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Verified\\\\[4pt]\n\\ref{WR15} & Coordinate certificate and arbitrary basis change & Verified\\\\[4pt]\n\\ref{WR16} & Consistent inexact stopping rule & Verified\\\\\n\\bottomrule\\end{longtable}\n\n\\subsection{Bounded realization and budget geometry}\\label{app:wr-source-map}\nThe formal bounded-realization theorem concerns bounded linear maps between real Hilbert spaces. It gives an exact preimage within the same norm bound as the approximating corrections. Its proof uses compactness of a dual ball and the Riesz isometric equivalence. The printed proof uses the weak-subsequence lemma, Lemma~\\ref{lem:weak-subsequence}, whose particular proof route is not separately formalized here. Neither route requires separability or closed range. Neither asserts strong convergence of the original approximating sequence.\n\nThe nested-constraint theorem constructs the minimum-norm points of nonempty closed affine sets. It proves the exact energy identity, norm monotonicity, and the equivalence between a nonempty intersection and bounded minimum norms. Under that bound, the corrections converge strongly to the unique minimum-norm point of the intersection. If the intersection is empty, their norms tend to infinity. The affine assumption is retained: for general convex sets the corresponding energy statement is an inequality.\n\n\\begin{longtable}{@{}p{.38\\linewidth}p{.55\\linewidth}@{}}\n\\toprule Namespace & Representative declarations\\\\\\midrule\\endhead\n\\path{BoundedApproximateRealization} & \\path{exists_solution_of_bounded_residual_tendsto_zero}\\newline\\path{mem_range_iff_bounded_approximation}\\newline\\path{norm_tendsto_atTop_of_not_mem_range}\\\\[5pt]\n\\path{NestedMinimumCorrections} & \\path{nested_affine_energy}\\newline\\path{nested_affine_intersection_iff_bounded}\\newline\\path{nested_affine_strong_realization}\\newline\\path{nested_affine_escape}\\\\[5pt]\n\\path{ProtectedBoundedRealization} & \\path{exists_exact_repair_of_bounded_approximation}\\\\\n\\bottomrule\\end{longtable}\nAll namespaces in this appendix have the prefix \\path{KomorebiSpace}. The protected application uses the actual subtype $\\ker p$ with its inherited Hilbert norm. It constructs a repaired candidate preserving the declared observable and the original correction budget.\n\nThe budget-profile theorems concern bounded operators between real complete Hilbert spaces. The budget is nonnegative and may be zero. Attainment of a residual minimum over a bounded correction ball does not require the entire operator range to be closed. The sharp support formula supplies an actual maximizing witness, expressed formally by \\code{IsGreatest}. It does not infer attainment from the mere existence of a supremum.\n\n\\begin{longtable}{@{}p{.38\\linewidth}p{.55\\linewidth}@{}}\n\\toprule Namespace & Representative declarations\\\\\\midrule\\endhead\n\\path{BudgetProfile} & \\path{exists_budget_minimizer}\\newline\\path{beta_eq_iInf}\\newline\\path{beta_tendsto_floor}\\newline\\path{beta_eq_floor_iff}\\newline\\path{beta_eq_zero_iff}\\\\[5pt]\n\\path{BudgetProfileGeometry} & \\path{beta_antitone}\\newline\\path{beta_zero}\\newline\\path{beta_convex_combination}\\newline\\path{beta_profile_bounds}\\\\[5pt]\n\\path{BudgetDuality} & \\path{dualValue_le_beta}\\newline\\path{exists_attaining_dual_witness}\\newline\\path{sharp_budget_duality}\\newline\\path{strict_exclusion_certificate}\\\\[5pt]\n\\path{BudgetRobustness} & \\path{candidate_residual_le}\\newline\\path{beta_abs_sub_le}\\newline\\path{nominal_dual_lower_certificate}\\newline\\path{excludes_budget_of_nominal_dual_certificate}\\\\\n\\bottomrule\\end{longtable}\n\nThe robustness comparison uses the same correction space for both problems. Its bound $\\delta+\\eta M$ controls data and operator uncertainty at budget $M$. A changing protected kernel is a different problem and is not covered by that estimate.\n\n\\subsection{Regularization, noise, and stopping}\nFor positive parameter $\\alpha$, the regularized normal operator is constructed as a continuous linear equivalence. Its inverse defines the correction. The formal development proves uniqueness, the energy identity, parameter monotonicity, budget optimality, and the two data-sensitivity estimates. The fixed-space protected application uses the restricted operator and its actual adjoint.\n\nThe limit theory distinguishes projection onto the closure of the range from membership in the range itself. Fitted values converge to the projected datum and residual norms converge to the irreducible floor. If that projected datum is attained, the corrections converge in norm to its unique minimum-norm preimage. Otherwise their norms tend to infinity. The data and stopping theorems use this limit theorem together with their explicit parameter, noise, and solver-tolerance assumptions.\n\n\\begin{longtable}{@{}p{.38\\linewidth}p{.55\\linewidth}@{}}\n\\toprule Namespace & Representative declarations\\\\\\midrule\\endhead\n\\path{RegularizedHilbertRepair} & \\path{exists_unique_minimizer}\\newline\\path{energy_identity}\\newline\\path{residual_best_at_budget}\\newline\\path{parameter_norm_antitone}\\newline\\path{parameter_residual_monotone}\\newline\\path{data_noise_bound}\\newline\\path{data_fitted_bound}\\\\[5pt]\n\\path{RegularizationLimits} & \\path{fitted_tendsto_projectedDatum}\\newline\\path{residual_norm_tendsto_floor}\\newline\\path{weighted_penalty_tendsto_zero}\\newline\\path{norm_tendsto_atTop_of_not_mem_range}\\\\[5pt]\n\\path{RegularizationStrongLimit} & \\path{exists_unique_minimum_and_limit}\\\\[5pt]\n\\path{RegularizationNoiseLimits} & \\path{data_noise_coefficient_attained}\\newline\\path{noisy_parameter_choice}\\newline\\path{inexact_parameter_choice}\\newline\\path{stopping_rule_sequence}\\\\\n\\bottomrule\\end{longtable}\n\nThe stopping theorem compares proposals, exact regularized corrections, and the intended minimum correction in one fixed Hilbert model. It includes the pointwise three-term bound and the convergence conclusion. Protection is exact at every index. The explicit diagonal schedule at the end of Section~\\ref{sec:stopping-rules} combines this solver bound with the formalized noisy rate in Corollary~\\ref{WR12}. The declaration \\path{DiagonalRegularization.diagonal_inexact_rounded_rate} proves this specialization with the literal ceiling, the data-noise hypothesis, and the actual measured-data normal-residual bound. The exact-data bias estimate is derived rather than assumed.\n\n\\subsection{Certificates and their coordinate metric}\nThe solver certificate controls correction error, fitted error, and objective excess through the actual Hilbert normal residual. The primal and dual identity supplies objective bracketing and a correction-distance bound for an arbitrary dual witness. 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It is a family satisfying every prescribed transition equation. The first task is to construct the space of such families and to keep its possible emptiness visible.}\n\\section{Views and coherent realizations}\\label{sec:views}\n\\subsection{The data of a view system}\nFix a small category \\(\\V\\), or work in fixed universes so that the products below exist in the chosen ambient universe. A \\emph{view} is an object of \\(\\V\\). It can encode a resolution level, a coordinate presentation, or a context in which part of an object is visible. These interpretations are mathematical choices, not additional axioms.","isfying every prescribed transition equation. The first task is to construct the space of such families and to keep its possible emptiness visible.}\n\\section{Views and coherent realizations}\\label{sec:views}\n\\subsection{The data of a view system}\nFix a small category \\(\\V\\), or work in fixed universes so that the products below exist in the chosen ambient universe. A \\emph{view} is an object of \\(\\V\\). It can encode a resolution level, a coordinate presentation, or a context in which part of an object is visible. These interpretations are mathematical choices, not additional axioms.","\\begin{definition}[View system]\nA view system consists of sets \\(F_v\\) and maps\n\\[\n f^*:F_w\\longrightarrow F_v\\qquad(f:v\\longrightarrow w)\n\\]\nsatisfying \\((1_v)^*=\\id\\) and \\((g\\circ f)^*=f^*g^*\\). Equivalently, it is a functor \\(F:\\V^{\\mathrm{op}}\\to\\Set\\).\n\\end{definition}\nThe direction matters. An arrow \\(v\\to w\\) induces a restriction from the fiber at \\(w\\) to the fiber at \\(v\\). A restriction need not be invertible. A category with several arrows between two views can retain different routes between their descriptions. Replacing it by a preorder can lose those routes and the constraints they impose.","\\begin{definition}[Strict Komorebi realization]\nThe ambient realization set is\n\\begin{equation}\\label{eq:ambient}\n \\mathfrak K(F)=\\Gamma(F)=\n \\left\\{x\\in\\prod_{v\\in\\Ob\\V}F_v:\n f^*(x_w)=x_v\\text{ for every }f:v\\to w\\right\\}.\n\\end{equation}\nA point of \\(\\mathfrak K(F)\\) is a whole coherent family. Its value in view \\(v\\) is the evaluation \\(\\operatorname{ev}_v(x)=x_v\\).\n\\end{definition}\nThe raw product contains all families, including inconsistent ones. Equation~\\eqref{eq:ambient} cuts out the exact families. No nonemptiness hypothesis is built into the definition.","\\begin{theorem}[Joint determination and the universal property]\\label{KS01}\nThe evaluations jointly determine a realization. If a set \\(T\\) carries maps \\(c_v:T\\to F_v\\) satisfying \\(f^*c_w=c_v\\), there is a unique map \\(c:T\\to\\Gamma(F)\\) satisfying \\(\\operatorname{ev}_v c=c_v\\) for every view.\n\\end{theorem}\n\\begin{proof}\nWe first verify that the evaluation maps themselves form a compatible family. If $x\\in\\Gamma(F)$ and $f:v\\to w$, the defining equation for $x$ gives\n\\[\n f^*(\\operatorname{ev}_w x)=f^*(x_w)=x_v=\\operatorname{ev}_v x.\n\\]\nIf $x,y\\in\\Gamma(F)$ have equal evaluations, then $x_v=y_v$ for every $v$. Equality of functions into a dependent product gives equality of their underlying families. Membership in $\\Gamma(F)$ is a property of a family, not a further choice of coordinate. Thus $x=y$.","{theorem}[Joint determination and the universal property]\\label{KS01}\nThe evaluations jointly determine a realization. If a set \\(T\\) carries maps \\(c_v:T\\to F_v\\) satisfying \\(f^*c_w=c_v\\), there is a unique map \\(c:T\\to\\Gamma(F)\\) satisfying \\(\\operatorname{ev}_v c=c_v\\) for every view.\n\\end{theorem}\n\\begin{proof}\nWe first verify that the evaluation maps themselves form a compatible family. If $x\\in\\Gamma(F)$ and $f:v\\to w$, the defining equation for $x$ gives\n\\[\n f^*(\\operatorname{ev}_w x)=f^*(x_w)=x_v=\\operatorname{ev}_v x.\n\\]\nIf $x,y\\in\\Gamma(F)$ have equal evaluations, then $x_v=y_v$ for every $v$. Equality of functions into a dependent product gives equality of their underlying families. Membership in $\\Gamma(F)$ is a property of a family, not a further choice of coordinate. Thus $x=y$.","rmination and the universal property]\\label{KS01}\nThe evaluations jointly determine a realization. If a set \\(T\\) carries maps \\(c_v:T\\to F_v\\) satisfying \\(f^*c_w=c_v\\), there is a unique map \\(c:T\\to\\Gamma(F)\\) satisfying \\(\\operatorname{ev}_v c=c_v\\) for every view.\n\\end{theorem}\n\\begin{proof}\nWe first verify that the evaluation maps themselves form a compatible family. If $x\\in\\Gamma(F)$ and $f:v\\to w$, the defining equation for $x$ gives\n\\[\n f^*(\\operatorname{ev}_w x)=f^*(x_w)=x_v=\\operatorname{ev}_v x.\n\\]\nIf $x,y\\in\\Gamma(F)$ have equal evaluations, then $x_v=y_v$ for every $v$. Equality of functions into a dependent product gives equality of their underlying families. Membership in $\\Gamma(F)$ is a property of a family, not a further choice of coordinate. Thus $x=y$.","label{KS01}\nThe evaluations jointly determine a realization. If a set \\(T\\) carries maps \\(c_v:T\\to F_v\\) satisfying \\(f^*c_w=c_v\\), there is a unique map \\(c:T\\to\\Gamma(F)\\) satisfying \\(\\operatorname{ev}_v c=c_v\\) for every view.\n\\end{theorem}\n\\begin{proof}\nWe first verify that the evaluation maps themselves form a compatible family. If $x\\in\\Gamma(F)$ and $f:v\\to w$, the defining equation for $x$ gives\n\\[\n f^*(\\operatorname{ev}_w x)=f^*(x_w)=x_v=\\operatorname{ev}_v x.\n\\]\nIf $x,y\\in\\Gamma(F)$ have equal evaluations, then $x_v=y_v$ for every $v$. Equality of functions into a dependent product gives equality of their underlying families. Membership in $\\Gamma(F)$ is a property of a family, not a further choice of coordinate. Thus $x=y$.","The evaluations jointly determine a realization. If a set \\(T\\) carries maps \\(c_v:T\\to F_v\\) satisfying \\(f^*c_w=c_v\\), there is a unique map \\(c:T\\to\\Gamma(F)\\) satisfying \\(\\operatorname{ev}_v c=c_v\\) for every view.\n\\end{theorem}\n\\begin{proof}\nWe first verify that the evaluation maps themselves form a compatible family. If $x\\in\\Gamma(F)$ and $f:v\\to w$, the defining equation for $x$ gives\n\\[\n f^*(\\operatorname{ev}_w x)=f^*(x_w)=x_v=\\operatorname{ev}_v x.\n\\]\nIf $x,y\\in\\Gamma(F)$ have equal evaluations, then $x_v=y_v$ for every $v$. Equality of functions into a dependent product gives equality of their underlying families. Membership in $\\Gamma(F)$ is a property of a family, not a further choice of coordinate. Thus $x=y$.","maps \\(c_v:T\\to F_v\\) satisfying \\(f^*c_w=c_v\\), there is a unique map \\(c:T\\to\\Gamma(F)\\) satisfying \\(\\operatorname{ev}_v c=c_v\\) for every view.\n\\end{theorem}\n\\begin{proof}\nWe first verify that the evaluation maps themselves form a compatible family. If $x\\in\\Gamma(F)$ and $f:v\\to w$, the defining equation for $x$ gives\n\\[\n f^*(\\operatorname{ev}_w x)=f^*(x_w)=x_v=\\operatorname{ev}_v x.\n\\]\nIf $x,y\\in\\Gamma(F)$ have equal evaluations, then $x_v=y_v$ for every $v$. Equality of functions into a dependent product gives equality of their underlying families. Membership in $\\Gamma(F)$ is a property of a family, not a further choice of coordinate. Thus $x=y$.","), there is a unique map \\(c:T\\to\\Gamma(F)\\) satisfying \\(\\operatorname{ev}_v c=c_v\\) for every view.\n\\end{theorem}\n\\begin{proof}\nWe first verify that the evaluation maps themselves form a compatible family. If $x\\in\\Gamma(F)$ and $f:v\\to w$, the defining equation for $x$ gives\n\\[\n f^*(\\operatorname{ev}_w x)=f^*(x_w)=x_v=\\operatorname{ev}_v x.\n\\]\nIf $x,y\\in\\Gamma(F)$ have equal evaluations, then $x_v=y_v$ for every $v$. Equality of functions into a dependent product gives equality of their underlying families. Membership in $\\Gamma(F)$ is a property of a family, not a further choice of coordinate. Thus $x=y$.","map \\(c:T\\to\\Gamma(F)\\) satisfying \\(\\operatorname{ev}_v c=c_v\\) for every view.\n\\end{theorem}\n\\begin{proof}\nWe first verify that the evaluation maps themselves form a compatible family. If $x\\in\\Gamma(F)$ and $f:v\\to w$, the defining equation for $x$ gives\n\\[\n f^*(\\operatorname{ev}_w x)=f^*(x_w)=x_v=\\operatorname{ev}_v x.\n\\]\nIf $x,y\\in\\Gamma(F)$ have equal evaluations, then $x_v=y_v$ for every $v$. Equality of functions into a dependent product gives equality of their underlying families. Membership in $\\Gamma(F)$ is a property of a family, not a further choice of coordinate. Thus $x=y$.","e first verify that the evaluation maps themselves form a compatible family. If $x\\in\\Gamma(F)$ and $f:v\\to w$, the defining equation for $x$ gives\n\\[\n f^*(\\operatorname{ev}_w x)=f^*(x_w)=x_v=\\operatorname{ev}_v x.\n\\]\nIf $x,y\\in\\Gamma(F)$ have equal evaluations, then $x_v=y_v$ for every $v$. Equality of functions into a dependent product gives equality of their underlying families. Membership in $\\Gamma(F)$ is a property of a family, not a further choice of coordinate. Thus $x=y$.","$x\\in\\Gamma(F)$ and $f:v\\to w$, the defining equation for $x$ gives\n\\[\n f^*(\\operatorname{ev}_w x)=f^*(x_w)=x_v=\\operatorname{ev}_v x.\n\\]\nIf $x,y\\in\\Gamma(F)$ have equal evaluations, then $x_v=y_v$ for every $v$. Equality of functions into a dependent product gives equality of their underlying families. Membership in $\\Gamma(F)$ is a property of a family, not a further choice of coordinate. Thus $x=y$.","Now fix the compatible maps $c_v:T\\to F_v$. For each $t\\in T$ define a family by $x(t)_v=c_v(t)$. For every arrow,\n\\[\n f^*x(t)_w=f^*c_w(t)=c_v(t)=x(t)_v.\n\\]\nThis calculation verifies membership in $\\Gamma(F)$, so the rule $c(t)=x(t)$ defines an actual function $c:T\\to\\Gamma(F)$. Evaluating it gives $\\operatorname{ev}_vc(t)=c_v(t)$, as required.","To prove uniqueness, let $b:T\\to\\Gamma(F)$ be another such function. At every $t$ and $v$,\n\\[\n (b(t))_v=\\operatorname{ev}_v b(t)=c_v(t)=\\operatorname{ev}_v c(t).\n\\]\nJoint determination first yields $b(t)=c(t)$, and function extensionality then yields $b=c$. Nothing in this argument assumes that $T$ or $\\Gamma(F)$ is inhabited. For example, when $T$ is empty there is the unique empty function. When $T$ is inhabited, existence of the compatible family $(c_v)$ is already substantive data.\n\\end{proof}\n\\source{KS01}\nThus the strict construction is precisely the ordinary limit of the view diagram, with the usual limit interpretation in the sequential case~\\cite{stacks-limits}. This identifies its mathematical content and its universal pr","To prove uniqueness, let $b:T\\to\\Gamma(F)$ be another such function. At every $t$ and $v$,\n\\[\n (b(t))_v=\\operatorname{ev}_v b(t)=c_v(t)=\\operatorname{ev}_v c(t).\n\\]\nJoint determination first yields $b(t)=c(t)$, and function extensionality then yields $b=c$. Nothing in this argument assumes that $T$ or $\\Gamma(F)$ is inhabited. For example, when $T$ is empty there is the unique empty function. When $T$ is inhabited, existence of the compatible family $(c_v)$ is already substantive data.\n\\end{proof}\n\\source{KS01}\nThus the strict construction is precisely the ordinary limit of the view diagram, with the usual limit interpretation in the sequential case~\\cite{stacks-limits}. This identifies its mathematical content and its universal property. It does not supply a universal","To prove uniqueness, let $b:T\\to\\Gamma(F)$ be another such function. At every $t$ and $v$,\n\\[\n (b(t))_v=\\operatorname{ev}_v b(t)=c_v(t)=\\operatorname{ev}_v c(t).\n\\]\nJoint determination first yields $b(t)=c(t)$, and function extensionality then yields $b=c$. Nothing in this argument assumes that $T$ or $\\Gamma(F)$ is inhabited. For example, when $T$ is empty there is the unique empty function. When $T$ is inhabited, existence of the compatible family $(c_v)$ is already substantive data.\n\\end{proof}\n\\source{KS01}\nThus the strict construction is precisely the ordinary limit of the view diagram, with the usual limit interpretation in the sequential case~\\cite{stacks-limits}. This identifies its mathematical content and its universal property. It does not supply a universal view dia","To prove uniqueness, let $b:T\\to\\Gamma(F)$ be another such function. At every $t$ and $v$,\n\\[\n (b(t))_v=\\operatorname{ev}_v b(t)=c_v(t)=\\operatorname{ev}_v c(t).\n\\]\nJoint determination first yields $b(t)=c(t)$, and function extensionality then yields $b=c$. Nothing in this argument assumes that $T$ or $\\Gamma(F)$ is inhabited. For example, when $T$ is empty there is the unique empty function. When $T$ is inhabited, existence of the compatible family $(c_v)$ is already substantive data.\n\\end{proof}\n\\source{KS01}\nThus the strict construction is precisely the ordinary limit of the view diagram, with the usual limit interpretation in the sequential case~\\cite{stacks-limits}. This identifies its mathematical content and its universal property. It does not supply a universal view diagram for every possible mathematical question.","prove uniqueness, let $b:T\\to\\Gamma(F)$ be another such function. At every $t$ and $v$,\n\\[\n (b(t))_v=\\operatorname{ev}_v b(t)=c_v(t)=\\operatorname{ev}_v c(t).\n\\]\nJoint determination first yields $b(t)=c(t)$, and function extensionality then yields $b=c$. Nothing in this argument assumes that $T$ or $\\Gamma(F)$ is inhabited. For example, when $T$ is empty there is the unique empty function. When $T$ is inhabited, existence of the compatible family $(c_v)$ is already substantive data.\n\\end{proof}\n\\source{KS01}\nThus the strict construction is precisely the ordinary limit of the view diagram, with the usual limit interpretation in the sequential case~\\cite{stacks-limits}. This identifies its mathematical content and its universal property. It does not supply a universal view diagram for every possible mathematical question.","uch function. At every $t$ and $v$,\n\\[\n (b(t))_v=\\operatorname{ev}_v b(t)=c_v(t)=\\operatorname{ev}_v c(t).\n\\]\nJoint determination first yields $b(t)=c(t)$, and function extensionality then yields $b=c$. Nothing in this argument assumes that $T$ or $\\Gamma(F)$ is inhabited. For example, when $T$ is empty there is the unique empty function. When $T$ is inhabited, existence of the compatible family $(c_v)$ is already substantive data.\n\\end{proof}\n\\source{KS01}\nThus the strict construction is precisely the ordinary limit of the view diagram, with the usual limit interpretation in the sequential case~\\cite{stacks-limits}. This identifies its mathematical content and its universal property. It does not supply a universal view diagram for every possible mathematical question.","\\subsection{Maps of realizations}\nLet \\(\\alpha:F\\to G\\) consist of maps \\(\\alpha_v:F_v\\to G_v\\) with\n\\[\n \\alpha_v f_F^*=f_G^*\\alpha_w.\n\\]\nThe induced map on realizations is \\(\\Gamma(\\alpha)(x)_v=\\alpha_v(x_v)\\). The naturality square is exactly what is needed to show that the image family is coherent.","\\begin{proposition}[Functorial transport]\\label{KS02}\nThese induced maps preserve identities and composition:\n\\[\n \\Gamma(\\id_F)=\\id_{\\Gamma(F)},\\qquad\n \\Gamma(\\beta\\alpha)=\\Gamma(\\beta)\\Gamma(\\alpha).\n\\]\n\\end{proposition}\n\\begin{proof}\nEvaluate each side at an arbitrary view and apply the corresponding identity for the coordinate maps. Theorem~\\ref{KS01} then identifies the maps on the whole realization set.\n\\end{proof}\n\\source{KS02}\nAn arbitrary collection of fiber maps need not define a map on realizations. Its compatibility equations are part of the theorem, not a convention that can be omitted after a notation has been chosen.","\\subsection{A global view and a useful warning}\n\\begin{proposition}[Terminal evaluation]\\label{KS03}\nIf \\(t\\) is terminal in \\(\\V\\), evaluation induces an equivalence \\(\\Gamma(F)\\simeq F_t\\).\n\\end{proposition}\n\\begin{proof}\nFor \\(a\\in F_t\\), take its restriction along the unique arrow \\(v\\to t\\) at each view. Uniqueness of the arrow to \\(t\\), together with the functor laws, proves coherence. Evaluation at \\(t\\) recovers \\(a\\). Conversely, any coherent family is the restriction of its coordinate at \\(t\\).\n\\end{proof}\n\\source{KS03}\nThis proposition is both useful and diagnostic. If the supposedly reconstructed global object has already been placed in a terminal fiber, the limit recovers that fiber. The proposition does not prove that arbitrary partial data have a preimage there.","{A global view and a useful warning}\n\\begin{proposition}[Terminal evaluation]\\label{KS03}\nIf \\(t\\) is terminal in \\(\\V\\), evaluation induces an equivalence \\(\\Gamma(F)\\simeq F_t\\).\n\\end{proposition}\n\\begin{proof}\nFor \\(a\\in F_t\\), take its restriction along the unique arrow \\(v\\to t\\) at each view. Uniqueness of the arrow to \\(t\\), together with the functor laws, proves coherence. Evaluation at \\(t\\) recovers \\(a\\). Conversely, any coherent family is the restriction of its coordinate at \\(t\\).\n\\end{proof}\n\\source{KS03}\nThis proposition is both useful and diagnostic. If the supposedly reconstructed global object has already been placed in a terminal fiber, the limit recovers that fiber. The proposition does not prove that arbitrary partial data have a preimage there.","and a useful warning}\n\\begin{proposition}[Terminal evaluation]\\label{KS03}\nIf \\(t\\) is terminal in \\(\\V\\), evaluation induces an equivalence \\(\\Gamma(F)\\simeq F_t\\).\n\\end{proposition}\n\\begin{proof}\nFor \\(a\\in F_t\\), take its restriction along the unique arrow \\(v\\to t\\) at each view. Uniqueness of the arrow to \\(t\\), together with the functor laws, proves coherence. Evaluation at \\(t\\) recovers \\(a\\). Conversely, any coherent family is the restriction of its coordinate at \\(t\\).\n\\end{proof}\n\\source{KS03}\nThis proposition is both useful and diagnostic. If the supposedly reconstructed global object has already been placed in a terminal fiber, the limit recovers that fiber. The proposition does not prove that arbitrary partial data have a preimage there.","(t\\) is terminal in \\(\\V\\), evaluation induces an equivalence \\(\\Gamma(F)\\simeq F_t\\).\n\\end{proposition}\n\\begin{proof}\nFor \\(a\\in F_t\\), take its restriction along the unique arrow \\(v\\to t\\) at each view. Uniqueness of the arrow to \\(t\\), together with the functor laws, proves coherence. Evaluation at \\(t\\) recovers \\(a\\). Conversely, any coherent family is the restriction of its coordinate at \\(t\\).\n\\end{proof}\n\\source{KS03}\nThis proposition is both useful and diagnostic. If the supposedly reconstructed global object has already been placed in a terminal fiber, the limit recovers that fiber. The proposition does not prove that arbitrary partial data have a preimage there.","The category of elements \\(\\int_\\V F\\) and \\(\\Gamma(F)\\) also have different roles. The former organizes individual appearances and their transitions. The latter selects coherent families across all views. The category of elements serves here as explanatory notation, while the Lean formalization directly represents families and their coherence.","\\chap{Partial data, gluing, and correspondence fibers}\\label{sec:glue}\n\\subsection{Prescribed views}\nFor \\(U\\subseteq\\Ob\\V\\), use the full subcategory on \\(U\\). A partial section \\(s\\in\\Gamma_U(F)\\) satisfies all transition equations whose endpoints lie in \\(U\\). Define\n\\begin{equation}\\label{eq:extension}\n \\Ext_U(s)=\\{x\\in\\Gamma(F):x|_U=s\\}.\n\\end{equation}\nThis fiber can be empty, a singleton, or much larger. A partial section is already coherent on its prescribed domain. A raw assignment that fails an internal transition equation is not a partial section of this kind.","Let \\((U_i)_{i\\in I}\\) be full patches. A matching family consists of \\(s_i\\in\\Gamma_{U_i}(F)\\) with equal values on every overlap. The relevant covering condition is\n\\begin{equation}\\label{eq:transitioncover}\n \\text{for every }f:v\\to w,\\quad\n \\text{some }U_i\\text{ contains both }v\\text{ and }w.\n\\end{equation}\nIdentities imply that this condition also covers every object.","\\begin{theorem}[Strict gluing]\\label{KS04}\nUnder \\eqref{eq:transitioncover}, restriction induces an equivalence\n\\[\n \\Gamma(F)\\simeq\\Match(F,(U_i)).\n\\]\nEvery fixed matching family has a unique global glue.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: every coordinate can be selected.} Apply the cover condition to the identity arrow of $v$. It gives an index containing $v$. Choose one such index $i(v)$ and put\n\\[\n x_v=(s_{i(v)})_v.\n\\]\nThe choice is an existence construction. It is not an algorithm selecting patches in a preferred order. If $v\\in U_j$, matching on the overlap $U_{i(v)}\\cap U_j$ gives\n\\[\n x_v=(s_j)_v.                                                   \\tag{*}\n\\]\nIn particular the value is independent of the selected index.","rem}[Strict gluing]\\label{KS04}\nUnder \\eqref{eq:transitioncover}, restriction induces an equivalence\n\\[\n \\Gamma(F)\\simeq\\Match(F,(U_i)).\n\\]\nEvery fixed matching family has a unique global glue.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: every coordinate can be selected.} Apply the cover condition to the identity arrow of $v$. It gives an index containing $v$. Choose one such index $i(v)$ and put\n\\[\n x_v=(s_{i(v)})_v.\n\\]\nThe choice is an existence construction. It is not an algorithm selecting patches in a preferred order. If $v\\in U_j$, matching on the overlap $U_{i(v)}\\cap U_j$ gives\n\\[\n x_v=(s_j)_v.                                                   \\tag{*}\n\\]\nIn particular the value is independent of the selected index.","_i)).\n\\]\nEvery fixed matching family has a unique global glue.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: every coordinate can be selected.} Apply the cover condition to the identity arrow of $v$. It gives an index containing $v$. Choose one such index $i(v)$ and put\n\\[\n x_v=(s_{i(v)})_v.\n\\]\nThe choice is an existence construction. It is not an algorithm selecting patches in a preferred order. If $v\\in U_j$, matching on the overlap $U_{i(v)}\\cap U_j$ gives\n\\[\n x_v=(s_j)_v.                                                   \\tag{*}\n\\]\nIn particular the value is independent of the selected index.","has a unique global glue.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: every coordinate can be selected.} Apply the cover condition to the identity arrow of $v$. It gives an index containing $v$. Choose one such index $i(v)$ and put\n\\[\n x_v=(s_{i(v)})_v.\n\\]\nThe choice is an existence construction. It is not an algorithm selecting patches in a preferred order. If $v\\in U_j$, matching on the overlap $U_{i(v)}\\cap U_j$ gives\n\\[\n x_v=(s_j)_v.                                                   \\tag{*}\n\\]\nIn particular the value is independent of the selected index.","\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: every coordinate can be selected.} Apply the cover condition to the identity arrow of $v$. It gives an index containing $v$. Choose one such index $i(v)$ and put\n\\[\n x_v=(s_{i(v)})_v.\n\\]\nThe choice is an existence construction. It is not an algorithm selecting patches in a preferred order. If $v\\in U_j$, matching on the overlap $U_{i(v)}\\cap U_j$ gives\n\\[\n x_v=(s_j)_v.                                                   \\tag{*}\n\\]\nIn particular the value is independent of the selected index.","\\emph{Step 2: the selected coordinates satisfy every arrow equation.} For $f:v\\to w$, choose a single index $j$ with $v,w\\in U_j$. Formula (*) applies to both endpoints. Since the patch is full, it contains this actual arrow, even when other arrows have the same endpoints. Coherence of $s_j$ gives\n\\[\n f^*x_w=f^*(s_j)_w=(s_j)_v=x_v.\n\\]\nThus $x\\in\\Gamma(F)$. Object coverage alone would not supply the common patch used in this step.","\\emph{Step 3: assembly and restriction are inverse.} Formula (*) proves $x|_{U_j}=s_j$ for every $j$. Conversely, start with $y\\in\\Gamma(F)$ and assemble its restrictions. The resulting coordinate at $v$ is $y_v$, regardless of which containing patch is chosen. Joint determination gives the original $y$. These two identities are exactly the inverse identities for the displayed equivalence.","For uniqueness with a fixed matching family, any other glue has coordinate $(s_{i(v)})_v=x_v$ at every $v$, hence equals $x$. If the view category is empty, its product has one empty family and the same conclusion follows without making any choices.\n\\end{proof}\n\\source{KS04}\nThe proof explains why object coverage alone is insufficient. Coherence is an equation attached to an arrow, so existence of a glue must see that equation somewhere in the local data.","\\begin{example}[An object cover with no glue]\\label{KS05}\nTake two views \\(0\\to1\\), Boolean fibers, and the identity restriction. The singleton patches cover all objects. A local value false at \\(0\\) and true at \\(1\\) agrees on the empty overlap, yet no global section realizes both values. A global family would have to assign the same Boolean to both views.\n\\end{example}\n\\source{KS05}\nObject coverage does prove uniqueness of a possible glue. It does not prove its existence. This distinction persists when sets are replaced by objects with symmetries.","\\subsection{Correspondences must also be coherent}\nSuppose natural maps form a span \\(F\\xleftarrow{p}M\\xrightarrow{q}G\\), and fix \\(a\\in\\Gamma(F)\\), \\(b\\in\\Gamma(G)\\). At each view define the ordinary fiber\n\\[\n B_{a,b}(v)=\\{m\\in M_v:p_v(m)=a_v,\\ q_v(m)=b_v\\}.\n\\]\nNaturality and coherence provide its restrictions. A global bridge is an element of\n\\[\n \\mathfrak B(a,b)=\\{z\\in\\Gamma(M):\\Gamma(p)z=a,\\ \\Gamma(q)z=b\\}.\n\\]","\\begin{theorem}[Fiber and section compatibility]\\label{KS06}\nThere is a coordinate preserving equivalence\n\\[\n \\Gamma(B_{a,b})\\simeq\\mathfrak B(a,b).\n\\]\nUnder a transition cover, matching local bridge sections glue to a unique global bridge with both prescribed projections unchanged.\n\\end{theorem}\n\\begin{proof}\nWe must first check that the displayed fibers really form a view system. Let $m\\in B_{a,b}(w)$ and $f:v\\to w$. Naturality of $p$ and coherence of $a$ give\n\\[\n p_v(f_M^*m)=f_F^*p_w(m)=f_F^*a_w=a_v.\n\\]\nThe same calculation with $q,b$ gives $q_v(f_M^*m)=b_v$. Therefore $f_M^*m$ belongs to $B_{a,b}(v)$. Identity and composition of these restricted maps are inherited from $M$, because equality in a subset is equality of its underlying elements.","em}[Fiber and section compatibility]\\label{KS06}\nThere is a coordinate preserving equivalence\n\\[\n \\Gamma(B_{a,b})\\simeq\\mathfrak B(a,b).\n\\]\nUnder a transition cover, matching local bridge sections glue to a unique global bridge with both prescribed projections unchanged.\n\\end{theorem}\n\\begin{proof}\nWe must first check that the displayed fibers really form a view system. Let $m\\in B_{a,b}(w)$ and $f:v\\to w$. Naturality of $p$ and coherence of $a$ give\n\\[\n p_v(f_M^*m)=f_F^*p_w(m)=f_F^*a_w=a_v.\n\\]\nThe same calculation with $q,b$ gives $q_v(f_M^*m)=b_v$. Therefore $f_M^*m$ belongs to $B_{a,b}(v)$. Identity and composition of these restricted maps are inherited from $M$, because equality in a subset is equality of its underlying elements.","atibility]\\label{KS06}\nThere is a coordinate preserving equivalence\n\\[\n \\Gamma(B_{a,b})\\simeq\\mathfrak B(a,b).\n\\]\nUnder a transition cover, matching local bridge sections glue to a unique global bridge with both prescribed projections unchanged.\n\\end{theorem}\n\\begin{proof}\nWe must first check that the displayed fibers really form a view system. Let $m\\in B_{a,b}(w)$ and $f:v\\to w$. Naturality of $p$ and coherence of $a$ give\n\\[\n p_v(f_M^*m)=f_F^*p_w(m)=f_F^*a_w=a_v.\n\\]\nThe same calculation with $q,b$ gives $q_v(f_M^*m)=b_v$. Therefore $f_M^*m$ belongs to $B_{a,b}(v)$. Identity and composition of these restricted maps are inherited from $M$, because equality in a subset is equality of its underlying elements.","}\nThere is a coordinate preserving equivalence\n\\[\n \\Gamma(B_{a,b})\\simeq\\mathfrak B(a,b).\n\\]\nUnder a transition cover, matching local bridge sections glue to a unique global bridge with both prescribed projections unchanged.\n\\end{theorem}\n\\begin{proof}\nWe must first check that the displayed fibers really form a view system. Let $m\\in B_{a,b}(w)$ and $f:v\\to w$. Naturality of $p$ and coherence of $a$ give\n\\[\n p_v(f_M^*m)=f_F^*p_w(m)=f_F^*a_w=a_v.\n\\]\nThe same calculation with $q,b$ gives $q_v(f_M^*m)=b_v$. Therefore $f_M^*m$ belongs to $B_{a,b}(v)$. Identity and composition of these restricted maps are inherited from $M$, because equality in a subset is equality of its underlying elements.","a coordinate preserving equivalence\n\\[\n \\Gamma(B_{a,b})\\simeq\\mathfrak B(a,b).\n\\]\nUnder a transition cover, matching local bridge sections glue to a unique global bridge with both prescribed projections unchanged.\n\\end{theorem}\n\\begin{proof}\nWe must first check that the displayed fibers really form a view system. Let $m\\in B_{a,b}(w)$ and $f:v\\to w$. Naturality of $p$ and coherence of $a$ give\n\\[\n p_v(f_M^*m)=f_F^*p_w(m)=f_F^*a_w=a_v.\n\\]\nThe same calculation with $q,b$ gives $q_v(f_M^*m)=b_v$. Therefore $f_M^*m$ belongs to $B_{a,b}(v)$. Identity and composition of these restricted maps are inherited from $M$, because equality in a subset is equality of its underlying elements.","k B(a,b).\n\\]\nUnder a transition cover, matching local bridge sections glue to a unique global bridge with both prescribed projections unchanged.\n\\end{theorem}\n\\begin{proof}\nWe must first check that the displayed fibers really form a view system. Let $m\\in B_{a,b}(w)$ and $f:v\\to w$. Naturality of $p$ and coherence of $a$ give\n\\[\n p_v(f_M^*m)=f_F^*p_w(m)=f_F^*a_w=a_v.\n\\]\nThe same calculation with $q,b$ gives $q_v(f_M^*m)=b_v$. Therefore $f_M^*m$ belongs to $B_{a,b}(v)$. Identity and composition of these restricted maps are inherited from $M$, because equality in a subset is equality of its underlying elements.","Under a transition cover, matching local bridge sections glue to a unique global bridge with both prescribed projections unchanged.\n\\end{theorem}\n\\begin{proof}\nWe must first check that the displayed fibers really form a view system. Let $m\\in B_{a,b}(w)$ and $f:v\\to w$. Naturality of $p$ and coherence of $a$ give\n\\[\n p_v(f_M^*m)=f_F^*p_w(m)=f_F^*a_w=a_v.\n\\]\nThe same calculation with $q,b$ gives $q_v(f_M^*m)=b_v$. Therefore $f_M^*m$ belongs to $B_{a,b}(v)$. Identity and composition of these restricted maps are inherited from $M$, because equality in a subset is equality of its underlying elements.","tching local bridge sections glue to a unique global bridge with both prescribed projections unchanged.\n\\end{theorem}\n\\begin{proof}\nWe must first check that the displayed fibers really form a view system. Let $m\\in B_{a,b}(w)$ and $f:v\\to w$. Naturality of $p$ and coherence of $a$ give\n\\[\n p_v(f_M^*m)=f_F^*p_w(m)=f_F^*a_w=a_v.\n\\]\nThe same calculation with $q,b$ gives $q_v(f_M^*m)=b_v$. Therefore $f_M^*m$ belongs to $B_{a,b}(v)$. Identity and composition of these restricted maps are inherited from $M$, because equality in a subset is equality of its underlying elements.","sections glue to a unique global bridge with both prescribed projections unchanged.\n\\end{theorem}\n\\begin{proof}\nWe must first check that the displayed fibers really form a view system. Let $m\\in B_{a,b}(w)$ and $f:v\\to w$. Naturality of $p$ and coherence of $a$ give\n\\[\n p_v(f_M^*m)=f_F^*p_w(m)=f_F^*a_w=a_v.\n\\]\nThe same calculation with $q,b$ gives $q_v(f_M^*m)=b_v$. Therefore $f_M^*m$ belongs to $B_{a,b}(v)$. Identity and composition of these restricted maps are inherited from $M$, because equality in a subset is equality of its underlying elements.","\\end{theorem}\n\\begin{proof}\nWe must first check that the displayed fibers really form a view system. Let $m\\in B_{a,b}(w)$ and $f:v\\to w$. Naturality of $p$ and coherence of $a$ give\n\\[\n p_v(f_M^*m)=f_F^*p_w(m)=f_F^*a_w=a_v.\n\\]\nThe same calculation with $q,b$ gives $q_v(f_M^*m)=b_v$. Therefore $f_M^*m$ belongs to $B_{a,b}(v)$. Identity and composition of these restricted maps are inherited from $M$, because equality in a subset is equality of its underlying elements.","Given $u\\in\\Gamma(B_{a,b})$, forget each fiber-membership condition and call the underlying family $z_v\\in M_v$. Coherence of $u$ is precisely coherence of $z$, so $z\\in\\Gamma(M)$. The equations $p_vz_v=a_v$ hold at every coordinate. Joint determination in $\\Gamma(F)$ yields $\\Gamma(p)z=a$, and the equations for $q$ similarly yield $\\Gamma(q)z=b$. Thus $z$ is a global bridge.","In the opposite direction, let $z\\in\\mathfrak B(a,b)$. Evaluate its two global projection equations at $v$. They show that $z_v$ belongs to $B_{a,b}(v)$, and its original coherence shows that these elements form a section of the fiber system. Forgetting and reattaching membership conditions preserves every coordinate, so the two constructions are inverse.","Finally apply strict gluing to this specific fiber system. The glued section retains its membership in every $B_{a,b}(v)$, which is why both prescribed projections survive gluing. Uniqueness is for the fixed matching bridge family. Different matching families for the same $(a,b)$ can still yield different global bridges.\n\\end{proof}\n\\source{KS06}\nThese are ordinary fibers. They do not retain comparison isomorphisms or automorphisms. Section~\\ref{sec:relative} constructs the appropriate groupoid version when this information matters.","\\subsection{Locally inhabited does not mean globally inhabited}\n\\begin{example}[The loop with no coherent choice]\\label{KS07}\nLet the view category have one object and automorphism group \\(C_2\\). Let its nonidentity arrow act on a set with two elements by swapping the elements. The fiber is nonempty and every transport is bijective. Nevertheless a global section would be a fixed point of the swap, and there is none.\n\\end{example}\n\\source{KS07}\nThe problem is not missing local elements or noninvertible transition maps. It is the absence of an element satisfying the loop equation. Inserting a point in each fiber does not certify a coherent selection of those points.","\\subsection{The compatibility differential}\nAssume each fiber \\(A_v\\) is an abelian group and each restriction is additive. Set\n\\[\n E=\\prod_v A_v,\\qquad R=\\prod_{f:v\\to w}A_v,\n \\qquad (\\delta x)_f=f^*(x_w)-x_v.\n\\]\nThe group \\(R\\) retains an entry for every arrow. It is not silently identified with a cohomology group.","\\begin{proposition}[Differential identities]\\label{KS08}\nThe map \\(\\delta:E\\to R\\) is additive, \\(\\Gamma(A)=\\ker\\delta\\), and\n\\[\n (\\delta x)_{1_v}=0,\\qquad\n (\\delta x)_{g\\circ f}=f^*((\\delta x)_g)+(\\delta x)_f.\n\\]\n\\end{proposition}\n\\begin{proof}\nAdditivity follows coordinate by coordinate. Vanishing is exactly the defining coherence equation. For composition, add and subtract \\(f^*x_w\\):\n\\[\n f^*g^*x_z-x_v\n =f^*(g^*x_z-x_w)+(f^*x_w-x_v).\n\\]\nThe identity equation uses \\((1_v)^*=\\id\\).\n\\end{proof}\n\\source{KS08}\nThis exact formula is more informative than a scalar error score. It specifies how defects transform and combine. Norms can be added later, but they do not replace the typed residual.",":E\\to R\\) is additive, \\(\\Gamma(A)=\\ker\\delta\\), and\n\\[\n (\\delta x)_{1_v}=0,\\qquad\n (\\delta x)_{g\\circ f}=f^*((\\delta x)_g)+(\\delta x)_f.\n\\]\n\\end{proposition}\n\\begin{proof}\nAdditivity follows coordinate by coordinate. Vanishing is exactly the defining coherence equation. For composition, add and subtract \\(f^*x_w\\):\n\\[\n f^*g^*x_z-x_v\n =f^*(g^*x_z-x_w)+(f^*x_w-x_v).\n\\]\nThe identity equation uses \\((1_v)^*=\\id\\).\n\\end{proof}\n\\source{KS08}\nThis exact formula is more informative than a scalar error score. It specifies how defects transform and combine. Norms can be added later, but they do not replace the typed residual.","\\subsection{The admissible corrections}\nFix \\(U\\subseteq\\Ob\\V\\). Write \\(p_U:E\\to\\prod_{v\\in U}A_v\\) for raw restriction. A correction preserving the prescribed values belongs to \\(\\ker p_U\\). Its possible defects form the subgroup\n\\[\n B_U=\\delta(\\ker p_U)\\le R.\n\\]\nFor a partial section \\(s\\), choose any raw extension \\(\\widetilde s\\), for example zero off \\(U\\), and define\n\\begin{equation}\\label{eq:obstruction}\n Q_U=R/B_U,\\qquad\n \\mathfrak o_U(s)=[\\delta\\widetilde s]\\in Q_U.\n\\end{equation}\nIf another raw extension is chosen, their difference vanishes on \\(U\\), so its differential belongs to \\(B_U\\). Thus the class is independent of the lift.","lues belongs to \\(\\ker p_U\\). Its possible defects form the subgroup\n\\[\n B_U=\\delta(\\ker p_U)\\le R.\n\\]\nFor a partial section \\(s\\), choose any raw extension \\(\\widetilde s\\), for example zero off \\(U\\), and define\n\\begin{equation}\\label{eq:obstruction}\n Q_U=R/B_U,\\qquad\n \\mathfrak o_U(s)=[\\delta\\widetilde s]\\in Q_U.\n\\end{equation}\nIf another raw extension is chosen, their difference vanishes on \\(U\\), so its differential belongs to \\(B_U\\). Thus the class is independent of the lift.","ible defects form the subgroup\n\\[\n B_U=\\delta(\\ker p_U)\\le R.\n\\]\nFor a partial section \\(s\\), choose any raw extension \\(\\widetilde s\\), for example zero off \\(U\\), and define\n\\begin{equation}\\label{eq:obstruction}\n Q_U=R/B_U,\\qquad\n \\mathfrak o_U(s)=[\\delta\\widetilde s]\\in Q_U.\n\\end{equation}\nIf another raw extension is chosen, their difference vanishes on \\(U\\), so its differential belongs to \\(B_U\\). Thus the class is independent of the lift.","\\begin{theorem}[Extension criterion]\\label{KS09}\nFor every coherent partial section,\n\\[\n \\mathfrak o_U(s)=0\\quad\\Longleftrightarrow\\quad\\Ext_U(s)\\ne\\varnothing.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\pi_U:R\\to R/B_U$ be the quotient homomorphism. By the definition of a quotient by a subgroup,\n\\[\n \\pi_U(r)=0\\quad\\Longleftrightarrow\\quad r\\in B_U.\n\\]\nHere $B_U$ is the \\emph{image} of $\\ker p_U$ under $\\delta$. Consequently membership of $\\delta\\widetilde s$ means the existence of a specific $h\\in E$ satisfying both $p_Uh=0$ and $\\delta h=\\delta\\widetilde s$. The subgroup need not be closed in any topology. No topology has been introduced.","begin{theorem}[Extension criterion]\\label{KS09}\nFor every coherent partial section,\n\\[\n \\mathfrak o_U(s)=0\\quad\\Longleftrightarrow\\quad\\Ext_U(s)\\ne\\varnothing.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\pi_U:R\\to R/B_U$ be the quotient homomorphism. By the definition of a quotient by a subgroup,\n\\[\n \\pi_U(r)=0\\quad\\Longleftrightarrow\\quad r\\in B_U.\n\\]\nHere $B_U$ is the \\emph{image} of $\\ker p_U$ under $\\delta$. Consequently membership of $\\delta\\widetilde s$ means the existence of a specific $h\\in E$ satisfying both $p_Uh=0$ and $\\delta h=\\delta\\widetilde s$. The subgroup need not be closed in any topology. No topology has been introduced.","abel{KS09}\nFor every coherent partial section,\n\\[\n \\mathfrak o_U(s)=0\\quad\\Longleftrightarrow\\quad\\Ext_U(s)\\ne\\varnothing.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\pi_U:R\\to R/B_U$ be the quotient homomorphism. By the definition of a quotient by a subgroup,\n\\[\n \\pi_U(r)=0\\quad\\Longleftrightarrow\\quad r\\in B_U.\n\\]\nHere $B_U$ is the \\emph{image} of $\\ker p_U$ under $\\delta$. Consequently membership of $\\delta\\widetilde s$ means the existence of a specific $h\\in E$ satisfying both $p_Uh=0$ and $\\delta h=\\delta\\widetilde s$. The subgroup need not be closed in any topology. No topology has been introduced.","hfrak o_U(s)=0\\quad\\Longleftrightarrow\\quad\\Ext_U(s)\\ne\\varnothing.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\pi_U:R\\to R/B_U$ be the quotient homomorphism. By the definition of a quotient by a subgroup,\n\\[\n \\pi_U(r)=0\\quad\\Longleftrightarrow\\quad r\\in B_U.\n\\]\nHere $B_U$ is the \\emph{image} of $\\ker p_U$ under $\\delta$. Consequently membership of $\\delta\\widetilde s$ means the existence of a specific $h\\in E$ satisfying both $p_Uh=0$ and $\\delta h=\\delta\\widetilde s$. The subgroup need not be closed in any topology. No topology has been introduced.","eftrightarrow\\quad\\Ext_U(s)\\ne\\varnothing.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\pi_U:R\\to R/B_U$ be the quotient homomorphism. By the definition of a quotient by a subgroup,\n\\[\n \\pi_U(r)=0\\quad\\Longleftrightarrow\\quad r\\in B_U.\n\\]\nHere $B_U$ is the \\emph{image} of $\\ker p_U$ under $\\delta$. Consequently membership of $\\delta\\widetilde s$ means the existence of a specific $h\\in E$ satisfying both $p_Uh=0$ and $\\delta h=\\delta\\widetilde s$. The subgroup need not be closed in any topology. No topology has been introduced.","\\quad\\Ext_U(s)\\ne\\varnothing.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\pi_U:R\\to R/B_U$ be the quotient homomorphism. By the definition of a quotient by a subgroup,\n\\[\n \\pi_U(r)=0\\quad\\Longleftrightarrow\\quad r\\in B_U.\n\\]\nHere $B_U$ is the \\emph{image} of $\\ker p_U$ under $\\delta$. Consequently membership of $\\delta\\widetilde s$ means the existence of a specific $h\\in E$ satisfying both $p_Uh=0$ and $\\delta h=\\delta\\widetilde s$. The subgroup need not be closed in any topology. No topology has been introduced.","If such $h$ exists, set $x=\\widetilde s-h$. Additivity gives the two equations separately:\n\\[\n \\delta x=\\delta\\widetilde s-\\delta h=0,\n \\qquad p_Ux=p_U\\widetilde s-p_Uh=s.\n\\]\nThe first equation puts $x$ in $\\Gamma(A)$, and the second puts that coherent family in $\\Ext_U(s)$.","Conversely, take $x\\in\\Ext_U(s)$ and set $h=\\widetilde s-x$. Since both raw families restrict to $s$, one has $p_Uh=0$. Since $\\delta x=0$, one has $\\delta h=\\delta\\widetilde s$. This witnesses $\\delta\\widetilde s\\in B_U$, hence vanishing of its quotient class.","The proof is independent of the selected raw lift. Indeed, if $\\widetilde s'$ is another lift, $k=\\widetilde s'-\\widetilde s$ lies in $\\ker p_U$, and\n\\[\n \\delta\\widetilde s'-\\delta\\widetilde s=\\delta k\\in B_U.\n\\]\nThus the two quotient classes agree before either feasibility direction is invoked. A raw lift always exists in this additive model by using zero outside $U$. Constructing a raw lift does not solve the coherence equations.\n\\end{proof}\n\\source{KS09}\nThe quotient is constructed before the solvability test. The theorem therefore does not define obstruction circularly by saying that it vanishes when a solution exists. It also does not give a decision procedure for membership in an arbitrary subgroup of an infinite product.","\\subsection{Existence is not uniqueness}\nDefine the ambiguity group\n\\[\n H_U=\\ker\\delta\\cap\\ker p_U.\n\\]\nIt consists of coherent changes that are invisible on the prescribed domain.","\\begin{theorem}[Classification of a nonempty extension fiber]\\label{KS10}\nIf \\(x_0\\in\\Ext_U(s)\\), translation gives an equivalence\n\\[\n H_U\\simeq\\Ext_U(s),\\qquad h\\longmapsto x_0+h.\n\\]\nConsequently the extension is unique exactly when \\(\\mathfrak o_U(s)=0\\) and \\(H_U=0\\).\n\\end{theorem}\n\\begin{proof}\nDefine $\\Theta:H_U\\to\\Ext_U(s)$ by $\\Theta(h)=x_0+h$. It lands in the extension set because\n\\[\n \\delta(x_0+h)=\\delta x_0+\\delta h=0,\n \\qquad p_U(x_0+h)=s+0=s.\n\\]\nDefine $\\Psi:\\Ext_U(s)\\to H_U$ by $\\Psi(x)=x-x_0$. Its differential is zero, and its restriction is $s-s=0$, so the claimed target is correct. The formulas give\n\\[\n \\Psi\\Theta(h)=x_0+h-x_0=h,\\qquad\n \\Theta\\Psi(x)=x_0+(x-x_0)=x.\n\\]\nThis proves the equivalence rather than only a comparison of cardinalities.","10}\nIf \\(x_0\\in\\Ext_U(s)\\), translation gives an equivalence\n\\[\n H_U\\simeq\\Ext_U(s),\\qquad h\\longmapsto x_0+h.\n\\]\nConsequently the extension is unique exactly when \\(\\mathfrak o_U(s)=0\\) and \\(H_U=0\\).\n\\end{theorem}\n\\begin{proof}\nDefine $\\Theta:H_U\\to\\Ext_U(s)$ by $\\Theta(h)=x_0+h$. It lands in the extension set because\n\\[\n \\delta(x_0+h)=\\delta x_0+\\delta h=0,\n \\qquad p_U(x_0+h)=s+0=s.\n\\]\nDefine $\\Psi:\\Ext_U(s)\\to H_U$ by $\\Psi(x)=x-x_0$. Its differential is zero, and its restriction is $s-s=0$, so the claimed target is correct. The formulas give\n\\[\n \\Psi\\Theta(h)=x_0+h-x_0=h,\\qquad\n \\Theta\\Psi(x)=x_0+(x-x_0)=x.\n\\]\nThis proves the equivalence rather than only a comparison of cardinalities.","translation gives an equivalence\n\\[\n H_U\\simeq\\Ext_U(s),\\qquad h\\longmapsto x_0+h.\n\\]\nConsequently the extension is unique exactly when \\(\\mathfrak o_U(s)=0\\) and \\(H_U=0\\).\n\\end{theorem}\n\\begin{proof}\nDefine $\\Theta:H_U\\to\\Ext_U(s)$ by $\\Theta(h)=x_0+h$. It lands in the extension set because\n\\[\n \\delta(x_0+h)=\\delta x_0+\\delta h=0,\n \\qquad p_U(x_0+h)=s+0=s.\n\\]\nDefine $\\Psi:\\Ext_U(s)\\to H_U$ by $\\Psi(x)=x-x_0$. Its differential is zero, and its restriction is $s-s=0$, so the claimed target is correct. The formulas give\n\\[\n \\Psi\\Theta(h)=x_0+h-x_0=h,\\qquad\n \\Theta\\Psi(x)=x_0+(x-x_0)=x.\n\\]\nThis proves the equivalence rather than only a comparison of cardinalities.","longmapsto x_0+h.\n\\]\nConsequently the extension is unique exactly when \\(\\mathfrak o_U(s)=0\\) and \\(H_U=0\\).\n\\end{theorem}\n\\begin{proof}\nDefine $\\Theta:H_U\\to\\Ext_U(s)$ by $\\Theta(h)=x_0+h$. It lands in the extension set because\n\\[\n \\delta(x_0+h)=\\delta x_0+\\delta h=0,\n \\qquad p_U(x_0+h)=s+0=s.\n\\]\nDefine $\\Psi:\\Ext_U(s)\\to H_U$ by $\\Psi(x)=x-x_0$. Its differential is zero, and its restriction is $s-s=0$, so the claimed target is correct. The formulas give\n\\[\n \\Psi\\Theta(h)=x_0+h-x_0=h,\\qquad\n \\Theta\\Psi(x)=x_0+(x-x_0)=x.\n\\]\nThis proves the equivalence rather than only a comparison of cardinalities.","tly when \\(\\mathfrak o_U(s)=0\\) and \\(H_U=0\\).\n\\end{theorem}\n\\begin{proof}\nDefine $\\Theta:H_U\\to\\Ext_U(s)$ by $\\Theta(h)=x_0+h$. It lands in the extension set because\n\\[\n \\delta(x_0+h)=\\delta x_0+\\delta h=0,\n \\qquad p_U(x_0+h)=s+0=s.\n\\]\nDefine $\\Psi:\\Ext_U(s)\\to H_U$ by $\\Psi(x)=x-x_0$. Its differential is zero, and its restriction is $s-s=0$, so the claimed target is correct. The formulas give\n\\[\n \\Psi\\Theta(h)=x_0+h-x_0=h,\\qquad\n \\Theta\\Psi(x)=x_0+(x-x_0)=x.\n\\]\nThis proves the equivalence rather than only a comparison of cardinalities.","If $H_U=\\{0\\}$, the parameter of every extension is zero, so every extension equals $x_0$. Conversely, if $x_0$ is the only extension, then for any $h\\in H_U$ the equality $x_0+h=x_0$ forces $h=0$ by cancellation. Combining this observation with the existence criterion proves the stated necessary and sufficient condition for a unique extension.","The choice of $x_0$ only supplies an origin for the parameters. If $x_1=x_0+h_1$ is used instead, the parameter of $x$ changes from $x-x_0$ to $x-x_1=(x-x_0)-h_1$. The affine set of answers itself has not changed.\n\\end{proof}\n\\source{KS10}\nAn extension fiber is therefore an affine copy, or torsor, of the invisible coherent group. Selecting one solution gives coordinates on the fiber. Without a selected solution, the affine structure does not designate an origin.","\\begin{example}[Parity and an independent ambiguity]\nFor two views with \\(f^*(m)=2m\\) on \\(\\Z\\), a prescribed value \\(k\\) at the coarse view extends exactly when \\(k\\) is even. The extension is \\((k,k/2)\\) and is unique. In particular \\(1\\) is obstructed and \\(2\\) extends nontrivially. The parity statement is an instance of Theorem~\\ref{KS09}. If a separate, disconnected integer view is added and left unprescribed, every extendable seed has infinitely many completions. This last variant is a direct application of Theorem~\\ref{KS10}, not a claim that the parity obstruction has changed.\n\\end{example}\n\\source{KS11}","\\subsection{The exact sequence behind the criterion}\nThe additive structures make restriction a homomorphism\n\\(r_U:\\Gamma(A)\\to\\Gamma_U(A)\\). Using the zero extension, or lift independence, \\(\\mathfrak o_U\\) is also a homomorphism.","\\begin{proposition}[Exactness at partial sections]\\label{KS12}\nThe sequence\n\\[\n \\Gamma(A)\\xrightarrow{r_U}\\Gamma_U(A)\n \\xrightarrow{\\mathfrak o_U}Q_U\n\\]\nis exact at \\(\\Gamma_U(A)\\). Thus \\(\\ker\\mathfrak o_U=\\im r_U\\), and restriction is onto precisely when \\(\\mathfrak o_U\\) is the zero map.\n\\end{proposition}\n\\begin{proof}\nFirst verify that $\\mathfrak o_U$ is a homomorphism, since exactness is being asserted in an additive setting. Given $s,t\\in\\Gamma_U(A)$, choose raw lifts $\\widetilde s,\\widetilde t$. Their sum lifts $s+t$. Lift independence and additivity imply\n\\[\n \\mathfrak o_U(s+t)\n =[\\delta(\\widetilde s+\\widetilde t)]\n =[\\delta\\widetilde s]+[\\delta\\widetilde t]\n =\\mathfrak o_U(s)+\\mathfrak o_U(t).\n\\]\nThe zero lift gives $\\mathfrak o_U(0)=0$. Restriction $r_U$ is a homomorphism because addition of global sections is coordinatewise.","ections]\\label{KS12}\nThe sequence\n\\[\n \\Gamma(A)\\xrightarrow{r_U}\\Gamma_U(A)\n \\xrightarrow{\\mathfrak o_U}Q_U\n\\]\nis exact at \\(\\Gamma_U(A)\\). Thus \\(\\ker\\mathfrak o_U=\\im r_U\\), and restriction is onto precisely when \\(\\mathfrak o_U\\) is the zero map.\n\\end{proposition}\n\\begin{proof}\nFirst verify that $\\mathfrak o_U$ is a homomorphism, since exactness is being asserted in an additive setting. Given $s,t\\in\\Gamma_U(A)$, choose raw lifts $\\widetilde s,\\widetilde t$. Their sum lifts $s+t$. Lift independence and additivity imply\n\\[\n \\mathfrak o_U(s+t)\n =[\\delta(\\widetilde s+\\widetilde t)]\n =[\\delta\\widetilde s]+[\\delta\\widetilde t]\n =\\mathfrak o_U(s)+\\mathfrak o_U(t).\n\\]\nThe zero lift gives $\\mathfrak o_U(0)=0$. Restriction $r_U$ is a homomorphism because addition of global sections is coordinatewise.","\\Gamma(A)\\xrightarrow{r_U}\\Gamma_U(A)\n \\xrightarrow{\\mathfrak o_U}Q_U\n\\]\nis exact at \\(\\Gamma_U(A)\\). Thus \\(\\ker\\mathfrak o_U=\\im r_U\\), and restriction is onto precisely when \\(\\mathfrak o_U\\) is the zero map.\n\\end{proposition}\n\\begin{proof}\nFirst verify that $\\mathfrak o_U$ is a homomorphism, since exactness is being asserted in an additive setting. Given $s,t\\in\\Gamma_U(A)$, choose raw lifts $\\widetilde s,\\widetilde t$. Their sum lifts $s+t$. Lift independence and additivity imply\n\\[\n \\mathfrak o_U(s+t)\n =[\\delta(\\widetilde s+\\widetilde t)]\n =[\\delta\\widetilde s]+[\\delta\\widetilde t]\n =\\mathfrak o_U(s)+\\mathfrak o_U(t).\n\\]\nThe zero lift gives $\\mathfrak o_U(0)=0$. Restriction $r_U$ is a homomorphism because addition of global sections is coordinatewise.","tarrow{\\mathfrak o_U}Q_U\n\\]\nis exact at \\(\\Gamma_U(A)\\). Thus \\(\\ker\\mathfrak o_U=\\im r_U\\), and restriction is onto precisely when \\(\\mathfrak o_U\\) is the zero map.\n\\end{proposition}\n\\begin{proof}\nFirst verify that $\\mathfrak o_U$ is a homomorphism, since exactness is being asserted in an additive setting. Given $s,t\\in\\Gamma_U(A)$, choose raw lifts $\\widetilde s,\\widetilde t$. Their sum lifts $s+t$. Lift independence and additivity imply\n\\[\n \\mathfrak o_U(s+t)\n =[\\delta(\\widetilde s+\\widetilde t)]\n =[\\delta\\widetilde s]+[\\delta\\widetilde t]\n =\\mathfrak o_U(s)+\\mathfrak o_U(t).\n\\]\nThe zero lift gives $\\mathfrak o_U(0)=0$. Restriction $r_U$ is a homomorphism because addition of global sections is coordinatewise.","a_U(A)\\). Thus \\(\\ker\\mathfrak o_U=\\im r_U\\), and restriction is onto precisely when \\(\\mathfrak o_U\\) is the zero map.\n\\end{proposition}\n\\begin{proof}\nFirst verify that $\\mathfrak o_U$ is a homomorphism, since exactness is being asserted in an additive setting. Given $s,t\\in\\Gamma_U(A)$, choose raw lifts $\\widetilde s,\\widetilde t$. Their sum lifts $s+t$. Lift independence and additivity imply\n\\[\n \\mathfrak o_U(s+t)\n =[\\delta(\\widetilde s+\\widetilde t)]\n =[\\delta\\widetilde s]+[\\delta\\widetilde t]\n =\\mathfrak o_U(s)+\\mathfrak o_U(t).\n\\]\nThe zero lift gives $\\mathfrak o_U(0)=0$. Restriction $r_U$ is a homomorphism because addition of global sections is coordinatewise.","that $\\mathfrak o_U$ is a homomorphism, since exactness is being asserted in an additive setting. Given $s,t\\in\\Gamma_U(A)$, choose raw lifts $\\widetilde s,\\widetilde t$. Their sum lifts $s+t$. Lift independence and additivity imply\n\\[\n \\mathfrak o_U(s+t)\n =[\\delta(\\widetilde s+\\widetilde t)]\n =[\\delta\\widetilde s]+[\\delta\\widetilde t]\n =\\mathfrak o_U(s)+\\mathfrak o_U(t).\n\\]\nThe zero lift gives $\\mathfrak o_U(0)=0$. Restriction $r_U$ is a homomorphism because addition of global sections is coordinatewise.","phism, since exactness is being asserted in an additive setting. Given $s,t\\in\\Gamma_U(A)$, choose raw lifts $\\widetilde s,\\widetilde t$. Their sum lifts $s+t$. Lift independence and additivity imply\n\\[\n \\mathfrak o_U(s+t)\n =[\\delta(\\widetilde s+\\widetilde t)]\n =[\\delta\\widetilde s]+[\\delta\\widetilde t]\n =\\mathfrak o_U(s)+\\mathfrak o_U(t).\n\\]\nThe zero lift gives $\\mathfrak o_U(0)=0$. Restriction $r_U$ is a homomorphism because addition of global sections is coordinatewise.","If $s=r_Ux$ for a global section $x$, use $x$ itself as a raw lift. Its differential vanishes, so $\\mathfrak o_U(s)=0$. This proves $\\im r_U\\subseteq\\ker\\mathfrak o_U$. If $s\\in\\ker\\mathfrak o_U$, the extension criterion constructs a global $x$ restricting to $s$, proving the reverse inclusion.","Finally, $r_U$ is surjective exactly when its image is all of $\\Gamma_U(A)$. By the equality just proved, this is equivalent to every partial section having zero obstruction, which says that $\\mathfrak o_U$ is the zero homomorphism. Nothing here claims that its image is all of $Q_U$.\n\\end{proof}\n\\source{KS12}\nNo surjectivity onto the entire residual quotient is asserted. The residual group contains many entries that need not arise as the differential of a partial lift.","\\subsection{Extension in stages}\nFor \\(U\\subseteq W\\subseteq\\Ob\\V\\), an extension first to \\(W\\) and then globally must retain the intermediate choice. There is an explicit equivalence\n\\begin{equation}\\label{eq:stages}\n \\Ext_U(s)\\simeq\n \\coprod_{\\substack{t\\in\\Gamma_W(A)","\\subsection{Extension in stages}\nFor \\(U\\subseteq W\\subseteq\\Ob\\V\\), an extension first to \\(W\\) and then globally must retain the intermediate choice. There is an explicit equivalence\n\\begin{equation}\\label{eq:stages}\n \\Ext_U(s)\\simeq\n \\coprod_{\\substack{t\\in\\Gamma_W(A)\\\\t|_U=s}}\\Ext_W(t).\n\\end{equation}\nIt sends a global extension to its restriction to \\(W\\), together with itself as an extension of that restriction. The inverse forgets the intermediate coordinate. These operations are inverse because the required intermediate coordinate of a global section is its actual restriction. A coherent intermediate choice can still be obstructed, even when a different intermediate choice completes. Equation~\\eqref{eq:stages} is why extension by stages is a dependent sum rather than a statement that every branch succeeds.\n\\source{KS13}","t|_U=s}}\\Ext_W(t).\n\\end{equation}\nIt sends a global extension to its restriction to \\(W\\), together with itself as an extension of that restriction. The inverse forgets the intermediate coordinate. These operations are inverse because the required intermediate coordinate of a global section is its actual restriction. A coherent intermediate choice can still be obstructed, even when a different intermediate choice completes. Equation~\\eqref{eq:stages} is why extension by stages is a dependent sum rather than a statement that every branch succeeds.\n\\source{KS13}","\\chap{Protected correction as a general principle}\\label{sec:protected}\nThe additive extension theorem is a particular instance of a useful general construction. Let \\(E,R,S\\) be abelian groups, with additive maps \\(d:E\\to R\\) and \\(p:E\\to S\\). The residual \\(d\\) measures compatibility, and \\(p\\) specifies the information that repair must preserve.","\\begin{theorem}[Semantics preserving correction]\\label{KS14}\nFor \\(x\\in E\\), the following conditions are equivalent:\n\\[\n\\begin{aligned}\n &[dx]=0\\text{ in }R/d(\\ker p),","\\begin{theorem}[Semantics preserving correction]\\label{KS14}\nFor \\(x\\in E\\), the following conditions are equivalent:\n\\[\n\\begin{aligned}\n &[dx]=0\\text{ in }R/d(\\ker p),\\\\\n &\\exists h\\in E:\\ p(h)=0,\\quad d(h)=d(x),\\\\\n &\\exists y\\in E:\\ d(y)=0,\\quad p(y)=p(x).\n\\end{aligned}\n\\]\nThe class of \\(dx\\) is unchanged when an admissible correction is subtracted from \\(x\\).\n\\end{theorem}\n\\begin{proof}\nThe image $d(\\ker p)$ is a subgroup of $R$. The defining property of the quotient says\n\\[\n [dx]=0\\quad\\Longleftrightarrow\\quad\n dx\\in d(\\ker p)\n \\quad\\Longleftrightarrow\\quad\n \\exists h\\in E:\\ ph=0,\\ dh=dx.\n\\]\nThis is the first equivalence with its witness made explicit.","&\\exists y\\in E:\\ d(y)=0,\\quad p(y)=p(x).\n\\end{aligned}\n\\]\nThe class of \\(dx\\) is unchanged when an admissible correction is subtracted from \\(x\\).\n\\end{theorem}\n\\begin{proof}\nThe image $d(\\ker p)$ is a subgroup of $R$. The defining property of the quotient says\n\\[\n [dx]=0\\quad\\Longleftrightarrow\\quad\n dx\\in d(\\ker p)\n \\quad\\Longleftrightarrow\\quad\n \\exists h\\in E:\\ ph=0,\\ dh=dx.\n\\]\nThis is the first equivalence with its witness made explicit.","For the second, a permitted $h$ gives $y=x-h$. Then\n\\[\n dy=dx-dh=0,\\qquad py=px-ph=px.\n\\]\nConversely an exact $y$ with $py=px$ gives $h=x-y$, which satisfies $ph=0$ and $dh=dx$. The two witness transformations are inverse: subtracting $h$ from $x$ and then subtracting that answer from $x$ recovers $h$, and conversely.","To verify invariance, fix any $h$ with $ph=0$. Then $d(x-h)=dx-dh$, and $dh$ lies in the subgroup divided out. Therefore $[d(x-h)]=[dx]$. This does not assert that the residual itself is unchanged. It asserts that permitted changes can alter the residual only within its already correctable coset.\n\\end{proof}\n\\source{KS14}\nHere \\emph{semantics} means the declared mathematical observable \\(p\\). It is not an assertion that one homomorphism captures every meaning of every object. In a tower, \\(p\\) might preserve a seed operator and every new output component. In a graph, it might preserve designated vertex values. Different choices give different problems.","up divided out. Therefore $[d(x-h)]=[dx]$. This does not assert that the residual itself is unchanged. It asserts that permitted changes can alter the residual only within its already correctable coset.\n\\end{proof}\n\\source{KS14}\nHere \\emph{semantics} means the declared mathematical observable \\(p\\). It is not an assertion that one homomorphism captures every meaning of every object. In a tower, \\(p\\) might preserve a seed operator and every new output component. In a graph, it might preserve designated vertex values. Different choices give different problems.","\\begin{proposition}[A certificate of impossibility]\\label{KS15}\nLet \\(L:R\\to D\\) be additive. If \\(L\\) vanishes on \\(d(\\ker p)\\) but \\(L(dx)\\ne0\\), no exact \\(y\\) can preserve \\(p(x)\\).\n\\end{proposition}\n\\begin{proof}\nSuppose, for contradiction, that $dy=0$ and $py=px$. Then $h=x-y$ satisfies $ph=0$ and $dh=dx$. By the annihilation hypothesis,\n\\[\n L(dx)=L(dh)=0,\n\\]\ncontrary to the assumed nonzero value. This already proves impossibility directly.","Equivalently, if $r-r'\\in d(\\ker p)$, then $L(r-r')=0$, so $L(r)=L(r')$. Hence $\\overline L([r])=L(r)$ is a well-defined homomorphism on $R/d(\\ker p)$. Its value on $[dx]$ is nonzero. This supplies the quotient interpretation of the same contradiction, without assuming that every nonzero quotient class can be detected by a map into one fixed preferred coefficient group.\n\\end{proof}\n\\source{KS15}\nThis is often the most economical negative proof. One need not classify every possible correction. It is enough to find an invariant annihilating all admissible corrections but not the observed defect.","\\subsection{Why unrestricted cancellation is empty information}\nIf every correction is permitted, taking \\(h=x\\) leaves the exact candidate zero. Thus \\(\\exists h:d(x-h)=0\\) is automatic. It says nothing about recovering an object of interest. A repair theorem becomes informative only when it preserves something, is quantitatively controlled, or satisfies a comparison property that excludes arbitrary replacement.","\\begin{example}[A rigid obstruction]\nTake \\(E=R=S=\\Z\\), \\(d=p=\\id\\), and \\(x=1\\). The only admissible correction is zero, while the residual is nonzero. Hence there is no exact answer preserving the observed value. This is not a failure of the formalism. It is the correct certificate for an incompatible demand.\n\\end{example}","\\subsection{What an obstruction does not yet provide}\nAn algebraic quotient does not automatically carry a useful norm. A zero infimum of distances to its boundary subgroup does not imply membership unless closure is controlled. Nor does vanishing supply a preferred correction when \\(\\ker d\\cap\\ker p\\) is nontrivial. These two issues lead to different parts of the book: ambiguity is treated algebraically and through groupoids, while closed range and condition numbers enter the analytic theory.","\\begin{insight}\nA repair problem is not determined by its defect alone. It is determined by the defect \\emph{and} the admissible changes. Keeping the pair \\((d,p)\\) explicit prevents a successful cancellation from being mistaken for a successful reconstruction.\n\\end{insight}","\\subsection{Three maps that must not be confused}\nKeep the additive view system of Section~\\ref{sec:obstruction}. For a prescribed domain \\(U\\), there are three distinct maps:\n\\[\n p_U:E\\longrightarrow\\prod_{u\\in U}A_u,\\qquad\n r_U:\\Gamma(A)\\longrightarrow\\Gamma_U(A),\\qquad\n \\mathfrak o_U:\\Gamma_U(A)\\longrightarrow Q_U.\n\\]\nThe first restricts a raw family, whether or not it is coherent. The second restricts a global coherent family. The third detects whether an already coherent partial family has a global completion. Their domains are different. In particular, surjectivity of the raw coordinate projection does not imply surjectivity of \\(r_U\\).","rightarrow\\Gamma_U(A),\\qquad\n \\mathfrak o_U:\\Gamma_U(A)\\longrightarrow Q_U.\n\\]\nThe first restricts a raw family, whether or not it is coherent. The second restricts a global coherent family. The third detects whether an already coherent partial family has a global completion. Their domains are different. In particular, surjectivity of the raw coordinate projection does not imply surjectivity of \\(r_U\\).","The obstruction group is \\(Q_U=R/B_U\\), where \\(B_U=\\delta(\\ker p_U)\\). The ambiguity group is \\(H_U=\\ker\\delta\\cap\\ker p_U\\). Both depend on what has been fixed. Enlarging the prescription reduces the permitted correction space and the invisible exact space at the same time.","\\subsection{Forgetting a prescription gives a quotient map}\nLet \\(U\\subseteq W\\). Then\n\\[\n \\ker p_W\\subseteq\\ker p_U,\\qquad B_W\\subseteq B_U,\n \\qquad H_W\\subseteq H_U.\n\\]\nA correction invisible on all of \\(W\\) is certainly invisible on \\(U\\). The direction of the quotient map is consequently\n\\[\n \\pi_{W,U}:Q_W\\longrightarrow Q_U,\\qquad\n [r]_{B_W}\\longmapsto[r]_{B_U}.\n\\]\nThe larger prescribed domain appears in the source. More protection means a smaller subgroup is divided out, so more obstruction information can remain.","\\begin{theorem}[Functorial forgetting of protected views]\\label{KS59}\nThe homomorphism \\(\\pi_{W,U}\\) is well defined and surjective. For \\(U\\subseteq W\\subseteq Z\\), it satisfies\n\\[\n \\pi_{U,U}=I,\\qquad \\pi_{W,U}\\pi_{Z,W}=\\pi_{Z,U}.\n\\]\nFor every \\(s\\in\\Gamma_W(A)\\),\n\\begin{equation}\\label{eq:prescribed-naturality}\n \\pi_{W,U}\\mathfrak o_W(s)=\\mathfrak o_U(s|_U).\n\\end{equation}\nIn particular an obstructed restriction remains obstructed when more compatible prescribed data are retained. A successful extension on the larger prescribed domain gives a successful extension of its restriction.\n\\end{theorem}\n\\begin{proof}\nIf \\(r-r'\\in B_W\\), then \\(r-r'\\in B_U\\), proving well-definedness. Every class modulo \\(B_U\\) has a representative \\(r\\), and the class of that same representative modulo \\(B_W\\","\\begin{theorem}[Functorial forgetting of protected views]\\label{KS59}\nThe homomorphism \\(\\pi_{W,U}\\) is well defined and surjective. For \\(U\\subseteq W\\subseteq Z\\), it satisfies\n\\[\n \\pi_{U,U}=I,\\qquad \\pi_{W,U}\\pi_{Z,W}=\\pi_{Z,U}.\n\\]\nFor every \\(s\\in\\Gamma_W(A)\\),\n\\begin{equation}\\label{eq:prescribed-naturality}\n \\pi_{W,U}\\mathfrak o_W(s)=\\mathfrak o_U(s|_U).\n\\end{equation}\nIn particular an obstructed restriction remains obstructed when more compatible prescribed data are retained. A successful extension on the larger prescribed domain gives a successful extension of its restriction.\n\\end{theorem}\n\\begin{proof}\nIf \\(r-r'\\in B_W\\), then \\(r-r'\\in B_U\\), proving well-definedness. Every class modulo \\(B_U\\) has a representative \\(r\\), and the class of that same representative modulo \\(B_W\\) maps to it. This proves surjectivity. Identity and composition","\\begin{theorem}[Functorial forgetting of protected views]\\label{KS59}\nThe homomorphism \\(\\pi_{W,U}\\) is well defined and surjective. For \\(U\\subseteq W\\subseteq Z\\), it satisfies\n\\[\n \\pi_{U,U}=I,\\qquad \\pi_{W,U}\\pi_{Z,W}=\\pi_{Z,U}.\n\\]\nFor every \\(s\\in\\Gamma_W(A)\\),\n\\begin{equation}\\label{eq:prescribed-naturality}\n \\pi_{W,U}\\mathfrak o_W(s)=\\mathfrak o_U(s|_U).\n\\end{equation}\nIn particular an obstructed restriction remains obstructed when more compatible prescribed data are retained. A successful extension on the larger prescribed domain gives a successful extension of its restriction.\n\\end{theorem}\n\\begin{proof}\nIf \\(r-r'\\in B_W\\), then \\(r-r'\\in B_U\\), proving well-definedness. Every class modulo \\(B_U\\) has a representative \\(r\\), and the class of that same representative modulo \\(B_W\\) maps to it. This proves surjectivity. Identity and composition follow by applying the formulas to a representative.","{Z,U}.\n\\]\nFor every \\(s\\in\\Gamma_W(A)\\),\n\\begin{equation}\\label{eq:prescribed-naturality}\n \\pi_{W,U}\\mathfrak o_W(s)=\\mathfrak o_U(s|_U).\n\\end{equation}\nIn particular an obstructed restriction remains obstructed when more compatible prescribed data are retained. A successful extension on the larger prescribed domain gives a successful extension of its restriction.\n\\end{theorem}\n\\begin{proof}\nIf \\(r-r'\\in B_W\\), then \\(r-r'\\in B_U\\), proving well-definedness. Every class modulo \\(B_U\\) has a representative \\(r\\), and the class of that same representative modulo \\(B_W\\) maps to it. This proves surjectivity. Identity and composition follow by applying the formulas to a representative.","s\\in\\Gamma_W(A)\\),\n\\begin{equation}\\label{eq:prescribed-naturality}\n \\pi_{W,U}\\mathfrak o_W(s)=\\mathfrak o_U(s|_U).\n\\end{equation}\nIn particular an obstructed restriction remains obstructed when more compatible prescribed data are retained. A successful extension on the larger prescribed domain gives a successful extension of its restriction.\n\\end{theorem}\n\\begin{proof}\nIf \\(r-r'\\in B_W\\), then \\(r-r'\\in B_U\\), proving well-definedness. Every class modulo \\(B_U\\) has a representative \\(r\\), and the class of that same representative modulo \\(B_W\\) maps to it. This proves surjectivity. Identity and composition follow by applying the formulas to a representative.","lity}\n \\pi_{W,U}\\mathfrak o_W(s)=\\mathfrak o_U(s|_U).\n\\end{equation}\nIn particular an obstructed restriction remains obstructed when more compatible prescribed data are retained. A successful extension on the larger prescribed domain gives a successful extension of its restriction.\n\\end{theorem}\n\\begin{proof}\nIf \\(r-r'\\in B_W\\), then \\(r-r'\\in B_U\\), proving well-definedness. Every class modulo \\(B_U\\) has a representative \\(r\\), and the class of that same representative modulo \\(B_W\\) maps to it. This proves surjectivity. Identity and composition follow by applying the formulas to a representative.","(s)=\\mathfrak o_U(s|_U).\n\\end{equation}\nIn particular an obstructed restriction remains obstructed when more compatible prescribed data are retained. A successful extension on the larger prescribed domain gives a successful extension of its restriction.\n\\end{theorem}\n\\begin{proof}\nIf \\(r-r'\\in B_W\\), then \\(r-r'\\in B_U\\), proving well-definedness. Every class modulo \\(B_U\\) has a representative \\(r\\), and the class of that same representative modulo \\(B_W\\) maps to it. This proves surjectivity. Identity and composition follow by applying the formulas to a representative.","obstructed restriction remains obstructed when more compatible prescribed data are retained. A successful extension on the larger prescribed domain gives a successful extension of its restriction.\n\\end{theorem}\n\\begin{proof}\nIf \\(r-r'\\in B_W\\), then \\(r-r'\\in B_U\\), proving well-definedness. Every class modulo \\(B_U\\) has a representative \\(r\\), and the class of that same representative modulo \\(B_W\\) maps to it. This proves surjectivity. Identity and composition follow by applying the formulas to a representative.","Choose a raw extension \\(\\widetilde s\\) of the partial section on \\(W\\). It is also a raw extension of \\(s|_U\\). Lift independence of the two extension obstructions therefore identifies both sides of \\eqref{eq:prescribed-naturality} with the class of \\(\\delta\\widetilde s\\) modulo \\(B_U\\). A homomorphism sends zero to zero. Thus nonvanishing on the smaller domain rules out vanishing on the larger one, while vanishing on the larger domain implies vanishing after restriction.\n\\end{proof}\n\\source{KS59}\nThe converse implication does not hold. The quotient map can annihilate the obstruction created by a newly fixed value. Surjectivity of \\(\\pi_{W,U}\\) is also not a theorem that arbitrary larger partial sections extend. It is a statem","Choose a raw extension \\(\\widetilde s\\) of the partial section on \\(W\\). It is also a raw extension of \\(s|_U\\). Lift independence of the two extension obstructions therefore identifies both sides of \\eqref{eq:prescribed-naturality} with the class of \\(\\delta\\widetilde s\\) modulo \\(B_U\\). A homomorphism sends zero to zero. Thus nonvanishing on the smaller domain rules out vanishing on the larger one, while vanishing on the larger domain implies vanishing after restriction.\n\\end{proof}\n\\source{KS59}\nThe converse implication does not hold. The quotient map can annihilate the obstruction created by a newly fixed value. Surjectivity of \\(\\pi_{W,U}\\) is also not a theorem that arbitrary larger partial sections extend. It is a statement about representatives in two","Choose a raw extension \\(\\widetilde s\\) of the partial section on \\(W\\). It is also a raw extension of \\(s|_U\\). Lift independence of the two extension obstructions therefore identifies both sides of \\eqref{eq:prescribed-naturality} with the class of \\(\\delta\\widetilde s\\) modulo \\(B_U\\). A homomorphism sends zero to zero. Thus nonvanishing on the smaller domain rules out vanishing on the larger one, while vanishing on the larger domain implies vanishing after restriction.\n\\end{proof}\n\\source{KS59}\nThe converse implication does not hold. The quotient map can annihilate the obstruction created by a newly fixed value. Surjectivity of \\(\\pi_{W,U}\\) is also not a theorem that arbitrary larger partial sections extend. It is a statement about representatives in two obstruction groups, not about the image","Choose a raw extension \\(\\widetilde s\\) of the partial section on \\(W\\). It is also a raw extension of \\(s|_U\\). Lift independence of the two extension obstructions therefore identifies both sides of \\eqref{eq:prescribed-naturality} with the class of \\(\\delta\\widetilde s\\) modulo \\(B_U\\). A homomorphism sends zero to zero. Thus nonvanishing on the smaller domain rules out vanishing on the larger one, while vanishing on the larger domain implies vanishing after restriction.\n\\end{proof}\n\\source{KS59}\nThe converse implication does not hold. The quotient map can annihilate the obstruction created by a newly fixed value. Surjectivity of \\(\\pi_{W,U}\\) is also not a theorem that arbitrary larger partial sections extend. It is a statement about representatives in two obstruction groups, not about the image of restriction from global objects.","class of \\(\\delta\\widetilde s\\) modulo \\(B_U\\). A homomorphism sends zero to zero. Thus nonvanishing on the smaller domain rules out vanishing on the larger one, while vanishing on the larger domain implies vanishing after restriction.\n\\end{proof}\n\\source{KS59}\nThe converse implication does not hold. The quotient map can annihilate the obstruction created by a newly fixed value. Surjectivity of \\(\\pi_{W,U}\\) is also not a theorem that arbitrary larger partial sections extend. It is a statement about representatives in two obstruction groups, not about the image of restriction from global objects.","\\subsection{An initially free parameter can become an obstruction}\nConsider the chain of views \\(0\\to1\\to2\\), with fiber \\(\\Z^2\\) at each view. Write the maps in the direction in which data are restricted:\n\\[\n \\rho_{0,1}(a,b)=(a,0),\\qquad\n \\rho_{1,2}(a,b)=(a,2b),\\qquad\n \\rho_{0,2}(a,b)=(a,0).\n\\]\nThe composite identity \\(\\rho_{0,1}\\rho_{1,2}=\\rho_{0,2}\\) is literal. A global family is exactly\n\\begin{equation}\\label{eq:branch-family}\n x_0=(k,0),\\qquad x_1=(k,2n),\\qquad x_2=(k,n),\n \\qquad k,n\\in\\Z.\n\\end{equation}\nIndeed the first coordinate must agree at every view. The second coordinate at view zero is forced to vanish, and the second coordinate at view one must be twice the second coordinate at view two. These conditions also suffice.","{An initially free parameter can become an obstruction}\nConsider the chain of views \\(0\\to1\\to2\\), with fiber \\(\\Z^2\\) at each view. Write the maps in the direction in which data are restricted:\n\\[\n \\rho_{0,1}(a,b)=(a,0),\\qquad\n \\rho_{1,2}(a,b)=(a,2b),\\qquad\n \\rho_{0,2}(a,b)=(a,0).\n\\]\nThe composite identity \\(\\rho_{0,1}\\rho_{1,2}=\\rho_{0,2}\\) is literal. A global family is exactly\n\\begin{equation}\\label{eq:branch-family}\n x_0=(k,0),\\qquad x_1=(k,2n),\\qquad x_2=(k,n),\n \\qquad k,n\\in\\Z.\n\\end{equation}\nIndeed the first coordinate must agree at every view. The second coordinate at view zero is forced to vanish, and the second coordinate at view one must be twice the second coordinate at view two. These conditions also suffice.","First prescribe only \\(U=\\{0\\}\\), with seed \\((k,0)\\). Every integer \\(n\\) in \\eqref{eq:branch-family} gives a completion. Now prescribe \\(W=\\{0,1\\}\\), with values \\((k,0)\\) and \\((k,m)\\). This is a coherent partial section for every integer \\(m\\), because \\(\\rho_{0,1}\\) forgets its second coordinate. Global extendability is a stronger requirement: \\(m\\) must be even.","\\begin{proposition}[A coherent branch can be a dead end]\\label{KS60}\nEvery initial seed \\((k,0)\\) in this model has a completion set in bijection with \\(\\Z\\). A fixed intermediate value \\((k,m)\\) extends precisely when \\(m\\in2\\Z\\), and then has a unique completion. In particular the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes fo","\\begin{proposition}[A coherent branch can be a dead end]\\label{KS60}\nEvery initial seed \\((k,0)\\) in this model has a completion set in bijection with \\(\\Z\\). A fixed intermediate value \\((k,m)\\) extends precisely when \\(m\\in2\\Z\\), and then has a unique completion. In particular the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qqu","\\begin{proposition}[A coherent branch can be a dead end]\\label{KS60}\nEvery initial seed \\((k,0)\\) in this model has a completion set in bijection with \\(\\Z\\). A fixed intermediate value \\((k,m)\\) extends precisely when \\(m\\in2\\Z\\), and then has a unique completion. In particular the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Re","\\begin{proposition}[A coherent branch can be a dead end]\\label{KS60}\nEvery initial seed \\((k,0)\\) in this model has a completion set in bijection with \\(\\Z\\). A fixed intermediate value \\((k,m)\\) extends precisely when \\(m\\in2\\Z\\), and then has a unique completion. In particular the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view o","\\begin{proposition}[A coherent branch can be a dead end]\\label{KS60}\nEvery initial seed \\((k,0)\\) in this model has a completion set in bijection with \\(\\Z\\). A fixed intermediate value \\((k,m)\\) extends precisely when \\(m\\in2\\Z\\), and then has a unique completion. In particular the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambigu","\\begin{proposition}[A coherent branch can be a dead end]\\label{KS60}\nEvery initial seed \\((k,0)\\) in this model has a completion set in bijection with \\(\\Z\\). A fixed intermediate value \\((k,m)\\) extends precisely when \\(m\\in2\\Z\\), and then has a unique completion. In particular the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuc","seed \\((k,0)\\) in this model has a completion set in bijection with \\(\\Z\\). A fixed intermediate value \\((k,m)\\) extends precisely when \\(m\\in2\\Z\\), and then has a unique completion. In particular the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","completion set in bijection with \\(\\Z\\). A fixed intermediate value \\((k,m)\\) extends precisely when \\(m\\in2\\Z\\), and then has a unique completion. In particular the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","(k,m)\\) extends precisely when \\(m\\in2\\Z\\), and then has a unique completion. In particular the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","a unique completion. In particular the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","lar the nonzero initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","o initial seed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","ed \\((7,0)\\) extends, while the coherent intermediate choice \\((7,1)\\) does not.\n\\end{proposition}\n\\begin{proof}\nEquation~\\eqref{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","ef{eq:branch-family} supplies a completion for each \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","ach \\(n\\), and the second coordinate of \\(x_2\\) recovers the parameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","ameter. Fixing the intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","he intermediate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","ate value imposes \\(2n=m\\). This equation has no integer solution for odd \\(m\\), and exactly one for even \\(m\\). Taking \\(k=7\\) and \\(m=1\\) proves the final assertion.\n\\end{proof}\n\\source{KS60}\nThe invisible coherent changes for the initial prescription are\n\\[\n ((0,0),(0,2n),(0,n)),\\qquad n\\in\\Z.\n\\]\nThus \\(H_U\\simeq\\Z\\). Requiring invisibility also at view one forces \\(2n=0\\), hence \\(n=0\\), so \\(H_W=0\\). More observation removes all ambiguity in the successful branches. It does not turn the unsuccessful branches into solutions.","\\subsection{Why the extension space is a dependent sum}\nLet \\(s\\) be the initial partial section. The stage theorem gives\n\\[\n \\Ext_U(s)\\simeq\n \\coprod_{\\substack{t\\in\\Gamma_W(A)","\\subsection{Why the extension space is a dependent sum}\nLet \\(s\\) be the initial partial section. The stage theorem gives\n\\[\n \\Ext_U(s)\\simeq\n \\coprod_{\\substack{t\\in\\Gamma_W(A)\\\\t|_U=s}}\\Ext_W(t).\n\\]\nIn the present example the index \\(t\\) is the integer \\(m\\). The summand is empty for odd \\(m\\) and a singleton for even \\(m\\). Taking their disjoint union gives one integer's worth of completions again. No contradiction appears between many initial completions and unique completion after a successful intermediate choice.","t|_U=s}}\\Ext_W(t).\n\\]\nIn the present example the index \\(t\\) is the integer \\(m\\). The summand is empty for odd \\(m\\) and a singleton for even \\(m\\). Taking their disjoint union gives one integer's worth of completions again. No contradiction appears between many initial completions and unique completion after a successful intermediate choice.","\\subsection{Changing the coefficients changes both questions}\nThe equations behind the example can be read over a general abelian coefficient group \\(T\\). With the initial first coordinate fixed, the possible completions are parametrized by \\(n\\in T\\). Prescribing the middle value \\(m\\) gives the equation \\(2n=m\\). Direct substitution therefore gives\n\\[\n \\text{feasible middle values}=2T,\\qquad\n \\text{ambiguity after prescribing the middle}=\\ker(2:T\\to T).\n\\]\nThis calculation specializes as follows.\n\\begin{center}\n\\begin{tabular}{@{}lll@{}}\n\\toprule\nCoefficient group & Which middle values extend? & Completions of an extendable value","\\midrule\n\\(\\Z\\) & Even integers & One","\\(\\Q\\) & Every rational & One","\\(\\mathbb F_2\\) & Only zero & Two","\\(\\mathbb F_3\\) & Every element & One","A solver that changes the coefficient domain while leaving its user-facing description untouched changes the mathematical problem. The rational answer \\(1/2\\) is not an integral completion of the odd middle value one. Conversely, a failure over \\(\\Z\\) is not evidence of a failure over \\(\\Q\\). The coefficient structure belongs beside the views, restrictions, and preserved values in the problem specification.","Resolution, repair, and retained detail}{Resolution gives a concrete family of views. Its compatibility differential connects exact maps, protected information, and residual coordinates. Recovering every finite coordinate is then separated from convergence in a chosen norm.}\n\\section{Resolution towers and exact transport}\\label{sec:towers}\n\\subsection{The sequential model}\nA tower in a category \\(\\mathcal C\\) is a sequence\n\\[\n \\cdots\\xrightarrow{\\rho_2}X_2\\xrightarrow{\\rho_1}X_1\n \\xrightarrow{\\rho_0}X_0.\n\\]\nFor sets, a point is a sequence \\(x_n\\in X_n\\) with \\(\\rho_n(x_{n+1})=x_n\\). Composite restrictions turn the tower into a view system on \\(\\N\\), with an arrow \\(i\\to j\\) when \\(i\\le j\\). Its section space is the same coherent point type, with the same coordinates.","ned detail}{Resolution gives a concrete family of views. Its compatibility differential connects exact maps, protected information, and residual coordinates. Recovering every finite coordinate is then separated from convergence in a chosen norm.}\n\\section{Resolution towers and exact transport}\\label{sec:towers}\n\\subsection{The sequential model}\nA tower in a category \\(\\mathcal C\\) is a sequence\n\\[\n \\cdots\\xrightarrow{\\rho_2}X_2\\xrightarrow{\\rho_1}X_1\n \\xrightarrow{\\rho_0}X_0.\n\\]\nFor sets, a point is a sequence \\(x_n\\in X_n\\) with \\(\\rho_n(x_{n+1})=x_n\\). Composite restrictions turn the tower into a view system on \\(\\N\\), with an arrow \\(i\\to j\\) when \\(i\\le j\\). Its section space is the same coherent point type, with the same coordinates.","information, and residual coordinates. Recovering every finite coordinate is then separated from convergence in a chosen norm.}\n\\section{Resolution towers and exact transport}\\label{sec:towers}\n\\subsection{The sequential model}\nA tower in a category \\(\\mathcal C\\) is a sequence\n\\[\n \\cdots\\xrightarrow{\\rho_2}X_2\\xrightarrow{\\rho_1}X_1\n \\xrightarrow{\\rho_0}X_0.\n\\]\nFor sets, a point is a sequence \\(x_n\\in X_n\\) with \\(\\rho_n(x_{n+1})=x_n\\). Composite restrictions turn the tower into a view system on \\(\\N\\), with an arrow \\(i\\to j\\) when \\(i\\le j\\). Its section space is the same coherent point type, with the same coordinates.","sidual coordinates. Recovering every finite coordinate is then separated from convergence in a chosen norm.}\n\\section{Resolution towers and exact transport}\\label{sec:towers}\n\\subsection{The sequential model}\nA tower in a category \\(\\mathcal C\\) is a sequence\n\\[\n \\cdots\\xrightarrow{\\rho_2}X_2\\xrightarrow{\\rho_1}X_1\n \\xrightarrow{\\rho_0}X_0.\n\\]\nFor sets, a point is a sequence \\(x_n\\in X_n\\) with \\(\\rho_n(x_{n+1})=x_n\\). Composite restrictions turn the tower into a view system on \\(\\N\\), with an arrow \\(i\\to j\\) when \\(i\\le j\\). Its section space is the same coherent point type, with the same coordinates.","ence in a chosen norm.}\n\\section{Resolution towers and exact transport}\\label{sec:towers}\n\\subsection{The sequential model}\nA tower in a category \\(\\mathcal C\\) is a sequence\n\\[\n \\cdots\\xrightarrow{\\rho_2}X_2\\xrightarrow{\\rho_1}X_1\n \\xrightarrow{\\rho_0}X_0.\n\\]\nFor sets, a point is a sequence \\(x_n\\in X_n\\) with \\(\\rho_n(x_{n+1})=x_n\\). Composite restrictions turn the tower into a view system on \\(\\N\\), with an arrow \\(i\\to j\\) when \\(i\\le j\\). Its section space is the same coherent point type, with the same coordinates.","}\n\\section{Resolution towers and exact transport}\\label{sec:towers}\n\\subsection{The sequential model}\nA tower in a category \\(\\mathcal C\\) is a sequence\n\\[\n \\cdots\\xrightarrow{\\rho_2}X_2\\xrightarrow{\\rho_1}X_1\n \\xrightarrow{\\rho_0}X_0.\n\\]\nFor sets, a point is a sequence \\(x_n\\in X_n\\) with \\(\\rho_n(x_{n+1})=x_n\\). Composite restrictions turn the tower into a view system on \\(\\N\\), with an arrow \\(i\\to j\\) when \\(i\\le j\\). Its section space is the same coherent point type, with the same coordinates.","ution towers and exact transport}\\label{sec:towers}\n\\subsection{The sequential model}\nA tower in a category \\(\\mathcal C\\) is a sequence\n\\[\n \\cdots\\xrightarrow{\\rho_2}X_2\\xrightarrow{\\rho_1}X_1\n \\xrightarrow{\\rho_0}X_0.\n\\]\nFor sets, a point is a sequence \\(x_n\\in X_n\\) with \\(\\rho_n(x_{n+1})=x_n\\). Composite restrictions turn the tower into a view system on \\(\\N\\), with an arrow \\(i\\to j\\) when \\(i\\le j\\). Its section space is the same coherent point type, with the same coordinates.","A candidate transport from \\(X\\) to \\(Y\\) consists of maps \\(f_n:X_n\\to Y_n\\). It is exact when\n\\begin{equation}\\label{eq:exacttower}\n \\rho^Y_n f_{n+1}=f_n\\rho^X_n\\quad\\text{for all }n.\n\\end{equation}","\\begin{proposition}[The exact tower category]\\label{KS16}\nIdentity maps and composites of exact transports are exact. An exact transport induces a map of coherent point spaces by \\(x\\mapsto(f_nx_n)_n\\), preserving identities and composition.\n\\end{proposition}\n\\begin{proof}\nFor two exact transports, substitute \\(\\rho^Z_ng_{n+1}=g_n\\rho^Y_n\\) and \\(\\rho^Y_nf_{n+1}=f_n\\rho^X_n\\) into the composite square. For a coherent point,\n\\[\n \\rho^Y_n f_{n+1}x_{n+1}\n =f_n\\rho^X_nx_{n+1}=f_nx_n.\n\\]\nIdentity, associativity, and composition on points are checked coordinatewise.\n\\end{proof}\n\\source{KS16}\nThis is a category on a fixed schedule. A full procategory allows additional reindexing and equivalence relations on maps. Those constructions must be supplied rather than identified with the fixed schedule by terminology.","tower category]\\label{KS16}\nIdentity maps and composites of exact transports are exact. An exact transport induces a map of coherent point spaces by \\(x\\mapsto(f_nx_n)_n\\), preserving identities and composition.\n\\end{proposition}\n\\begin{proof}\nFor two exact transports, substitute \\(\\rho^Z_ng_{n+1}=g_n\\rho^Y_n\\) and \\(\\rho^Y_nf_{n+1}=f_n\\rho^X_n\\) into the composite square. For a coherent point,\n\\[\n \\rho^Y_n f_{n+1}x_{n+1}\n =f_n\\rho^X_nx_{n+1}=f_nx_n.\n\\]\nIdentity, associativity, and composition on points are checked coordinatewise.\n\\end{proof}\n\\source{KS16}\nThis is a category on a fixed schedule. A full procategory allows additional reindexing and equivalence relations on maps. Those constructions must be supplied rather than identified with the fixed schedule by terminology.","\\subsection{The actual residual of a transport}\nAssume \\(\\mathcal C\\) is preadditive, so differences of maps make sense. The tower differential is\n\\[\n (\\partial f)_n=\\rho^Y_nf_{n+1}-f_n\\rho^X_n.\n\\]\nIts target has components \\(\\Hom(X_{n+1},Y_n)\\), not \\(\\Hom(X_n,Y_n)\\).","\\begin{proposition}[The transport Leibniz identity]\\label{KS17}\nFor composable candidates,\n\\[\n \\partial(gf)_n=(\\partial g)_nf_{n+1}+g_n(\\partial f)_n.\n\\]\nIn particular exactness is equivalent to vanishing of the differential.\n\\end{proposition}\n\\begin{proof}\nExpand the right side. The terms \\(-g_n\\rho^Y_nf_{n+1}\\) and \\(+g_n\\rho^Y_nf_{n+1}\\) cancel, leaving \\(\\rho^Z_ng_{n+1}f_{n+1}-g_nf_n\\rho^X_n\\). This is the differential of the composite. The exactness statement is \\eqref{eq:exacttower}.\n\\end{proof}\n\\source{KS17}\nThere is no additional mixed defect term in this identity. Such a term would require a different definition of the defect or a different expansion. Typed cancellation is a useful safeguard against importing an attractive but incorrect error formula.","\\subsection{Cofinal views preserve points, not every measurement}\nLet \\(\\phi:\\N\\to\\N\\) be monotone and unbounded. The sampled tower uses levels \\(X_{\\phi(n)}\\) and their composite restrictions.","\\begin{theorem}[Cofinal reconstruction of points]\\label{KS18}\nSampling induces an equivalence between the coherent points of \\(X\\) and of its sampled tower.\n\\end{theorem}\n\\begin{proof}\nWrite $\\rho_{i,j}:X_j\\to X_i$ for the composite restriction when $i\\le j$, with $\\rho_{i,i}=I$. These maps satisfy $\\rho_{i,j}\\rho_{j,k}=\\rho_{i,k}$. A coherent sampled point obeys\n\\[\n \\rho_{\\phi(n),\\phi(m)}y_m=y_n\\qquad(n\\le m),\n\\]\nby iterating its adjacent coherence equations.","For each $k$, unboundedness gives an index $n$ with $k\\le\\phi(n)$. Set $x_k=\\rho_{k,\\phi(n)}y_n$. If $m$ is another permissible index, choose $\\ell\\ge n,m$. Monotonicity permits both composite restrictions, and sampled coherence gives\n\\[\n \\rho_{k,\\phi(n)}y_n\n =\\rho_{k,\\phi(n)}\\rho_{\\phi(n),\\phi(\\ell)}y_\\ell\n =\\rho_{k,\\phi(\\ell)}y_\\ell\n =\\rho_{k,\\phi(m)}y_m.\n\\]\nThus $x_k$ is independent of the choice, including when the schedule repeats a level.","For the adjacent equation at $k$, choose one $n$ with $k+1\\le\\phi(n)$. Then\n\\[\n \\rho_{k,k+1}x_{k+1}\n =\\rho_{k,k+1}\\rho_{k+1,\\phi(n)}y_n\n =\\rho_{k,\\phi(n)}y_n=x_k.\n\\]\nThe recovered family is therefore a point of the original tower.","Sampling a full point and recovering its $k$th coordinate gives $\\rho_{k,\\phi(n)}x_{\\phi(n)}=x_k$ by its original coherence. Recovering a sampled point and evaluating at $k=\\phi(n)$ allows the index $n$ itself, giving $\\rho_{\\phi(n),\\phi(n)}y_n=y_n$. These are both inverse identities. No surjectivity of the bonding maps or of the point evaluations was used.\n\\end{proof}\n\\source{KS18}\nThis theorem is about exact coherent points. Raw residual size is a different quantity. On identity scalar towers, \\(f_n=(-1)^n\\) has nonzero fine differences, but its even subsequence is constant. The coarse residual vanishes because adjacent fine defects cancel. A cofinal equivalence of limits does not imply equality of residual norms or an absence of a","Sampling a full point and recovering its $k$th coordinate gives $\\rho_{k,\\phi(n)}x_{\\phi(n)}=x_k$ by its original coherence. Recovering a sampled point and evaluating at $k=\\phi(n)$ allows the index $n$ itself, giving $\\rho_{\\phi(n),\\phi(n)}y_n=y_n$. These are both inverse identities. No surjectivity of the bonding maps or of the point evaluations was used.\n\\end{proof}\n\\source{KS18}\nThis theorem is about exact coherent points. Raw residual size is a different quantity. On identity scalar towers, \\(f_n=(-1)^n\\) has nonzero fine differences, but its even subsequence is constant. The coarse residual vanishes because adjacent fine defects cancel. A cofinal equivalence of limits does not imply equality of residual norms or an absence of aliasing in uncorrected data.","s original coherence. Recovering a sampled point and evaluating at $k=\\phi(n)$ allows the index $n$ itself, giving $\\rho_{\\phi(n),\\phi(n)}y_n=y_n$. These are both inverse identities. No surjectivity of the bonding maps or of the point evaluations was used.\n\\end{proof}\n\\source{KS18}\nThis theorem is about exact coherent points. Raw residual size is a different quantity. On identity scalar towers, \\(f_n=(-1)^n\\) has nonzero fine differences, but its even subsequence is constant. The coarse residual vanishes because adjacent fine defects cancel. A cofinal equivalence of limits does not imply equality of residual norms or an absence of aliasing in uncorrected data.","\\chap{Banach repair and constructed split coordinates}\\label{sec:split}\n\\subsection{An actual complete candidate space}\nLet \\(X_n,Y_n\\) be Banach spaces over a complete nontrivially normed field, with continuous linear bonding maps and uniform bounds \\(\\norm{\\rho^X_n}\\le B_X\\), \\(\\norm{\\rho^Y_n}\\le B_Y\\). Use the spaces of bounded operator families\n\\[\n E=\\ell^\\infty_n\\mathcal L(X_n,Y_n),\\qquad\n R=\\ell^\\infty_n\\mathcal L(X_{n+1},Y_n).\n\\]\nThe source formalization derives the required completeness from the operator spaces and their bounded product. The same differential satisfies\n\\[\n \\norm{\\partial f}_\\infty\\le(B_X+B_Y)\\norm f_\\infty.\n\\]\nThus it is a continuous linear map, and evaluation at a fixed level is continuous. These facts connect algebraic compatibility to an analytic limit argument.","ed field, with continuous linear bonding maps and uniform bounds \\(\\norm{\\rho^X_n}\\le B_X\\), \\(\\norm{\\rho^Y_n}\\le B_Y\\). Use the spaces of bounded operator families\n\\[\n E=\\ell^\\infty_n\\mathcal L(X_n,Y_n),\\qquad\n R=\\ell^\\infty_n\\mathcal L(X_{n+1},Y_n).\n\\]\nThe source formalization derives the required completeness from the operator spaces and their bounded product. The same differential satisfies\n\\[\n \\norm{\\partial f}_\\infty\\le(B_X+B_Y)\\norm f_\\infty.\n\\]\nThus it is a continuous linear map, and evaluation at a fixed level is continuous. These facts connect algebraic compatibility to an analytic limit argument.","\\subsection{From a correction operator to a projection}\nThe next statement is general. Let \\(E\\) be Banach, let \\(d:E\\to R\\), \\(p:E\\to S\\), and \\(A:R\\to E\\) be continuous linear maps. Suppose \\(pA=0\\) and, for \\(0\\le q<1\\),\n\\begin{equation}\\label{eq:contraction}\n \\norm{df-dA(df)}\\le q\\norm{df}\\qquad(f\\in E).\n\\end{equation}\nThe hypothesis concerns residuals of actual candidates. It need not hold for every independent vector in \\(R\\).","erator to a projection}\nThe next statement is general. Let \\(E\\) be Banach, let \\(d:E\\to R\\), \\(p:E\\to S\\), and \\(A:R\\to E\\) be continuous linear maps. Suppose \\(pA=0\\) and, for \\(0\\le q<1\\),\n\\begin{equation}\\label{eq:contraction}\n \\norm{df-dA(df)}\\le q\\norm{df}\\qquad(f\\in E).\n\\end{equation}\nThe hypothesis concerns residuals of actual candidates. It need not hold for every independent vector in \\(R\\).","\\begin{theorem}[Constructed geometric repair]\\label{KS19}\nStarting from \\(f^{(0)}=f\\), define\n\\[\n f^{(m+1)}=f^{(m)}-A(df^{(m)}).\n\\]\nThere is a bounded linear projection \\(P_A:E\\to\\ker d\\) with\n\\[\n f^{(m)}\\longrightarrow P_Af,\\qquad p(P_Af)=p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle","\\begin{theorem}[Constructed geometric repair]\\label{KS19}\nStarting from \\(f^{(0)}=f\\), define\n\\[\n f^{(m+1)}=f^{(m)}-A(df^{(m)}).\n\\]\nThere is a bounded linear projection \\(P_A:E\\to\\ker d\\) with\n\\[\n f^{(m)}\\longrightarrow P_Af,\\qquad p(P_Af)=p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f","\\begin{theorem}[Constructed geometric repair]\\label{KS19}\nStarting from \\(f^{(0)}=f\\), define\n\\[\n f^{(m+1)}=f^{(m)}-A(df^{(m)}).\n\\]\nThere is a bounded linear projection \\(P_A:E\\to\\ker d\\) with\n\\[\n f^{(m)}\\longrightarrow P_Af,\\qquad p(P_Af)=p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\l","\\begin{theorem}[Constructed geometric repair]\\label{KS19}\nStarting from \\(f^{(0)}=f\\), define\n\\[\n f^{(m+1)}=f^{(m)}-A(df^{(m)}).\n\\]\nThere is a bounded linear projection \\(P_A:E\\to\\ker d\\) with\n\\[\n f^{(m)}\\longrightarrow P_Af,\\qquad p(P_Af)=p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\n","\\begin{theorem}[Constructed geometric repair]\\label{KS19}\nStarting from \\(f^{(0)}=f\\), define\n\\[\n f^{(m+1)}=f^{(m)}-A(df^{(m)}).\n\\]\nThere is a bounded linear projection \\(P_A:E\\to\\ker d\\) with\n\\[\n f^{(m)}\\longrightarrow P_Af,\\qquad p(P_Af)=p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$","\\begin{theorem}[Constructed geometric repair]\\label{KS19}\nStarting from \\(f^{(0)}=f\\), define\n\\[\n f^{(m+1)}=f^{(m)}-A(df^{(m)}).\n\\]\nThere is a bounded linear projection \\(P_A:E\\to\\ker d\\) with\n\\[\n f^{(m)}\\longrightarrow P_Af,\\qquad p(P_Af)=p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality …","bel{KS19}\nStarting from \\(f^{(0)}=f\\), define\n\\[\n f^{(m+1)}=f^{(m)}-A(df^{(m)}).\n\\]\nThere is a bounded linear projection \\(P_A:E\\to\\ker d\\) with\n\\[\n f^{(m)}\\longrightarrow P_Af,\\qquad p(P_Af)=p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","0)}=f\\), define\n\\[\n f^{(m+1)}=f^{(m)}-A(df^{(m)}).\n\\]\nThere is a bounded linear projection \\(P_A:E\\to\\ker d\\) with\n\\[\n f^{(m)}\\longrightarrow P_Af,\\qquad p(P_Af)=p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","fine\n\\[\n f^{(m+1)}=f^{(m)}-A(df^{(m)}).\n\\]\nThere is a bounded linear projection \\(P_A:E\\to\\ker d\\) with\n\\[\n f^{(m)}\\longrightarrow P_Af,\\qquad p(P_Af)=p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","f^{(m)}).\n\\]\nThere is a bounded linear projection \\(P_A:E\\to\\ker d\\) with\n\\[\n f^{(m)}\\longrightarrow P_Af,\\qquad p(P_Af)=p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","p(f),\n\\]\nand\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","nd\n\\[\n \\norm{f^{(m)}-P_Af}\n \\le\\frac{\\norm A}{1-q}\\norm{df^{(m)}}.\n\\]\nIt fixes every exact candidate and satisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","tisfies\n\\[\n \\norm{P_A}\\le1+\\frac{\\norm A\\norm d}{1-q}.\n\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","\\]\n\\end{theorem}\n\\begin{proof}\nPut $r_m=df^{(m)}$. Additivity of $d$ gives\n\\[\n r_{m+1}=r_m-dAr_m.\n\\]\nEach $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","n. Induction yields, for $j\\ge0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","0$,\n\\[\n \\norm{r_{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","{m+j}}\\le q^j\\norm{r_m},\\qquad\n \\norm{f^{(m+j+1)}-f^{(m+j)}}\n \\le\\norm A\\,q^j\\norm{r_m}.\n\\]\nFor $n>m$, the finite triangle inequality therefore gives\n\\[\n \\norm{f^{(n)}-f^{(m)}}\n \\le\\norm A\\,\\norm{r_m}\\sum_{j=0}^{n-m-1}q^j\n \\le\\frac{\\norm A}{1-q}\\norm{r_m}.\n\\]\nSince $\\norm{r_m}\\le q^m\\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\\to\\infty$ in the same norm inequality proves the required tail estimate.","Continuity of $d$ implies $d(P_Af)=\\lim_m r_m=0$. Protection holds before taking a limit: $pA=0$ gives $pf^{(m+1)}=pf^{(m)}$, hence $pf^{(m)}=pf$ for every $m$. Continuity of $p$ gives $p(P_Af)=pf$. This explains why continuity of the observable belongs to this analytic statement.","For linearity, write $L=I-Ad$, so $f^{(m)}=L^mf$. For fixed $f,g$ and scalars $\\lambda,\\mu$,\n\\[\n L^m(\\lambda f+\\mu g)=\\lambda L^mf+\\mu L^mg.\n\\]\nContinuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis","For linearity, write $L=I-Ad$, so $f^{(m)}=L^mf$. For fixed $f,g$ and scalars $\\lambda,\\mu$,\n\\[\n L^m(\\lambda f+\\mu g)=\\lambda L^mf+\\mu L^mg.\n\\]\nContinuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs","For linearity, write $L=I-Ad$, so $f^{(m)}=L^mf$. For fixed $f,g$ and scalars $\\lambda,\\mu$,\n\\[\n L^m(\\lambda f+\\mu g)=\\lambda L^mf+\\mu L^mg.\n\\]\nContinuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics fro","For linearity, write $L=I-Ad$, so $f^{(m)}=L^mf$. For fixed $f,g$ and scalars $\\lambda,\\mu$,\n\\[\n L^m(\\lambda f+\\mu g)=\\lambda L^mf+\\mu L^mg.\n\\]\nContinuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not const","For linearity, write $L=I-Ad$, so $f^{(m)}=L^mf$. For fixed $f,g$ and scalars $\\lambda,\\mu$,\n\\[\n L^m(\\lambda f+\\mu g)=\\lambda L^mf+\\mu L^mg.\n\\]\nContinuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not construct a suitable \\(A\\) for every residual problem. T","For linearity, write $L=I-Ad$, so $f^{(m)}=L^mf$. For fixed $f,g$ and scalars $\\lambda,\\mu$,\n\\[\n L^m(\\lambda f+\\mu g)=\\lambda L^mf+\\mu L^mg.\n\\]\nContinuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not construct a suitable \\(A\\) for every residual problem. That missing input can, however, be derived for a useful class of actual towers.","$f^{(m)}=L^mf$. For fixed $f,g$ and scalars $\\lambda,\\mu$,\n\\[\n L^m(\\lambda f+\\mu g)=\\lambda L^mf+\\mu L^mg.\n\\]\nContinuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not construct a suitable \\(A\\) for every residual problem. That missing input can, however, be derived for a useful class of actual towers.",",g$ and scalars $\\lambda,\\mu$,\n\\[\n L^m(\\lambda f+\\mu g)=\\lambda L^mf+\\mu L^mg.\n\\]\nContinuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not construct a suitable \\(A\\) for every residual problem. That missing input can, however, be derived for a useful class of actual towers.","lars $\\lambda,\\mu$,\n\\[\n L^m(\\lambda f+\\mu g)=\\lambda L^mf+\\mu L^mg.\n\\]\nContinuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not construct a suitable \\(A\\) for every residual problem. That missing input can, however, be derived for a useful class of actual towers.","f+\\mu g)=\\lambda L^mf+\\mu L^mg.\n\\]\nContinuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not construct a suitable \\(A\\) for every residual problem. That missing input can, however, be derived for a useful class of actual towers.","n allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not construct a suitable \\(A\\) for every residual problem. That missing input can, however, be derived for a useful class of actual towers.","to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives\n\\[\n \\norm{P_Af}\\le\\norm f+\\frac{\\norm A}{1-q}\\norm{df}\n \\le\\left(1+\\frac{\\norm A\\norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not construct a suitable \\(A\\) for every residual problem. That missing input can, however, be derived for a useful class of actual towers.","norm d}{1-q}\\right)\\norm f,\n\\]\nso the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not construct a suitable \\(A\\) for every residual problem. That missing input can, however, be derived for a useful class of actual towers.","near map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.\n\\end{proof}\n\\source{KS19}\nThis theorem constructs the dynamics from \\(A\\). It does not construct a suitable \\(A\\) for every residual problem. That missing input can, however, be derived for a useful class of actual towers.","\\subsection{A right inverse from target splittings}\nSuppose each target bonding map has a continuous linear section\n\\[\n s_n:Y_n\\to Y_{n+1},\\qquad \\rho^Y_ns_n=I,\\qquad \\norm{s_n}\\le S.\n\\]\nAssume the propagation gap\n\\begin{equation}\\label{eq:splitgap}\n SB_X<1,\\qquad K=\\frac{S}{1-SB_X}.\n\\end{equation}\nFor \\(r\\in R\\), recursively set\n\\begin{equation}\\label{eq:greenrecurrence}\n (Ar)_0=0,\\qquad\n (Ar)_{n+1}=s_n\\bigl(r_n+(Ar)_n\\rho^X_n\\bigr).\n\\end{equation}","\\begin{theorem}[Split tower correction]\\label{KS20}\nThe recurrence defines a continuous linear map \\(A:R\\to E\\) with\n\\[\n \\partial A=I_R,\\qquad \\norm A\\le K.\n\\]\nThe operator \\(P=I-A\\partial\\) is an exact projection and\n\\[\n (Pf)_0=f_0,\\qquad\n \\norm{f-Pf}_\\infty\\le K\\norm{\\partial f}_\\infty.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\theta=SB_X<1$. The number $K=S/(1-\\theta)$ is nonnegative and satisfies\n\\[\n S+SB_XK=K.\n\\]\nFix $r\\in R$ and define its coordinates recursively as in \\eqref{eq:greenrecurrence}. The initial coordinate has norm zero. If $\\norm{(Ar)_n}\\le K\\norm r_\\infty$, submultiplicativity of operator norms gives\n\\[\n \\begin{aligned}\n \\norm{(Ar)_{n+1}}\n &\\le\\norm{s_n}\\bigl(\\norm{r_n}+\\norm{(Ar)_n}\\norm{\\rho^X_n}\\bigr)","begin{theorem}[Split tower correction]\\label{KS20}\nThe recurrence defines a continuous linear map \\(A:R\\to E\\) with\n\\[\n \\partial A=I_R,\\qquad \\norm A\\le K.\n\\]\nThe operator \\(P=I-A\\partial\\) is an exact projection and\n\\[\n (Pf)_0=f_0,\\qquad\n \\norm{f-Pf}_\\infty\\le K\\norm{\\partial f}_\\infty.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\theta=SB_X<1$. The number $K=S/(1-\\theta)$ is nonnegative and satisfies\n\\[\n S+SB_XK=K.\n\\]\nFix $r\\in R$ and define its coordinates recursively as in \\eqref{eq:greenrecurrence}. The initial coordinate has norm zero. If $\\norm{(Ar)_n}\\le K\\norm r_\\infty$, submultiplicativity of operator norms gives\n\\[\n \\begin{aligned}\n \\norm{(Ar)_{n+1}}\n &\\le\\norm{s_n}\\bigl(\\norm{r_n}+\\norm{(Ar)_n}\\norm{\\rho^X_n}\\bigr)","tower correction]\\label{KS20}\nThe recurrence defines a continuous linear map \\(A:R\\to E\\) with\n\\[\n \\partial A=I_R,\\qquad \\norm A\\le K.\n\\]\nThe operator \\(P=I-A\\partial\\) is an exact projection and\n\\[\n (Pf)_0=f_0,\\qquad\n \\norm{f-Pf}_\\infty\\le K\\norm{\\partial f}_\\infty.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\theta=SB_X<1$. The number $K=S/(1-\\theta)$ is nonnegative and satisfies\n\\[\n S+SB_XK=K.\n\\]\nFix $r\\in R$ and define its coordinates recursively as in \\eqref{eq:greenrecurrence}. The initial coordinate has norm zero. If $\\norm{(Ar)_n}\\le K\\norm r_\\infty$, submultiplicativity of operator norms gives\n\\[\n \\begin{aligned}\n \\norm{(Ar)_{n+1}}\n &\\le\\norm{s_n}\\bigl(\\norm{r_n}+\\norm{(Ar)_n}\\norm{\\rho^X_n}\\bigr)","orm A\\le K.\n\\]\nThe operator \\(P=I-A\\partial\\) is an exact projection and\n\\[\n (Pf)_0=f_0,\\qquad\n \\norm{f-Pf}_\\infty\\le K\\norm{\\partial f}_\\infty.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\theta=SB_X<1$. The number $K=S/(1-\\theta)$ is nonnegative and satisfies\n\\[\n S+SB_XK=K.\n\\]\nFix $r\\in R$ and define its coordinates recursively as in \\eqref{eq:greenrecurrence}. The initial coordinate has norm zero. If $\\norm{(Ar)_n}\\le K\\norm r_\\infty$, submultiplicativity of operator norms gives\n\\[\n \\begin{aligned}\n \\norm{(Ar)_{n+1}}\n &\\le\\norm{s_n}\\bigl(\\norm{r_n}+\\norm{(Ar)_n}\\norm{\\rho^X_n}\\bigr)","_0=f_0,\\qquad\n \\norm{f-Pf}_\\infty\\le K\\norm{\\partial f}_\\infty.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $\\theta=SB_X<1$. The number $K=S/(1-\\theta)$ is nonnegative and satisfies\n\\[\n S+SB_XK=K.\n\\]\nFix $r\\in R$ and define its coordinates recursively as in \\eqref{eq:greenrecurrence}. The initial coordinate has norm zero. If $\\norm{(Ar)_n}\\le K\\norm r_\\infty$, submultiplicativity of operator norms gives\n\\[\n \\begin{aligned}\n \\norm{(Ar)_{n+1}}\n &\\le\\norm{s_n}\\bigl(\\norm{r_n}+\\norm{(Ar)_n}\\norm{\\rho^X_n}\\bigr)\\\\\n &\\le S(1+KB_X)\\norm r_\\infty=K\\norm r_\\infty.\n \\end{aligned}\n\\]\nInduction proves the estimate at every coordinate, so the recursively defined family actually belongs to the bounded product $E$.","&\\le S(1+KB_X)\\norm r_\\infty=K\\norm r_\\infty.\n \\end{aligned}\n\\]\nInduction proves the estimate at every coordinate, so the recursively defined family actually belongs to the bounded product $E$.","The coordinate recurrence preserves addition and scalar multiplication, starting from the zero initial coordinate. Induction therefore proves these laws for the family map $A$. Its uniform estimate proves $\\norm A\\le K$, giving a continuous linear map with the claimed domain and codomain.","For every $n$, apply the target restriction to the recurrence. The equation $\\rho^Y_ns_n=I$ gives\n\\[\n \\rho^Y_n(Ar)_{n+1}=r_n+(Ar)_n\\rho^X_n,\n \\qquad (\\partial Ar)_n=r_n.\n\\]\nCoordinate equality yields the operator identity $\\partial A=I_R$. In particular this identity holds for every bounded residual, not just residuals already known to lie in the image.","Set $P=I-A\\partial$. Then\n\\[\n \\partial P=\\partial-\\partial A\\partial=0,\\qquad\n P^2=I-2A\\partial+A(\\partial A)\\partial=I-A\\partial=P.\n\\]\nThe expression $2A\\partial$ abbreviates addition of the same operator twice and uses no division by two. Every exact $f$ is fixed because $\\partial f=0$. Since $(Ar)_0=0$, the seed of $Pf$ is $f_0$. Finally $f-Pf=A\\partial f$, and the bound on $A$ gives the asserted residual estimate. The denominators above are justified exactly by the strict propagation gap.\n\\end{proof}\n\\source{KS20}\nThe surjectivity is onto all bounded residual families. It is stronger than the assertion that some exact output can be assigned to each candidate. The gap and the chosen splittings are genuine","Set $P=I-A\\partial$. Then\n\\[\n \\partial P=\\partial-\\partial A\\partial=0,\\qquad\n P^2=I-2A\\partial+A(\\partial A)\\partial=I-A\\partial=P.\n\\]\nThe expression $2A\\partial$ abbreviates addition of the same operator twice and uses no division by two. Every exact $f$ is fixed because $\\partial f=0$. Since $(Ar)_0=0$, the seed of $Pf$ is $f_0$. Finally $f-Pf=A\\partial f$, and the bound on $A$ gives the asserted residual estimate. The denominators above are justified exactly by the strict propagation gap.\n\\end{proof}\n\\source{KS20}\nThe surjectivity is onto all bounded residual families. It is stronger than the assertion that some exact output can be assigned to each candidate. The gap and the chosen splittings are genuine inputs.","\\subsection{Protected information beyond the seed}\nWhen the target grows, its new kernel directions are independent information. Define\n\\[\n N_n(f)=(I-s_n\\rho^Y_n)f_{n+1},\\qquad\n (p_Lf)_0=f_0,\\quad(p_Lf)_{n+1}=N_n(f).\n\\]\nEach \\(N_n(f)\\) lands in \\(\\ker\\rho^Y_n\\). The observable \\(p_L\\) retains both the seed and every new output component, relative to the chosen sections.","\\begin{theorem}[Split coordinate theorem]\\label{KS21}\nThe observable is a bounded projection with \\(p_LA=0\\) and \\(p_LP=p_L\\). It separates exact candidates. If\n\\[\n \\mathcal S_L=\\{s\\in E:p_Ls=s\\},\n\\]\nthen\n\\[\n \\Phi:E\\simeq_L\\mathcal S_L\\times R,\n \\qquad \\Phi(f)=(p_Lf,\\partial f),\n \\qquad \\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and","\\begin{theorem}[Split coordinate theorem]\\label{KS21}\nThe observable is a bounded projection with \\(p_LA=0\\) and \\(p_LP=p_L\\). It separates exact candidates. If\n\\[\n \\mathcal S_L=\\{s\\in E:p_Ls=s\\},\n\\]\nthen\n\\[\n \\Phi:E\\simeq_L\\mathcal S_L\\times R,\n \\qquad \\Phi(f)=(p_Lf,\\partial f),\n \\qquad \\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+","\\begin{theorem}[Split coordinate theorem]\\label{KS21}\nThe observable is a bounded projection with \\(p_LA=0\\) and \\(p_LP=p_L\\). It separates exact candidates. If\n\\[\n \\mathcal S_L=\\{s\\in E:p_Ls=s\\},\n\\]\nthen\n\\[\n \\Phi:E\\simeq_L\\mathcal S_L\\times R,\n \\qquad \\Phi(f)=(p_Lf,\\partial f),\n \\qquad \\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, no","\\begin{theorem}[Split coordinate theorem]\\label{KS21}\nThe observable is a bounded projection with \\(p_LA=0\\) and \\(p_LP=p_L\\). It separates exact candidates. If\n\\[\n \\mathcal S_L=\\{s\\in E:p_Ls=s\\},\n\\]\nthen\n\\[\n \\Phi:E\\simeq_L\\mathcal S_L\\times R,\n \\qquad \\Phi(f)=(p_Lf,\\partial f),\n \\qquad \\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","orem]\\label{KS21}\nThe observable is a bounded projection with \\(p_LA=0\\) and \\(p_LP=p_L\\). It separates exact candidates. If\n\\[\n \\mathcal S_L=\\{s\\in E:p_Ls=s\\},\n\\]\nthen\n\\[\n \\Phi:E\\simeq_L\\mathcal S_L\\times R,\n \\qquad \\Phi(f)=(p_Lf,\\partial f),\n \\qquad \\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","ble is a bounded projection with \\(p_LA=0\\) and \\(p_LP=p_L\\). It separates exact candidates. If\n\\[\n \\mathcal S_L=\\{s\\in E:p_Ls=s\\},\n\\]\nthen\n\\[\n \\Phi:E\\simeq_L\\mathcal S_L\\times R,\n \\qquad \\Phi(f)=(p_Lf,\\partial f),\n \\qquad \\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","\\(p_LP=p_L\\). It separates exact candidates. If\n\\[\n \\mathcal S_L=\\{s\\in E:p_Ls=s\\},\n\\]\nthen\n\\[\n \\Phi:E\\simeq_L\\mathcal S_L\\times R,\n \\qquad \\Phi(f)=(p_Lf,\\partial f),\n \\qquad \\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","=\\{s\\in E:p_Ls=s\\},\n\\]\nthen\n\\[\n \\Phi:E\\simeq_L\\mathcal S_L\\times R,\n \\qquad \\Phi(f)=(p_Lf,\\partial f),\n \\qquad \\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","\\]\nthen\n\\[\n \\Phi:E\\simeq_L\\mathcal S_L\\times R,\n \\qquad \\Phi(f)=(p_Lf,\\partial f),\n \\qquad \\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","eq_L\\mathcal S_L\\times R,\n \\qquad \\Phi(f)=(p_Lf,\\partial f),\n \\qquad \\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","\\Phi^{-1}(s,r)=Ps+Ar.\n\\]\nRepair becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","r becomes \\((s,r)\\mapsto(s,0)\\). It is the unique exact repair preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","ir preserving \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","g \\(p_Lf\\).\n\\end{theorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","heorem}\n\\begin{proof}\nPut \\(B_n=I-s_n\\rho^Y_n\\). Since \\(\\rho^Y_ns_n=I\\), direct multiplication gives\n\\[\n B_n^2=B_n,\\qquad \\rho^Y_nB_n=0,\\qquad B_ns_n=0.\n\\]\nMoreover, \\(B_ny=y\\) whenever \\(\\rho^Y_ny=0\\). Thus \\(B_n\\) really projects onto the target kernel. Its norm is at most \\(1+SB_Y\\). Coordinate zero of \\(p_L\\) is the identity, and every later coordinate applies \\(B_n\\), so \\(p_L^2=p_L\\) and \\(\\norm{p_L}\\le1+SB_Y\\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.","The recurrence for \\(Ar\\) gives zero seed and places \\((Ar)_{n+1}\\) in the image of \\(s_n\\). The identities just proved make all its innovations zero. Therefore \\(p_LA=0\\), and\n\\[\n p_LP=p_L(I-A\\partial)=p_L.\n\\]\nWe next prove separation on the exact space. If \\(f\\) is exact, its restriction equation rewrites as\n\\[\n f_{n+1}=s_n\\rho^Y_nf_{n+1}+B_nf_{n+1}\n          =s_nf_n\\rho^X_n+N_n(f).\n\\]\nTwo exact families with the same seed and innovations agree first at level zero and then at every later level by this recurrence. More generally, if two arbitrary families have the same observable and differential, their difference is exact with zero observable, so separation on the exact space makes their difference zero.","The proposed chart has the claimed target because \\(p_L(p_Lf)=p_Lf\\). Given \\(s\\in\\mathcal S_L\\) and \\(r\\in R\\), set \\(f=Ps+Ar\\). The preceding identities and \\(\\partial A=I\\) give\n\\[\n p_Lf=p_Ls=s,\\qquad \\partial f=r.\n\\]\nThis proves one round trip. For the other, \\(f\\) and \\(P(p_Lf)+A\\partial f\\) have the same observable and residual, so the joint separation just proved identifies them. All maps are linear and bounded on their indicated subspaces and finite products, establishing a continuous linear equivalence.","Finally, \\(Pf\\) has coordinates \\((p_Lf,0)\\). Any exact answer with the same observable has those same two coordinates, hence equals \\(Pf\\). This uniqueness is relative to the complete observable of seed and innovations. It is not obtained by assuming that the seed alone determines an exact growing-target family.\n\\end{proof}\n\\source{KS21}\nThis is an actual analytic coordinate equivalence, not generally an isometry. The protected coordinate depends on the declared splitting. It should not be called an intrinsic semantic coordinate without a further comparison theorem.","\\subsection{A growing example and the gap boundary}\nUse \\(X_n=Y_n=\\R^{n+1}\\) with maximum norms. Let \\(\\pi_n\\) delete the last coordinate, take \\(\\rho^X_n=\\tfrac12\\pi_n\\), \\(\\rho^Y_n=\\pi_n\\), and let \\(s_n\\) append zero. Then \\(S=B_Y=1\\), \\(B_X=1/2\\), and \\(K=2\\).","\\begin{example}[Repair of the identity family]\\label{KS22}\nThe bounded candidate \\(f_n=I\\) is not exact. Its repair is\n\\[\n (Pf)_n=\\operatorname{diag}(2^{-n},2^{-(n-1)},\\ldots,2^{-1},1).\n\\]\nThe new last output row is retained at every stage while inherited rows are propagated by the factor \\(1/2\\).\n\\end{example}\n\\begin{proof}\nAt level zero the value is one. Exactness forces every inherited row at the next level to be one half of its previous value. Preservation of the innovation forces the new row to remain the corresponding identity row. Induction gives the displayed diagonal.\n\\end{proof}\n\\source{KS22}\nFor constant real scalar identity towers, by contrast, \\((\\partial f)_n=f_{n+1}-f_n\\). The constant residual \\(r_n=1\\) has no bounded primitive, since any primitive satisfies \\(f","\\begin{example}[Repair of the identity family]\\label{KS22}\nThe bounded candidate \\(f_n=I\\) is not exact. Its repair is\n\\[\n (Pf)_n=\\operatorname{diag}(2^{-n},2^{-(n-1)},\\ldots,2^{-1},1).\n\\]\nThe new last output row is retained at every stage while inherited rows are propagated by the factor \\(1/2\\).\n\\end{example}\n\\begin{proof}\nAt level zero the value is one. Exactness forces every inherited row at the next level to be one half of its previous value. Preservation of the innovation forces the new row to remain the corresponding identity row. Induction gives the displayed diagonal.\n\\end{proof}\n\\source{KS22}\nFor constant real scalar identity towers, by contrast, \\((\\partial f)_n=f_{n+1}-f_n\\). The constant residual \\(r_n=1\\) has no bounded primitive, since any primitive satisfies \\(f_n=f_0+n\\). Here \\(SB_X=1\\). This counterexample excludes discarding the gap indiscriminately. It","\\begin{example}[Repair of the identity family]\\label{KS22}\nThe bounded candidate \\(f_n=I\\) is not exact. Its repair is\n\\[\n (Pf)_n=\\operatorname{diag}(2^{-n},2^{-(n-1)},\\ldots,2^{-1},1).\n\\]\nThe new last output row is retained at every stage while inherited rows are propagated by the factor \\(1/2\\).\n\\end{example}\n\\begin{proof}\nAt level zero the value is one. Exactness forces every inherited row at the next level to be one half of its previous value. Preservation of the innovation forces the new row to remain the corresponding identity row. Induction gives the displayed diagonal.\n\\end{proof}\n\\source{KS22}\nFor constant real scalar identity towers, by contrast, \\((\\partial f)_n=f_{n+1}-f_n\\). The constant residual \\(r_n=1\\) has no bounded primitive, since any primitive satisfies \\(f_n=f_0+n\\). Here \\(SB_X=1\\). This counterexample excludes discarding the gap indiscriminately. It does not prove that the gap is necessary for every individual tower. Exact repair preserving the initial coor","opagated by the factor \\(1/2\\).\n\\end{example}\n\\begin{proof}\nAt level zero the value is one. Exactness forces every inherited row at the next level to be one half of its previous value. Preservation of the innovation forces the new row to remain the corresponding identity row. Induction gives the displayed diagonal.\n\\end{proof}\n\\source{KS22}\nFor constant real scalar identity towers, by contrast, \\((\\partial f)_n=f_{n+1}-f_n\\). The constant residual \\(r_n=1\\) has no bounded primitive, since any primitive satisfies \\(f_n=f_0+n\\). Here \\(SB_X=1\\). This counterexample excludes discarding the gap indiscriminately. It does not prove that the gap is necessary for every individual tower. Exact repair preserving the initial coordinate still exists in this constant example, by replacing a candidate with its initial constant family. Surjectivity onto all bounded residuals does not follow from that repair.\n\\source{KS23}","roof}\nAt level zero the value is one. Exactness forces every inherited row at the next level to be one half of its previous value. Preservation of the innovation forces the new row to remain the corresponding identity row. Induction gives the displayed diagonal.\n\\end{proof}\n\\source{KS22}\nFor constant real scalar identity towers, by contrast, \\((\\partial f)_n=f_{n+1}-f_n\\). The constant residual \\(r_n=1\\) has no bounded primitive, since any primitive satisfies \\(f_n=f_0+n\\). Here \\(SB_X=1\\). This counterexample excludes discarding the gap indiscriminately. It does not prove that the gap is necessary for every individual tower. Exact repair preserving the initial coordinate still exists in this constant example, by replacing a candidate with its initial constant family. Surjectivity onto all bounded residuals does not follow from that repair.\n\\source{KS23}","ited row at the next level to be one half of its previous value. Preservation of the innovation forces the new row to remain the corresponding identity row. Induction gives the displayed diagonal.\n\\end{proof}\n\\source{KS22}\nFor constant real scalar identity towers, by contrast, \\((\\partial f)_n=f_{n+1}-f_n\\). The constant residual \\(r_n=1\\) has no bounded primitive, since any primitive satisfies \\(f_n=f_0+n\\). Here \\(SB_X=1\\). This counterexample excludes discarding the gap indiscriminately. It does not prove that the gap is necessary for every individual tower. Exact repair preserving the initial coordinate still exists in this constant example, by replacing a candidate with its initial constant family. Surjectivity onto all bounded residuals does not follow from that repair.\n\\source{KS23}","level to be one half of its previous value. Preservation of the innovation forces the new row to remain the corresponding identity row. Induction gives the displayed diagonal.\n\\end{proof}\n\\source{KS22}\nFor constant real scalar identity towers, by contrast, \\((\\partial f)_n=f_{n+1}-f_n\\). The constant residual \\(r_n=1\\) has no bounded primitive, since any primitive satisfies \\(f_n=f_0+n\\). Here \\(SB_X=1\\). This counterexample excludes discarding the gap indiscriminately. It does not prove that the gap is necessary for every individual tower. Exact repair preserving the initial coordinate still exists in this constant example, by replacing a candidate with its initial constant family. Surjectivity onto all bounded residuals does not follow from that repair.\n\\source{KS23}","\\subsection{What is canonical}\nFor fixed splitting and protected observable, the repair above is unique. Changing the splitting can change both the observable and the repaired answer. In the growing example, replace \\(s_n(y)=(y,0)\\) by \\(s_n'(y)=(y,y_0)\\). At the first positive level, repair of the identity becomes\n\\[\n \\begin{pmatrix}1/2&0","\\subsection{What is canonical}\nFor fixed splitting and protected observable, the repair above is unique. Changing the splitting can change both the observable and the repaired answer. In the growing example, replace \\(s_n(y)=(y,0)\\) by \\(s_n'(y)=(y,y_0)\\). At the first positive level, repair of the identity becomes\n\\[\n \\begin{pmatrix}1/2&0\\\\-1/2&1\\end{pmatrix}\n \\quad\\text{instead of}\\quad\n \\begin{pmatrix}1/2&0\\\\0&1\\end{pmatrix}.\n\\]\nBoth are exact continuations for their respective innovation profiles and have the same seed. Thus a unique answer for a declared preservation problem is not an answer independent of all preservation choices.\n\\source{KS24}","\\chap{Multiscale residuals and the limits of reconstruction}\\label{sec:cascade}\n\\subsection{A split coarsener retains its missing detail}\nLet \\(C:R\\to\\overline R\\) and \\(J:\\overline R\\to R\\) be linear maps with \\(CJ=I\\). Put \\(H=I-JC\\). Then\n\\[\n r=JCr+Hr,\\qquad CH=0,\\qquad H^2=H.\n\\]\nThe two coordinates \\((Cr,Hr)\\) retain all information. The coarse coordinate alone generally does not.","The actual dyadic tower coarsener has the typed formula\n\\begin{equation}\\label{eq:dyadic}\n (Cr)_n=\\rho^Y_{2n}r_{2n+1}+r_{2n}\\rho^X_{2n+1}.\n\\end{equation}\nIt is obtained by telescoping two adjacent compatibility squares. With target sections, a lift is\n\\[\n (Jc)_{2n}=0,\\qquad (Jc)_{2n+1}=s_{2n}c_n.\n\\]\nThe section identity gives \\(CJ=I\\), so the missing detail is explicitly \\(Hr\\), not an unspecified reconstruction error. The normed construction imposes the bounds needed to make these maps bounded on its operator family spaces.\n\\source{KS25}","\\subsection{A whole cascade}\nAt scale \\(j\\), let \\(C_j:R_j\\to R_{j+1}\\) and \\(J_j:R_{j+1}\\to R_j\\) be bounded linear maps with \\(C_jJ_j=I\\). For \\(x\\in R_0\\), define\n\\[\n x_0=x,\\quad x_{j+1}=C_jx_j,\\quad\n L_0=I,\\quad L_N=J_0\\cdots J_{N-1},\n\\]\n\\[\n h_j=x_j-J_jx_{j+1},\\qquad Q_Nx=L_Nx_N,\n \\qquad S_N=I-Q_N.\n\\]\nThe map \\(Q_N\\) restores the surviving coarse remainder to the original space. It is this restored remainder, not just \\(x_N\\), that controls reconstruction.","At scale \\(j\\), let \\(C_j:R_j\\to R_{j+1}\\) and \\(J_j:R_{j+1}\\to R_j\\) be bounded linear maps with \\(C_jJ_j=I\\). For \\(x\\in R_0\\), define\n\\[\n x_0=x,\\quad x_{j+1}=C_jx_j,\\quad\n L_0=I,\\quad L_N=J_0\\cdots J_{N-1},\n\\]\n\\[\n h_j=x_j-J_jx_{j+1},\\qquad Q_Nx=L_Nx_N,\n \\qquad S_N=I-Q_N.\n\\]\nThe map \\(Q_N\\) restores the surviving coarse remainder to the original space. It is this restored remainder, not just \\(x_N\\), that controls reconstruction.","\\begin{theorem}[Finite reconstruction and nested remainders]\\label{KS26}\nFor every \\(N\\),\n\\[\n x=Q_Nx+\\sum_{j<N}L_jh_j,\n \\qquad S_Nx=\\sum_{j<N}L_jh_j.\n\\]\nEach \\(Q_N\\) is a projection, and for \\(N\\le M\\),\n\\[\n Q_NQ_M=Q_MQ_N=Q_M.\n\\]\nThe vectors with all details zero are exactly the persistent subspace\n\\[\n \\mathcal P_\\infty=\\{x:Q_Nx=x\\text{ for every }N\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $D_0=I$ and $D_N=C_{N-1}\\cdots C_0:R_0\\to R_N$. Thus $x_N=D_Nx$, $Q_N=L_ND_N$, and repeated use of $C_jJ_j=I$ gives\n\\[\n D_NL_N=I_{R_N}.\n\\]\nThis cancellation takes place between adjacent maps with matching domains. No identification of different residual spaces is needed.","egin{theorem}[Finite reconstruction and nested remainders]\\label{KS26}\nFor every \\(N\\),\n\\[\n x=Q_Nx+\\sum_{j<N}L_jh_j,\n \\qquad S_Nx=\\sum_{j<N}L_jh_j.\n\\]\nEach \\(Q_N\\) is a projection, and for \\(N\\le M\\),\n\\[\n Q_NQ_M=Q_MQ_N=Q_M.\n\\]\nThe vectors with all details zero are exactly the persistent subspace\n\\[\n \\mathcal P_\\infty=\\{x:Q_Nx=x\\text{ for every }N\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $D_0=I$ and $D_N=C_{N-1}\\cdots C_0:R_0\\to R_N$. Thus $x_N=D_Nx$, $Q_N=L_ND_N$, and repeated use of $C_jJ_j=I$ gives\n\\[\n D_NL_N=I_{R_N}.\n\\]\nThis cancellation takes place between adjacent maps with matching domains. No identification of different residual spaces is needed.","ested remainders]\\label{KS26}\nFor every \\(N\\),\n\\[\n x=Q_Nx+\\sum_{j<N}L_jh_j,\n \\qquad S_Nx=\\sum_{j<N}L_jh_j.\n\\]\nEach \\(Q_N\\) is a projection, and for \\(N\\le M\\),\n\\[\n Q_NQ_M=Q_MQ_N=Q_M.\n\\]\nThe vectors with all details zero are exactly the persistent subspace\n\\[\n \\mathcal P_\\infty=\\{x:Q_Nx=x\\text{ for every }N\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $D_0=I$ and $D_N=C_{N-1}\\cdots C_0:R_0\\to R_N$. Thus $x_N=D_Nx$, $Q_N=L_ND_N$, and repeated use of $C_jJ_j=I$ gives\n\\[\n D_NL_N=I_{R_N}.\n\\]\nThis cancellation takes place between adjacent maps with matching domains. No identification of different residual spaces is needed.","ders]\\label{KS26}\nFor every \\(N\\),\n\\[\n x=Q_Nx+\\sum_{j<N}L_jh_j,\n \\qquad S_Nx=\\sum_{j<N}L_jh_j.\n\\]\nEach \\(Q_N\\) is a projection, and for \\(N\\le M\\),\n\\[\n Q_NQ_M=Q_MQ_N=Q_M.\n\\]\nThe vectors with all details zero are exactly the persistent subspace\n\\[\n \\mathcal P_\\infty=\\{x:Q_Nx=x\\text{ for every }N\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $D_0=I$ and $D_N=C_{N-1}\\cdots C_0:R_0\\to R_N$. Thus $x_N=D_Nx$, $Q_N=L_ND_N$, and repeated use of $C_jJ_j=I$ gives\n\\[\n D_NL_N=I_{R_N}.\n\\]\nThis cancellation takes place between adjacent maps with matching domains. No identification of different residual spaces is needed.","\\[\n x=Q_Nx+\\sum_{j<N}L_jh_j,\n \\qquad S_Nx=\\sum_{j<N}L_jh_j.\n\\]\nEach \\(Q_N\\) is a projection, and for \\(N\\le M\\),\n\\[\n Q_NQ_M=Q_MQ_N=Q_M.\n\\]\nThe vectors with all details zero are exactly the persistent subspace\n\\[\n \\mathcal P_\\infty=\\{x:Q_Nx=x\\text{ for every }N\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $D_0=I$ and $D_N=C_{N-1}\\cdots C_0:R_0\\to R_N$. Thus $x_N=D_Nx$, $Q_N=L_ND_N$, and repeated use of $C_jJ_j=I$ gives\n\\[\n D_NL_N=I_{R_N}.\n\\]\nThis cancellation takes place between adjacent maps with matching domains. No identification of different residual spaces is needed.","j<N}L_jh_j,\n \\qquad S_Nx=\\sum_{j<N}L_jh_j.\n\\]\nEach \\(Q_N\\) is a projection, and for \\(N\\le M\\),\n\\[\n Q_NQ_M=Q_MQ_N=Q_M.\n\\]\nThe vectors with all details zero are exactly the persistent subspace\n\\[\n \\mathcal P_\\infty=\\{x:Q_Nx=x\\text{ for every }N\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $D_0=I$ and $D_N=C_{N-1}\\cdots C_0:R_0\\to R_N$. Thus $x_N=D_Nx$, $Q_N=L_ND_N$, and repeated use of $C_jJ_j=I$ gives\n\\[\n D_NL_N=I_{R_N}.\n\\]\nThis cancellation takes place between adjacent maps with matching domains. No identification of different residual spaces is needed.","Since $L_{j+1}=L_jJ_j$ and $x_{j+1}=C_jx_j$,\n\\[\n L_jh_j=L_j(x_j-J_jx_{j+1})=Q_jx-Q_{j+1}x.\n\\]\nSumming from $j=0$ to $N-1$ leaves $x-Q_Nx$ because $Q_0=I$. This proves both finite reconstruction identities, including $N=0$, when the detail sum is empty.","Idempotence follows from $D_NL_N=I$:\n\\[\n Q_N^2=L_N(D_NL_N)D_N=Q_N.\n\\]\nFor $M\\ge N$, put $C_{N,M}=C_{M-1}\\cdots C_N$ and $J_{N,M}=J_N\\cdots J_{M-1}$, using identities for $M=N$. Then\n\\[\n D_M=C_{N,M}D_N,\\quad L_M=L_NJ_{N,M},\\quad C_{N,M}J_{N,M}=I.\n\\]\nThe identities $D_NL_M=J_{N,M}$ and $D_ML_N=C_{N,M}$ now give\n\\[\n Q_NQ_M=L_NJ_{N,M}D_M=Q_M,\\qquad\n Q_MQ_N=L_MC_{N,M}D_N=Q_M.\n\\]\nBoth orders have been checked, rather than inferred from one range inclusion.","If all $h_j$ vanish, the finite telescope gives $Q_Nx=x$ for every $N$. Conversely, if every $Q_Nx=x$, then $L_jh_j=Q_jx-Q_{j+1}x=0$. Apply the left inverse $D_j$ of $L_j$ to obtain $h_j=0$. The injectivity of the restoration composite is the step that recovers the original detail from its transported value.\n\\end{proof}\n\\source{KS26}","\\subsection{Convergence has its own theorem}\nCall \\(x\\) norm recoverable when \\(Q_Nx\\to0\\). This is equivalent to \\(S_Nx\\to x\\), but it does not follow merely from vanishing persistent subspace. Let\n\\[\n \\mathcal F=\\bigcup_N\\ker Q_N\n\\]\nbe the finite core.","\\begin{theorem}[Density criterion for norm recovery]\\label{KS27}\nSuppose \\(\\sup_N\\norm{Q_N}\\le K<\\infty\\). Then the norm recoverable vectors are exactly \\(\\overline{\\mathcal F}\\). In particular all vectors are norm recoverable if and only if the finite core is dense.\n\\end{theorem}\n\\begin{proof}\nThe nested identities first imply that the kernels increase: if $Q_My=0$ and $N\\ge M$, then $Q_Ny=Q_NQ_My=0$. Thus the finite core really is an increasing union of linear subspaces.","Suppose $Q_Nx\\to0$. Put $y_N=S_Nx=x-Q_Nx$. Idempotence gives $Q_Ny_N=Q_Nx-Q_N^2x=0$, so $y_N\\in\\mathcal F$. Since $x-y_N=Q_Nx\\to0$, these are finite-core approximants converging to $x$. Hence $x\\in\\overline{\\mathcal F}$.","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach sp","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary li","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathc","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace i","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\l","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\pr","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\prod_{j<N}\\norm","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\prod_{j<N}\\norm{J_j}\\right)\\norm{x_","Conversely, fix $x\\in\\overline{\\mathcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\prod_{j<N}\\norm{J_j}\\right)\\norm{x_N}\n\\]\ngives a direct sufficient decay test. A decreasing coarse norm by itself is not enough when restoration gains g…","thcal F}$ and $\\epsilon>0$. If $K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\prod_{j<N}\\norm{J_j}\\right)\\norm{x_N}\n\\]\ngives a direct sufficient decay test. A decreasing coarse norm by itself is not enough when restoration gains grow.","$K=0$, the common operator bound gives $\\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\prod_{j<N}\\norm{J_j}\\right)\\norm{x_N}\n\\]\ngives a direct sufficient decay test. A decreasing coarse norm by itself is not enough when restoration gains grow.","\\in\\mathcal F$ satisfying $\\norm{x-y}<\\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\\ge M$, nesting and the common operator bound give\n\\[\n \\norm{Q_Nx}=\\norm{Q_N(x-y)}\\le K\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\prod_{j<N}\\norm{J_j}\\right)\\norm{x_N}\n\\]\ngives a direct sufficient decay test. A decreasing coarse norm by itself is not enough when restoration gains grow.","\\norm{x-y}\n <\\epsilon\\frac{K}{K+1}\\le\\epsilon.\n\\]\nThe approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\prod_{j<N}\\norm{J_j}\\right)\\norm{x_N}\n\\]\ngives a direct sufficient decay test. A decreasing coarse norm by itself is not enough when restoration gains grow.","mation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\prod_{j<N}\\norm{J_j}\\right)\\norm{x_N}\n\\]\ngives a direct sufficient decay test. A decreasing coarse norm by itself is not enough when restoration gains grow.","order in norm convergence. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\prod_{j<N}\\norm{J_j}\\right)\\norm{x_N}\n\\]\ngives a direct sufficient decay test. A decreasing coarse norm by itself is not enough when restoration gains grow.","nce. This proves equality of the two sets. Applying it to every $x\\in R_0$ gives the density criterion.\n\\end{proof}\n\\source{KS27}\nThere are other sufficient conditions. If the transported details are absolutely summable in the Banach space \\(R_0\\), their sum exists and the remaining boundary lies in \\(\\mathcal P_\\infty\\). If that persistent subspace is zero, the sum equals \\(x\\). Alternatively,\n\\[\n \\norm{x-S_Nx}\\le\\left(\\prod_{j<N}\\norm{J_j}\\right)\\norm{x_N}\n\\]\ngives a direct sufficient decay test. A decreasing coarse norm by itself is not enough when restoration gains grow.","\\begin{theorem}[Dyadic determination and finite coordinate recovery]\\label{KS28}\nIn the actual dyadic system, all detail coordinates jointly determine the residual. Moreover,\n\\[\n k+1<2^N\\quad\\Longrightarrow\\quad\n (Q_Nr)_k=0,\\qquad(S_Nr)_k=r_k.\n\\]\nThus every fixed residual coordinate is recovered exactly at a sufficiently large finite depth. The statement does not assert convergence in the supremum norm over all coordinates.\n\\end{theorem}\n\\begin{proof}\nWrite \\(R_j\\) for the residual space at scale \\(j\\). Its lift has the actual coordinate formula\n\\[\n (J_jz)_{2m}=0,\\qquad (J_jz)_{2m+1}=s^{(j)}_{2m}z_m,\n\\]\nwhere \\(s^{(j)}_{2m}\\) is the corresponding target section at that scale. The expression on the right is composition of the section with the residual operator. Its source and target are those of the odd coordinate at scale \\(j\\).","ery]\\label{KS28}\nIn the actual dyadic system, all detail coordinates jointly determine the residual. Moreover,\n\\[\n k+1<2^N\\quad\\Longrightarrow\\quad\n (Q_Nr)_k=0,\\qquad(S_Nr)_k=r_k.\n\\]\nThus every fixed residual coordinate is recovered exactly at a sufficiently large finite depth. The statement does not assert convergence in the supremum norm over all coordinates.\n\\end{theorem}\n\\begin{proof}\nWrite \\(R_j\\) for the residual space at scale \\(j\\). Its lift has the actual coordinate formula\n\\[\n (J_jz)_{2m}=0,\\qquad (J_jz)_{2m+1}=s^{(j)}_{2m}z_m,\n\\]\nwhere \\(s^{(j)}_{2m}\\) is the corresponding target section at that scale. The expression on the right is composition of the section with the residual operator. Its source and target are those of the odd coordinate at scale \\(j\\).","stem, all detail coordinates jointly determine the residual. Moreover,\n\\[\n k+1<2^N\\quad\\Longrightarrow\\quad\n (Q_Nr)_k=0,\\qquad(S_Nr)_k=r_k.\n\\]\nThus every fixed residual coordinate is recovered exactly at a sufficiently large finite depth. The statement does not assert convergence in the supremum norm over all coordinates.\n\\end{theorem}\n\\begin{proof}\nWrite \\(R_j\\) for the residual space at scale \\(j\\). Its lift has the actual coordinate formula\n\\[\n (J_jz)_{2m}=0,\\qquad (J_jz)_{2m+1}=s^{(j)}_{2m}z_m,\n\\]\nwhere \\(s^{(j)}_{2m}\\) is the corresponding target section at that scale. The expression on the right is composition of the section with the residual operator. Its source and target are those of the odd coordinate at scale \\(j\\).","residual. Moreover,\n\\[\n k+1<2^N\\quad\\Longrightarrow\\quad\n (Q_Nr)_k=0,\\qquad(S_Nr)_k=r_k.\n\\]\nThus every fixed residual coordinate is recovered exactly at a sufficiently large finite depth. The statement does not assert convergence in the supremum norm over all coordinates.\n\\end{theorem}\n\\begin{proof}\nWrite \\(R_j\\) for the residual space at scale \\(j\\). Its lift has the actual coordinate formula\n\\[\n (J_jz)_{2m}=0,\\qquad (J_jz)_{2m+1}=s^{(j)}_{2m}z_m,\n\\]\nwhere \\(s^{(j)}_{2m}\\) is the corresponding target section at that scale. The expression on the right is composition of the section with the residual operator. Its source and target are those of the odd coordinate at scale \\(j\\).","_k=0,\\qquad(S_Nr)_k=r_k.\n\\]\nThus every fixed residual coordinate is recovered exactly at a sufficiently large finite depth. The statement does not assert convergence in the supremum norm over all coordinates.\n\\end{theorem}\n\\begin{proof}\nWrite \\(R_j\\) for the residual space at scale \\(j\\). Its lift has the actual coordinate formula\n\\[\n (J_jz)_{2m}=0,\\qquad (J_jz)_{2m+1}=s^{(j)}_{2m}z_m,\n\\]\nwhere \\(s^{(j)}_{2m}\\) is the corresponding target section at that scale. The expression on the right is composition of the section with the residual operator. Its source and target are those of the odd coordinate at scale \\(j\\).","orem}\n\\begin{proof}\nWrite \\(R_j\\) for the residual space at scale \\(j\\). Its lift has the actual coordinate formula\n\\[\n (J_jz)_{2m}=0,\\qquad (J_jz)_{2m+1}=s^{(j)}_{2m}z_m,\n\\]\nwhere \\(s^{(j)}_{2m}\\) is the corresponding target section at that scale. The expression on the right is composition of the section with the residual operator. Its source and target are those of the odd coordinate at scale \\(j\\).","Set \\(L_{j,0}=I_{R_j}\\) and \\(L_{j,N+1}=J_jL_{j+1,N}\\). We prove, simultaneously for all starting scales \\(j\\), that\n\\[\n k+1<2^N\\quad\\Longrightarrow\\quad (L_{j,N}z)_k=0\n \\qquad(z\\in R_{j+N}).\n\\]\nFor \\(N=0\\), the inequality \\(k+1<1\\) has no solution. Suppose the assertion holds at depth \\(N\\). At depth \\(N+1\\), an even coordinate vanishes by the displayed lift formula. For an odd coordinate \\(k=2m+1\\), the hypothesis gives\n\\[\n 2(m+1)<2^{N+1},\\qquad m+1<2^N.\n\\]\nThe induction hypothesis at starting scale \\(j+1\\) makes \\((L_{j+1,N}z)_m\\) zero. Composing that zero operator with \\(s^{(j)}_{2m}\\) gives zero at coordinate \\(k\\). This completes the induction without identifying residual spaces at different scales.","\\(L_{j,N+1}=J_jL_{j+1,N}\\). We prove, simultaneously for all starting scales \\(j\\), that\n\\[\n k+1<2^N\\quad\\Longrightarrow\\quad (L_{j,N}z)_k=0\n \\qquad(z\\in R_{j+N}).\n\\]\nFor \\(N=0\\), the inequality \\(k+1<1\\) has no solution. Suppose the assertion holds at depth \\(N\\). At depth \\(N+1\\), an even coordinate vanishes by the displayed lift formula. For an odd coordinate \\(k=2m+1\\), the hypothesis gives\n\\[\n 2(m+1)<2^{N+1},\\qquad m+1<2^N.\n\\]\nThe induction hypothesis at starting scale \\(j+1\\) makes \\((L_{j+1,N}z)_m\\) zero. Composing that zero operator with \\(s^{(j)}_{2m}\\) gives zero at coordinate \\(k\\). This completes the induction without identifying residual spaces at different scales.","), that\n\\[\n k+1<2^N\\quad\\Longrightarrow\\quad (L_{j,N}z)_k=0\n \\qquad(z\\in R_{j+N}).\n\\]\nFor \\(N=0\\), the inequality \\(k+1<1\\) has no solution. Suppose the assertion holds at depth \\(N\\). At depth \\(N+1\\), an even coordinate vanishes by the displayed lift formula. For an odd coordinate \\(k=2m+1\\), the hypothesis gives\n\\[\n 2(m+1)<2^{N+1},\\qquad m+1<2^N.\n\\]\nThe induction hypothesis at starting scale \\(j+1\\) makes \\((L_{j+1,N}z)_m\\) zero. Composing that zero operator with \\(s^{(j)}_{2m}\\) gives zero at coordinate \\(k\\). This completes the induction without identifying residual spaces at different scales.","1<2^N\\quad\\Longrightarrow\\quad (L_{j,N}z)_k=0\n \\qquad(z\\in R_{j+N}).\n\\]\nFor \\(N=0\\), the inequality \\(k+1<1\\) has no solution. Suppose the assertion holds at depth \\(N\\). At depth \\(N+1\\), an even coordinate vanishes by the displayed lift formula. For an odd coordinate \\(k=2m+1\\), the hypothesis gives\n\\[\n 2(m+1)<2^{N+1},\\qquad m+1<2^N.\n\\]\nThe induction hypothesis at starting scale \\(j+1\\) makes \\((L_{j+1,N}z)_m\\) zero. Composing that zero operator with \\(s^{(j)}_{2m}\\) gives zero at coordinate \\(k\\). This completes the induction without identifying residual spaces at different scales.","R_{j+N}).\n\\]\nFor \\(N=0\\), the inequality \\(k+1<1\\) has no solution. Suppose the assertion holds at depth \\(N\\). At depth \\(N+1\\), an even coordinate vanishes by the displayed lift formula. For an odd coordinate \\(k=2m+1\\), the hypothesis gives\n\\[\n 2(m+1)<2^{N+1},\\qquad m+1<2^N.\n\\]\nThe induction hypothesis at starting scale \\(j+1\\) makes \\((L_{j+1,N}z)_m\\) zero. Composing that zero operator with \\(s^{(j)}_{2m}\\) gives zero at coordinate \\(k\\). This completes the induction without identifying residual spaces at different scales.","(k+1<1\\) has no solution. Suppose the assertion holds at depth \\(N\\). At depth \\(N+1\\), an even coordinate vanishes by the displayed lift formula. For an odd coordinate \\(k=2m+1\\), the hypothesis gives\n\\[\n 2(m+1)<2^{N+1},\\qquad m+1<2^N.\n\\]\nThe induction hypothesis at starting scale \\(j+1\\) makes \\((L_{j+1,N}z)_m\\) zero. Composing that zero operator with \\(s^{(j)}_{2m}\\) gives zero at coordinate \\(k\\). This completes the induction without identifying residual spaces at different scales.","Now \\(Q_Nr=L_{0,N}x_N\\), so the support assertion applies to the restored remainder. Since \\(S_Nr=r-Q_Nr\\), it gives the reconstruction equation. Finally, if two residuals have identical details at every scale, the finite telescope makes their \\(S_N\\) reconstructions equal for every \\(N\\). At any fixed coordinate choose \\(N\\) with \\(k+1<2^N\\). The reconstruction equation then makes their original coordinates equal. Hence the residuals are equal.\n\\end{proof}\n\\source{KS28}\nThe strict inequality matters: the next coordinate, \\(k=2^N-1\\), can still have a nonzero remainder. There is also a sharper, nonconsecutive description of the support.","ction equation. Finally, if two residuals have identical details at every scale, the finite telescope makes their \\(S_N\\) reconstructions equal for every \\(N\\). At any fixed coordinate choose \\(N\\) with \\(k+1<2^N\\). The reconstruction equation then makes their original coordinates equal. Hence the residuals are equal.\n\\end{proof}\n\\source{KS28}\nThe strict inequality matters: the next coordinate, \\(k=2^N-1\\), can still have a nonzero remainder. There is also a sharper, nonconsecutive description of the support.","\\begin{proposition}[The dyadic support pattern]\\label{KS55}\nFor the same lifts,\n\\[\n 2^N\\nmid(k+1)\\quad\\Longrightarrow\\quad(Q_Nr)_k=0.\n\\]\nThus coordinate \\(k\\) is certainly recovered once \\(N>\\nu_2(k+1)\\), where \\(\\nu_2(m)\\) is the largest power-of-two exponent dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The fo","\\begin{proposition}[The dyadic support pattern]\\label{KS55}\nFor the same lifts,\n\\[\n 2^N\\nmid(k+1)\\quad\\Longrightarrow\\quad(Q_Nr)_k=0.\n\\]\nThus coordinate \\(k\\) is certainly recovered once \\(N>\\nu_2(k+1)\\), where \\(\\nu_2(m)\\) is the largest power-of-two exponent dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift","\\begin{proposition}[The dyadic support pattern]\\label{KS55}\nFor the same lifts,\n\\[\n 2^N\\nmid(k+1)\\quad\\Longrightarrow\\quad(Q_Nr)_k=0.\n\\]\nThus coordinate \\(k\\) is certainly recovered once \\(N>\\nu_2(k+1)\\), where \\(\\nu_2(m)\\) is the largest power-of-two exponent dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_","\\begin{proposition}[The dyadic support pattern]\\label{KS55}\nFor the same lifts,\n\\[\n 2^N\\nmid(k+1)\\quad\\Longrightarrow\\quad(Q_Nr)_k=0.\n\\]\nThus coordinate \\(k\\) is certainly recovered once \\(N>\\nu_2(k+1)\\), where \\(\\nu_2(m)\\) is the largest power-of-two exponent dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\en","\\begin{proposition}[The dyadic support pattern]\\label{KS55}\nFor the same lifts,\n\\[\n 2^N\\nmid(k+1)\\quad\\Longrightarrow\\quad(Q_Nr)_k=0.\n\\]\nThus coordinate \\(k\\) is certainly recovered once \\(N>\\nu_2(k+1)\\), where \\(\\nu_2(m)\\) is the largest power-of-two exponent dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform i","\\begin{proposition}[The dyadic support pattern]\\label{KS55}\nFor the same lifts,\n\\[\n 2^N\\nmid(k+1)\\quad\\Longrightarrow\\quad(Q_Nr)_k=0.\n\\]\nThus coordinate \\(k\\) is certainly recovered once \\(N>\\nu_2(k+1)\\), where \\(\\nu_2(m)\\) is the largest power-of-two exponent dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\c","\\begin{proposition}[The dyadic support pattern]\\label{KS55}\nFor the same lifts,\n\\[\n 2^N\\nmid(k+1)\\quad\\Longrightarrow\\quad(Q_Nr)_k=0.\n\\]\nThus coordinate \\(k\\) is certainly recovered once \\(N>\\nu_2(k+1)\\), where \\(\\nu_2(m)\\) is the largest power-of-two exponent dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\code{DyadicSupport.lean}. It asserts where a rema…",",\n\\[\n 2^N\\nmid(k+1)\\quad\\Longrightarrow\\quad(Q_Nr)_k=0.\n\\]\nThus coordinate \\(k\\) is certainly recovered once \\(N>\\nu_2(k+1)\\), where \\(\\nu_2(m)\\) is the largest power-of-two exponent dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\code{DyadicSupport.lean}. It asserts where a remainder \\emph{may} survive, not that every allowed coordinate is nonzero. In the…","\\quad(Q_Nr)_k=0.\n\\]\nThus coordinate \\(k\\) is certainly recovered once \\(N>\\nu_2(k+1)\\), where \\(\\nu_2(m)\\) is the largest power-of-two exponent dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\code{DyadicSupport.lean}. It asserts where a remainder \\emph{may} survive, not that every allowed coordinate is nonzero. In the scalar identity-bonding model with \\(C…","\\) is the largest power-of-two exponent dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\code{DyadicSupport.lean}. It asserts where a remainder \\emph{may} survive, not that every allowed coordinate is nonzero. In the scalar identity-bonding model with \\(C(a,b)=a+b\\), the residual constantly equal to one has value \\(2^N\\) at every surviving position. That ex…","nt dividing the positive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\code{DyadicSupport.lean}. It asserts where a remainder \\emph{may} survive, not that every allowed coordinate is nonzero. In the scalar identity-bonding model with \\(C(a,b)=a+b\\), the residual constantly equal to one has value \\(2^N\\) at every surviving position. That example proves sharpness of the uniform…","ositive integer \\(m\\). The only possible unrecovered coordinates at depth \\(N\\) are\n\\[\n 2^N-1,\\quad 2\\cdot2^N-1,\\quad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\code{DyadicSupport.lean}. It asserts where a remainder \\emph{may} survive, not that every allowed coordinate is nonzero. In the scalar identity-bonding model with \\(C(a,b)=a+b\\), the residual constantly equal to one has value \\(2^N\\) at every surviving position. That example proves sharpness of the uniform prefix cutoff.","ad3\\cdot2^N-1,\\quad\\ldots.\n\\]\n\\end{proposition}\n\\begin{proof}\nProve the assertion for \\(L_{j,N}\\) by induction, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\code{DyadicSupport.lean}. It asserts where a remainder \\emph{may} survive, not that every allowed coordinate is nonzero. In the scalar identity-bonding model with \\(C(a,b)=a+b\\), the residual constantly equal to one has value \\(2^N\\) at every surviving position. That example proves sharpness of the uniform prefix cutoff.","n, uniformly in \\(j\\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \\(N+1\\), even coordinates vanish. If \\(k=2m+1\\), then\n\\[\n 2^{N+1}\\mid(k+1)\\quad\\Longleftrightarrow\\quad2^N\\mid(m+1).\n\\]\nWhen divisibility fails, the inner coordinate in \\(J_jL_{j+1,N}\\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\code{DyadicSupport.lean}. It asserts where a remainder \\emph{may} survive, not that every allowed coordinate is nonzero. In the scalar identity-bonding model with \\(C(a,b)=a+b\\), the residual constantly equal to one has value \\(2^N\\) at every surviving position. That example proves sharpness of the uniform prefix cutoff.","induction. The formula for the odd lift finishes the proof. Apply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\code{DyadicSupport.lean}. It asserts where a remainder \\emph{may} survive, not that every allowed coordinate is nonzero. In the scalar identity-bonding model with \\(C(a,b)=a+b\\), the residual constantly equal to one has value \\(2^N\\) at every surviving position. That example proves sharpness of the uniform prefix cutoff.","ply the result to \\(Q_Nr=L_{0,N}x_N\\).\n\\end{proof}\n\\source{KS55}\nThis refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \\code{DyadicSupport.lean}. It asserts where a remainder \\emph{may} survive, not that every allowed coordinate is nonzero. In the scalar identity-bonding model with \\(C(a,b)=a+b\\), the residual constantly equal to one has value \\(2^N\\) at every surviving position. That example proves sharpness of the uniform prefix cutoff.","\\plate{16_dyadic_support}{The possible support of the restored dyadic remainder is sparse and periodic. At depth \\(N\\), only positions whose index plus one is divisible by \\(2^N\\) can remain. Open cells are guaranteed recovered. Marked cells are allowed to survive, not asserted nonzero for every residual.}{fig:dyadicsupport}","\\subsection{Two boundary examples}\nOn \\(\\ell^\\infty(\\N,\\R)\\), let \\(C\\) remove the first coordinate and let \\(J\\) prepend zero. Then \\(Q_Nx\\) is the tail of \\(x\\), placed at its original coordinates with zeros before \\(N\\). Every fixed coordinate is eventually zero in the remainder, and \\(\\norm{Q_N}\\le1\\). Nevertheless the constant sequence \\(\\mathbf1\\) satisfies \\(\\norm{Q_N\\mathbf1}=1\\) for every \\(N\\).","bsection{Two boundary examples}\nOn \\(\\ell^\\infty(\\N,\\R)\\), let \\(C\\) remove the first coordinate and let \\(J\\) prepend zero. Then \\(Q_Nx\\) is the tail of \\(x\\), placed at its original coordinates with zeros before \\(N\\). Every fixed coordinate is eventually zero in the remainder, and \\(\\norm{Q_N}\\le1\\). Nevertheless the constant sequence \\(\\mathbf1\\) satisfies \\(\\norm{Q_N\\mathbf1}=1\\) for every \\(N\\).","\\begin{proposition}[The recovery domain is \\(c_0\\)]\\label{KS29}\nFor this shift cascade,\n\\[\n S_Nx\\to x\\text{ in }\\ell^\\infty\n \\quad\\Longleftrightarrow\\quad x_n\\to0.\n\\]\nFor \\(x_n=q^n\\), \\(0\\le q<1\\), the error is exactly \\(q^N\\).\n\\end{proposition}\n\\begin{proof}\nThe error norm is \\(\\sup_{n\\ge N}|x_n|\\). It tends to zero precisely when the sequence tends to zero. For a geometric sequence its largest tail value is the first one, \\(q^N\\).\n\\end{proof}\n\\source{KS29}\nAs a second example, take scalar spaces with \\(C=\\tfrac12I\\), \\(J=2I\\). The coarse value \\(x_N=2^{-N}x\\) tends to zero, but restoration gives \\(Q_Nx=x\\). Every detail is zero and the entire input is persistent. The example belongs to this scalar cascade, not to the actual dyadic hiera","\\begin{proposition}[The recovery domain is \\(c_0\\)]\\label{KS29}\nFor this shift cascade,\n\\[\n S_Nx\\to x\\text{ in }\\ell^\\infty\n \\quad\\Longleftrightarrow\\quad x_n\\to0.\n\\]\nFor \\(x_n=q^n\\), \\(0\\le q<1\\), the error is exactly \\(q^N\\).\n\\end{proposition}\n\\begin{proof}\nThe error norm is \\(\\sup_{n\\ge N}|x_n|\\). It tends to zero precisely when the sequence tends to zero. For a geometric sequence its largest tail value is the first one, \\(q^N\\).\n\\end{proof}\n\\source{KS29}\nAs a second example, take scalar spaces with \\(C=\\tfrac12I\\), \\(J=2I\\). The coarse value \\(x_N=2^{-N}x\\) tends to zero, but restoration gives \\(Q_Nx=x\\). Every detail is zero and the entire input is persistent. The example belongs to this scalar cascade, not to the actual dyadic hierarchy, whose persistent subspace was proved to vanish.","ry domain is \\(c_0\\)]\\label{KS29}\nFor this shift cascade,\n\\[\n S_Nx\\to x\\text{ in }\\ell^\\infty\n \\quad\\Longleftrightarrow\\quad x_n\\to0.\n\\]\nFor \\(x_n=q^n\\), \\(0\\le q<1\\), the error is exactly \\(q^N\\).\n\\end{proposition}\n\\begin{proof}\nThe error norm is \\(\\sup_{n\\ge N}|x_n|\\). It tends to zero precisely when the sequence tends to zero. For a geometric sequence its largest tail value is the first one, \\(q^N\\).\n\\end{proof}\n\\source{KS29}\nAs a second example, take scalar spaces with \\(C=\\tfrac12I\\), \\(J=2I\\). The coarse value \\(x_N=2^{-N}x\\) tends to zero, but restoration gives \\(Q_Nx=x\\). Every detail is zero and the entire input is persistent. The example belongs to this scalar cascade, not to the actual dyadic hierarchy, whose persistent subspace was proved to vanish.","}\\ell^\\infty\n \\quad\\Longleftrightarrow\\quad x_n\\to0.\n\\]\nFor \\(x_n=q^n\\), \\(0\\le q<1\\), the error is exactly \\(q^N\\).\n\\end{proposition}\n\\begin{proof}\nThe error norm is \\(\\sup_{n\\ge N}|x_n|\\). It tends to zero precisely when the sequence tends to zero. For a geometric sequence its largest tail value is the first one, \\(q^N\\).\n\\end{proof}\n\\source{KS29}\nAs a second example, take scalar spaces with \\(C=\\tfrac12I\\), \\(J=2I\\). The coarse value \\(x_N=2^{-N}x\\) tends to zero, but restoration gives \\(Q_Nx=x\\). Every detail is zero and the entire input is persistent. The example belongs to this scalar cascade, not to the actual dyadic hierarchy, whose persistent subspace was proved to vanish.","uad\\Longleftrightarrow\\quad x_n\\to0.\n\\]\nFor \\(x_n=q^n\\), \\(0\\le q<1\\), the error is exactly \\(q^N\\).\n\\end{proposition}\n\\begin{proof}\nThe error norm is \\(\\sup_{n\\ge N}|x_n|\\). It tends to zero precisely when the sequence tends to zero. For a geometric sequence its largest tail value is the first one, \\(q^N\\).\n\\end{proof}\n\\source{KS29}\nAs a second example, take scalar spaces with \\(C=\\tfrac12I\\), \\(J=2I\\). The coarse value \\(x_N=2^{-N}x\\) tends to zero, but restoration gives \\(Q_Nx=x\\). Every detail is zero and the entire input is persistent. The example belongs to this scalar cascade, not to the actual dyadic hierarchy, whose persistent subspace was proved to vanish.","x_n\\to0.\n\\]\nFor \\(x_n=q^n\\), \\(0\\le q<1\\), the error is exactly \\(q^N\\).\n\\end{proposition}\n\\begin{proof}\nThe error norm is \\(\\sup_{n\\ge N}|x_n|\\). It tends to zero precisely when the sequence tends to zero. For a geometric sequence its largest tail value is the first one, \\(q^N\\).\n\\end{proof}\n\\source{KS29}\nAs a second example, take scalar spaces with \\(C=\\tfrac12I\\), \\(J=2I\\). The coarse value \\(x_N=2^{-N}x\\) tends to zero, but restoration gives \\(Q_Nx=x\\). Every detail is zero and the entire input is persistent. The example belongs to this scalar cascade, not to the actual dyadic hierarchy, whose persistent subspace was proved to vanish.","is exactly \\(q^N\\).\n\\end{proposition}\n\\begin{proof}\nThe error norm is \\(\\sup_{n\\ge N}|x_n|\\). It tends to zero precisely when the sequence tends to zero. For a geometric sequence its largest tail value is the first one, \\(q^N\\).\n\\end{proof}\n\\source{KS29}\nAs a second example, take scalar spaces with \\(C=\\tfrac12I\\), \\(J=2I\\). The coarse value \\(x_N=2^{-N}x\\) tends to zero, but restoration gives \\(Q_Nx=x\\). Every detail is zero and the entire input is persistent. The example belongs to this scalar cascade, not to the actual dyadic hierarchy, whose persistent subspace was proved to vanish.","\\subsection{An output level may require a finer input level}\nLet \\(X\\) and \\(Y\\) be sequential set-valued resolutions. Write \\(\\rho^X_{i,j}:X_j\\to X_i\\) for composite restriction. A reindexed transport consists of a monotone schedule \\(\\phi:\\N\\to\\N\\) and maps\n\\[\n f_n:X_{\\phi(n)}\\longrightarrow Y_n\n\\]\nsatisfying, for \\(i\\le j\\),\n\\begin{equation}\\label{eq:scheduled-naturality}\n \\rho^Y_{i,j}f_j=f_i\\rho^X_{\\phi(i),\\phi(j)}.\n\\end{equation}\nThe schedule records how much input is used for each output level. Monotonicity means that asking for more output never forces the representation to discard input already available. It does not require strict growth, a bounded look-ahead, or an efficiency estimate.","The realized map is\n\\[\n |f|:\\operatorname{Point}(X)\\to\\operatorname{Point}(Y),\\qquad\n |f|(x)_n=f_n(x_{\\phi(n)}).\n\\]\nEquation~\\eqref{eq:scheduled-naturality} proves that this is a coherent output. A second transport \\(g_n:Y_{\\psi(n)}\\to Z_n\\) composes with \\(f\\) using schedule \\(\\phi\\circ\\psi\\) and coordinate map \\(g_n f_{\\psi(n)}\\). The order of the schedules is fixed by the types, not by the order in which their letters are written in prose.","\\subsection{Equality is witnessed at finite stages}\nTwo different schedules can represent the same transformation. To compare \\((\\phi,f)\\) and \\((\\psi,g)\\), require that for every output level \\(n\\) there be a finite level \\(k\\ge\\phi(n),\\psi(n)\\) with\n\\begin{equation}\\label{eq:eventual-equality}\n f_n\\rho^X_{\\phi(n),k}=g_n\\rho^X_{\\psi(n),k}\n \\quad\\text{as functions on all of }X_k.\n\\end{equation}\nCall this eventual equality. The witnessing \\(k\\) may depend on \\(n\\). Equality is required on the whole finite-level fiber, not only on states that happen to extend globally.","stages}\nTwo different schedules can represent the same transformation. To compare \\((\\phi,f)\\) and \\((\\psi,g)\\), require that for every output level \\(n\\) there be a finite level \\(k\\ge\\phi(n),\\psi(n)\\) with\n\\begin{equation}\\label{eq:eventual-equality}\n f_n\\rho^X_{\\phi(n),k}=g_n\\rho^X_{\\psi(n),k}\n \\quad\\text{as functions on all of }X_k.\n\\end{equation}\nCall this eventual equality. The witnessing \\(k\\) may depend on \\(n\\). Equality is required on the whole finite-level fiber, not only on states that happen to extend globally.","t schedules can represent the same transformation. To compare \\((\\phi,f)\\) and \\((\\psi,g)\\), require that for every output level \\(n\\) there be a finite level \\(k\\ge\\phi(n),\\psi(n)\\) with\n\\begin{equation}\\label{eq:eventual-equality}\n f_n\\rho^X_{\\phi(n),k}=g_n\\rho^X_{\\psi(n),k}\n \\quad\\text{as functions on all of }X_k.\n\\end{equation}\nCall this eventual equality. The witnessing \\(k\\) may depend on \\(n\\). Equality is required on the whole finite-level fiber, not only on states that happen to extend globally.","ntual transport category]\\label{KS61}\nEventual equality is an equivalence relation and is preserved by composition. Its quotient therefore gives a category of resolutions and eventual transports. Realization descends to a functor into sets, preserving identities and composition.\n\\end{theorem}\n\\begin{proof}\nA finite-stage equality remains true after moving to a later stage. Indeed, if it holds on $X_k$ and $\\ell\\ge k$, precompose both sides with $\\rho^X_{k,\\ell}$ and use the composite-restriction laws. Reflexivity uses $k=\\phi(n)$, symmetry reverses the equality, and transitivity moves two witnesses to their maximum. The shared middle presentation then cancels as a literal function on that common fiber. This proves an equivalence relation without referring to global points.","sport category]\\label{KS61}\nEventual equality is an equivalence relation and is preserved by composition. Its quotient therefore gives a category of resolutions and eventual transports. Realization descends to a functor into sets, preserving identities and composition.\n\\end{theorem}\n\\begin{proof}\nA finite-stage equality remains true after moving to a later stage. Indeed, if it holds on $X_k$ and $\\ell\\ge k$, precompose both sides with $\\rho^X_{k,\\ell}$ and use the composite-restriction laws. Reflexivity uses $k=\\phi(n)$, symmetry reverses the equality, and transitivity moves two witnesses to their maximum. The shared middle presentation then cancels as a literal function on that common fiber. This proves an equivalence relation without referring to global points.","ality is an equivalence relation and is preserved by composition. Its quotient therefore gives a category of resolutions and eventual transports. Realization descends to a functor into sets, preserving identities and composition.\n\\end{theorem}\n\\begin{proof}\nA finite-stage equality remains true after moving to a later stage. Indeed, if it holds on $X_k$ and $\\ell\\ge k$, precompose both sides with $\\rho^X_{k,\\ell}$ and use the composite-restriction laws. Reflexivity uses $k=\\phi(n)$, symmetry reverses the equality, and transitivity moves two witnesses to their maximum. The shared middle presentation then cancels as a literal function on that common fiber. This proves an equivalence relation without referring to global points.","ent therefore gives a category of resolutions and eventual transports. Realization descends to a functor into sets, preserving identities and composition.\n\\end{theorem}\n\\begin{proof}\nA finite-stage equality remains true after moving to a later stage. Indeed, if it holds on $X_k$ and $\\ell\\ge k$, precompose both sides with $\\rho^X_{k,\\ell}$ and use the composite-restriction laws. Reflexivity uses $k=\\phi(n)$, symmetry reverses the equality, and transitivity moves two witnesses to their maximum. The shared middle presentation then cancels as a literal function on that common fiber. This proves an equivalence relation without referring to global points.","We check compatibility with composition with all indices displayed. Write $f,f'$ with schedules $\\phi,\\phi'$ and $g,g'$ with schedules $\\psi,\\psi'$. Fix an output level $n$. Choose $j\\ge\\psi(n),\\psi'(n)$ with\n\\[\n g_n\\rho^Y_{\\psi(n),j}=g'_n\\rho^Y_{\\psi'(n),j}.\n\\]\nNext choose $k\\ge\\phi(j),\\phi'(j)$ with\n\\[\n f_j\\rho^X_{\\phi(j),k}=f'_j\\rho^X_{\\phi'(j),k}.\n\\]\nMonotonicity implies that $k$ is also at least both composite input levels. Applying the naturality equations and then the two witness equations gives\n\\[\n\\begin{aligned}\n g_nf_{\\psi(n)}\\rho^X_{\\phi(\\psi(n)),k}\n &=g_n\\rho^Y_{\\psi(n),j}f_j\\rho^X_{\\phi(j),k}","nd $g,g'$ with schedules $\\psi,\\psi'$. Fix an output level $n$. Choose $j\\ge\\psi(n),\\psi'(n)$ with\n\\[\n g_n\\rho^Y_{\\psi(n),j}=g'_n\\rho^Y_{\\psi'(n),j}.\n\\]\nNext choose $k\\ge\\phi(j),\\phi'(j)$ with\n\\[\n f_j\\rho^X_{\\phi(j),k}=f'_j\\rho^X_{\\phi'(j),k}.\n\\]\nMonotonicity implies that $k$ is also at least both composite input levels. Applying the naturality equations and then the two witness equations gives\n\\[\n\\begin{aligned}\n g_nf_{\\psi(n)}\\rho^X_{\\phi(\\psi(n)),k}\n &=g_n\\rho^Y_{\\psi(n),j}f_j\\rho^X_{\\phi(j),k}\\\\\n &=g'_n\\rho^Y_{\\psi'(n),j}f_j\\rho^X_{\\phi(j),k}\\\\\n &=g'_n\\rho^Y_{\\psi'(n),j}f'_j\\rho^X_{\\phi'(j),k}\\\\\n &=g'_nf'_{\\psi'(n)}\\rho^X_{\\phi'(\\psi'(n)),k}.\n\\end{aligned}\n\\]\nThis is the required equality on every element of $X_k$. Thus composition descends to classes. The representative composition is associative because function composition and schedule composition are associative. The schedule $n\\mapsto n$ with identity coordinate maps gives both identity laws. Hence the quotient has the claimed category structure.","&=g'_nf'_{\\psi'(n)}\\rho^X_{\\phi'(\\psi'(n)),k}.\n\\end{aligned}\n\\]\nThis is the required equality on every element of $X_k$. Thus composition descends to classes. The representative composition is associative because function composition and schedule composition are associative. The schedule $n\\mapsto n$ with identity coordinate maps gives both identity laws. Hence the quotient has the claimed category structure.","Finally, evaluate a witness equation at the $k$th coordinate of a coherent point $x$. Its two restricted arguments become $x_{\\phi(n)}$ and $x_{\\psi(n)}$, proving equality of the realized outputs. Thus realization is well defined on the quotient. For a composite, direct evaluation gives\n\\[\n |g\\circ f|(x)_n=g_n\\bigl(f_{\\psi(n)}(x_{\\phi(\\psi(n))})\\bigr)\n =(|g|(|f|(x)))_n.\n\\]\nThe identity law is equally coordinatewise. Joint determination of output points proves the functor laws.\n\\end{proof}\n\\source{KS61}\nThis is an actual quotient category in the formal source. No identification with the full procategory of an arbitrary base category is needed or claimed. The objects here are sequential set-valued resolutions, and the equivalence relation","Finally, evaluate a witness equation at the $k$th coordinate of a coherent point $x$. Its two restricted arguments become $x_{\\phi(n)}$ and $x_{\\psi(n)}$, proving equality of the realized outputs. Thus realization is well defined on the quotient. For a composite, direct evaluation gives\n\\[\n |g\\circ f|(x)_n=g_n\\bigl(f_{\\psi(n)}(x_{\\phi(\\psi(n))})\\bigr)\n =(|g|(|f|(x)))_n.\n\\]\nThe identity law is equally coordinatewise. Joint determination of output points proves the functor laws.\n\\end{proof}\n\\source{KS61}\nThis is an actual quotient category in the formal source. No identification with the full procategory of an arbitrary base category is needed or claimed. The objects here are sequential set-valued resolutions, and the equivalence relation is the finite-stage relation just de","Finally, evaluate a witness equation at the $k$th coordinate of a coherent point $x$. Its two restricted arguments become $x_{\\phi(n)}$ and $x_{\\psi(n)}$, proving equality of the realized outputs. Thus realization is well defined on the quotient. For a composite, direct evaluation gives\n\\[\n |g\\circ f|(x)_n=g_n\\bigl(f_{\\psi(n)}(x_{\\phi(\\psi(n))})\\bigr)\n =(|g|(|f|(x)))_n.\n\\]\nThe identity law is equally coordinatewise. Joint determination of output points proves the functor laws.\n\\end{proof}\n\\source{KS61}\nThis is an actual quotient category in the formal source. No identification with the full procategory of an arbitrary base category is needed or claimed. The objects here are sequential set-valued resolutions, and the equivalence relation is the finite-stage relation just defined.","\\subsection{When global behavior determines the finite presentation}\nSay that \\(X\\) has surjective point projections when\n\\[\n \\forall k\\quad \\operatorname{ev}_k:\\operatorname{Point}(X)\\to X_k\n \\quad\\text{is onto}.\n\\]\nThis means that every finite input state really occurs in some coherent global input. It is stronger than nonemptiness of the global space.","\\begin{theorem}[Faithfulness on an observable source]\\label{KS62}\nIf \\(X\\) has surjective point projections, two reindexed maps out of \\(X\\) are eventually equal exactly when they induce the same realized function. Surjective adjacent bonding maps are a sufficient condition for surjective point projections.\n\\end{theorem}\n\\begin{proof}\nRealized equality follows from eventual equality by the preceding theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of t","\\begin{theorem}[Faithfulness on an observable source]\\label{KS62}\nIf \\(X\\) has surjective point projections, two reindexed maps out of \\(X\\) are eventually equal exactly when they induce the same realized function. Surjective adjacent bonding maps are a sufficient condition for surjective point projections.\n\\end{theorem}\n\\begin{proof}\nRealized equality follows from eventual equality by the preceding theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]","]\\label{KS62}\nIf \\(X\\) has surjective point projections, two reindexed maps out of \\(X\\) are eventually equal exactly when they induce the same realized function. Surjective adjacent bonding maps are a sufficient condition for surjective point projections.\n\\end{theorem}\n\\begin{proof}\nRealized equality follows from eventual equality by the preceding theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\…","ive point projections, two reindexed maps out of \\(X\\) are eventually equal exactly when they induce the same realized function. Surjective adjacent bonding maps are a sufficient condition for surjective point projections.\n\\end{theorem}\n\\begin{proof}\nRealized equality follows from eventual equality by the preceding theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves","ctions, two reindexed maps out of \\(X\\) are eventually equal exactly when they induce the same realized function. Surjective adjacent bonding maps are a sufficient condition for surjective point projections.\n\\end{theorem}\n\\begin{proof}\nRealized equality follows from eventual equality by the preceding theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","n they induce the same realized function. Surjective adjacent bonding maps are a sufficient condition for surjective point projections.\n\\end{theorem}\n\\begin{proof}\nRealized equality follows from eventual equality by the preceding theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","nction. Surjective adjacent bonding maps are a sufficient condition for surjective point projections.\n\\end{theorem}\n\\begin{proof}\nRealized equality follows from eventual equality by the preceding theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","re a sufficient condition for surjective point projections.\n\\end{theorem}\n\\begin{proof}\nRealized equality follows from eventual equality by the preceding theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","jections.\n\\end{theorem}\n\\begin{proof}\nRealized equality follows from eventual equality by the preceding theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","end{theorem}\n\\begin{proof}\nRealized equality follows from eventual equality by the preceding theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","ing theorem, so only its converse remains. Suppose \\(|f|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","|=|g|\\). Fix the output level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","level \\(n\\), and let \\(k=\\max\\{\\phi(n),\\psi(n)\\}\\). To prove equality of two maps defined on \\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","\\(X_k\\), fix an arbitrary element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","element \\(a\\in X_k\\). The assumed surjectivity of \\(\\operatorname{ev}_k\\) supplies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","lies a coherent point \\(x\\) with \\(x_k=a\\). Equality of the two realized maps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","ps gives equality of their output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","output coordinates:\n\\[\n f_n(x_{\\phi(n)})=g_n(x_{\\psi(n)}).\n\\]\nCoherence of \\(x\\) rewrites this as\n\\[\n f_n\\bigl(\\rho^X_{\\phi(n),k}(a)\\bigr)\n =g_n\\bigl(\\rho^X_{\\psi(n),k}(a)\\bigr).\n\\]\nThe index \\(k\\) was selected before \\(a\\), so this is one common finite-stage equality for \\emph{all} elements of \\(X_k\\). It is exactly the witness required by eventual equality at output \\(n\\). Repeating for each \\(n\\) proves faithfulness.","For the sufficient condition on bonding maps, fix \\(k\\) and \\(a\\in X_k\\). Define coordinates below \\(k\\) by restriction of \\(a\\). At each later stage use surjectivity of the adjacent bonding map to choose a preimage of the coordinate already chosen. Recursion with classical choice constructs the upper tail. Adjacent equations hold by construction, and induction on the interval length gives every composite equation. This produces a point evaluating to \\(a\\). If a fiber is empty, the statement that each of its elements lifts is vacuous, as required by surjectivity onto that fiber. Without the projection-surjectivity assumption, equality on realizable inputs alone need not identify finite maps on values that never occur in a global point.\n\\end{pr","For the sufficient condition on bonding maps, fix \\(k\\) and \\(a\\in X_k\\). Define coordinates below \\(k\\) by restriction of \\(a\\). At each later stage use surjectivity of the adjacent bonding map to choose a preimage of the coordinate already chosen. Recursion with classical choice constructs the upper tail. Adjacent equations hold by construction, and induction on the interval length gives every composite equation. This produces a point evaluating to \\(a\\). If a fiber is empty, the statement that each of its elements lifts is vacuous, as required by surjectivity onto that fiber. Without the projection-surjectivity assumption, equality on realizable inputs alone need not identify finite maps on values that never occur in a global point.\n\\end{proof}\n\\source{KS62","For the sufficient condition on bonding maps, fix \\(k\\) and \\(a\\in X_k\\). Define coordinates below \\(k\\) by restriction of \\(a\\). At each later stage use surjectivity of the adjacent bonding map to choose a preimage of the coordinate already chosen. Recursion with classical choice constructs the upper tail. Adjacent equations hold by construction, and induction on the interval length gives every composite equation. This produces a point evaluating to \\(a\\). If a fiber is empty, the statement that each of its elements lifts is vacuous, as required by surjectivity onto that fiber. Without the projection-surjectivity assumption, equality on realizable inputs alone need not identify finite maps on values that never occur in a global point.\n\\end{proof}\n\\source{KS62}\nThe hypothesis is sufficient,","For the sufficient condition on bonding maps, fix \\(k\\) and \\(a\\in X_k\\). Define coordinates below \\(k\\) by restriction of \\(a\\). At each later stage use surjectivity of the adjacent bonding map to choose a preimage of the coordinate already chosen. Recursion with classical choice constructs the upper tail. Adjacent equations hold by construction, and induction on the interval length gives every composite equation. This produces a point evaluating to \\(a\\). If a fiber is empty, the statement that each of its elements lifts is vacuous, as required by surjectivity onto that fiber. Without the projection-surjectivity assumption, equality on realizable inputs alone need not identify finite maps on values that never occur in a global point.\n\\end{proof}\n\\source{KS62}\nThe hypothesis is sufficient, not presented as a chara","nd \\(a\\in X_k\\). Define coordinates below \\(k\\) by restriction of \\(a\\). At each later stage use surjectivity of the adjacent bonding map to choose a preimage of the coordinate already chosen. Recursion with classical choice constructs the upper tail. Adjacent equations hold by construction, and induction on the interval length gives every composite equation. This produces a point evaluating to \\(a\\). If a fiber is empty, the statement that each of its elements lifts is vacuous, as required by surjectivity onto that fiber. Without the projection-surjectivity assumption, equality on realizable inputs alone need not identify finite maps on values that never occur in a global point.\n\\end{proof}\n\\source{KS62}\nThe hypothesis is sufficient, not presented as a characterization of every source on which faithfulness happens to hold. Its role in the proof is precise: it converts a statement about globally realized inputs into a statement about all finite inputs.","Without it, global behavior can lose information. For example, use \\(X_n=\\{m\\in\\N:m\\ge n\\}\\) with inclusion bonds and a constant Boolean output tower. The source point space is empty. The two transports that always output false and always output true induce the same function from the empty set, but disagree at every finite stage. Every \\(X_k\\) is nonempty, so no eventual equality witness exists. This calculation uses the finite-prefix boundary of Example~\\ref{KS54} and does not identify empty global behavior with equality of finite presentations.","\\subsection{The exact image of realization}\nFor a function \\(F:\\operatorname{Point}(X)\\to\\operatorname{Point}(Y)\\), define finite observation by the quantifiers\n\\begin{equation}\\label{eq:finite-observation}\n \\forall n\\ \\exists k\\ \\exists a:X_k\\to Y_n\\ \\forall x:\\quad\n F(x)_n=a(x_k).\n\\end{equation}\nThe input level is uniform over all global inputs for each fixed output level. It may vary with the output level. This is not the weaker condition that each individual input can be inspected to some input-dependent finite depth.","\\begin{theorem}[Characterization of realized transformations]\\label{KS63}\nIf \\(X\\) has surjective point projections, realization induces a bijection\n\\[\n \\operatorname{EventualHom}(X,Y)\n \\simeq\n \\{F:\\operatorname{Point}(X)\\to\\operatorname{Point}(Y):F\n \\text{ is finitely observed}\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a reindexed map $(\\phi,f)$, the formula $|f|(x)_n=f_n(x_{\\phi(n)})$ immediately supplies a finite-observation witness for each $n$. Eventual equality does not affect the realized function by Theorem~\\ref{KS61}, so this defines a map from the quotient to the stated set of functions.","ransformations]\\label{KS63}\nIf \\(X\\) has surjective point projections, realization induces a bijection\n\\[\n \\operatorname{EventualHom}(X,Y)\n \\simeq\n \\{F:\\operatorname{Point}(X)\\to\\operatorname{Point}(Y):F\n \\text{ is finitely observed}\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a reindexed map $(\\phi,f)$, the formula $|f|(x)_n=f_n(x_{\\phi(n)})$ immediately supplies a finite-observation witness for each $n$. Eventual equality does not affect the realized function by Theorem~\\ref{KS61}, so this defines a map from the quotient to the stated set of functions.","For surjectivity, let $F$ be finitely observed. Choose one pair $(k_n,a_n)$ at each output level, satisfying $F(x)_n=a_n(x_{k_n})$ for every coherent input. Define\n\\[\n \\phi(n)=\\max_{0\\le i\\le n}k_i,\n \\qquad f_n=a_n\\rho^X_{k_n,\\phi(n)}.\n\\]\nThe maximum is of a nonempty finite set. It gives a monotone schedule with $k_n\\le\\phi(n)$. For every point $x$, coherence in $X$ gives\n\\[\n f_n(x_{\\phi(n)})\n =a_n\\bigl(\\rho^X_{k_n,\\phi(n)}x_{\\phi(n)}\\bigr)\n =a_n(x_{k_n})=F(x)_n.                                      \\tag{*}\n\\]","The remaining obligation is naturality on \\emph{all} finite states. Fix $i\\le j$ and $u\\in X_{\\phi(j)}$. By surjectivity of that point projection, choose $x$ with $x_{\\phi(j)}=u$. Equation (*) and output coherence yield\n\\[\n\\begin{aligned}\n \\rho^Y_{i,j}f_j(u)&=\\rho^Y_{i,j}F(x)_j=F(x)_i,","The remaining obligation is naturality on \\emph{all} finite states. Fix $i\\le j$ and $u\\in X_{\\phi(j)}$. By surjectivity of that point projection, choose $x$ with $x_{\\phi(j)}=u$. Equation (*) and output coherence yield\n\\[\n\\begin{aligned}\n \\rho^Y_{i,j}f_j(u)&=\\rho^Y_{i,j}F(x)_j=F(x)_i,\\\\\n f_i\\rho^X_{\\phi(i),\\phi(j)}(u)&=f_i(x_{\\phi(i)})=F(x)_i.\n\\end{aligned}\n\\]\nThe state $u$ was arbitrary. Hence the two finite-level functions are equal, which proves scheduled naturality. The resulting transport realizes $F$ by (*).","f_i\\rho^X_{\\phi(i),\\phi(j)}(u)&=f_i(x_{\\phi(i)})=F(x)_i.\n\\end{aligned}\n\\]\nThe state $u$ was arbitrary. Hence the two finite-level functions are equal, which proves scheduled naturality. The resulting transport realizes $F$ by (*).","Different witness choices can give different schedules and coordinate maps, but the realized function is still $F$. By Theorem~\\ref{KS62} these transports are eventually equal, so their quotient class is independent of all choices. The same faithfulness result says that two classes with equal realized functions are identical. We have proved both surjectivity and injectivity, including the well-definedness of the inverse choice construction. No bound on the growth of $\\phi$ follows from these quantifiers.\n\\end{proof}\n\\source{KS63}\nThe finite-level compatibility is derived in this proof. It is not inserted as an additional field in the hypothesis that \\(F\\) is finitely observed. This is one of the useful bridges between a global description of a transformation and an actual presentation in the resolution","ps, but the realized function is still $F$. By Theorem~\\ref{KS62} these transports are eventually equal, so their quotient class is independent of all choices. The same faithfulness result says that two classes with equal realized functions are identical. We have proved both surjectivity and injectivity, including the well-definedness of the inverse choice construction. No bound on the growth of $\\phi$ follows from these quantifiers.\n\\end{proof}\n\\source{KS63}\nThe finite-level compatibility is derived in this proof. It is not inserted as an additional field in the hypothesis that \\(F\\) is finitely observed. This is one of the useful bridges between a global description of a transformation and an actual presentation in the resolution calculus.","says that two classes with equal realized functions are identical. We have proved both surjectivity and injectivity, including the well-definedness of the inverse choice construction. No bound on the growth of $\\phi$ follows from these quantifiers.\n\\end{proof}\n\\source{KS63}\nThe finite-level compatibility is derived in this proof. It is not inserted as an additional field in the hypothesis that \\(F\\) is finitely observed. This is one of the useful bridges between a global description of a transformation and an actual presentation in the resolution calculus.","\\subsection{Binary prefixes: a complete interpretation}\nLet \\(X_n\\) consist of binary words of length \\(n\\), with prefix restriction. A coherent point is exactly an infinite binary sequence: its length-\\(n\\) view is its first \\(n\\) bits. Every prefix extends, for example by appending zeros. Hence the source satisfies the hypothesis of the characterization theorem.","Bitwise complementation is represented at level \\(n\\) by complementing the first \\(n\\) bits. The left shift \\(F(x)_i=x_{i+1}\\) uses the schedule \\(\\phi(n)=n+1\\). A nearest-neighbor difference over \\(\\mathbb F_2\\),\n\\[\n G(x)_i=x_i+x_{i+1},\n\\]\nalso uses \\(n+1\\) input bits for \\(n\\) output bits. Applying \\(G\\) twice uses \\(n+2\\) bits, and the middle terms cancel:\n\\[\n G(G(x))_i=x_i+x_{i+2}.\n\\]\nThe compatible finite maps make each statement meaningful on the entire coherent sequence, not merely on one chosen truncation. These are worked instances of the scheduled formulas. The finite regression suite checks their prefix compatibility independently.","Not every set function on these points is finitely observed. Consider the Boolean output that is one exactly when every input bit is zero, regarded as a point of the constant Boolean tower. Given any proposed finite level \\(k\\), compare the all-zero sequence with a sequence whose first one occurs at position \\(k\\). Their first \\(k\\) bits agree, but the outputs differ. Therefore no finite witness \\(a:X_k\\to\\{0,1\\}\\) can compute even that one output coordinate. This rules out a reindexed presentation by the forward direction of the theorem.","ot every set function on these points is finitely observed. Consider the Boolean output that is one exactly when every input bit is zero, regarded as a point of the constant Boolean tower. Given any proposed finite level \\(k\\), compare the all-zero sequence with a sequence whose first one occurs at position \\(k\\). Their first \\(k\\) bits agree, but the outputs differ. Therefore no finite witness \\(a:X_k\\to\\{0,1\\}\\) can compute even that one output coordinate. This rules out a reindexed presentation by the forward direction of the theorem.","\\subsection{Cofinal changes also preserve the morphism calculus}\nA monotone unbounded schedule \\(\\theta\\) gives the sampled resolution \\(X_\\theta\\). There is a transport from \\(X\\) to \\(X_\\theta\\) that reads the selected level. For the reverse direction choose, for each \\(n\\), a level \\(j(n)\\) with \\(n\\le\\theta(j(n))\\), and restrict the available value down to \\(X_n\\). Taking a least such index gives a monotone schedule.","\\begin{theorem}[Cofinal invariance of eventual transports]\\label{KS64}\nSampling and recovery are inverse in the eventual transport category. Consequently, for monotone unbounded schedules \\(\\theta\\) and \\(\\eta\\), there is a bijection\n\\[\n \\operatorname{EventualHom}(X,Y)\n \\simeq\n \\operatorname{EventualHom}(X_\\theta,Y_\\eta).\n\\]\nOn point spaces it agrees with the previously constructed cofinal realization equivalences.\n\\end{theorem}\n\\begin{proof}\nLet \\(s:X\\to X_\\theta\\) be sampling, with schedule \\(\\theta\\) and identity level maps. For recovery choose the least \\(j(n)\\) with \\(n\\le\\theta(j(n))\\). Unboundedness makes this set nonempty. Increasing \\(n\\) only strengthens the inequality, so \\(j\\) is monotone. The level map of \\(r:X_\\theta\\to X\\) is \\(\\rho^X_{n,\\theta(j(n))}\\). Composite restriction proves its naturality.","orts]\\label{KS64}\nSampling and recovery are inverse in the eventual transport category. Consequently, for monotone unbounded schedules \\(\\theta\\) and \\(\\eta\\), there is a bijection\n\\[\n \\operatorname{EventualHom}(X,Y)\n \\simeq\n \\operatorname{EventualHom}(X_\\theta,Y_\\eta).\n\\]\nOn point spaces it agrees with the previously constructed cofinal realization equivalences.\n\\end{theorem}\n\\begin{proof}\nLet \\(s:X\\to X_\\theta\\) be sampling, with schedule \\(\\theta\\) and identity level maps. For recovery choose the least \\(j(n)\\) with \\(n\\le\\theta(j(n))\\). Unboundedness makes this set nonempty. Increasing \\(n\\) only strengthens the inequality, so \\(j\\) is monotone. The level map of \\(r:X_\\theta\\to X\\) is \\(\\rho^X_{n,\\theta(j(n))}\\). Composite restriction proves its naturality.","erse in the eventual transport category. Consequently, for monotone unbounded schedules \\(\\theta\\) and \\(\\eta\\), there is a bijection\n\\[\n \\operatorname{EventualHom}(X,Y)\n \\simeq\n \\operatorname{EventualHom}(X_\\theta,Y_\\eta).\n\\]\nOn point spaces it agrees with the previously constructed cofinal realization equivalences.\n\\end{theorem}\n\\begin{proof}\nLet \\(s:X\\to X_\\theta\\) be sampling, with schedule \\(\\theta\\) and identity level maps. For recovery choose the least \\(j(n)\\) with \\(n\\le\\theta(j(n))\\). Unboundedness makes this set nonempty. Increasing \\(n\\) only strengthens the inequality, so \\(j\\) is monotone. The level map of \\(r:X_\\theta\\to X\\) is \\(\\rho^X_{n,\\theta(j(n))}\\). Composite restriction proves its naturality.","e unbounded schedules \\(\\theta\\) and \\(\\eta\\), there is a bijection\n\\[\n \\operatorname{EventualHom}(X,Y)\n \\simeq\n \\operatorname{EventualHom}(X_\\theta,Y_\\eta).\n\\]\nOn point spaces it agrees with the previously constructed cofinal realization equivalences.\n\\end{theorem}\n\\begin{proof}\nLet \\(s:X\\to X_\\theta\\) be sampling, with schedule \\(\\theta\\) and identity level maps. For recovery choose the least \\(j(n)\\) with \\(n\\le\\theta(j(n))\\). Unboundedness makes this set nonempty. Increasing \\(n\\) only strengthens the inequality, so \\(j\\) is monotone. The level map of \\(r:X_\\theta\\to X\\) is \\(\\rho^X_{n,\\theta(j(n))}\\). Composite restriction proves its naturality.","hedules \\(\\theta\\) and \\(\\eta\\), there is a bijection\n\\[\n \\operatorname{EventualHom}(X,Y)\n \\simeq\n \\operatorname{EventualHom}(X_\\theta,Y_\\eta).\n\\]\nOn point spaces it agrees with the previously constructed cofinal realization equivalences.\n\\end{theorem}\n\\begin{proof}\nLet \\(s:X\\to X_\\theta\\) be sampling, with schedule \\(\\theta\\) and identity level maps. For recovery choose the least \\(j(n)\\) with \\(n\\le\\theta(j(n))\\). Unboundedness makes this set nonempty. Increasing \\(n\\) only strengthens the inequality, so \\(j\\) is monotone. The level map of \\(r:X_\\theta\\to X\\) is \\(\\rho^X_{n,\\theta(j(n))}\\). Composite restriction proves its naturality.","EventualHom}(X,Y)\n \\simeq\n \\operatorname{EventualHom}(X_\\theta,Y_\\eta).\n\\]\nOn point spaces it agrees with the previously constructed cofinal realization equivalences.\n\\end{theorem}\n\\begin{proof}\nLet \\(s:X\\to X_\\theta\\) be sampling, with schedule \\(\\theta\\) and identity level maps. For recovery choose the least \\(j(n)\\) with \\(n\\le\\theta(j(n))\\). Unboundedness makes this set nonempty. Increasing \\(n\\) only strengthens the inequality, so \\(j\\) is monotone. The level map of \\(r:X_\\theta\\to X\\) is \\(\\rho^X_{n,\\theta(j(n))}\\). Composite restriction proves its naturality.","ventualHom}(X_\\theta,Y_\\eta).\n\\]\nOn point spaces it agrees with the previously constructed cofinal realization equivalences.\n\\end{theorem}\n\\begin{proof}\nLet \\(s:X\\to X_\\theta\\) be sampling, with schedule \\(\\theta\\) and identity level maps. For recovery choose the least \\(j(n)\\) with \\(n\\le\\theta(j(n))\\). Unboundedness makes this set nonempty. Increasing \\(n\\) only strengthens the inequality, so \\(j\\) is monotone. The level map of \\(r:X_\\theta\\to X\\) is \\(\\rho^X_{n,\\theta(j(n))}\\). Composite restriction proves its naturality.","point spaces it agrees with the previously constructed cofinal realization equivalences.\n\\end{theorem}\n\\begin{proof}\nLet \\(s:X\\to X_\\theta\\) be sampling, with schedule \\(\\theta\\) and identity level maps. For recovery choose the least \\(j(n)\\) with \\(n\\le\\theta(j(n))\\). Unboundedness makes this set nonempty. Increasing \\(n\\) only strengthens the inequality, so \\(j\\) is monotone. The level map of \\(r:X_\\theta\\to X\\) is \\(\\rho^X_{n,\\theta(j(n))}\\). Composite restriction proves its naturality.","with the previously constructed cofinal realization equivalences.\n\\end{theorem}\n\\begin{proof}\nLet \\(s:X\\to X_\\theta\\) be sampling, with schedule \\(\\theta\\) and identity level maps. For recovery choose the least \\(j(n)\\) with \\(n\\le\\theta(j(n))\\). Unboundedness makes this set nonempty. Increasing \\(n\\) only strengthens the inequality, so \\(j\\) is monotone. The level map of \\(r:X_\\theta\\to X\\) is \\(\\rho^X_{n,\\theta(j(n))}\\). Composite restriction proves its naturality.","The composite \\(rs:X\\to X\\) reads level \\(\\theta(j(n))\\) and restricts to \\(n\\). At that very stage its map agrees with the identity transport after restriction from \\(\\theta(j(n))\\) to \\(n\\). Thus \\(rs\\) is eventually equal to the identity.","For \\(sr:X_\\theta\\to X_\\theta\\), the schedule at output \\(n\\) is \\(j(\\theta(n))\\), and the level map is \\(\\rho^X_{\\theta(n),\\theta(j(\\theta(n)))}\\). Compare it with the identity at a sampled index\n\\[\n k\\ge\\max\\{n,j(\\theta(n))\\}.\n\\]\nRestricting either route from level \\(k\\) of the sampled system gives the same original restriction \\(\\rho^X_{\\theta(n),\\theta(k)}\\). This proves eventual equality even when \\(\\theta\\) has repeated values. Hence the two classes are inverse isomorphisms, without requiring them to be literal inverse schedules.",":X_\\theta\\to X_\\theta\\), the schedule at output \\(n\\) is \\(j(\\theta(n))\\), and the level map is \\(\\rho^X_{\\theta(n),\\theta(j(\\theta(n)))}\\). Compare it with the identity at a sampled index\n\\[\n k\\ge\\max\\{n,j(\\theta(n))\\}.\n\\]\nRestricting either route from level \\(k\\) of the sampled system gives the same original restriction \\(\\rho^X_{\\theta(n),\\theta(k)}\\). This proves eventual equality even when \\(\\theta\\) has repeated values. Hence the two classes are inverse isomorphisms, without requiring them to be literal inverse schedules.","Write \\(s_X,r_X\\) and \\(s_Y,r_Y\\) for the sampling and recovery isomorphisms of source and target. The asserted Hom map is\n\\[\n [f]\\longmapsto s_Y[f]r_X,\n\\]\nwith inverse \\([g]\\mapsto r_Y[g]s_X\\). Associativity and the two inverse equations cancel the inserted pairs, proving both round trips. Applying the realization functor transports these equations to point spaces, where sampling and recovery are the cofinal point equivalences already constructed. No norm invariance or surjectivity of source point projections is needed for this categorical conjugation argument.\n\\end{proof}\n\\source{KS64}\nThis strengthens cofinal invariance from points to a specified category of transformations. It still does not make raw r","Write \\(s_X,r_X\\) and \\(s_Y,r_Y\\) for the sampling and recovery isomorphisms of source and target. The asserted Hom map is\n\\[\n [f]\\longmapsto s_Y[f]r_X,\n\\]\nwith inverse \\([g]\\mapsto r_Y[g]s_X\\). Associativity and the two inverse equations cancel the inserted pairs, proving both round trips. Applying the realization functor transports these equations to point spaces, where sampling and recovery are the cofinal point equivalences already constructed. No norm invariance or surjectivity of source point projections is needed for this categorical conjugation argument.\n\\end{proof}\n\\source{KS64}\nThis strengthens cofinal invariance from points to a specified category of transformations. It still does not make raw residual norms in","Write \\(s_X,r_X\\) and \\(s_Y,r_Y\\) for the sampling and recovery isomorphisms of source and target. The asserted Hom map is\n\\[\n [f]\\longmapsto s_Y[f]r_X,\n\\]\nwith inverse \\([g]\\mapsto r_Y[g]s_X\\). Associativity and the two inverse equations cancel the inserted pairs, proving both round trips. Applying the realization functor transports these equations to point spaces, where sampling and recovery are the cofinal point equivalences already constructed. No norm invariance or surjectivity of source point projections is needed for this categorical conjugation argument.\n\\end{proof}\n\\source{KS64}\nThis strengthens cofinal invariance from points to a specified category of transformations. It still does not make raw residual norms invariant, identify every global set function with a finite presentation, or preserve the numerical conditioning of a repair","Write \\(s_X,r_X\\) and \\(s_Y,r_Y\\) for the sampling and recovery isomorphisms of source and target. The asserted Hom map is\n\\[\n [f]\\longmapsto s_Y[f]r_X,\n\\]\nwith inverse \\([g]\\mapsto r_Y[g]s_X\\). Associativity and the two inverse equations cancel the inserted pairs, proving both round trips. Applying the realization functor transports these equations to point spaces, where sampling and recovery are the cofinal point equivalences already constructed. No norm invariance or surjectivity of source point projections is needed for this categorical conjugation argument.\n\\end{proof}\n\\source{KS64}\nThis strengthens cofinal invariance from points to a specified category of transformations. It still does not make raw residual norms invariant, identify every global set function with a finite presentation, or preserve the numerical conditioning of a repair. Those are separate questions, addres","\\subsection{A section is a choice of residual coordinates}\nLet \\(d:E\\to R\\) be a continuous linear map between normed spaces over a nontrivially normed field. Suppose it has a continuous linear right inverse \\(A:R\\to E\\), so \\(dA=I_R\\). Set\n\\[\n K=\\ker d,\\qquad P_A=I_E-Ad.\n\\]\nThen \\(dP_A=0\\), \\(P_A\\) fixes \\(K\\), and \\(P_A^2=P_A\\). The right inverse chooses where each residual is represented inside the candidate space. Different right inverses can choose different representatives of the same residual.","\\begin{theorem}[The split defect chart]\\label{KS65}\nThe maps\n\\[\n \\chi_A:E\\simeq_L K\\times R,\\qquad\n \\chi_A(x)=(P_Ax,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition fo","\\begin{theorem}[The split defect chart]\\label{KS65}\nThe maps\n\\[\n \\chi_A:E\\simeq_L K\\times R,\\qquad\n \\chi_A(x)=(P_Ax,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they de","\\begin{theorem}[The split defect chart]\\label{KS65}\nThe maps\n\\[\n \\chi_A:E\\simeq_L K\\times R,\\qquad\n \\chi_A(x)=(P_Ax,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuou","\\begin{theorem}[The split defect chart]\\label{KS65}\nThe maps\n\\[\n \\chi_A:E\\simeq_L K\\times R,\\qquad\n \\chi_A(x)=(P_Ax,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equi","\\begin{theorem}[The split defect chart]\\label{KS65}\nThe maps\n\\[\n \\chi_A:E\\simeq_L K\\times R,\\qquad\n \\chi_A(x)=(P_Ax,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on","\\begin{theorem}[The split defect chart]\\label{KS65}\nThe maps\n\\[\n \\chi_A:E\\simeq_L K\\times R,\\qquad\n \\chi_A(x)=(P_Ax,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied t","\\begin{theorem}[The split defect chart]\\label{KS65}\nThe maps\n\\[\n \\chi_A:E\\simeq_L K\\times R,\\qquad\n \\chi_A(x)=(P_Ax,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart","\\begin{theorem}[The split defect chart]\\label{KS65}\nThe maps\n\\[\n \\chi_A:E\\simeq_L K\\times R,\\qquad\n \\chi_A(x)=(P_Ax,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the","5}\nThe maps\n\\[\n \\chi_A:E\\simeq_L K\\times R,\\qquad\n \\chi_A(x)=(P_Ax,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","\\chi_A(x)=(P_Ax,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","x,dx),\\qquad\n \\chi_A^{-1}(u,r)=u+Ar\n\\]\nare inverse continuous linear equivalences. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","nces. In these coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","hese coordinates, \\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","\\(P_A\\) sends \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","s \\((u,r)\\) to \\((u,0)\\). If \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","f \\(B\\) is another right inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","inverse, then \\(A-B\\) takes values in \\(K\\), and\n\\[\n P_Bx-P_Ax=(A-B)dx.\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.",".\n\\]\nThe projections commute if and only if they are equal, and they are equal if and only if \\(A=B\\).\n\\end{theorem}\n\\begin{proof}\nThe identity \\(dA=I\\) gives \\(dP_A=d-dAd=0\\), so the proposed first chart coordinate actually belongs to \\(K\\). If \\(u\\in K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","K\\), then \\(du=0\\) and \\(P_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","_Au=u\\). Therefore, for \\((u,r)\\in K\\times R\\),\n\\[\n d(u+Ar)=r,\\qquad P_A(u+Ar)=u+Ar-A(r)=u.\n\\]\nIn the other direction, \\(P_Ax+A(dx)=x-Adx+Adx=x\\). These are both inverse equations, including the membership condition for the kernel coordinate. All component maps are bounded linear, so they define a continuous linear equivalence on the standard finite product. Applied to \\(P_Ax\\), the chart gives \\((P_Ax,0)\\), proving the repair formula.","For a second right inverse \\(B\\), the equation \\(d(A-B)=I-I=0\\) places its difference from \\(A\\) in \\(K\\). Direct subtraction gives\n\\[\n P_B-P_A=(I-Bd)-(I-Ad)=(A-B)d.\n\\]\nBoth projections have range \\(K\\) and fix every point of \\(K\\). Thus applying \\(P_A\\) after \\(P_B\\) gives \\(P_B\\), while applying \\(P_B\\) after \\(P_A\\) gives \\(P_A\\). These two compositions are equal if and only if the projections are equal.","If \\(P_A=P_B\\), evaluate their difference at \\(Ar\\). Since \\(dAr=r\\), the difference formula yields \\((A-B)r=0\\) for every \\(r\\), hence \\(A=B\\). Equality of sections conversely makes their projection formulas identical. The argument uses the right inverse to test every residual. It does not infer equality of sections merely from agreement on some restricted set of observed residuals.\n\\end{proof}\n\\source{KS65}\nThis is a linear splitting, not necessarily an orthogonal one. The exact coordinate \\(u\\) depends on the chosen section. The residual coordinate \\(r=dx\\) does not. A specified Hilbert geometry can make one section distinguished through orthogonality, but the algebraic split alone does not do so.","\\((A-B)r=0\\) for every \\(r\\), hence \\(A=B\\). Equality of sections conversely makes their projection formulas identical. The argument uses the right inverse to test every residual. It does not infer equality of sections merely from agreement on some restricted set of observed residuals.\n\\end{proof}\n\\source{KS65}\nThis is a linear splitting, not necessarily an orthogonal one. The exact coordinate \\(u\\) depends on the chosen section. The residual coordinate \\(r=dx\\) does not. A specified Hilbert geometry can make one section distinguished through orthogonality, but the algebraic split alone does not do so.","tions conversely makes their projection formulas identical. The argument uses the right inverse to test every residual. It does not infer equality of sections merely from agreement on some restricted set of observed residuals.\n\\end{proof}\n\\source{KS65}\nThis is a linear splitting, not necessarily an orthogonal one. The exact coordinate \\(u\\) depends on the chosen section. The residual coordinate \\(r=dx\\) does not. A specified Hilbert geometry can make one section distinguished through orthogonality, but the algebraic split alone does not do so.","\\subsection{A change of representation has two components}\nLet \\(D:F\\to T\\) be a second continuous linear defect map. A compatible change consists of bounded linear maps\n\\[\n S:E\\to F,\\qquad C:R\\to T,\\qquad DS=Cd.\n\\]\nThe equation is the commuting defect square. It says that transforming a candidate and then measuring its defect agrees with transforming its actual defect. No sections have yet been used.","In particular, \\(S\\) carries \\(\\ker d\\) into \\(\\ker D\\). It may still have a kernel, lose information, or fail to be onto. Calling it a change of representation does not assert an equivalence unless inverse maps have also been supplied.","Choose sections \\(A\\) of \\(d\\) and \\(B\\) of \\(D\\). Define\n\\begin{equation}\\label{eq:repair-mismatch}\n M=SA-BC:R\\longrightarrow F.\n\\end{equation}\nThe commuting square and section identities give \\(DM=0\\). Thus this mismatch takes residual information into an exact direction of the target.","\\begin{theorem}[Triangular transport and repair mismatch]\\label{KS66}\nIn split defect coordinates, the change of representation is\n\\begin{equation}\\label{eq:triangular-chart}\n \\chi_BS\\chi_A^{-1}(u,r)=(Su+Mr,Cr).\n\\end{equation}\nThe two repair routes differ by\n\\begin{equation}\\label{eq:repair-change}\n P_BSx-SP_Ax=M(dx).\n\\end{equation}\nConsequently they agree on exact inputs, and\n\\[\n \\norm{P_BSx-SP_Ax}\\le\\norm M\\norm{dx},\n \\qquad \\norm M\\le\\norm S\\norm A+\\norm B\\norm C.\n\\]\nThey agree on all inputs if and only if \\(SA=BC\\).","esentation is\n\\begin{equation}\\label{eq:triangular-chart}\n \\chi_BS\\chi_A^{-1}(u,r)=(Su+Mr,Cr).\n\\end{equation}\nThe two repair routes differ by\n\\begin{equation}\\label{eq:repair-change}\n P_BSx-SP_Ax=M(dx).\n\\end{equation}\nConsequently they agree on exact inputs, and\n\\[\n \\norm{P_BSx-SP_Ax}\\le\\norm M\\norm{dx},\n \\qquad \\norm M\\le\\norm S\\norm A+\\norm B\\norm C.\n\\]\nThey agree on all inputs if and only if \\(SA=BC\\).","For two consecutive compatible changes \\((S_1,C_1)\\), \\((S_2,C_2)\\), with compatible choices of sections at the three ends, their mismatches satisfy\n\\[\n M_{21}=S_2M_1+M_2C_1.\n\\]\nIf \\(dx_j\\to0\\) in norm for a fixed change and fixed sections, the repair discrepancy tends to zero.\n\\end{theorem}\n\\begin{proof}\nFirst \\(DS=Cd\\) sends \\(\\ker d\\) into \\(\\ker D\\), and\n\\[\n DM=DSA-DBC=CdA-C=C-C=0.\n\\]\nThus both terms in the proposed exact coordinate have the correct codomain. Given \\(u\\in\\ker d\\), compute the residual and exact coordinates of \\(S(u+Ar)\\) separately:\n\\[\n DS(u+Ar)=Cr,\\qquad\n S(u+Ar)-BCr=Su+(SA-BC)r.\n\\]\nThis proves the triangular formula as an equality in \\(\\ker D\\times T\\).","mpatible choices of sections at the three ends, their mismatches satisfy\n\\[\n M_{21}=S_2M_1+M_2C_1.\n\\]\nIf \\(dx_j\\to0\\) in norm for a fixed change and fixed sections, the repair discrepancy tends to zero.\n\\end{theorem}\n\\begin{proof}\nFirst \\(DS=Cd\\) sends \\(\\ker d\\) into \\(\\ker D\\), and\n\\[\n DM=DSA-DBC=CdA-C=C-C=0.\n\\]\nThus both terms in the proposed exact coordinate have the correct codomain. Given \\(u\\in\\ker d\\), compute the residual and exact coordinates of \\(S(u+Ar)\\) separately:\n\\[\n DS(u+Ar)=Cr,\\qquad\n S(u+Ar)-BCr=Su+(SA-BC)r.\n\\]\nThis proves the triangular formula as an equality in \\(\\ker D\\times T\\).","es satisfy\n\\[\n M_{21}=S_2M_1+M_2C_1.\n\\]\nIf \\(dx_j\\to0\\) in norm for a fixed change and fixed sections, the repair discrepancy tends to zero.\n\\end{theorem}\n\\begin{proof}\nFirst \\(DS=Cd\\) sends \\(\\ker d\\) into \\(\\ker D\\), and\n\\[\n DM=DSA-DBC=CdA-C=C-C=0.\n\\]\nThus both terms in the proposed exact coordinate have the correct codomain. Given \\(u\\in\\ker d\\), compute the residual and exact coordinates of \\(S(u+Ar)\\) separately:\n\\[\n DS(u+Ar)=Cr,\\qquad\n S(u+Ar)-BCr=Su+(SA-BC)r.\n\\]\nThis proves the triangular formula as an equality in \\(\\ker D\\times T\\).","2M_1+M_2C_1.\n\\]\nIf \\(dx_j\\to0\\) in norm for a fixed change and fixed sections, the repair discrepancy tends to zero.\n\\end{theorem}\n\\begin{proof}\nFirst \\(DS=Cd\\) sends \\(\\ker d\\) into \\(\\ker D\\), and\n\\[\n DM=DSA-DBC=CdA-C=C-C=0.\n\\]\nThus both terms in the proposed exact coordinate have the correct codomain. Given \\(u\\in\\ker d\\), compute the residual and exact coordinates of \\(S(u+Ar)\\) separately:\n\\[\n DS(u+Ar)=Cr,\\qquad\n S(u+Ar)-BCr=Su+(SA-BC)r.\n\\]\nThis proves the triangular formula as an equality in \\(\\ker D\\times T\\).","ds to zero.\n\\end{theorem}\n\\begin{proof}\nFirst \\(DS=Cd\\) sends \\(\\ker d\\) into \\(\\ker D\\), and\n\\[\n DM=DSA-DBC=CdA-C=C-C=0.\n\\]\nThus both terms in the proposed exact coordinate have the correct codomain. Given \\(u\\in\\ker d\\), compute the residual and exact coordinates of \\(S(u+Ar)\\) separately:\n\\[\n DS(u+Ar)=Cr,\\qquad\n S(u+Ar)-BCr=Su+(SA-BC)r.\n\\]\nThis proves the triangular formula as an equality in \\(\\ker D\\times T\\).","For a general input, use the same commuting square to obtain\n\\[\n \\begin{aligned}\n P_BSx-SP_Ax\n &=Sx-BDSx-Sx+SAdx\\\\\n &=(SA-BC)dx=M(dx).\n \\end{aligned}\n\\]\nIts norm is at most \\(\\norm M\\norm{dx}\\). The triangle inequality and submultiplicativity give \\(\\norm M\\le\\norm S\\norm A+\\norm B\\norm C\\). Exact inputs have \\(dx=0\\), so the two routes agree there regardless of the section choices. If they agree on \\emph{every} input, substitute \\(x=Ar\\). Then \\(dAr=r\\) makes \\(Mr=0\\) for each residual, hence \\(SA=BC\\). That equality conversely makes the discrepancy vanish everywhere.","&=(SA-BC)dx=M(dx).\n \\end{aligned}\n\\]\nIts norm is at most \\(\\norm M\\norm{dx}\\). The triangle inequality and submultiplicativity give \\(\\norm M\\le\\norm S\\norm A+\\norm B\\norm C\\). Exact inputs have \\(dx=0\\), so the two routes agree there regardless of the section choices. If they agree on \\emph{every} input, substitute \\(x=Ar\\). Then \\(dAr=r\\) makes \\(Mr=0\\) for each residual, hence \\(SA=BC\\). That equality conversely makes the discrepancy vanish everywhere.","For composition, denote the three sections by \\(A,B,A_2\\). The composite residual map is \\(C_2C_1\\), so its mismatch is \\(S_2S_1A-A_2C_2C_1\\). Insert and subtract \\(S_2BC_1\\):\n\\[\n \\begin{aligned}\n S_2S_1A-A_2C_2C_1\n &=S_2(S_1A-BC_1)+(S_2B-A_2C_2)C_1","For composition, denote the three sections by \\(A,B,A_2\\). The composite residual map is \\(C_2C_1\\), so its mismatch is \\(S_2S_1A-A_2C_2C_1\\). Insert and subtract \\(S_2BC_1\\):\n\\[\n \\begin{aligned}\n S_2S_1A-A_2C_2C_1\n &=S_2(S_1A-BC_1)+(S_2B-A_2C_2)C_1\\\\\n &=S_2M_1+M_2C_1.\n \\end{aligned}\n\\]\nAll terms map the first residual space into the last candidate space, in the written order. Finally, for fixed bounded \\(M\\), the estimate \\(\\norm{M(dx_j)}\\le\\norm M\\norm{dx_j}\\) proves convergence of the discrepancy when \\(dx_j\\to0\\). Allowing the section or representation norms to grow with \\(j\\) would require a separate uniform bound.\n\\end{proof}\n\\source{KS66}\nThe off-diagonal term in \\eqref{eq:triangular-chart} has a specific meaning. A raw residual can become part of the target's chosen exact coordinate after a change of presentation. Deleting the residual before changing presentation removes that contribution. Deleting it afterwards does not. This is not a failure of exactness prese","&=S_2M_1+M_2C_1.\n \\end{aligned}\n\\]\nAll terms map the first residual space into the last candidate space, in the written order. Finally, for fixed bounded \\(M\\), the estimate \\(\\norm{M(dx_j)}\\le\\norm M\\norm{dx_j}\\) proves convergence of the discrepancy when \\(dx_j\\to0\\). Allowing the section or representation norms to grow with \\(j\\) would require a separate uniform bound.\n\\end{proof}\n\\source{KS66}\nThe off-diagonal term in \\eqref{eq:triangular-chart} has a specific meaning. A raw residual can become part of the target's chosen exact coordinate after a change of presentation. Deleting the residual before changing presentation removes that contribution. Deleting it afterwards does not. This is not a failure of exactness preservation. Both routes end in the target exact space, but they can end at different points in it.","\\subsection{A two-coordinate model in which the routes differ}\nUse \\(E=F=\\R^2\\), write candidates as \\((u,r)\\), and take\n\\[\n d(u,r)=D(u,r)=r,\\qquad A(r)=B(r)=(0,r).\n\\]\nThe exact space is the horizontal axis and both repairs are \\(P(u,r)=(u,0)\\). For a real parameter \\(c\\), set\n\\[\n S_c(u,r)=(u+cr,r),\\qquad C=I.\n\\]\nThe defect square commutes because the second coordinate is unchanged. The mismatch is \\(M_c(r)=(cr,0)\\). Hence\n\\[\n PS_c(u,r)=(u+cr,0),\\qquad S_cP(u,r)=(u,0).\n\\]\nWith \\(c=2\\) and input \\((0,1)\\), one route gives \\((2,0)\\) and the other gives \\((0,0)\\). Both have zero residual. A report containing only their final residuals would miss the entire difference.","section{A two-coordinate model in which the routes differ}\nUse \\(E=F=\\R^2\\), write candidates as \\((u,r)\\), and take\n\\[\n d(u,r)=D(u,r)=r,\\qquad A(r)=B(r)=(0,r).\n\\]\nThe exact space is the horizontal axis and both repairs are \\(P(u,r)=(u,0)\\). For a real parameter \\(c\\), set\n\\[\n S_c(u,r)=(u+cr,r),\\qquad C=I.\n\\]\nThe defect square commutes because the second coordinate is unchanged. The mismatch is \\(M_c(r)=(cr,0)\\). Hence\n\\[\n PS_c(u,r)=(u+cr,0),\\qquad S_cP(u,r)=(u,0).\n\\]\nWith \\(c=2\\) and input \\((0,1)\\), one route gives \\((2,0)\\) and the other gives \\((0,0)\\). Both have zero residual. A report containing only their final residuals would miss the entire difference.","model in which the routes differ}\nUse \\(E=F=\\R^2\\), write candidates as \\((u,r)\\), and take\n\\[\n d(u,r)=D(u,r)=r,\\qquad A(r)=B(r)=(0,r).\n\\]\nThe exact space is the horizontal axis and both repairs are \\(P(u,r)=(u,0)\\). For a real parameter \\(c\\), set\n\\[\n S_c(u,r)=(u+cr,r),\\qquad C=I.\n\\]\nThe defect square commutes because the second coordinate is unchanged. The mismatch is \\(M_c(r)=(cr,0)\\). Hence\n\\[\n PS_c(u,r)=(u+cr,0),\\qquad S_cP(u,r)=(u,0).\n\\]\nWith \\(c=2\\) and input \\((0,1)\\), one route gives \\((2,0)\\) and the other gives \\((0,0)\\). Both have zero residual. A report containing only their final residuals would miss the entire difference.","te candidates as \\((u,r)\\), and take\n\\[\n d(u,r)=D(u,r)=r,\\qquad A(r)=B(r)=(0,r).\n\\]\nThe exact space is the horizontal axis and both repairs are \\(P(u,r)=(u,0)\\). For a real parameter \\(c\\), set\n\\[\n S_c(u,r)=(u+cr,r),\\qquad C=I.\n\\]\nThe defect square commutes because the second coordinate is unchanged. The mismatch is \\(M_c(r)=(cr,0)\\). Hence\n\\[\n PS_c(u,r)=(u+cr,0),\\qquad S_cP(u,r)=(u,0).\n\\]\nWith \\(c=2\\) and input \\((0,1)\\), one route gives \\((2,0)\\) and the other gives \\((0,0)\\). Both have zero residual. A report containing only their final residuals would miss the entire difference.","((u,r)\\), and take\n\\[\n d(u,r)=D(u,r)=r,\\qquad A(r)=B(r)=(0,r).\n\\]\nThe exact space is the horizontal axis and both repairs are \\(P(u,r)=(u,0)\\). For a real parameter \\(c\\), set\n\\[\n S_c(u,r)=(u+cr,r),\\qquad C=I.\n\\]\nThe defect square commutes because the second coordinate is unchanged. The mismatch is \\(M_c(r)=(cr,0)\\). Hence\n\\[\n PS_c(u,r)=(u+cr,0),\\qquad S_cP(u,r)=(u,0).\n\\]\nWith \\(c=2\\) and input \\((0,1)\\), one route gives \\((2,0)\\) and the other gives \\((0,0)\\). Both have zero residual. A report containing only their final residuals would miss the entire difference.","r)=D(u,r)=r,\\qquad A(r)=B(r)=(0,r).\n\\]\nThe exact space is the horizontal axis and both repairs are \\(P(u,r)=(u,0)\\). For a real parameter \\(c\\), set\n\\[\n S_c(u,r)=(u+cr,r),\\qquad C=I.\n\\]\nThe defect square commutes because the second coordinate is unchanged. The mismatch is \\(M_c(r)=(cr,0)\\). Hence\n\\[\n PS_c(u,r)=(u+cr,0),\\qquad S_cP(u,r)=(u,0).\n\\]\nWith \\(c=2\\) and input \\((0,1)\\), one route gives \\((2,0)\\) and the other gives \\((0,0)\\). Both have zero residual. A report containing only their final residuals would miss the entire difference.","0,r).\n\\]\nThe exact space is the horizontal axis and both repairs are \\(P(u,r)=(u,0)\\). For a real parameter \\(c\\), set\n\\[\n S_c(u,r)=(u+cr,r),\\qquad C=I.\n\\]\nThe defect square commutes because the second coordinate is unchanged. The mismatch is \\(M_c(r)=(cr,0)\\). Hence\n\\[\n PS_c(u,r)=(u+cr,0),\\qquad S_cP(u,r)=(u,0).\n\\]\nWith \\(c=2\\) and input \\((0,1)\\), one route gives \\((2,0)\\) and the other gives \\((0,0)\\). Both have zero residual. A report containing only their final residuals would miss the entire difference.","\\plate{21_mismatch}{The shear preserves the residual coordinate but mixes it into the chosen exact coordinate. Repair after the shear reaches \\((2,0)\\). Repair before the shear reaches \\((0,0)\\). Their difference is the exact-valued mismatch applied to the original residual. The example uses the ordinary Euclidean drawing, while the algebraic identity does not depend on that drawing.}{fig:mismatch}","Two shears compose by adding their parameters. In the mismatch law this appears as \\(M_{c_2+c_1}=S_{c_2}M_{c_1}+M_{c_2}\\), and \\(S_{c_2}\\) fixes the exact horizontal axis. The identity is therefore a simple instance of the same composition rule, not an unrelated coordinate trick.","\\subsection{Actual tower resampling gives a defect map}\nReturn to Banach towers with contractive source and target bonds. Let \\(\\phi\\) be a monotone schedule. Candidate sampling is\n\\[\n (S_\\phi f)_n=f_{\\phi(n)}.\n\\]\nThe sampled bonding map is the actual composite restriction between the selected levels. The residual map \\(C_\\phi\\) is not pointwise sampling of a scalar error score. It accumulates transported fine defects across each omitted block.","For a block from \\(a\\) to \\(b\\), its value on a residual family \\(r\\) is\n\\begin{equation}\\label{eq:block-defect-sum}\n (C_{a,b}r)=\\sum_{k=a}^{b-1}\n \\rho^Y_{a,k}\\,r_k\\,\\rho^X_{k+1,b}.\n\\end{equation}\nFor the schedule \\(\\phi\\), set \\((C_\\phi r)_n=C_{\\phi(n),\\phi(n+1)}r\\). Every summand maps \\(X_b\\) into \\(Y_a\\). These surrounding restrictions are necessary when the spaces change with the level. For an actual differential, the sum telescopes to\n\\[\n C_{a,b}(\\partial f)=\\rho^Y_{a,b}f_b-f_a\\rho^X_{a,b}.\n\\]\nThe dyadic coarsener in Section~\\ref{sec:cascade} is the case \\(b=a+2\\).","a residual family \\(r\\) is\n\\begin{equation}\\label{eq:block-defect-sum}\n (C_{a,b}r)=\\sum_{k=a}^{b-1}\n \\rho^Y_{a,k}\\,r_k\\,\\rho^X_{k+1,b}.\n\\end{equation}\nFor the schedule \\(\\phi\\), set \\((C_\\phi r)_n=C_{\\phi(n),\\phi(n+1)}r\\). Every summand maps \\(X_b\\) into \\(Y_a\\). These surrounding restrictions are necessary when the spaces change with the level. For an actual differential, the sum telescopes to\n\\[\n C_{a,b}(\\partial f)=\\rho^Y_{a,b}f_b-f_a\\rho^X_{a,b}.\n\\]\nThe dyadic coarsener in Section~\\ref{sec:cascade} is the case \\(b=a+2\\).","ho^X_{k+1,b}.\n\\end{equation}\nFor the schedule \\(\\phi\\), set \\((C_\\phi r)_n=C_{\\phi(n),\\phi(n+1)}r\\). Every summand maps \\(X_b\\) into \\(Y_a\\). These surrounding restrictions are necessary when the spaces change with the level. For an actual differential, the sum telescopes to\n\\[\n C_{a,b}(\\partial f)=\\rho^Y_{a,b}f_b-f_a\\rho^X_{a,b}.\n\\]\nThe dyadic coarsener in Section~\\ref{sec:cascade} is the case \\(b=a+2\\).","\\begin{theorem}[Bounded-gap resampling and selected repair]\\label{KS67}\nSuppose \\(\\phi(n+1)-\\phi(n)\\le m\\) for a fixed natural number \\(m\\). Then the actual sampling maps are bounded and satisfy\n\\[\n \\norm{S_\\phi}\\le1,\\qquad \\norm{C_\\phi}\\le m,\n \\qquad \\partial_\\phi S_\\phi=C_\\phi\\partial.\n\\]\nExact candidates remain exact under every monotone common resampling, even without a gap bound. If \\(A\\) and \\(B\\) are bounded right inverses for the fine and sampled differentials, then\n\\[\n \\norm{P_BS_\\phi f-S_\\phi P_Af}\n \\le(\\norm A+m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) a","\\begin{theorem}[Bounded-gap resampling and selected repair]\\label{KS67}\nSuppose \\(\\phi(n+1)-\\phi(n)\\le m\\) for a fixed natural number \\(m\\). Then the actual sampling maps are bounded and satisfy\n\\[\n \\norm{S_\\phi}\\le1,\\qquad \\norm{C_\\phi}\\le m,\n \\qquad \\partial_\\phi S_\\phi=C_\\phi\\partial.\n\\]\nExact candidates remain exact under every monotone common resampling, even without a gap bound. If \\(A\\) and \\(B\\) are bounded right inverses for the fine and sampled differentials, then\n\\[\n \\norm{P_BS_\\phi f-S_\\phi P_Af}\n \\le(\\norm A+m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\in","\\begin{theorem}[Bounded-gap resampling and selected repair]\\label{KS67}\nSuppose \\(\\phi(n+1)-\\phi(n)\\le m\\) for a fixed natural number \\(m\\). Then the actual sampling maps are bounded and satisfy\n\\[\n \\norm{S_\\phi}\\le1,\\qquad \\norm{C_\\phi}\\le m,\n \\qquad \\partial_\\phi S_\\phi=C_\\phi\\partial.\n\\]\nExact candidates remain exact under every monotone common resampling, even without a gap bound. If \\(A\\) and \\(B\\) are bounded right inverses for the fine and sampled differentials, then\n\\[\n \\norm{P_BS_\\phi f-S_\\phi P_Af}\n \\le(\\norm A+m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\","\\begin{theorem}[Bounded-gap resampling and selected repair]\\label{KS67}\nSuppose \\(\\phi(n+1)-\\phi(n)\\le m\\) for a fixed natural number \\(m\\). Then the actual sampling maps are bounded and satisfy\n\\[\n \\norm{S_\\phi}\\le1,\\qquad \\norm{C_\\phi}\\le m,\n \\qquad \\partial_\\phi S_\\phi=C_\\phi\\partial.\n\\]\nExact candidates remain exact under every monotone common resampling, even without a gap bound. If \\(A\\) and \\(B\\) are bounded right inverses for the fine and sampled differentials, then\n\\[\n \\norm{P_BS_\\phi f-S_\\phi P_Af}\n \\le(\\norm A+m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\infty\\). Taking the supremum over blocks proves that \\(C_\\phi\\) maps into the bounded residual product an","gin{theorem}[Bounded-gap resampling and selected repair]\\label{KS67}\nSuppose \\(\\phi(n+1)-\\phi(n)\\le m\\) for a fixed natural number \\(m\\). Then the actual sampling maps are bounded and satisfy\n\\[\n \\norm{S_\\phi}\\le1,\\qquad \\norm{C_\\phi}\\le m,\n \\qquad \\partial_\\phi S_\\phi=C_\\phi\\partial.\n\\]\nExact candidates remain exact under every monotone common resampling, even without a gap bound. If \\(A\\) and \\(B\\) are bounded right inverses for the fine and sampled differentials, then\n\\[\n \\norm{P_BS_\\phi f-S_\\phi P_Af}\n \\le(\\norm A+m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\infty\\). Taking the supremum over blocks proves that \\(C_\\phi\\) maps into the bounded residual product and has the","(\\phi(n+1)-\\phi(n)\\le m\\) for a fixed natural number \\(m\\). Then the actual sampling maps are bounded and satisfy\n\\[\n \\norm{S_\\phi}\\le1,\\qquad \\norm{C_\\phi}\\le m,\n \\qquad \\partial_\\phi S_\\phi=C_\\phi\\partial.\n\\]\nExact candidates remain exact under every monotone common resampling, even without a gap bound. If \\(A\\) and \\(B\\) are bounded right inverses for the fine and sampled differentials, then\n\\[\n \\norm{P_BS_\\phi f-S_\\phi P_Af}\n \\le(\\norm A+m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\infty\\). Taking the supremum over blocks proves that \\(C_\\phi\\) maps into the bounded residual product and has the asserted operator norm. If a block has length zero its sum is zero, in agreem…","uad \\norm{C_\\phi}\\le m,\n \\qquad \\partial_\\phi S_\\phi=C_\\phi\\partial.\n\\]\nExact candidates remain exact under every monotone common resampling, even without a gap bound. If \\(A\\) and \\(B\\) are bounded right inverses for the fine and sampled differentials, then\n\\[\n \\norm{P_BS_\\phi f-S_\\phi P_Af}\n \\le(\\norm A+m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\infty\\). Taking the supremum over blocks proves that \\(C_\\phi\\) maps into the bounded residual product and has the asserted operator norm. If a block has length zero its sum is zero, in agreement with the identity restriction.",". If \\(A\\) and \\(B\\) are bounded right inverses for the fine and sampled differentials, then\n\\[\n \\norm{P_BS_\\phi f-S_\\phi P_Af}\n \\le(\\norm A+m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\infty\\). Taking the supremum over blocks proves that \\(C_\\phi\\) maps into the bounded residual product and has the asserted operator norm. If a block has length zero its sum is zero, in agreement with the identity restriction.","and \\(B\\) are bounded right inverses for the fine and sampled differentials, then\n\\[\n \\norm{P_BS_\\phi f-S_\\phi P_Af}\n \\le(\\norm A+m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\infty\\). Taking the supremum over blocks proves that \\(C_\\phi\\) maps into the bounded residual product and has the asserted operator norm. If a block has length zero its sum is zero, in agreement with the identity restriction.","nverses for the fine and sampled differentials, then\n\\[\n \\norm{P_BS_\\phi f-S_\\phi P_Af}\n \\le(\\norm A+m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\infty\\). Taking the supremum over blocks proves that \\(C_\\phi\\) maps into the bounded residual product and has the asserted operator norm. If a block has length zero its sum is zero, in agreement with the identity restriction.","m\\norm B)\\norm{\\partial f}.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\infty\\). Taking the supremum over blocks proves that \\(C_\\phi\\) maps into the bounded residual product and has the asserted operator norm. If a block has length zero its sum is zero, in agreement with the identity restriction.","proof}\nFor a bounded candidate family, \\(\\sup_n\\norm{f_{\\phi(n)}}\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\infty\\). Taking the supremum over blocks proves that \\(C_\\phi\\) maps into the bounded residual product and has the asserted operator norm. If a block has length zero its sum is zero, in agreement with the identity restriction.","\\le\\sup_k\\norm{f_k}\\), proving the sampling bound. For a bounded residual family, each summand in \\eqref{eq:block-defect-sum} is a map from \\(X_b\\) to \\(Y_a\\) and has norm at most \\(\\norm r_\\infty\\), because every composite bond is contractive. A selected block has length at most \\(m\\), so the triangle inequality bounds its sum by \\(m\\norm r_\\infty\\). Taking the supremum over blocks proves that \\(C_\\phi\\) maps into the bounded residual product and has the asserted operator norm. If a block has length zero its sum is zero, in agreement with the identity restriction.","To verify the commuting square, insert\n\\(r_k=\\rho^Y_kf_{k+1}-f_k\\rho^X_k\\). The \\(k\\)-th term becomes\n\\[\n \\rho^Y_{a,k+1}f_{k+1}\\rho^X_{k+1,b}\n -\\rho^Y_{a,k}f_k\\rho^X_{k,b}.\n\\]\nSuccessive internal terms cancel in the finite sum, leaving \\(\\rho^Y_{a,b}f_b-f_a\\rho^X_{a,b}\\). This is exactly the sampled differential. No infinite series interchange is involved.","When the two bounded sections exist, Theorem~\\ref{KS66} applies to the commuting square just established. Substitute \\(\\norm{S_\\phi}\\le1\\) and \\(\\norm{C_\\phi}\\le m\\) in its estimate to get \\((\\norm A+m\\norm B)\\norm{\\partial f}\\). Existence of those sections is an additional hypothesis, not a consequence of the sampling estimate.\n\\end{proof}\n\\source{KS67}\nThe existence of the two right inverses is not inferred merely from contractivity of the bonds. The split-tower theorem provides them under its additional section and propagation-gap conditions. Keeping these hypotheses separate prevents a resampling theorem from being mistaken for an unconditional reconstruction algorithm.","\\subsection{Two distinct losses under coarsening}\nOn scalar identity towers, a fine residual block \\((1,-1)\\) has zero two-step sum. Thus \\(C_\\phi r=0\\) need not imply \\(r=0\\), even for a bounded-gap schedule. This is cancellation. Retaining an independent detail coordinate remedies that information loss in a split coarsener, as the next section shows.","Unbounded gaps cause a different issue. With \\(\\phi(n)=n^2\\), the bounded residual family \\(r_k=1\\) has block sums \\(2n+1\\). These are unbounded. Therefore block accumulation cannot be a map from every bounded residual family to a bounded sampled residual family in this example. The constant residual is not claimed to be the differential of a bounded candidate on the identity tower. It is a witness about the proposed operator on the entire residual space.","\\partnote{Symmetry and effective descent}{A set of solutions forgets the witnesses relating equivalent solutions. The next layer retains those witnesses, their composition, and their automorphisms. Gluing then has to respect the overlap identifications themselves.}\n\\section{Nonabelian coherence and relative trivialization}\\label{sec:gauge}\n\\subsection{Coherent transition labels}\nFix groups \\(G_v\\) and restriction homomorphisms \\(f^*:G_w\\to G_v\\), with the same identity and composition laws as before. The groups need not be abelian and the homomorphisms need not be bijective.","descent}{A set of solutions forgets the witnesses relating equivalent solutions. The next layer retains those witnesses, their composition, and their automorphisms. Gluing then has to respect the overlap identifications themselves.}\n\\section{Nonabelian coherence and relative trivialization}\\label{sec:gauge}\n\\subsection{Coherent transition labels}\nFix groups \\(G_v\\) and restriction homomorphisms \\(f^*:G_w\\to G_v\\), with the same identity and composition laws as before. The groups need not be abelian and the homomorphisms need not be bijective.","\\begin{definition}[Cocycle]\nA cocycle is a family \\(c_f\\in G_v\\) for every \\(f:v\\to w\\), satisfying\n\\[\n c_{1_v}=1,\\qquad c_{g\\circ f}=c_f f^*(c_g).\n\\]\nIt determines a twisted view system with maps\n\\[\n T_c(f)(x)=c_f f^*(x).\n\\]\nA section of the twisted system satisfies \\(x_v=c_f f^*(x_w)\\).\n\\end{definition}\nThe cocycle law says that the transition rules compose coherently. It does not assert that the twisted system has a section. The coherent transition system and a global trivialization of that system are different objects.","For a family \\(a=(a_v)\\in\\prod_vG_v\\), define\n\\begin{equation}\\label{eq:gauge}\n (a\\cdot c)_f=a_vc_f[f^*(a_w)]^{-1}.\n\\end{equation}","\\begin{proposition}[Gauge action and its groupoid]\\label{KS30}\nEquation~\\eqref{eq:gauge} defines an action on cocycles. A gauge carrying \\(c\\) to \\(d\\) induces an equivalence of twisted section sets by \\(x_v\\mapsto a_vx_v\\). The cocycles and actual gauges form a groupoid: an arrow \\(a:c\\to d\\) followed by \\(b:d\\to e\\) has label \\(ba\\).\n\\end{proposition}\n\\begin{proof}\nFirst, $T_c$ really is a view system. For $f:v\\to w$, $g:w\\to z$, and $x\\in G_z$, the homomorphism and cocycle laws give\n\\[\n T_c(f)T_c(g)(x)=c_ff^*(c_gg^*x)\n =c_ff^*(c_g)f^*g^*x=c_{g\\circ f}(g\\circ f)^*x.\n\\]\nThe identity equation follows from $c_{1_v}=1$. No rearrangement of factors is involved.","ition}[Gauge action and its groupoid]\\label{KS30}\nEquation~\\eqref{eq:gauge} defines an action on cocycles. A gauge carrying \\(c\\) to \\(d\\) induces an equivalence of twisted section sets by \\(x_v\\mapsto a_vx_v\\). The cocycles and actual gauges form a groupoid: an arrow \\(a:c\\to d\\) followed by \\(b:d\\to e\\) has label \\(ba\\).\n\\end{proposition}\n\\begin{proof}\nFirst, $T_c$ really is a view system. For $f:v\\to w$, $g:w\\to z$, and $x\\in G_z$, the homomorphism and cocycle laws give\n\\[\n T_c(f)T_c(g)(x)=c_ff^*(c_gg^*x)\n =c_ff^*(c_g)f^*g^*x=c_{g\\circ f}(g\\circ f)^*x.\n\\]\nThe identity equation follows from $c_{1_v}=1$. No rearrangement of factors is involved.","action and its groupoid]\\label{KS30}\nEquation~\\eqref{eq:gauge} defines an action on cocycles. A gauge carrying \\(c\\) to \\(d\\) induces an equivalence of twisted section sets by \\(x_v\\mapsto a_vx_v\\). The cocycles and actual gauges form a groupoid: an arrow \\(a:c\\to d\\) followed by \\(b:d\\to e\\) has label \\(ba\\).\n\\end{proposition}\n\\begin{proof}\nFirst, $T_c$ really is a view system. For $f:v\\to w$, $g:w\\to z$, and $x\\in G_z$, the homomorphism and cocycle laws give\n\\[\n T_c(f)T_c(g)(x)=c_ff^*(c_gg^*x)\n =c_ff^*(c_g)f^*g^*x=c_{g\\circ f}(g\\circ f)^*x.\n\\]\nThe identity equation follows from $c_{1_v}=1$. No rearrangement of factors is involved.","groupoid]\\label{KS30}\nEquation~\\eqref{eq:gauge} defines an action on cocycles. A gauge carrying \\(c\\) to \\(d\\) induces an equivalence of twisted section sets by \\(x_v\\mapsto a_vx_v\\). The cocycles and actual gauges form a groupoid: an arrow \\(a:c\\to d\\) followed by \\(b:d\\to e\\) has label \\(ba\\).\n\\end{proposition}\n\\begin{proof}\nFirst, $T_c$ really is a view system. For $f:v\\to w$, $g:w\\to z$, and $x\\in G_z$, the homomorphism and cocycle laws give\n\\[\n T_c(f)T_c(g)(x)=c_ff^*(c_gg^*x)\n =c_ff^*(c_g)f^*g^*x=c_{g\\circ f}(g\\circ f)^*x.\n\\]\nThe identity equation follows from $c_{1_v}=1$. No rearrangement of factors is involved.","nes an action on cocycles. A gauge carrying \\(c\\) to \\(d\\) induces an equivalence of twisted section sets by \\(x_v\\mapsto a_vx_v\\). The cocycles and actual gauges form a groupoid: an arrow \\(a:c\\to d\\) followed by \\(b:d\\to e\\) has label \\(ba\\).\n\\end{proposition}\n\\begin{proof}\nFirst, $T_c$ really is a view system. For $f:v\\to w$, $g:w\\to z$, and $x\\in G_z$, the homomorphism and cocycle laws give\n\\[\n T_c(f)T_c(g)(x)=c_ff^*(c_gg^*x)\n =c_ff^*(c_g)f^*g^*x=c_{g\\circ f}(g\\circ f)^*x.\n\\]\nThe identity equation follows from $c_{1_v}=1$. No rearrangement of factors is involved.","a_vx_v\\). The cocycles and actual gauges form a groupoid: an arrow \\(a:c\\to d\\) followed by \\(b:d\\to e\\) has label \\(ba\\).\n\\end{proposition}\n\\begin{proof}\nFirst, $T_c$ really is a view system. For $f:v\\to w$, $g:w\\to z$, and $x\\in G_z$, the homomorphism and cocycle laws give\n\\[\n T_c(f)T_c(g)(x)=c_ff^*(c_gg^*x)\n =c_ff^*(c_g)f^*g^*x=c_{g\\circ f}(g\\circ f)^*x.\n\\]\nThe identity equation follows from $c_{1_v}=1$. No rearrangement of factors is involved.","Put $d=a\\cdot c$. Its identity label is $a_v1a_v^{-1}=1$. For a composite, expand all factors in their original order:\n\\[\n\\begin{aligned}\n d_f f^*(d_g)\n &=a_vc_f[f^*(a_w)]^{-1}\n   f^*(a_wc_g[g^*(a_z)]^{-1})","Put $d=a\\cdot c$. Its identity label is $a_v1a_v^{-1}=1$. For a composite, expand all factors in their original order:\n\\[\n\\begin{aligned}\n d_f f^*(d_g)\n &=a_vc_f[f^*(a_w)]^{-1}\n   f^*(a_wc_g[g^*(a_z)]^{-1})\\\\\n &=a_vc_f[f^*(a_w)]^{-1}f^*(a_w)f^*(c_g)\n       [f^*g^*(a_z)]^{-1}\\\\\n &=a_vc_{g\\circ f}[(g\\circ f)^*(a_z)]^{-1}\n =d_{g\\circ f}.\n\\end{aligned}\n\\]\nThus the gauge formula preserves the cocycle condition. For two gauges $a,b$, substitution gives\n\\[\n (a\\cdot(b\\cdot c))_f\n =a_vb_vc_f[f^*(b_w)]^{-1}[f^*(a_w)]^{-1}\n =((ab)\\cdot c)_f.\n\\]\nThe reversed order in the two inverse factors is exactly the formula for the inverse of a product. The identity gauge acts trivially.","&=a_vc_{g\\circ f}[(g\\circ f)^*(a_z)]^{-1}\n =d_{g\\circ f}.\n\\end{aligned}\n\\]\nThus the gauge formula preserves the cocycle condition. For two gauges $a,b$, substitution gives\n\\[\n (a\\cdot(b\\cdot c))_f\n =a_vb_vc_f[f^*(b_w)]^{-1}[f^*(a_w)]^{-1}\n =((ab)\\cdot c)_f.\n\\]\nThe reversed order in the two inverse factors is exactly the formula for the inverse of a product. The identity gauge acts trivially.","If $d=a\\cdot c$ and $x$ is a section of $T_c$, put $y_v=a_vx_v$. Then\n\\[\n d_ff^*(y_w)\n =a_vc_f[f^*(a_w)]^{-1}f^*(a_w)f^*(x_w)\n =a_vx_v=y_v.\n\\]\nThe inverse map is multiplication by $a_v^{-1}$ at each view, proving an equivalence of the actual section sets.","For the groupoid, an arrow $a:c\\to d$ means the equation $a\\cdot c=d$. If $b:d\\to e$, then $(ba)\\cdot c=b\\cdot(a\\cdot c)=e$, giving composition. The identity label is $1$ and the inverse label is $a^{-1}$. Associativity and the two inverse identities now follow from the corresponding multiplication laws in the product group. The label $ba$ is therefore forced by the order of application, not an arbitrary notational convention.\n\\end{proof}\n\\source{KS30}\nThe action groupoid \\(Z^1(\\V;G)//\\prod_vG_v\\) retains comparison witnesses. Its orbit set \\(Z^1(\\V;G)/\\prod_vG_v\\) retains only isomorphism classes. There is no reason for the orbit set to be a group in this nonabelian setting.","to e$, then $(ba)\\cdot c=b\\cdot(a\\cdot c)=e$, giving composition. The identity label is $1$ and the inverse label is $a^{-1}$. Associativity and the two inverse identities now follow from the corresponding multiplication laws in the product group. The label $ba$ is therefore forced by the order of application, not an arbitrary notational convention.\n\\end{proof}\n\\source{KS30}\nThe action groupoid \\(Z^1(\\V;G)//\\prod_vG_v\\) retains comparison witnesses. Its orbit set \\(Z^1(\\V;G)/\\prod_vG_v\\) retains only isomorphism classes. There is no reason for the orbit set to be a group in this nonabelian setting.","identity label is $1$ and the inverse label is $a^{-1}$. Associativity and the two inverse identities now follow from the corresponding multiplication laws in the product group. The label $ba$ is therefore forced by the order of application, not an arbitrary notational convention.\n\\end{proof}\n\\source{KS30}\nThe action groupoid \\(Z^1(\\V;G)//\\prod_vG_v\\) retains comparison witnesses. Its orbit set \\(Z^1(\\V;G)/\\prod_vG_v\\) retains only isomorphism classes. There is no reason for the orbit set to be a group in this nonabelian setting.","\\subsection{Automorphisms and ambiguity are not the same question}\nThe stabilizer of a cocycle is the group of families with\n\\[\n a_v=c_f f^*(a_w)c_f^{-1}.\n\\]\nIf one gauge \\(a:c\\to d\\) is known, all gauges from \\(c\\) to \\(d\\) are uniquely of the form \\(ah\\), with \\(h\\in\\Stab(c)\\). Indeed \\(a^{-1}b\\) stabilizes \\(c\\) for every other gauge \\(b\\). This is the ordinary Hom torsor principle in a groupoid.","\\subsection{The relative obstruction without commutativity}\nPrescribe a coherent partial section \\(s\\) of \\(T_c\\) on \\(U\\). The based gauge group is\n\\[\n \\mathcal G_U=\\{a\\in\\prod_vG_v:a_u=1\\text{ for }u\\in U\\}.\n\\]\nChoose a raw family \\(q\\) equal to \\(s\\) on \\(U\\). Identity elements can be used elsewhere. Normalize the labels by\n\\[\n c^q_f=q_v^{-1}c_f f^*(q_w).\n\\]\nIts labels on arrows internal to \\(U\\) are one. Define the pointed orbit obstruction\n\\[\n \\mathfrak o_U(c,s)=[c^q]\\in Z^1(\\V;G)/\\mathcal G_U,\n\\]\nwhose neutral point is the class of the trivial cocycle.","ativity}\nPrescribe a coherent partial section \\(s\\) of \\(T_c\\) on \\(U\\). The based gauge group is\n\\[\n \\mathcal G_U=\\{a\\in\\prod_vG_v:a_u=1\\text{ for }u\\in U\\}.\n\\]\nChoose a raw family \\(q\\) equal to \\(s\\) on \\(U\\). Identity elements can be used elsewhere. Normalize the labels by\n\\[\n c^q_f=q_v^{-1}c_f f^*(q_w).\n\\]\nIts labels on arrows internal to \\(U\\) are one. Define the pointed orbit obstruction\n\\[\n \\mathfrak o_U(c,s)=[c^q]\\in Z^1(\\V;G)/\\mathcal G_U,\n\\]\nwhose neutral point is the class of the trivial cocycle.","\\begin{theorem}[Nonabelian relative extension]\\label{KS31}\nThe obstruction is independent of the raw lift. It is neutral exactly when \\(s\\) extends to a global section of \\(T_c\\).\n\\end{theorem}\n\\begin{proof}\nThe normalized label $c^q$ is the gauge transform of $c$ by $q^{-1}$, so Proposition~\\ref{KS30} makes it a cocycle. If $f:v\\to w$ is internal to $U$, the partial-section equation says $q_v=c_ff^*(q_w)$, and therefore $c_f^q=1$ on that arrow.","relative extension]\\label{KS31}\nThe obstruction is independent of the raw lift. It is neutral exactly when \\(s\\) extends to a global section of \\(T_c\\).\n\\end{theorem}\n\\begin{proof}\nThe normalized label $c^q$ is the gauge transform of $c$ by $q^{-1}$, so Proposition~\\ref{KS30} makes it a cocycle. If $f:v\\to w$ is internal to $U$, the partial-section equation says $q_v=c_ff^*(q_w)$, and therefore $c_f^q=1$ on that arrow.","If $t$ is another raw lift, define $a_v=t_v^{-1}q_v$. On $U$ both lifts equal $s$, so $a_v=1$ and $a\\in\\mathcal G_U$. Its action is\n\\[\n\\begin{aligned}\n (a\\cdot c^q)_f\n &=t_v^{-1}q_v\\,q_v^{-1}c_ff^*(q_w)\n       [f^*(t_w^{-1}q_w)]^{-1}","If $t$ is another raw lift, define $a_v=t_v^{-1}q_v$. On $U$ both lifts equal $s$, so $a_v=1$ and $a\\in\\mathcal G_U$. Its action is\n\\[\n\\begin{aligned}\n (a\\cdot c^q)_f\n &=t_v^{-1}q_v\\,q_v^{-1}c_ff^*(q_w)\n       [f^*(t_w^{-1}q_w)]^{-1}\\\\\n &=t_v^{-1}c_ff^*(q_w)\n       [f^*(q_w)]^{-1}f^*(t_w)\n =t_v^{-1}c_ff^*(t_w)=c_f^t.\n\\end{aligned}\n\\]\nThus changing the raw lift leaves the based orbit unchanged.","Suppose the orbit is neutral. Equality of orbit classes means there exists an actual based gauge $b$ with $b\\cdot c^q=1$. At an arrow $f:v\\to w$, this equation is\n\\[\n b_vq_v^{-1}c_ff^*(q_w)[f^*(b_w)]^{-1}=1.\n\\]\nMultiply on the left by $q_vb_v^{-1}$. Homomorphic preservation of products then gives\n\\[\n c_ff^*(q_wb_w^{-1})=q_vb_v^{-1}.\n\\]\nTherefore $x_v=q_vb_v^{-1}$ is a twisted coherent family. On $U$, $b_v=1$ and $q_v=s_v$, so it extends the prescribed section.","se the orbit is neutral. Equality of orbit classes means there exists an actual based gauge $b$ with $b\\cdot c^q=1$. At an arrow $f:v\\to w$, this equation is\n\\[\n b_vq_v^{-1}c_ff^*(q_w)[f^*(b_w)]^{-1}=1.\n\\]\nMultiply on the left by $q_vb_v^{-1}$. Homomorphic preservation of products then gives\n\\[\n c_ff^*(q_wb_w^{-1})=q_vb_v^{-1}.\n\\]\nTherefore $x_v=q_vb_v^{-1}$ is a twisted coherent family. On $U$, $b_v=1$ and $q_v=s_v$, so it extends the prescribed section.","is neutral. Equality of orbit classes means there exists an actual based gauge $b$ with $b\\cdot c^q=1$. At an arrow $f:v\\to w$, this equation is\n\\[\n b_vq_v^{-1}c_ff^*(q_w)[f^*(b_w)]^{-1}=1.\n\\]\nMultiply on the left by $q_vb_v^{-1}$. Homomorphic preservation of products then gives\n\\[\n c_ff^*(q_wb_w^{-1})=q_vb_v^{-1}.\n\\]\nTherefore $x_v=q_vb_v^{-1}$ is a twisted coherent family. On $U$, $b_v=1$ and $q_v=s_v$, so it extends the prescribed section.","Conversely, suppose $x$ is a global twisted section extending $s$. Put $b_v=x_v^{-1}q_v$. This is based, and direct substitution gives\n\\[\n (b\\cdot c^q)_f\n =x_v^{-1}c_ff^*(x_w)=1,\n\\]\nwhere the last equality is precisely the section equation for $x$. Hence the obstruction is neutral. These arguments establish the result using group cancellation only. In particular they do not assume an additive quotient, a norm on labels, or a decision procedure for finding $b$.\n\\end{proof}\n\\source{KS31}\nThe orbit is defined from actual cocycles and an actual group action. It is not a Boolean test defined to be neutral precisely when an extension exists. Deciding orbit membership in arbitrary groups remains a separate problem.","$s$. Put $b_v=x_v^{-1}q_v$. This is based, and direct substitution gives\n\\[\n (b\\cdot c^q)_f\n =x_v^{-1}c_ff^*(x_w)=1,\n\\]\nwhere the last equality is precisely the section equation for $x$. Hence the obstruction is neutral. These arguments establish the result using group cancellation only. In particular they do not assume an additive quotient, a norm on labels, or a decision procedure for finding $b$.\n\\end{proof}\n\\source{KS31}\nThe orbit is defined from actual cocycles and an actual group action. It is not a Boolean test defined to be neutral precisely when an extension exists. Deciding orbit membership in arbitrary groups remains a separate problem.","Let\n\\[\n H_U=\\{h:f^*(h_w)=h_v\\text{ for every }f,\\ h|_U=1\\}.\n\\]","\\begin{theorem}[The nonabelian extension torsor]\\label{KS32}\nGiven one extension \\(x_0\\), the map \\(h\\mapsto x_0h\\) is an equivalence from \\(H_U\\) to the extension set. The extension is unique if and only if the obstruction is neutral and \\(H_U\\) is trivial.\n\\end{theorem}\n\\begin{proof}\nTake \\(h\\in H_U\\). Its pointwise product with the chosen extension \\(x_0\\) agrees with the prescribed data on \\(U\\), because \\(h|_U=1\\). For an arrow \\(f:v\\to w\\), the coefficient homomorphism preserves multiplication and the two coherence equations give\n\\[\n c_f f^*((x_0)_wh_w)\n =c_f f^*((x_0)_w)f^*(h_w)\n =(x_0)_vh_v.\n\\]\nThus \\(x_0h\\) is another twisted extension.","}[The nonabelian extension torsor]\\label{KS32}\nGiven one extension \\(x_0\\), the map \\(h\\mapsto x_0h\\) is an equivalence from \\(H_U\\) to the extension set. The extension is unique if and only if the obstruction is neutral and \\(H_U\\) is trivial.\n\\end{theorem}\n\\begin{proof}\nTake \\(h\\in H_U\\). Its pointwise product with the chosen extension \\(x_0\\) agrees with the prescribed data on \\(U\\), because \\(h|_U=1\\). For an arrow \\(f:v\\to w\\), the coefficient homomorphism preserves multiplication and the two coherence equations give\n\\[\n c_f f^*((x_0)_wh_w)\n =c_f f^*((x_0)_w)f^*(h_w)\n =(x_0)_vh_v.\n\\]\nThus \\(x_0h\\) is another twisted extension.","n torsor]\\label{KS32}\nGiven one extension \\(x_0\\), the map \\(h\\mapsto x_0h\\) is an equivalence from \\(H_U\\) to the extension set. The extension is unique if and only if the obstruction is neutral and \\(H_U\\) is trivial.\n\\end{theorem}\n\\begin{proof}\nTake \\(h\\in H_U\\). Its pointwise product with the chosen extension \\(x_0\\) agrees with the prescribed data on \\(U\\), because \\(h|_U=1\\). For an arrow \\(f:v\\to w\\), the coefficient homomorphism preserves multiplication and the two coherence equations give\n\\[\n c_f f^*((x_0)_wh_w)\n =c_f f^*((x_0)_w)f^*(h_w)\n =(x_0)_vh_v.\n\\]\nThus \\(x_0h\\) is another twisted extension.","\\) to the extension set. The extension is unique if and only if the obstruction is neutral and \\(H_U\\) is trivial.\n\\end{theorem}\n\\begin{proof}\nTake \\(h\\in H_U\\). Its pointwise product with the chosen extension \\(x_0\\) agrees with the prescribed data on \\(U\\), because \\(h|_U=1\\). For an arrow \\(f:v\\to w\\), the coefficient homomorphism preserves multiplication and the two coherence equations give\n\\[\n c_f f^*((x_0)_wh_w)\n =c_f f^*((x_0)_w)f^*(h_w)\n =(x_0)_vh_v.\n\\]\nThus \\(x_0h\\) is another twisted extension.","uction is neutral and \\(H_U\\) is trivial.\n\\end{theorem}\n\\begin{proof}\nTake \\(h\\in H_U\\). Its pointwise product with the chosen extension \\(x_0\\) agrees with the prescribed data on \\(U\\), because \\(h|_U=1\\). For an arrow \\(f:v\\to w\\), the coefficient homomorphism preserves multiplication and the two coherence equations give\n\\[\n c_f f^*((x_0)_wh_w)\n =c_f f^*((x_0)_w)f^*(h_w)\n =(x_0)_vh_v.\n\\]\nThus \\(x_0h\\) is another twisted extension.","Conversely, if \\(x\\) is a twisted extension of the same prescribed data, define \\(h_v=(x_0)_v^{-1}x_v\\). On \\(U\\) its value is one. For every arrow,\n\\[\n \\begin{aligned}\n h_v\n &=[c_f f^*((x_0)_w)]^{-1}[c_f f^*(x_w)]","Conversely, if \\(x\\) is a twisted extension of the same prescribed data, define \\(h_v=(x_0)_v^{-1}x_v\\). On \\(U\\) its value is one. For every arrow,\n\\[\n \\begin{aligned}\n h_v\n &=[c_f f^*((x_0)_w)]^{-1}[c_f f^*(x_w)]\\\\\n &=f^*((x_0)_w)^{-1}c_f^{-1}c_f f^*(x_w)\\\\\n &=f^*((x_0)_w^{-1}x_w)=f^*(h_w).\n \\end{aligned}\n\\]\nThe cancellation uses inverse multiplication in its actual order. No commuting factors have been moved past one another. Hence \\(h\\in H_U\\). Pointwise group cancellation shows that \\(h\\mapsto x_0h\\) and \\(x\\mapsto x_0^{-1}x\\) are inverse maps.","&=f^*((x_0)_w^{-1}x_w)=f^*(h_w).\n \\end{aligned}\n\\]\nThe cancellation uses inverse multiplication in its actual order. No commuting factors have been moved past one another. Hence \\(h\\in H_U\\). Pointwise group cancellation shows that \\(h\\mapsto x_0h\\) and \\(x\\mapsto x_0^{-1}x\\) are inverse maps.","The extension set is nonempty precisely when the earlier orbit obstruction is neutral. Once it is nonempty, the displayed bijection makes it a singleton precisely when \\(H_U\\) has only its identity element. Both conditions are needed for unique existence. Changing the chosen extension changes the parametrization, not the extension set or the criterion.\n\\end{proof}\n\\source{KS32}\nThe additive theory is recovered by applying a multiplicative type tag to an additive view system and choosing the trivial cocycle. The normalized label then becomes \\(f^*q_w-q_v\\). The source proves agreement of the actual extension sets and vanishing criteria. It does not identify the whole nonabelian orbit set with every element of the larger additive arrow residual quotient.","ton precisely when \\(H_U\\) has only its identity element. Both conditions are needed for unique existence. Changing the chosen extension changes the parametrization, not the extension set or the criterion.\n\\end{proof}\n\\source{KS32}\nThe additive theory is recovered by applying a multiplicative type tag to an additive view system and choosing the trivial cocycle. The normalized label then becomes \\(f^*q_w-q_v\\). The source proves agreement of the actual extension sets and vanishing criteria. It does not identify the whole nonabelian orbit set with every element of the larger additive arrow residual quotient.","\\subsection{A small noncommutative model}\n\\begin{example}[A \\(C_2\\) loop with \\(S_3\\) coefficients]\\label{KS33}\nTake one view with a loop \\(\\sigma\\) satisfying \\(\\sigma^2=1\\), and constant coefficients in a group \\(H\\). A cocycle is determined by an element \\(p\\in H\\) whose square is the identity. Gauge equivalence is conjugacy, and its automorphisms form the centralizer of \\(p\\).","For \\(H=S_3\\), these elements are the identity and the three transpositions. There are exactly two orbit classes. The identity cocycle has automorphism group \\(S_3\\), which is noncommutative. A nonidentity involution gives a coherent cocycle whose regular twisted section equation \\(x=px\\) has no solution.\n\\end{example}\n\\begin{proof}\nThe only nontrivial cocycle relation is \\(p^2=1\\). At a single view the gauge action is \\(p\\mapsto apa^{-1}\\). All three transpositions are conjugate and none is conjugate to the identity. The stabilizer is therefore the centralizer. Finally \\(x=px\\) implies \\(p=1\\) after multiplication by \\(x^{-1}\\).\n\\end{proof}\n\\source{KS33}\nThe example makes an important boundary visible","For \\(H=S_3\\), these elements are the identity and the three transpositions. There are exactly two orbit classes. The identity cocycle has automorphism group \\(S_3\\), which is noncommutative. A nonidentity involution gives a coherent cocycle whose regular twisted section equation \\(x=px\\) has no solution.\n\\end{example}\n\\begin{proof}\nThe only nontrivial cocycle relation is \\(p^2=1\\). At a single view the gauge action is \\(p\\mapsto apa^{-1}\\). All three transpositions are conjugate and none is conjugate to the identity. The stabilizer is therefore the centralizer. Finally \\(x=px\\) implies \\(p=1\\) after multiplication by \\(x^{-1}\\).\n\\end{proof}\n\\source{KS33}\nThe example makes an important boundary visible. A cocycle can exist and glue as an object even when it admits no global trivializing section. Stackhood will concern the former gluing problem, not a","For \\(H=S_3\\), these elements are the identity and the three transpositions. There are exactly two orbit classes. The identity cocycle has automorphism group \\(S_3\\), which is noncommutative. A nonidentity involution gives a coherent cocycle whose regular twisted section equation \\(x=px\\) has no solution.\n\\end{example}\n\\begin{proof}\nThe only nontrivial cocycle relation is \\(p^2=1\\). At a single view the gauge action is \\(p\\mapsto apa^{-1}\\). All three transpositions are conjugate and none is conjugate to the identity. The stabilizer is therefore the centralizer. Finally \\(x=px\\) implies \\(p=1\\) after multiplication by \\(x^{-1}\\).\n\\end{proof}\n\\source{KS33}\nThe example makes an important boundary visible. A cocycle can exist and glue as an object even when it admits no global trivializing section. Stackhood will concern the former gluing problem, not a claim that every cocycle becomes trivial.","ents are the identity and the three transpositions. There are exactly two orbit classes. The identity cocycle has automorphism group \\(S_3\\), which is noncommutative. A nonidentity involution gives a coherent cocycle whose regular twisted section equation \\(x=px\\) has no solution.\n\\end{example}\n\\begin{proof}\nThe only nontrivial cocycle relation is \\(p^2=1\\). At a single view the gauge action is \\(p\\mapsto apa^{-1}\\). All three transpositions are conjugate and none is conjugate to the identity. The stabilizer is therefore the centralizer. Finally \\(x=px\\) implies \\(p=1\\) after multiplication by \\(x^{-1}\\).\n\\end{proof}\n\\source{KS33}\nThe example makes an important boundary visible. A cocycle can exist and glue as an object even when it admits no global trivializing section. Stackhood will concern the former gluing problem, not a claim that every cocycle becomes trivial.","commutative. A nonidentity involution gives a coherent cocycle whose regular twisted section equation \\(x=px\\) has no solution.\n\\end{example}\n\\begin{proof}\nThe only nontrivial cocycle relation is \\(p^2=1\\). At a single view the gauge action is \\(p\\mapsto apa^{-1}\\). All three transpositions are conjugate and none is conjugate to the identity. The stabilizer is therefore the centralizer. Finally \\(x=px\\) implies \\(p=1\\) after multiplication by \\(x^{-1}\\).\n\\end{proof}\n\\source{KS33}\nThe example makes an important boundary visible. A cocycle can exist and glue as an object even when it admits no global trivializing section. Stackhood will concern the former gluing problem, not a claim that every cocycle becomes trivial.","nidentity involution gives a coherent cocycle whose regular twisted section equation \\(x=px\\) has no solution.\n\\end{example}\n\\begin{proof}\nThe only nontrivial cocycle relation is \\(p^2=1\\). At a single view the gauge action is \\(p\\mapsto apa^{-1}\\). All three transpositions are conjugate and none is conjugate to the identity. The stabilizer is therefore the centralizer. Finally \\(x=px\\) implies \\(p=1\\) after multiplication by \\(x^{-1}\\).\n\\end{proof}\n\\source{KS33}\nThe example makes an important boundary visible. A cocycle can exist and glue as an object even when it admits no global trivializing section. Stackhood will concern the former gluing problem, not a claim that every cocycle becomes trivial.","erent cocycle whose regular twisted section equation \\(x=px\\) has no solution.\n\\end{example}\n\\begin{proof}\nThe only nontrivial cocycle relation is \\(p^2=1\\). At a single view the gauge action is \\(p\\mapsto apa^{-1}\\). All three transpositions are conjugate and none is conjugate to the identity. The stabilizer is therefore the centralizer. Finally \\(x=px\\) implies \\(p=1\\) after multiplication by \\(x^{-1}\\).\n\\end{proof}\n\\source{KS33}\nThe example makes an important boundary visible. A cocycle can exist and glue as an object even when it admits no global trivializing section. Stackhood will concern the former gluing problem, not a claim that every cocycle becomes trivial.","\\plate{06_groupoid}{The orbit set sees two classes in the \\(C_2\\)-to-\\(S_3\\) example. The groupoid also retains the arrows between conjugate transpositions and all automorphisms. Collapsing the arrows loses information that matters in descent.}{fig:groupoid}","Let \\(U_i\\) be full patches in \\(\\V\\). Assume the composition cover condition\n\\begin{equation}\\label{eq:compositioncover}\n \\forall(v\\xrightarrow f w\\xrightarrow g z),\\quad\n \\exists i:\\ v,w,z\\in U_i.\n\\end{equation}\nBy inserting identity arrows, this implies coverage of objects and arrows.","A descent datum consists of local cocycles \\(c_i\\) and overlap gauges \\(a_{ij}(v)\\) with\n\\begin{align}\n c_j(f)&=a_{ij}(v)c_i(f)[f^*a_{ij}(w)]^{-1},\\label{eq:localgauge}","A descent datum consists of local cocycles \\(c_i\\) and overlap gauges \\(a_{ij}(v)\\) with\n\\begin{align}\n c_j(f)&=a_{ij}(v)c_i(f)[f^*a_{ij}(w)]^{-1},\\label{eq:localgauge}\\\\\n a_{ik}(v)&=a_{jk}(v)a_{ij}(v).\\label{eq:triple}\n\\end{align}\nThe triple equation implies \\(a_{ii}=1\\) and \\(a_{ji}=a_{ij}^{-1}\\). A morphism of descent data is a family of local gauges \\(r_i\\) satisfying\n\\[\n r_j(v)a_{ij}(v)=a'_{ij}(v)r_i(v).\n\\]\nThese objects and arrows form a groupoid.","a_{ik}(v)&=a_{jk}(v)a_{ij}(v).\\label{eq:triple}\n\\end{align}\nThe triple equation implies \\(a_{ii}=1\\) and \\(a_{ji}=a_{ij}^{-1}\\). A morphism of descent data is a family of local gauges \\(r_i\\) satisfying\n\\[\n r_j(v)a_{ij}(v)=a'_{ij}(v)r_i(v).\n\\]\nThese objects and arrows form a groupoid.","theorem}[Effective cocycle descent]\\label{KS34}\nUnder \\eqref{eq:compositioncover}, restriction is an equivalence between the global cocycle gauge groupoid and the groupoid of descent data. The overlap gauges supplied in the datum are realized, not discarded.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: normalize the supplied identifications.} The triple-overlap equation with all indices equal gives $a_{ii}=a_{ii}a_{ii}$, hence $a_{ii}=1$. With indices $i,j,i$ it gives $a_{ji}a_{ij}=1$, hence $a_{ji}=a_{ij}^{-1}$. Composition coverage implies object coverage by inserting identity arrows. Choose a containing patch $\\kappa(v)$ for each view and put\n\\[\n b_i(v)=a_{\\kappa(v),i}(v)\\quad(v\\in U_i).\n\\]\nAt a point of $U_i\\cap U_j$, the triple law, now on patches $\\kappa(v),i,j$, gives\n\\[\n b_j(v)=a_{ij}(v)b_i(v).                                    \\tag{*}\n\\]\nThe two endpoints of an arrow may have different reference patches. No common reference patch is being assumed.","scent]\\label{KS34}\nUnder \\eqref{eq:compositioncover}, restriction is an equivalence between the global cocycle gauge groupoid and the groupoid of descent data. The overlap gauges supplied in the datum are realized, not discarded.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: normalize the supplied identifications.} The triple-overlap equation with all indices equal gives $a_{ii}=a_{ii}a_{ii}$, hence $a_{ii}=1$. With indices $i,j,i$ it gives $a_{ji}a_{ij}=1$, hence $a_{ji}=a_{ij}^{-1}$. Composition coverage implies object coverage by inserting identity arrows. Choose a containing patch $\\kappa(v)$ for each view and put\n\\[\n b_i(v)=a_{\\kappa(v),i}(v)\\quad(v\\in U_i).\n\\]\nAt a point of $U_i\\cap U_j$, the triple law, now on patches $\\kappa(v),i,j$, gives\n\\[\n b_j(v)=a_{ij}(v)b_i(v).                                    \\tag{*}\n\\]\nThe two endpoints of an arrow may have different reference patches. No common reference patch is being assumed.","\\eqref{eq:compositioncover}, restriction is an equivalence between the global cocycle gauge groupoid and the groupoid of descent data. The overlap gauges supplied in the datum are realized, not discarded.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: normalize the supplied identifications.} The triple-overlap equation with all indices equal gives $a_{ii}=a_{ii}a_{ii}$, hence $a_{ii}=1$. With indices $i,j,i$ it gives $a_{ji}a_{ij}=1$, hence $a_{ji}=a_{ij}^{-1}$. Composition coverage implies object coverage by inserting identity arrows. Choose a containing patch $\\kappa(v)$ for each view and put\n\\[\n b_i(v)=a_{\\kappa(v),i}(v)\\quad(v\\in U_i).\n\\]\nAt a point of $U_i\\cap U_j$, the triple law, now on patches $\\kappa(v),i,j$, gives\n\\[\n b_j(v)=a_{ij}(v)b_i(v).                                    \\tag{*}\n\\]\nThe two endpoints of an arrow may have different reference patches. No common reference patch is being assumed.","tioncover}, restriction is an equivalence between the global cocycle gauge groupoid and the groupoid of descent data. The overlap gauges supplied in the datum are realized, not discarded.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: normalize the supplied identifications.} The triple-overlap equation with all indices equal gives $a_{ii}=a_{ii}a_{ii}$, hence $a_{ii}=1$. With indices $i,j,i$ it gives $a_{ji}a_{ij}=1$, hence $a_{ji}=a_{ij}^{-1}$. Composition coverage implies object coverage by inserting identity arrows. Choose a containing patch $\\kappa(v)$ for each view and put\n\\[\n b_i(v)=a_{\\kappa(v),i}(v)\\quad(v\\in U_i).\n\\]\nAt a point of $U_i\\cap U_j$, the triple law, now on patches $\\kappa(v),i,j$, gives\n\\[\n b_j(v)=a_{ij}(v)b_i(v).                                    \\tag{*}\n\\]\nThe two endpoints of an arrow may have different reference patches. No common reference patch is being assumed.","is an equivalence between the global cocycle gauge groupoid and the groupoid of descent data. The overlap gauges supplied in the datum are realized, not discarded.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: normalize the supplied identifications.} The triple-overlap equation with all indices equal gives $a_{ii}=a_{ii}a_{ii}$, hence $a_{ii}=1$. With indices $i,j,i$ it gives $a_{ji}a_{ij}=1$, hence $a_{ji}=a_{ij}^{-1}$. Composition coverage implies object coverage by inserting identity arrows. Choose a containing patch $\\kappa(v)$ for each view and put\n\\[\n b_i(v)=a_{\\kappa(v),i}(v)\\quad(v\\in U_i).\n\\]\nAt a point of $U_i\\cap U_j$, the triple law, now on patches $\\kappa(v),i,j$, gives\n\\[\n b_j(v)=a_{ij}(v)b_i(v).                                    \\tag{*}\n\\]\nThe two endpoints of an arrow may have different reference patches. No common reference patch is being assumed.","n the datum are realized, not discarded.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: normalize the supplied identifications.} The triple-overlap equation with all indices equal gives $a_{ii}=a_{ii}a_{ii}$, hence $a_{ii}=1$. With indices $i,j,i$ it gives $a_{ji}a_{ij}=1$, hence $a_{ji}=a_{ij}^{-1}$. Composition coverage implies object coverage by inserting identity arrows. Choose a containing patch $\\kappa(v)$ for each view and put\n\\[\n b_i(v)=a_{\\kappa(v),i}(v)\\quad(v\\in U_i).\n\\]\nAt a point of $U_i\\cap U_j$, the triple law, now on patches $\\kappa(v),i,j$, gives\n\\[\n b_j(v)=a_{ij}(v)b_i(v).                                    \\tag{*}\n\\]\nThe two endpoints of an arrow may have different reference patches. No common reference patch is being assumed.","rded.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: normalize the supplied identifications.} The triple-overlap equation with all indices equal gives $a_{ii}=a_{ii}a_{ii}$, hence $a_{ii}=1$. With indices $i,j,i$ it gives $a_{ji}a_{ij}=1$, hence $a_{ji}=a_{ij}^{-1}$. Composition coverage implies object coverage by inserting identity arrows. Choose a containing patch $\\kappa(v)$ for each view and put\n\\[\n b_i(v)=a_{\\kappa(v),i}(v)\\quad(v\\in U_i).\n\\]\nAt a point of $U_i\\cap U_j$, the triple law, now on patches $\\kappa(v),i,j$, gives\n\\[\n b_j(v)=a_{ij}(v)b_i(v).                                    \\tag{*}\n\\]\nThe two endpoints of an arrow may have different reference patches. No common reference patch is being assumed.","supplied identifications.} The triple-overlap equation with all indices equal gives $a_{ii}=a_{ii}a_{ii}$, hence $a_{ii}=1$. With indices $i,j,i$ it gives $a_{ji}a_{ij}=1$, hence $a_{ji}=a_{ij}^{-1}$. Composition coverage implies object coverage by inserting identity arrows. Choose a containing patch $\\kappa(v)$ for each view and put\n\\[\n b_i(v)=a_{\\kappa(v),i}(v)\\quad(v\\in U_i).\n\\]\nAt a point of $U_i\\cap U_j$, the triple law, now on patches $\\kappa(v),i,j$, gives\n\\[\n b_j(v)=a_{ij}(v)b_i(v).                                    \\tag{*}\n\\]\nThe two endpoints of an arrow may have different reference patches. No common reference patch is being assumed.","equation with all indices equal gives $a_{ii}=a_{ii}a_{ii}$, hence $a_{ii}=1$. With indices $i,j,i$ it gives $a_{ji}a_{ij}=1$, hence $a_{ji}=a_{ij}^{-1}$. Composition coverage implies object coverage by inserting identity arrows. Choose a containing patch $\\kappa(v)$ for each view and put\n\\[\n b_i(v)=a_{\\kappa(v),i}(v)\\quad(v\\in U_i).\n\\]\nAt a point of $U_i\\cap U_j$, the triple law, now on patches $\\kappa(v),i,j$, gives\n\\[\n b_j(v)=a_{ij}(v)b_i(v).                                    \\tag{*}\n\\]\nThe two endpoints of an arrow may have different reference patches. No common reference patch is being assumed.","s equal gives $a_{ii}=a_{ii}a_{ii}$, hence $a_{ii}=1$. With indices $i,j,i$ it gives $a_{ji}a_{ij}=1$, hence $a_{ji}=a_{ij}^{-1}$. Composition coverage implies object coverage by inserting identity arrows. Choose a containing patch $\\kappa(v)$ for each view and put\n\\[\n b_i(v)=a_{\\kappa(v),i}(v)\\quad(v\\in U_i).\n\\]\nAt a point of $U_i\\cap U_j$, the triple law, now on patches $\\kappa(v),i,j$, gives\n\\[\n b_j(v)=a_{ij}(v)b_i(v).                                    \\tag{*}\n\\]\nThe two endpoints of an arrow may have different reference patches. No common reference patch is being assumed.","\\emph{Step 2: obtain literal matching labels.} Define\n\\[\n \\widetilde c_i(f)=b_i(v)^{-1}c_i(f)f^*(b_i(w))\n \\quad(f:v\\to w\\text{ in }U_i).\n\\]\nThis is a gauge transform and hence a cocycle on $U_i$. If the arrow lies also in $U_j$, substitute the original overlap equation and (*) at its two endpoints:\n\\[\n\\begin{aligned}\n \\widetilde c_j(f)\n &=b_i(v)^{-1}a_{ij}(v)^{-1}\n   a_{ij}(v)c_i(f)[f^*a_{ij}(w)]^{-1}\n   f^*(a_{ij}(w)b_i(w))","\\emph{Step 2: obtain literal matching labels.} Define\n\\[\n \\widetilde c_i(f)=b_i(v)^{-1}c_i(f)f^*(b_i(w))\n \\quad(f:v\\to w\\text{ in }U_i).\n\\]\nThis is a gauge transform and hence a cocycle on $U_i$. If the arrow lies also in $U_j$, substitute the original overlap equation and (*) at its two endpoints:\n\\[\n\\begin{aligned}\n \\widetilde c_j(f)\n &=b_i(v)^{-1}a_{ij}(v)^{-1}\n   a_{ij}(v)c_i(f)[f^*a_{ij}(w)]^{-1}\n   f^*(a_{ij}(w)b_i(w))\\\\\n &=b_i(v)^{-1}c_i(f)f^*(b_i(w))\n =\\widetilde c_i(f).\n\\end{aligned}\n\\]\nEvery cancellation is between neighboring inverse factors. Commutativity is not required.","\\emph{Step 3: construct the global cocycle.} Every arrow is contained in a patch, again by inserting an identity in a composable pair. Define $c_f$ to be $\\widetilde c_i(f)$ in any such patch. Step 2 proves independence. For an identity arrow, choose a patch through its view to obtain $c_{1_v}=1$. For $f:v\\to w$ and $g:w\\to z$, composition coverage supplies a patch containing all three vertices. Fullness places $f$, $g$, and $g\\circ f$ in that same patch. Its cocycle law therefore proves\n\\[\n c_{g\\circ f}=\\widetilde c_i(g\\circ f)\n =\\widetilde c_i(f)f^*(\\widetilde c_i(g))=c_ff^*(c_g).\n\\]\nThis is the exact point where composition coverage, rather than arrow coverage alone, is used.","nstruct the global cocycle.} Every arrow is contained in a patch, again by inserting an identity in a composable pair. Define $c_f$ to be $\\widetilde c_i(f)$ in any such patch. Step 2 proves independence. For an identity arrow, choose a patch through its view to obtain $c_{1_v}=1$. For $f:v\\to w$ and $g:w\\to z$, composition coverage supplies a patch containing all three vertices. Fullness places $f$, $g$, and $g\\circ f$ in that same patch. Its cocycle law therefore proves\n\\[\n c_{g\\circ f}=\\widetilde c_i(g\\circ f)\n =\\widetilde c_i(f)f^*(\\widetilde c_i(g))=c_ff^*(c_g).\n\\]\nThis is the exact point where composition coverage, rather than arrow coverage alone, is used.","ct the global cocycle.} Every arrow is contained in a patch, again by inserting an identity in a composable pair. Define $c_f$ to be $\\widetilde c_i(f)$ in any such patch. Step 2 proves independence. For an identity arrow, choose a patch through its view to obtain $c_{1_v}=1$. For $f:v\\to w$ and $g:w\\to z$, composition coverage supplies a patch containing all three vertices. Fullness places $f$, $g$, and $g\\circ f$ in that same patch. Its cocycle law therefore proves\n\\[\n c_{g\\circ f}=\\widetilde c_i(g\\circ f)\n =\\widetilde c_i(f)f^*(\\widetilde c_i(g))=c_ff^*(c_g).\n\\]\nThis is the exact point where composition coverage, rather than arrow coverage alone, is used.","obal cocycle.} Every arrow is contained in a patch, again by inserting an identity in a composable pair. Define $c_f$ to be $\\widetilde c_i(f)$ in any such patch. Step 2 proves independence. For an identity arrow, choose a patch through its view to obtain $c_{1_v}=1$. For $f:v\\to w$ and $g:w\\to z$, composition coverage supplies a patch containing all three vertices. Fullness places $f$, $g$, and $g\\circ f$ in that same patch. Its cocycle law therefore proves\n\\[\n c_{g\\circ f}=\\widetilde c_i(g\\circ f)\n =\\widetilde c_i(f)f^*(\\widetilde c_i(g))=c_ff^*(c_g).\n\\]\nThis is the exact point where composition coverage, rather than arrow coverage alone, is used.","rting an identity in a composable pair. Define $c_f$ to be $\\widetilde c_i(f)$ in any such patch. Step 2 proves independence. For an identity arrow, choose a patch through its view to obtain $c_{1_v}=1$. For $f:v\\to w$ and $g:w\\to z$, composition coverage supplies a patch containing all three vertices. Fullness places $f$, $g$, and $g\\circ f$ in that same patch. Its cocycle law therefore proves\n\\[\n c_{g\\circ f}=\\widetilde c_i(g\\circ f)\n =\\widetilde c_i(f)f^*(\\widetilde c_i(g))=c_ff^*(c_g).\n\\]\nThis is the exact point where composition coverage, rather than arrow coverage alone, is used.","Undoing the normalization proves\n\\[\n b_i(v)c_f[f^*(b_i(w))]^{-1}=c_i(f).\n\\]\nTogether with (*), these gauges constitute an isomorphism from the restriction of $c$ to the supplied descent datum. They implement its original overlaps, proving essential surjectivity of restriction.","\\emph{Step 4: retain every morphism.} Let $c,d$ be global cocycles. Their canonical restrictions have identity overlap gauges. A morphism between those restrictions is consequently a family $r_i$ satisfying $r_j(v)=r_i(v)$ on each overlap. Assemble $r_v$ from any containing patch. For an arrow choose a patch containing both endpoints. Its gauge equation proves\n\\[\n d_f=r_vc_f[f^*(r_w)]^{-1}.\n\\]\nThus $r$ is a global gauge. Its restriction is the original family. Conversely, assembling the restrictions of a global gauge preserves every coordinate and returns that gauge. Restriction is therefore bijective on each Hom set, not merely injective on isomorphism classes. This is full faithfulness.","ph{Step 4: retain every morphism.} Let $c,d$ be global cocycles. Their canonical restrictions have identity overlap gauges. A morphism between those restrictions is consequently a family $r_i$ satisfying $r_j(v)=r_i(v)$ on each overlap. Assemble $r_v$ from any containing patch. For an arrow choose a patch containing both endpoints. Its gauge equation proves\n\\[\n d_f=r_vc_f[f^*(r_w)]^{-1}.\n\\]\nThus $r$ is a global gauge. Its restriction is the original family. Conversely, assembling the restrictions of a global gauge preserves every coordinate and returns that gauge. Restriction is therefore bijective on each Hom set, not merely injective on isomorphism classes. This is full faithfulness.","\\subsection{The correct uniqueness statement}\nSuppose \\((c,b_i)\\) and \\((c',b'_i)\\) realize the same descent datum. The only possible compatible comparison is\n\\[\n g_v=b'_i(v)^{-1}b_i(v),\n\\]\nusing any patch through \\(v\\). The overlap identities make it independent of the patch, and \\(b'_ig=b_i\\). It is therefore the unique global comparison respecting the fixed local identifications.","\\subsection{A failed triple}\nLet \\(t\\ne1\\) be an involution and ask for frames \\(b_0,b_1,b_2\\) satisfying\n\\[\n b_1=tb_0,\\qquad b_2=tb_1,\\qquad b_2=tb_0.\n\\]\nThe first two give \\(b_2=b_0\\), while the third forces \\(t=1\\). Each displayed transition is individually invertible, but they cannot be the overlap identifications of a global frame. The defect lies in the triple relation.","\\plate{07_cover}{Object coverage, arrow coverage, and composition coverage are distinct. The proper two-view patches on a three-view chain see every arrow but no patch sees the relation \\(0\\to1\\to2\\). The four proper three-view patches on a four-view chain do cover every composable pair.}{fig:cover}","\\chap{A genuine stack on the composition site}\\label{sec:stack}\nThe preceding theorem can be placed in a standard site, rather than called stack-like. We use the usual effective-descent meaning of a stack~\\cite{stacks-stack,mathlib-stack}. The base of this site is the inclusion poset of full patches\n\\[\n \\mathcal P(\\V)=(\\operatorname{Set}(\\Ob\\V),\\subseteq).\n\\]\nThis patch category is different from the original category of views.","\\subsection{The topology is defined without assuming effectivity}\nFor a composable pair \\(q=(v\\xrightarrow f w\\xrightarrow g z)\\), let its support be \\(|q|=\\{v,w,z\\}\\). A sieve \\(\\mathcal R\\) on \\(W\\) is covering when\n\\begin{equation}\\label{eq:site}\n |q|\\subseteq W\\quad\\Longrightarrow\\quad\n (|q|\\hookrightarrow W)\\in\\mathcal R\n \\quad\\text{for every composable }q.\n\\end{equation}\nThis condition depends only on the view category, not on the coefficient groups or on a promised global realization.","\\begin{proposition}[The composition topology]\\label{KS35}\nCondition~\\eqref{eq:site} defines a Grothendieck topology \\(J_{\\mathrm{comp}}\\). The sieve generated by a family of inclusions \\(U_i\\subseteq W\\) is covering exactly when the family covers every composition support in \\(W\\).\n\\end{proposition}\n\\begin{proof}\nIn the inclusion poset, identify a sieve on $W$ with a downward-closed collection of subsets of $W$. Pullback along $Z\\subseteq W$ consists of those subsets of $Z$ that were already members of the original sieve. We verify the three topology axioms using this description.","sition}[The composition topology]\\label{KS35}\nCondition~\\eqref{eq:site} defines a Grothendieck topology \\(J_{\\mathrm{comp}}\\). The sieve generated by a family of inclusions \\(U_i\\subseteq W\\) is covering exactly when the family covers every composition support in \\(W\\).\n\\end{proposition}\n\\begin{proof}\nIn the inclusion poset, identify a sieve on $W$ with a downward-closed collection of subsets of $W$. Pullback along $Z\\subseteq W$ consists of those subsets of $Z$ that were already members of the original sieve. We verify the three topology axioms using this description.","ion topology]\\label{KS35}\nCondition~\\eqref{eq:site} defines a Grothendieck topology \\(J_{\\mathrm{comp}}\\). The sieve generated by a family of inclusions \\(U_i\\subseteq W\\) is covering exactly when the family covers every composition support in \\(W\\).\n\\end{proposition}\n\\begin{proof}\nIn the inclusion poset, identify a sieve on $W$ with a downward-closed collection of subsets of $W$. Pullback along $Z\\subseteq W$ consists of those subsets of $Z$ that were already members of the original sieve. We verify the three topology axioms using this description.","q:site} defines a Grothendieck topology \\(J_{\\mathrm{comp}}\\). The sieve generated by a family of inclusions \\(U_i\\subseteq W\\) is covering exactly when the family covers every composition support in \\(W\\).\n\\end{proposition}\n\\begin{proof}\nIn the inclusion poset, identify a sieve on $W$ with a downward-closed collection of subsets of $W$. Pullback along $Z\\subseteq W$ consists of those subsets of $Z$ that were already members of the original sieve. We verify the three topology axioms using this description.","The maximal sieve contains every subset of $W$, hence every composition support. For stability under pullback, let $\\mathcal R$ cover $W$ and let $q$ be a composable pair whose support lies in $Z\\subseteq W$. The same pair lies in the full patch $W$, so $|q|\\in\\mathcal R$. Since $|q|\\subseteq Z$, it belongs to the pulled-back sieve on $Z$. Thus that sieve covers $Z$.","For transitivity, suppose $\\mathcal R$ covers $W$, and that for every $U\\in\\mathcal R$ the pullback of a sieve $\\mathcal T$ to $U$ covers $U$. Fix a composable pair $q$ in $W$. Its support $S=|q|$ belongs to $\\mathcal R$. Since patches are full, $q$ is also a composable pair in $S$. The covering condition for the pullback of $\\mathcal T$ to $S$ therefore puts the identity inclusion $S\\subseteq S$ in that pullback. By the definition of pullback, this says $S\\in\\mathcal T$. It holds for every $q$, so $\\mathcal T$ covers $W$.","for every $U\\in\\mathcal R$ the pullback of a sieve $\\mathcal T$ to $U$ covers $U$. Fix a composable pair $q$ in $W$. Its support $S=|q|$ belongs to $\\mathcal R$. Since patches are full, $q$ is also a composable pair in $S$. The covering condition for the pullback of $\\mathcal T$ to $S$ therefore puts the identity inclusion $S\\subseteq S$ in that pullback. By the definition of pullback, this says $S\\in\\mathcal T$. It holds for every $q$, so $\\mathcal T$ covers $W$.","ck of a sieve $\\mathcal T$ to $U$ covers $U$. Fix a composable pair $q$ in $W$. Its support $S=|q|$ belongs to $\\mathcal R$. Since patches are full, $q$ is also a composable pair in $S$. The covering condition for the pullback of $\\mathcal T$ to $S$ therefore puts the identity inclusion $S\\subseteq S$ in that pullback. By the definition of pullback, this says $S\\in\\mathcal T$. It holds for every $q$, so $\\mathcal T$ covers $W$.","sieve $\\mathcal T$ to $U$ covers $U$. Fix a composable pair $q$ in $W$. Its support $S=|q|$ belongs to $\\mathcal R$. Since patches are full, $q$ is also a composable pair in $S$. The covering condition for the pullback of $\\mathcal T$ to $S$ therefore puts the identity inclusion $S\\subseteq S$ in that pullback. By the definition of pullback, this says $S\\in\\mathcal T$. It holds for every $q$, so $\\mathcal T$ covers $W$.","o $U$ covers $U$. Fix a composable pair $q$ in $W$. Its support $S=|q|$ belongs to $\\mathcal R$. Since patches are full, $q$ is also a composable pair in $S$. The covering condition for the pullback of $\\mathcal T$ to $S$ therefore puts the identity inclusion $S\\subseteq S$ in that pullback. By the definition of pullback, this says $S\\in\\mathcal T$. It holds for every $q$, so $\\mathcal T$ covers $W$.","Finally, the sieve generated by $U_i\\subseteq W$ consists exactly of the subpatches $Z$ contained in at least one $U_i$. One inclusion follows because a sieve contains all restrictions of its generators. The reverse inclusion follows because this collection is itself downward closed and contains all generators, hence is the smallest such sieve. Consequently a support belongs to the generated sieve exactly when some $U_i$ contains all its endpoints. This proves the family characterization. When $W$ is empty, there are no composable pairs to test, so even the empty sieve is covering.\n\\end{proof}\n\\source{KS35}\nIntersections of a covering family with a subpatch remain covering. Replacing each member by a covering family again gives a cover.","Finally, the sieve generated by $U_i\\subseteq W$ consists exactly of the subpatches $Z$ contained in at least one $U_i$. One inclusion follows because a sieve contains all restrictions of its generators. The reverse inclusion follows because this collection is itself downward closed and contains all generators, hence is the smallest such sieve. Consequently a support belongs to the generated sieve exactly when some $U_i$ contains all its endpoints. This proves the family characterization. When $W$ is empty, there are no composable pairs to test, so even the empty sieve is covering.\n\\end{proof}\n\\source{KS35}\nIntersections of a covering family with a subpatch remain covering. Replacing each member by a covering family again gives a cover. Identities imply coverage of single vie","Finally, the sieve generated by $U_i\\subseteq W$ consists exactly of the subpatches $Z$ contained in at least one $U_i$. One inclusion follows because a sieve contains all restrictions of its generators. The reverse inclusion follows because this collection is itself downward closed and contains all generators, hence is the smallest such sieve. Consequently a support belongs to the generated sieve exactly when some $U_i$ contains all its endpoints. This proves the family characterization. When $W$ is empty, there are no composable pairs to test, so even the empty sieve is covering.\n\\end{proof}\n\\source{KS35}\nIntersections of a covering family with a subpatch remain covering. Replacing each member by a covering family again gives a cover. Identities imply coverage of single views and arrows. The empty patch a","enerated by $U_i\\subseteq W$ consists exactly of the subpatches $Z$ contained in at least one $U_i$. One inclusion follows because a sieve contains all restrictions of its generators. The reverse inclusion follows because this collection is itself downward closed and contains all generators, hence is the smallest such sieve. Consequently a support belongs to the generated sieve exactly when some $U_i$ contains all its endpoints. This proves the family characterization. When $W$ is empty, there are no composable pairs to test, so even the empty sieve is covering.\n\\end{proof}\n\\source{KS35}\nIntersections of a covering family with a subpatch remain covering. Replacing each member by a covering family again gives a cover. Identities imply coverage of single views and arrows. The empty patch admits the empty cover.","at least one $U_i$. One inclusion follows because a sieve contains all restrictions of its generators. The reverse inclusion follows because this collection is itself downward closed and contains all generators, hence is the smallest such sieve. Consequently a support belongs to the generated sieve exactly when some $U_i$ contains all its endpoints. This proves the family characterization. When $W$ is empty, there are no composable pairs to test, so even the empty sieve is covering.\n\\end{proof}\n\\source{KS35}\nIntersections of a covering family with a subpatch remain covering. Replacing each member by a covering family again gives a cover. Identities imply coverage of single views and arrows. The empty patch admits the empty cover.","\\subsection{The actual patch pseudofunctor}\nOn a patch \\(W\\), use\n\\[\n \\mathfrak K_G(W)=Z^1(\\V|_W;G|_W)//\\prod_{v\\in W}G_v.\n\\]\nRestriction keeps the existing cocycle values and gauge values on the smaller patch. It first gives a strict category valued functor on the opposite patch poset. Its standard pseudofunctor form has identity comparison isomorphisms, with the required coherence laws.","\\begin{theorem}[Stackhood on the composition site]\\label{KS36}\nThe patch pseudofunctor \\(\\mathfrak K_G\\) is a stack for \\(J_{\\mathrm{comp}}\\). It satisfies the standard condition of descent for morphisms and effective descent for objects on every covering sieve.\n\\end{theorem}\n\\begin{proof}\nWe use the equivalence-of-descent-categories formulation of stackhood for the already constructed patch pseudofunctor. Its restriction maps act on actual cocycle labels and gauge values. Their identity and composition laws have been proved, so the descent condition is applied to a specified pseudofunctor, not assumed as part of its definition.","\\emph{From standard descent data to overlap labels.} For a covering family $(U_i\\subseteq W)$, standard descent data provide local objects and comparison gauges on common subpatches, natural under further restriction. At a view $v\\in U_i\\cap U_j$, restrict the comparison to the singleton patch $\\{v\\}$ and call its value $a_{ij}(v)$. To verify the gauge equation along $f:v\\to w$ in an overlap, use the comparison on $U_i\\cap U_j$, which is a full patch containing the arrow. Naturality identifies its two endpoint values with the singleton values. Its own gauge equation then becomes \\eqref{eq:localgauge}. Evaluating the standard triple-comparison identity at a singleton gives \\eqref{eq:triple}. Thus these labels form the concrete descent datum of Theorem~\\ref{KS34}.","descent data to overlap labels.} For a covering family $(U_i\\subseteq W)$, standard descent data provide local objects and comparison gauges on common subpatches, natural under further restriction. At a view $v\\in U_i\\cap U_j$, restrict the comparison to the singleton patch $\\{v\\}$ and call its value $a_{ij}(v)$. To verify the gauge equation along $f:v\\to w$ in an overlap, use the comparison on $U_i\\cap U_j$, which is a full patch containing the arrow. Naturality identifies its two endpoint values with the singleton values. Its own gauge equation then becomes \\eqref{eq:localgauge}. Evaluating the standard triple-comparison identity at a singleton gives \\eqref{eq:triple}. Thus these labels form the concrete descent datum of Theorem~\\ref{KS34}.","\\emph{From the concrete realization back to standard data.} By Theorem~\\ref{KS34} there is a global cocycle $c$ and local gauges $b_i$ implementing these overlaps. On any common subpatch $Z\\subseteq U_i\\cap U_j$, the required equation between gauge arrows is an equality of their values at every $v\\in Z$. At such a view it is $b_j(v)=a_{ij}(v)b_i(v)$, already proved by the construction. Equality of coordinate functions proves equality of the arrows on $Z$. Thus every standard common-subpatch equation holds, not only the equations on singleton tests. This gives an isomorphism from the standard restriction of $c$ to the original standard descent object.","a.} By Theorem~\\ref{KS34} there is a global cocycle $c$ and local gauges $b_i$ implementing these overlaps. On any common subpatch $Z\\subseteq U_i\\cap U_j$, the required equation between gauge arrows is an equality of their values at every $v\\in Z$. At such a view it is $b_j(v)=a_{ij}(v)b_i(v)$, already proved by the construction. Equality of coordinate functions proves equality of the arrows on $Z$. Thus every standard common-subpatch equation holds, not only the equations on singleton tests. This gives an isomorphism from the standard restriction of $c$ to the original standard descent object.","e required equation between gauge arrows is an equality of their values at every $v\\in Z$. At such a view it is $b_j(v)=a_{ij}(v)b_i(v)$, already proved by the construction. Equality of coordinate functions proves equality of the arrows on $Z$. Thus every standard common-subpatch equation holds, not only the equations on singleton tests. This gives an isomorphism from the standard restriction of $c$ to the original standard descent object.","\\emph{Every covering sieve.} Let $\\mathcal R$ be an arbitrary covering sieve on $W$. Index a family by all its member inclusions $U\\subseteq W$. Each composition support belongs to this family by the covering definition, so it is a composition cover. Its generated sieve is exactly $\\mathcal R$: one inclusion uses that $\\mathcal R$ is downward closed, and the other uses each member as its own generator. The family argument therefore proves the descent equivalence on the original sieve. Applying it to every covering sieve proves both descent of morphisms and effectiveness of objects. These are the required standard stack conditions \\cite{stacks-stack,mathlib-stack}.\n\\end{proof}\n\\source{KS36}\nThe principal source declaration is \\code{Gro","\\emph{Every covering sieve.} Let $\\mathcal R$ be an arbitrary covering sieve on $W$. Index a family by all its member inclusions $U\\subseteq W$. Each composition support belongs to this family by the covering definition, so it is a composition cover. Its generated sieve is exactly $\\mathcal R$: one inclusion uses that $\\mathcal R$ is downward closed, and the other uses each member as its own generator. The family argument therefore proves the descent equivalence on the original sieve. Applying it to every covering sieve proves both descent of morphisms and effectiveness of objects. These are the required standard stack conditions \\cite{stacks-stack,mathlib-stack}.\n\\end{proof}\n\\source{KS36}\nThe principal source declaration is \\code{GroupViewSystem.patch_isStack}. Its targe","\\emph{Every covering sieve.} Let $\\mathcal R$ be an arbitrary covering sieve on $W$. Index a family by all its member inclusions $U\\subseteq W$. Each composition support belongs to this family by the covering definition, so it is a composition cover. Its generated sieve is exactly $\\mathcal R$: one inclusion uses that $\\mathcal R$ is downward closed, and the other uses each member as its own generator. The family argument therefore proves the descent equivalence on the original sieve. Applying it to every covering sieve proves both descent of morphisms and effectiveness of objects. These are the required standard stack conditions \\cite{stacks-stack,mathlib-stack}.\n\\end{proof}\n\\source{KS36}\nThe principal source declaration is \\code{GroupViewSystem.patch_isStack}. Its target is mathlib's \\code{Pseudofunctor.IsStack}, not a project p","\\emph{Every covering sieve.} Let $\\mathcal R$ be an arbitrary covering sieve on $W$. Index a family by all its member inclusions $U\\subseteq W$. Each composition support belongs to this family by the covering definition, so it is a composition cover. Its generated sieve is exactly $\\mathcal R$: one inclusion uses that $\\mathcal R$ is downward closed, and the other uses each member as its own generator. The family argument therefore proves the descent equivalence on the original sieve. Applying it to every covering sieve proves both descent of morphisms and effectiveness of objects. These are the required standard stack conditions \\cite{stacks-stack,mathlib-stack}.\n\\end{proof}\n\\source{KS36}\nThe principal source declaration is \\code{GroupViewSystem.patch_isStack}. Its target is mathlib's \\code{Pseudofunctor.IsStack}, not a project predicate that repeats the conclusion as an assumption.","On the three-view chain, the analogous proper patches cover each object and each arrow but miss the support of \\(0\\to1\\to2\\). They do not form a cover in this topology. Consequently the theorem is neither vacuous nor an unrestricted gluing principle.","\\chap{Equivalent presentations and identified boundaries}\\label{sec:relative}\n\\subsection{Transporting the same question}\nLet \\(H:\\V\\to\\W\\) be part of a category equivalence, with inverse \\(L\\) and counit \\(\\varepsilon:HL\\Rightarrow\\id_\\W\\). Pull a view system \\(F\\) on \\(\\W\\) back along \\(H\\). Sampling a global section gives \\((H^*x)_v=x_{Hv}\\).","\\begin{proposition}[Equivalence of section spaces]\\label{KS37}\nPullback induces \\(\\Gamma(F)\\simeq\\Gamma(H^*F)\\). If \\(U\\subseteq\\Ob\\W\\) is closed under isomorphism and \\(U'=H^{-1}U\\), it also induces\n\\[\n \\Ext_U^F(s)\\simeq\\Ext_{U'}^{H^*F}(H^*s).\n\\]\nExistence and uniqueness of the prescribed extension are preserved and reflected.\n\\end{proposition}\n\\begin{proof}\nWrite the unit of the equivalence as $\\eta:I_\\V\\Rightarrow LH$, and take the unit and counit with their triangle identities. Set $t_w=\\varepsilon_w^{-1}:w\\to HLw$. Given $y\\in\\Gamma(H^*F)$, define\n\\[\n (Ry)_w=F(t_w)y_{Lw}.\n\\]\nFor an arrow $f:w\\to w'$, naturality of the inverse counit states\n\\[\n HL(f)\\,t_w=t_{w'}f.\n\\]\nContravariance of $F$ and coherence of $y$ therefore give\n\\[\n\\begin{aligned}\n F(f)(Ry)_{w'}\n &=F(f)F(t_{w'})y_{Lw'}","simeq\\Gamma(H^*F)\\). If \\(U\\subseteq\\Ob\\W\\) is closed under isomorphism and \\(U'=H^{-1}U\\), it also induces\n\\[\n \\Ext_U^F(s)\\simeq\\Ext_{U'}^{H^*F}(H^*s).\n\\]\nExistence and uniqueness of the prescribed extension are preserved and reflected.\n\\end{proposition}\n\\begin{proof}\nWrite the unit of the equivalence as $\\eta:I_\\V\\Rightarrow LH$, and take the unit and counit with their triangle identities. Set $t_w=\\varepsilon_w^{-1}:w\\to HLw$. Given $y\\in\\Gamma(H^*F)$, define\n\\[\n (Ry)_w=F(t_w)y_{Lw}.\n\\]\nFor an arrow $f:w\\to w'$, naturality of the inverse counit states\n\\[\n HL(f)\\,t_w=t_{w'}f.\n\\]\nContravariance of $F$ and coherence of $y$ therefore give\n\\[\n\\begin{aligned}\n F(f)(Ry)_{w'}\n &=F(f)F(t_{w'})y_{Lw'}","\\) is closed under isomorphism and \\(U'=H^{-1}U\\), it also induces\n\\[\n \\Ext_U^F(s)\\simeq\\Ext_{U'}^{H^*F}(H^*s).\n\\]\nExistence and uniqueness of the prescribed extension are preserved and reflected.\n\\end{proposition}\n\\begin{proof}\nWrite the unit of the equivalence as $\\eta:I_\\V\\Rightarrow LH$, and take the unit and counit with their triangle identities. Set $t_w=\\varepsilon_w^{-1}:w\\to HLw$. Given $y\\in\\Gamma(H^*F)$, define\n\\[\n (Ry)_w=F(t_w)y_{Lw}.\n\\]\nFor an arrow $f:w\\to w'$, naturality of the inverse counit states\n\\[\n HL(f)\\,t_w=t_{w'}f.\n\\]\nContravariance of $F$ and coherence of $y$ therefore give\n\\[\n\\begin{aligned}\n F(f)(Ry)_{w'}\n &=F(f)F(t_{w'})y_{Lw'}","sm and \\(U'=H^{-1}U\\), it also induces\n\\[\n \\Ext_U^F(s)\\simeq\\Ext_{U'}^{H^*F}(H^*s).\n\\]\nExistence and uniqueness of the prescribed extension are preserved and reflected.\n\\end{proposition}\n\\begin{proof}\nWrite the unit of the equivalence as $\\eta:I_\\V\\Rightarrow LH$, and take the unit and counit with their triangle identities. Set $t_w=\\varepsilon_w^{-1}:w\\to HLw$. Given $y\\in\\Gamma(H^*F)$, define\n\\[\n (Ry)_w=F(t_w)y_{Lw}.\n\\]\nFor an arrow $f:w\\to w'$, naturality of the inverse counit states\n\\[\n HL(f)\\,t_w=t_{w'}f.\n\\]\nContravariance of $F$ and coherence of $y$ therefore give\n\\[\n\\begin{aligned}\n F(f)(Ry)_{w'}\n &=F(f)F(t_{w'})y_{Lw'}","[\n \\Ext_U^F(s)\\simeq\\Ext_{U'}^{H^*F}(H^*s).\n\\]\nExistence and uniqueness of the prescribed extension are preserved and reflected.\n\\end{proposition}\n\\begin{proof}\nWrite the unit of the equivalence as $\\eta:I_\\V\\Rightarrow LH$, and take the unit and counit with their triangle identities. Set $t_w=\\varepsilon_w^{-1}:w\\to HLw$. Given $y\\in\\Gamma(H^*F)$, define\n\\[\n (Ry)_w=F(t_w)y_{Lw}.\n\\]\nFor an arrow $f:w\\to w'$, naturality of the inverse counit states\n\\[\n HL(f)\\,t_w=t_{w'}f.\n\\]\nContravariance of $F$ and coherence of $y$ therefore give\n\\[\n\\begin{aligned}\n F(f)(Ry)_{w'}\n &=F(f)F(t_{w'})y_{Lw'}","ce and uniqueness of the prescribed extension are preserved and reflected.\n\\end{proposition}\n\\begin{proof}\nWrite the unit of the equivalence as $\\eta:I_\\V\\Rightarrow LH$, and take the unit and counit with their triangle identities. Set $t_w=\\varepsilon_w^{-1}:w\\to HLw$. Given $y\\in\\Gamma(H^*F)$, define\n\\[\n (Ry)_w=F(t_w)y_{Lw}.\n\\]\nFor an arrow $f:w\\to w'$, naturality of the inverse counit states\n\\[\n HL(f)\\,t_w=t_{w'}f.\n\\]\nContravariance of $F$ and coherence of $y$ therefore give\n\\[\n\\begin{aligned}\n F(f)(Ry)_{w'}\n &=F(f)F(t_{w'})y_{Lw'}","n are preserved and reflected.\n\\end{proposition}\n\\begin{proof}\nWrite the unit of the equivalence as $\\eta:I_\\V\\Rightarrow LH$, and take the unit and counit with their triangle identities. Set $t_w=\\varepsilon_w^{-1}:w\\to HLw$. Given $y\\in\\Gamma(H^*F)$, define\n\\[\n (Ry)_w=F(t_w)y_{Lw}.\n\\]\nFor an arrow $f:w\\to w'$, naturality of the inverse counit states\n\\[\n HL(f)\\,t_w=t_{w'}f.\n\\]\nContravariance of $F$ and coherence of $y$ therefore give\n\\[\n\\begin{aligned}\n F(f)(Ry)_{w'}\n &=F(f)F(t_{w'})y_{Lw'}","eflected.\n\\end{proposition}\n\\begin{proof}\nWrite the unit of the equivalence as $\\eta:I_\\V\\Rightarrow LH$, and take the unit and counit with their triangle identities. Set $t_w=\\varepsilon_w^{-1}:w\\to HLw$. Given $y\\in\\Gamma(H^*F)$, define\n\\[\n (Ry)_w=F(t_w)y_{Lw}.\n\\]\nFor an arrow $f:w\\to w'$, naturality of the inverse counit states\n\\[\n HL(f)\\,t_w=t_{w'}f.\n\\]\nContravariance of $F$ and coherence of $y$ therefore give\n\\[\n\\begin{aligned}\n F(f)(Ry)_{w'}\n &=F(f)F(t_{w'})y_{Lw'}","sition}\n\\begin{proof}\nWrite the unit of the equivalence as $\\eta:I_\\V\\Rightarrow LH$, and take the unit and counit with their triangle identities. Set $t_w=\\varepsilon_w^{-1}:w\\to HLw$. Given $y\\in\\Gamma(H^*F)$, define\n\\[\n (Ry)_w=F(t_w)y_{Lw}.\n\\]\nFor an arrow $f:w\\to w'$, naturality of the inverse counit states\n\\[\n HL(f)\\,t_w=t_{w'}f.\n\\]\nContravariance of $F$ and coherence of $y$ therefore give\n\\[\n\\begin{aligned}\n F(f)(Ry)_{w'}\n &=F(f)F(t_{w'})y_{Lw'}\\\\\n &=F(t_w)F(HL(f))y_{Lw'}\n =F(t_w)y_{Lw}=(Ry)_w.\n\\end{aligned}\n\\]\nThus recovery gives an actual coherent section.","If $x\\in\\Gamma(F)$, coherence along $t_w$ gives\n\\[\n (RH^*x)_w=F(t_w)x_{HLw}=x_w.\n\\]\nFor the other composite, the triangle identity says $t_{Hv}=H(\\eta_v)$. Coherence of $y$ along $\\eta_v:v\\to LHv$ yields\n\\[\n (H^*Ry)_v=F(t_{Hv})y_{LHv}=F(H\\eta_v)y_{LHv}=y_v.\n\\]\nThese coordinate equations prove both inverse identities.","For the relative claim, pullback clearly sends an extension of $s$ to an extension of $H^*s$ on $U'=H^{-1}U$. Conversely suppose $y$ extends $H^*s$. If $w\\in U$, the isomorphism $w\\cong HLw$ and isomorphism closure imply $HLw\\in U$. Hence $Lw\\in U'$ and $y_{Lw}=s_{HLw}$. Both endpoints of $t_w$ lie in $U$, so partial coherence gives\n\\[\n (Ry)_w=F(t_w)s_{HLw}=s_w.\n\\]\nRecovery therefore preserves the required boundary. The two inverse identities remain valid on the subsets of extensions. Bijections preserve nonemptiness and identify any two elements exactly when their images are equal, proving preservation and reflection of existence and uniqueness.\n\\end{proof}\n\\source{KS37}\nThe coefficients really are pulled back. This does not compare unrelated coefficient","For the relative claim, pullback clearly sends an extension of $s$ to an extension of $H^*s$ on $U'=H^{-1}U$. Conversely suppose $y$ extends $H^*s$. If $w\\in U$, the isomorphism $w\\cong HLw$ and isomorphism closure imply $HLw\\in U$. Hence $Lw\\in U'$ and $y_{Lw}=s_{HLw}$. Both endpoints of $t_w$ lie in $U$, so partial coherence gives\n\\[\n (Ry)_w=F(t_w)s_{HLw}=s_w.\n\\]\nRecovery therefore preserves the required boundary. The two inverse identities remain valid on the subsets of extensions. Bijections preserve nonemptiness and identify any two elements exactly when their images are equal, proving preservation and reflection of existence and uniqueness.\n\\end{proof}\n\\source{KS37}\nThe coefficients really are pulled back. This does not compare unrelated coefficient systems simply because th","For the relative claim, pullback clearly sends an extension of $s$ to an extension of $H^*s$ on $U'=H^{-1}U$. Conversely suppose $y$ extends $H^*s$. If $w\\in U$, the isomorphism $w\\cong HLw$ and isomorphism closure imply $HLw\\in U$. Hence $Lw\\in U'$ and $y_{Lw}=s_{HLw}$. Both endpoints of $t_w$ lie in $U$, so partial coherence gives\n\\[\n (Ry)_w=F(t_w)s_{HLw}=s_w.\n\\]\nRecovery therefore preserves the required boundary. The two inverse identities remain valid on the subsets of extensions. Bijections preserve nonemptiness and identify any two elements exactly when their images are equal, proving preservation and reflection of existence and uniqueness.\n\\end{proof}\n\\source{KS37}\nThe coefficients really are pulled back. This does not compare unrelated coefficient systems simply because their diagrams loo","For the relative claim, pullback clearly sends an extension of $s$ to an extension of $H^*s$ on $U'=H^{-1}U$. Conversely suppose $y$ extends $H^*s$. If $w\\in U$, the isomorphism $w\\cong HLw$ and isomorphism closure imply $HLw\\in U$. Hence $Lw\\in U'$ and $y_{Lw}=s_{HLw}$. Both endpoints of $t_w$ lie in $U$, so partial coherence gives\n\\[\n (Ry)_w=F(t_w)s_{HLw}=s_w.\n\\]\nRecovery therefore preserves the required boundary. The two inverse identities remain valid on the subsets of extensions. Bijections preserve nonemptiness and identify any two elements exactly when their images are equal, proving preservation and reflection of existence and uniqueness.\n\\end{proof}\n\\source{KS37}\nThe coefficients really are pulled back. This does not compare unrelated coefficient systems simply because their diagrams look alike. Isomorphism clo","For the relative claim, pullback clearly sends an extension of $s$ to an extension of $H^*s$ on $U'=H^{-1}U$. Conversely suppose $y$ extends $H^*s$. If $w\\in U$, the isomorphism $w\\cong HLw$ and isomorphism closure imply $HLw\\in U$. Hence $Lw\\in U'$ and $y_{Lw}=s_{HLw}$. Both endpoints of $t_w$ lie in $U$, so partial coherence gives\n\\[\n (Ry)_w=F(t_w)s_{HLw}=s_w.\n\\]\nRecovery therefore preserves the required boundary. The two inverse identities remain valid on the subsets of extensions. Bijections preserve nonemptiness and identify any two elements exactly when their images are equal, proving preservation and reflection of existence and uniqueness.\n\\end{proof}\n\\source{KS37}\nThe coefficients really are pulled back. This does not compare unrelated coefficient systems simply because their diagrams look alike. Isomorphism closure is a suffi","For the relative claim, pullback clearly sends an extension of $s$ to an extension of $H^*s$ on $U'=H^{-1}U$. Conversely suppose $y$ extends $H^*s$. If $w\\in U$, the isomorphism $w\\cong HLw$ and isomorphism closure imply $HLw\\in U$. Hence $Lw\\in U'$ and $y_{Lw}=s_{HLw}$. Both endpoints of $t_w$ lie in $U$, so partial coherence gives\n\\[\n (Ry)_w=F(t_w)s_{HLw}=s_w.\n\\]\nRecovery therefore preserves the required boundary. The two inverse identities remain valid on the subsets of extensions. Bijections preserve nonemptiness and identify any two elements exactly when their images are equal, proving preservation and reflection of existence and uniqueness.\n\\end{proof}\n\\source{KS37}\nThe coefficients really are pulled back. This does not compare unrelated coefficient systems simply because their diagrams look alike. Isomorphism closure is a sufficient general","For the relative claim, pullback clearly sends an extension of $s$ to an extension of $H^*s$ on $U'=H^{-1}U$. Conversely suppose $y$ extends $H^*s$. If $w\\in U$, the isomorphism $w\\cong HLw$ and isomorphism closure imply $HLw\\in U$. Hence $Lw\\in U'$ and $y_{Lw}=s_{HLw}$. Both endpoints of $t_w$ lie in $U$, so partial coherence gives\n\\[\n (Ry)_w=F(t_w)s_{HLw}=s_w.\n\\]\nRecovery therefore preserves the required boundary. The two inverse identities remain valid on the subsets of extensions. Bijections preserve nonemptiness and identify any two elements exactly when their images are equal, proving preservation and reflection of existence and uniqueness.\n\\end{proof}\n\\source{KS37}\nThe coefficients really are pulled back. This does not compare unrelated coefficient systems simply because their diagrams look alike. Isomorphism closure is a sufficient general hypothesis for the relative s","For the relative claim, pullback clearly sends an extension of $s$ to an extension of $H^*s$ on $U'=H^{-1}U$. Conversely suppose $y$ extends $H^*s$. If $w\\in U$, the isomorphism $w\\cong HLw$ and isomorphism closure imply $HLw\\in U$. Hence $Lw\\in U'$ and $y_{Lw}=s_{HLw}$. Both endpoints of $t_w$ lie in $U$, so partial coherence gives\n\\[\n (Ry)_w=F(t_w)s_{HLw}=s_w.\n\\]\nRecovery therefore preserves the required boundary. The two inverse identities remain valid on the subsets of extensions. Bijections preserve nonemptiness and identify any two elements exactly when their images are equal, proving preservation and reflection of existence and uniqueness.\n\\end{proof}\n\\source{KS37}\nThe coefficients really are pulled back. This does not compare unrelated coefficient systems simply because their diagrams look alike. Isomorphism closure is a sufficient general hypothesis for the relative statement. The theorem does not assert tha","For the relative claim, pullback clearly sends an extension of $s$ to an extension of $H^*s$ on $U'=H^{-1}U$. Conversely suppose $y$ extends $H^*s$. If $w\\in U$, the isomorphism $w\\cong HLw$ and isomorphism closure imply $HLw\\in U$. Hence $Lw\\in U'$ and $y_{Lw}=s_{HLw}$. Both endpoints of $t_w$ lie in $U$, so partial coherence gives\n\\[\n (Ry)_w=F(t_w)s_{HLw}=s_w.\n\\]\nRecovery therefore preserves the required boundary. The two inverse identities remain valid on the subsets of extensions. Bijections preserve nonemptiness and identify any two elements exactly when their images are equal, proving preservation and reflection of existence and uniqueness.\n\\end{proof}\n\\source{KS37}\nThe coefficients really are pulled back. This does not compare unrelated coefficient systems simply because their diagrams look alike. Isomorphism closure is a sufficient general hypothesis for the relative statement. The theorem does not assert that every unsaturate","For the relative claim, pullback clearly sends an extension of $s$ to an extension of $H^*s$ on $U'=H^{-1}U$. Conversely suppose $y$ extends $H^*s$. If $w\\in U$, the isomorphism $w\\cong HLw$ and isomorphism closure imply $HLw\\in U$. Hence $Lw\\in U'$ and $y_{Lw}=s_{HLw}$. Both endpoints of $t_w$ lie in $U$, so partial coherence gives\n\\[\n (Ry)_w=F(t_w)s_{HLw}=s_w.\n\\]\nRecovery therefore preserves the required boundary. The two inverse identities remain valid on the subsets of extensions. Bijections preserve nonemptiness and identify any two elements exactly when their images are equal, proving preservation and reflection of existence and uniqueness.\n\\end{proof}\n\\source{KS37}\nThe coefficients really are pulled back. This does not compare unrelated coefficient systems simply because their diagrams look alike. Isomorphism closure is a sufficient general hypothesis for the relative statement. The theorem does not assert that every unsaturated domain fails.","For group coefficients, the invisible coherent groups \\(H_U\\) are transported by a multiplicative equivalence, and the parameterizations \\(x_0h\\) commute with pullback. Thus feasibility and ambiguity are both retained. This is more than a comparison of the number of extensions. It is not an assertion that every possible based gauge orbit set is presentation independent.","\\subsection{Why a boundary must retain its identification}\nLet \\(R:\\mathcal C\\to\\mathcal D\\) be a functor between groupoids, and fix \\(b\\in\\mathcal D\\). A realization over \\(b\\) should include a chosen isomorphism to \\(b\\), rather than require literal equality of representatives.","\\begin{definition}[Marked realization groupoid]\nThe objects of \\(\\mathcal E_b(R)\\) are pairs \\((x,\\phi)\\), with \\(x\\in\\mathcal C\\) and \\(\\phi:Rx\\xrightarrow{\\sim}b\\). An arrow \\((x,\\phi)\\to(y,\\psi)\\) is an arrow \\(a:x\\to y\\) with \\(\\psi R(a)=\\phi\\).\n\\end{definition}\nComposition and inverses preserve this equation, so this is a groupoid. It is the explicit groupoid homotopy fiber construction, without invoking a higher categorical interface.","\\begin{proposition}[Existence and invisible automorphisms]\\label{KS38}\nThe groupoid \\(\\mathcal E_b(R)\\) is nonempty exactly when \\(b\\) is in the essential image of \\(R\\). An automorphism of \\((x,\\phi)\\) is an automorphism \\(a\\) of \\(x\\) satisfying \\(R(a)=1_{Rx}\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion is the object definition. For the second, the boundary equation is \\(\\phi R(a)=\\phi\\). Cancel the isomorphism \\(\\phi\\). Conversely \\(R(a)=1\\) makes the equation hold.\n\\end{proof}\n\\source{KS38}\nThis retains precisely the symmetries invisible at the identified boundary. It avoids freezing arbitrary representative coordinates and then mistaking the result for an invariant realization problem.","and invisible automorphisms]\\label{KS38}\nThe groupoid \\(\\mathcal E_b(R)\\) is nonempty exactly when \\(b\\) is in the essential image of \\(R\\). An automorphism of \\((x,\\phi)\\) is an automorphism \\(a\\) of \\(x\\) satisfying \\(R(a)=1_{Rx}\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion is the object definition. For the second, the boundary equation is \\(\\phi R(a)=\\phi\\). Cancel the isomorphism \\(\\phi\\). Conversely \\(R(a)=1\\) makes the equation hold.\n\\end{proof}\n\\source{KS38}\nThis retains precisely the symmetries invisible at the identified boundary. It avoids freezing arbitrary representative coordinates and then mistaking the result for an invariant realization problem.","rphisms]\\label{KS38}\nThe groupoid \\(\\mathcal E_b(R)\\) is nonempty exactly when \\(b\\) is in the essential image of \\(R\\). An automorphism of \\((x,\\phi)\\) is an automorphism \\(a\\) of \\(x\\) satisfying \\(R(a)=1_{Rx}\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion is the object definition. For the second, the boundary equation is \\(\\phi R(a)=\\phi\\). Cancel the isomorphism \\(\\phi\\). Conversely \\(R(a)=1\\) makes the equation hold.\n\\end{proof}\n\\source{KS38}\nThis retains precisely the symmetries invisible at the identified boundary. It avoids freezing arbitrary representative coordinates and then mistaking the result for an invariant realization problem.","\\subsection{Weakly commuting squares are enough}\nSuppose \\(E:\\mathcal C\\to\\mathcal C'\\) and \\(K:\\mathcal D\\to\\mathcal D'\\) are equivalences, with functors \\(R,R'\\) and a natural isomorphism\n\\[\n \\sigma:KR\\xRightarrow{\\sim}R'E.\n\\]\nThe square is not required to commute literally.","\\begin{theorem}[Invariance of the marked realization problem]\\label{KS39}\nThe map\n\\[\n (x,\\phi)\\longmapsto(Ex,K(\\phi)\\sigma_x^{-1}),\\qquad a\\longmapsto E(a)\n\\]\ndefines an equivalence \\(\\mathcal E_b(R)\\simeq\\mathcal E_{Kb}(R')\\).\n\\end{theorem}\n\\begin{proof}\nFor an object $(x,\\phi)$ put $\\beta_x=K(\\phi)\\sigma_x^{-1}:R'Ex\\to Kb$. This is an isomorphism of the required type. If $a:(x,\\phi)\\to(y,\\psi)$, naturality of $\\sigma$ says\n\\[\n R'E(a)\\sigma_x=\\sigma_yKR(a),\n \\qquad \\sigma_y^{-1}R'E(a)=KR(a)\\sigma_x^{-1}.\n\\]\nConsequently\n\\[\n \\beta_yR'E(a)\n =K(\\psi)KR(a)\\sigma_x^{-1}\n =K(\\psi R(a))\\sigma_x^{-1}\n =K(\\phi)\\sigma_x^{-1}=\\beta_x.\n\\]\nSo $E(a)$ is a marked arrow. Identity and composition are preserved because the underlying arrow map is that of the functor $E$.","Invariance of the marked realization problem]\\label{KS39}\nThe map\n\\[\n (x,\\phi)\\longmapsto(Ex,K(\\phi)\\sigma_x^{-1}),\\qquad a\\longmapsto E(a)\n\\]\ndefines an equivalence \\(\\mathcal E_b(R)\\simeq\\mathcal E_{Kb}(R')\\).\n\\end{theorem}\n\\begin{proof}\nFor an object $(x,\\phi)$ put $\\beta_x=K(\\phi)\\sigma_x^{-1}:R'Ex\\to Kb$. This is an isomorphism of the required type. If $a:(x,\\phi)\\to(y,\\psi)$, naturality of $\\sigma$ says\n\\[\n R'E(a)\\sigma_x=\\sigma_yKR(a),\n \\qquad \\sigma_y^{-1}R'E(a)=KR(a)\\sigma_x^{-1}.\n\\]\nConsequently\n\\[\n \\beta_yR'E(a)\n =K(\\psi)KR(a)\\sigma_x^{-1}\n =K(\\psi R(a))\\sigma_x^{-1}\n =K(\\phi)\\sigma_x^{-1}=\\beta_x.\n\\]\nSo $E(a)$ is a marked arrow. Identity and composition are preserved because the underlying arrow map is that of the functor $E$.","he marked realization problem]\\label{KS39}\nThe map\n\\[\n (x,\\phi)\\longmapsto(Ex,K(\\phi)\\sigma_x^{-1}),\\qquad a\\longmapsto E(a)\n\\]\ndefines an equivalence \\(\\mathcal E_b(R)\\simeq\\mathcal E_{Kb}(R')\\).\n\\end{theorem}\n\\begin{proof}\nFor an object $(x,\\phi)$ put $\\beta_x=K(\\phi)\\sigma_x^{-1}:R'Ex\\to Kb$. This is an isomorphism of the required type. If $a:(x,\\phi)\\to(y,\\psi)$, naturality of $\\sigma$ says\n\\[\n R'E(a)\\sigma_x=\\sigma_yKR(a),\n \\qquad \\sigma_y^{-1}R'E(a)=KR(a)\\sigma_x^{-1}.\n\\]\nConsequently\n\\[\n \\beta_yR'E(a)\n =K(\\psi)KR(a)\\sigma_x^{-1}\n =K(\\psi R(a))\\sigma_x^{-1}\n =K(\\phi)\\sigma_x^{-1}=\\beta_x.\n\\]\nSo $E(a)$ is a marked arrow. Identity and composition are preserved because the underlying arrow map is that of the functor $E$.","),\\qquad a\\longmapsto E(a)\n\\]\ndefines an equivalence \\(\\mathcal E_b(R)\\simeq\\mathcal E_{Kb}(R')\\).\n\\end{theorem}\n\\begin{proof}\nFor an object $(x,\\phi)$ put $\\beta_x=K(\\phi)\\sigma_x^{-1}:R'Ex\\to Kb$. This is an isomorphism of the required type. If $a:(x,\\phi)\\to(y,\\psi)$, naturality of $\\sigma$ says\n\\[\n R'E(a)\\sigma_x=\\sigma_yKR(a),\n \\qquad \\sigma_y^{-1}R'E(a)=KR(a)\\sigma_x^{-1}.\n\\]\nConsequently\n\\[\n \\beta_yR'E(a)\n =K(\\psi)KR(a)\\sigma_x^{-1}\n =K(\\psi R(a))\\sigma_x^{-1}\n =K(\\phi)\\sigma_x^{-1}=\\beta_x.\n\\]\nSo $E(a)$ is a marked arrow. Identity and composition are preserved because the underlying arrow map is that of the functor $E$.","}\n\\begin{proof}\nFor an object $(x,\\phi)$ put $\\beta_x=K(\\phi)\\sigma_x^{-1}:R'Ex\\to Kb$. This is an isomorphism of the required type. If $a:(x,\\phi)\\to(y,\\psi)$, naturality of $\\sigma$ says\n\\[\n R'E(a)\\sigma_x=\\sigma_yKR(a),\n \\qquad \\sigma_y^{-1}R'E(a)=KR(a)\\sigma_x^{-1}.\n\\]\nConsequently\n\\[\n \\beta_yR'E(a)\n =K(\\psi)KR(a)\\sigma_x^{-1}\n =K(\\psi R(a))\\sigma_x^{-1}\n =K(\\phi)\\sigma_x^{-1}=\\beta_x.\n\\]\nSo $E(a)$ is a marked arrow. Identity and composition are preserved because the underlying arrow map is that of the functor $E$.","uired type. If $a:(x,\\phi)\\to(y,\\psi)$, naturality of $\\sigma$ says\n\\[\n R'E(a)\\sigma_x=\\sigma_yKR(a),\n \\qquad \\sigma_y^{-1}R'E(a)=KR(a)\\sigma_x^{-1}.\n\\]\nConsequently\n\\[\n \\beta_yR'E(a)\n =K(\\psi)KR(a)\\sigma_x^{-1}\n =K(\\psi R(a))\\sigma_x^{-1}\n =K(\\phi)\\sigma_x^{-1}=\\beta_x.\n\\]\nSo $E(a)$ is a marked arrow. Identity and composition are preserved because the underlying arrow map is that of the functor $E$.","To prove full faithfulness, start with a marked arrow $u:Ex\\to Ey$ between the transported objects. Full faithfulness of $E$ gives a unique $a:x\\to y$ with $E(a)=u$. Its marked equation, rewritten by the naturality identity above, is\n\\[\n K(\\psi R(a))\\sigma_x^{-1}=K(\\phi)\\sigma_x^{-1}.\n\\]\nCancel the isomorphism on the right and use faithfulness of $K$ to obtain $\\psi R(a)=\\phi$. Thus the lifted arrow is marked. Uniqueness of the underlying lift also proves uniqueness as a marked arrow, giving a bijection on every Hom set.","For essential surjectivity, let $(y,\\psi)$ be a target realization. Choose $x$ and an isomorphism $i:Ex\\to y$, using essential surjectivity of $E$. Fullness of $K$ lifts the well-typed composite\n\\[\n KRx\\xrightarrow{\\sigma_x}R'Ex\n \\xrightarrow{R'(i)}R'y\\xrightarrow{\\psi}Kb\n\\]\nto an arrow $\\phi:Rx\\to b$. Since $\\mathcal D$ is a groupoid, $\\phi$ is automatically an isomorphism. Its image satisfies $K(\\phi)=\\psi R'(i)\\sigma_x$, hence $\\beta_x=\\psi R'(i)$. This is precisely the condition for $i$ to be an arrow from the transported realization to $(y,\\psi)$. It is an isomorphism because all categories involved are groupoids.",",\\psi)$ be a target realization. Choose $x$ and an isomorphism $i:Ex\\to y$, using essential surjectivity of $E$. Fullness of $K$ lifts the well-typed composite\n\\[\n KRx\\xrightarrow{\\sigma_x}R'Ex\n \\xrightarrow{R'(i)}R'y\\xrightarrow{\\psi}Kb\n\\]\nto an arrow $\\phi:Rx\\to b$. Since $\\mathcal D$ is a groupoid, $\\phi$ is automatically an isomorphism. Its image satisfies $K(\\phi)=\\psi R'(i)\\sigma_x$, hence $\\beta_x=\\psi R'(i)$. This is precisely the condition for $i$ to be an arrow from the transported realization to $(y,\\psi)$. It is an isomorphism because all categories involved are groupoids.","hism $i:Ex\\to y$, using essential surjectivity of $E$. Fullness of $K$ lifts the well-typed composite\n\\[\n KRx\\xrightarrow{\\sigma_x}R'Ex\n \\xrightarrow{R'(i)}R'y\\xrightarrow{\\psi}Kb\n\\]\nto an arrow $\\phi:Rx\\to b$. Since $\\mathcal D$ is a groupoid, $\\phi$ is automatically an isomorphism. Its image satisfies $K(\\phi)=\\psi R'(i)\\sigma_x$, hence $\\beta_x=\\psi R'(i)$. This is precisely the condition for $i$ to be an arrow from the transported realization to $(y,\\psi)$. It is an isomorphism because all categories involved are groupoids.","Fullness of $K$ lifts the well-typed composite\n\\[\n KRx\\xrightarrow{\\sigma_x}R'Ex\n \\xrightarrow{R'(i)}R'y\\xrightarrow{\\psi}Kb\n\\]\nto an arrow $\\phi:Rx\\to b$. Since $\\mathcal D$ is a groupoid, $\\phi$ is automatically an isomorphism. Its image satisfies $K(\\phi)=\\psi R'(i)\\sigma_x$, hence $\\beta_x=\\psi R'(i)$. This is precisely the condition for $i$ to be an arrow from the transported realization to $(y,\\psi)$. It is an isomorphism because all categories involved are groupoids.","We have verified full faithfulness and essential surjectivity of the specified transport functor. Choosing these preimages and lifting arrows through the Hom bijections constructs a quasi-inverse, as in Theorem~\\ref{KS34}, proving the claimed equivalence. No literal equality $KR=R'E$ was used.\n\\end{proof}\n\\source{KS39}\nApply this to the actual restriction functors between cocycle patch groupoids and the equivalences induced by an equivalent view presentation. The full realization problem, including comparison witnesses and automorphisms, is transported.","Replacing \\(b\\) by an isomorphic boundary \\(b'\\) also gives an equivalence, by composing the identification with \\(b\\to b'\\). For \\(R=\\id\\), the marked realization groupoid has center \\((b,1_b)\\), and between \\((x,\\phi)\\) and \\((y,\\psi)\\) there is exactly one compatible arrow \\(\\psi^{-1}\\phi\\). This is a useful example of contractible realization data, not a claim that the original groupoid has no symmetries.","\\subsection{Duplicate a view without adding information}\nLet the original view category have one object and only its identity arrow. Replace it by a category with two objects, \\(0\\) and \\(1\\), and exactly one arrow between each ordered pair. These arrows are inverse isomorphisms. The two categories are equivalent: selecting object zero and forgetting the duplicate give inverse equivalences up to the evident isomorphism.","Use a constant group \\(H\\) as coefficients, with identity pullback homomorphisms. A cocycle on the duplicated category is specified by its label \\(r\\in H\\) on the arrow \\(0\\to1\\). Its reverse label is \\(r^{-1}\\), and identity labels are one. A gauge \\((a_0,a_1)\\) carries \\(r\\) to\n\\begin{equation}\\label{eq:two-copy-gauge}\n r'=a_0ra_1^{-1}.\n\\end{equation}\nEvery \\(r\\) is isomorphic to the trivial cocycle, for example with gauge \\((1,r)\\). The automorphisms of \\(r\\) are the pairs\n\\[\n (a_0,r^{-1}a_0r),\\qquad a_0\\in H.\n\\]\nThus the unframed groupoid has one isomorphism class but retains a copy of \\(H\\) as the automorphism group of each object. This is compatible with the original one-view coefficient groupoid, which has o","Use a constant group \\(H\\) as coefficients, with identity pullback homomorphisms. A cocycle on the duplicated category is specified by its label \\(r\\in H\\) on the arrow \\(0\\to1\\). Its reverse label is \\(r^{-1}\\), and identity labels are one. A gauge \\((a_0,a_1)\\) carries \\(r\\) to\n\\begin{equation}\\label{eq:two-copy-gauge}\n r'=a_0ra_1^{-1}.\n\\end{equation}\nEvery \\(r\\) is isomorphic to the trivial cocycle, for example with gauge \\((1,r)\\). The automorphisms of \\(r\\) are the pairs\n\\[\n (a_0,r^{-1}a_0r),\\qquad a_0\\in H.\n\\]\nThus the unframed groupoid has one isomorphism class but retains a copy of \\(H\\) as the automorphism group of each object. This is compatible with the original one-view coefficient groupoid, which has one object and automorphism group \\(H\\).","oefficients, with identity pullback homomorphisms. A cocycle on the duplicated category is specified by its label \\(r\\in H\\) on the arrow \\(0\\to1\\). Its reverse label is \\(r^{-1}\\), and identity labels are one. A gauge \\((a_0,a_1)\\) carries \\(r\\) to\n\\begin{equation}\\label{eq:two-copy-gauge}\n r'=a_0ra_1^{-1}.\n\\end{equation}\nEvery \\(r\\) is isomorphic to the trivial cocycle, for example with gauge \\((1,r)\\). The automorphisms of \\(r\\) are the pairs\n\\[\n (a_0,r^{-1}a_0r),\\qquad a_0\\in H.\n\\]\nThus the unframed groupoid has one isomorphism class but retains a copy of \\(H\\) as the automorphism group of each object. This is compatible with the original one-view coefficient groupoid, which has one object and automorphism group \\(H\\).","ack homomorphisms. A cocycle on the duplicated category is specified by its label \\(r\\in H\\) on the arrow \\(0\\to1\\). Its reverse label is \\(r^{-1}\\), and identity labels are one. A gauge \\((a_0,a_1)\\) carries \\(r\\) to\n\\begin{equation}\\label{eq:two-copy-gauge}\n r'=a_0ra_1^{-1}.\n\\end{equation}\nEvery \\(r\\) is isomorphic to the trivial cocycle, for example with gauge \\((1,r)\\). The automorphisms of \\(r\\) are the pairs\n\\[\n (a_0,r^{-1}a_0r),\\qquad a_0\\in H.\n\\]\nThus the unframed groupoid has one isomorphism class but retains a copy of \\(H\\) as the automorphism group of each object. This is compatible with the original one-view coefficient groupoid, which has one object and automorphism group \\(H\\).","cocycle on the duplicated category is specified by its label \\(r\\in H\\) on the arrow \\(0\\to1\\). Its reverse label is \\(r^{-1}\\), and identity labels are one. A gauge \\((a_0,a_1)\\) carries \\(r\\) to\n\\begin{equation}\\label{eq:two-copy-gauge}\n r'=a_0ra_1^{-1}.\n\\end{equation}\nEvery \\(r\\) is isomorphic to the trivial cocycle, for example with gauge \\((1,r)\\). The automorphisms of \\(r\\) are the pairs\n\\[\n (a_0,r^{-1}a_0r),\\qquad a_0\\in H.\n\\]\nThus the unframed groupoid has one isomorphism class but retains a copy of \\(H\\) as the automorphism group of each object. This is compatible with the original one-view coefficient groupoid, which has one object and automorphism group \\(H\\).","s \\(r^{-1}\\), and identity labels are one. A gauge \\((a_0,a_1)\\) carries \\(r\\) to\n\\begin{equation}\\label{eq:two-copy-gauge}\n r'=a_0ra_1^{-1}.\n\\end{equation}\nEvery \\(r\\) is isomorphic to the trivial cocycle, for example with gauge \\((1,r)\\). The automorphisms of \\(r\\) are the pairs\n\\[\n (a_0,r^{-1}a_0r),\\qquad a_0\\in H.\n\\]\nThus the unframed groupoid has one isomorphism class but retains a copy of \\(H\\) as the automorphism group of each object. This is compatible with the original one-view coefficient groupoid, which has one object and automorphism group \\(H\\).","ry \\(r\\) is isomorphic to the trivial cocycle, for example with gauge \\((1,r)\\). The automorphisms of \\(r\\) are the pairs\n\\[\n (a_0,r^{-1}a_0r),\\qquad a_0\\in H.\n\\]\nThus the unframed groupoid has one isomorphism class but retains a copy of \\(H\\) as the automorphism group of each object. This is compatible with the original one-view coefficient groupoid, which has one object and automorphism group \\(H\\).","\\subsection{Freezing both representative gauges breaks that comparison}\nNow require a based gauge to be the identity at every view. On the duplicated category the only permitted gauge is \\((1,1)\\). Equation~\\eqref{eq:two-copy-gauge} then carries \\(r\\) only to itself. Distinct labels cannot be compared.","Restrict to the selected view zero. Every cocycle becomes the same one-view identity cocycle. The distinction between \\(r\\) and \\(1\\) has disappeared in the target, even though the fully framed source has no comparison arrow between them.","ing is not invariant]\\label{KS68}\nFor a nontrivial coefficient group, the restriction from the fully framed duplicated model to the fully framed single-view model need not be full, and hence need not be an equivalence. This failure occurs even though the underlying view categories are equivalent and the full prescribed domains are closed under isomorphism.\n\\end{proposition}\n\\begin{proof}\nChoose \\(r\\ne1\\). In the fully framed duplicated model there is no arrow from \\(r\\) to \\(1\\), because the only allowed gauge leaves \\(r\\) unchanged. After restriction to view zero, both become the same cocycle. There is an identity arrow between their images. That target arrow has no preimage, so the functor is not full. Fullness is necessary for an equivalence.\n\\end{proof}\n\\source{KS68}\nThe failure is not repaired by declaring the two objects equivalent as sets of labels. The issue is exactly the missing arrow. Nor does it contradict the theorem for marked relative realizations. That theorem does not freeze every arbitrary representative coordinate. It transports an actual boundary object together w…","he restriction from the fully framed duplicated model to the fully framed single-view model need not be full, and hence need not be an equivalence. This failure occurs even though the underlying view categories are equivalent and the full prescribed domains are closed under isomorphism.\n\\end{proposition}\n\\begin{proof}\nChoose \\(r\\ne1\\). In the fully framed duplicated model there is no arrow from \\(r\\) to \\(1\\), because the only allowed gauge leaves \\(r\\) unchanged. After restriction to view zero, both become the same cocycle. There is an identity arrow between their images. That target arrow has no preimage, so the functor is not full. Fullness is necessary for an equivalence.\n\\end{proof}\n\\source{KS68}\nThe failure is not repaired by declaring the two objects equivalent as sets of labels. The issue is exactly the missing arrow. Nor does it contradict the theorem for marked relative realizations. That theorem does not freeze every arbitrary representative coordinate. It transports an actual boundary object together with an identification to it.","ction from the fully framed duplicated model to the fully framed single-view model need not be full, and hence need not be an equivalence. This failure occurs even though the underlying view categories are equivalent and the full prescribed domains are closed under isomorphism.\n\\end{proposition}\n\\begin{proof}\nChoose \\(r\\ne1\\). In the fully framed duplicated model there is no arrow from \\(r\\) to \\(1\\), because the only allowed gauge leaves \\(r\\) unchanged. After restriction to view zero, both become the same cocycle. There is an identity arrow between their images. That target arrow has no preimage, so the functor is not full. Fullness is necessary for an equivalence.\n\\end{proof}\n\\source{KS68}\nThe failure is not repaired by declaring the two objects equivalent as sets of labels. The issue is exactly the missing arrow. Nor does it contradict the theorem for marked relative realizations. That theorem does not freeze every arbitrary representative coordinate. It transports an actual boundary object together with an identification to it.","to the fully framed single-view model need not be full, and hence need not be an equivalence. This failure occurs even though the underlying view categories are equivalent and the full prescribed domains are closed under isomorphism.\n\\end{proposition}\n\\begin{proof}\nChoose \\(r\\ne1\\). In the fully framed duplicated model there is no arrow from \\(r\\) to \\(1\\), because the only allowed gauge leaves \\(r\\) unchanged. After restriction to view zero, both become the same cocycle. There is an identity arrow between their images. That target arrow has no preimage, so the functor is not full. Fullness is necessary for an equivalence.\n\\end{proof}\n\\source{KS68}\nThe failure is not repaired by declaring the two objects equivalent as sets of labels. The issue is exactly the missing arrow. Nor does it contradict the theorem for marked relative realizations. That theorem does not freeze every arbitrary representative coordinate. It transports an actual boundary object together with an identification to it.","\\subsection{The marked fiber keeps the comparison that matters}\nIn the same example, let \\(R\\) restrict an unframed duplicated cocycle to view zero. Fix the unique boundary object. A marked realization has two pieces of data:\n\\[\n (r,\\phi),\\qquad r\\in H,\\quad\\phi\\in H.\n\\]\nThe element \\(\\phi\\) is the chosen identification of the restricted object with the fixed boundary. It is not discarded just because the boundary has only one object: that object still has many automorphisms.","A comparison from \\((r,\\phi)\\) to \\((r',\\psi)\\) is a gauge \\((a_0,a_1)\\) satisfying both the cocycle equation and the boundary equation\n\\[\n r'=a_0ra_1^{-1},\\qquad \\psi a_0=\\phi.\n\\]\nThese equations determine it uniquely:\n\\begin{equation}\\label{eq:marked-two-copy}\n a_0=\\psi^{-1}\\phi,\\qquad a_1=(r')^{-1}a_0r.\n\\end{equation}\nSubstituting the formulas proves both equations without commuting any factors. Thus in this example there is exactly one identification-preserving arrow between any two marked realizations. The marked groupoid is nonempty and equivalent to the terminal groupoid, although its raw object set has many elements.","This is a worked two-view model of the marked transport mechanism. The generic groupoid construction and its equivalence theorem are Proposition~\\ref{KS38} and Theorem~\\ref{KS39}. Formula~\\eqref{eq:marked-two-copy} is checked in the independent finite suite for \\(H=S_3\\), retaining the order of multiplication. This finite check is separate from the general formal theorem.","\\subsection{A two-bit phase law}\nWrite \\(\\mathbf F_2=\\Z/2\\Z\\) additively, and let\n\\[\n G=\\mathbf F_2^2,\\qquad x=(x_1,x_2),\\qquad\n c(x,y)=x_2y_1.\n\\]\nAll entries and operations in this subsection lie in \\(\\mathbf F_2\\). In particular minus and plus agree, although retaining a minus sign in the coboundary formula helps explain its origin.","\\begin{proposition}[A normalized associative phase]\\label{KS80}\nThe phase vanishes when either argument is zero and satisfies\n\\begin{equation}\\label{eq:phase-cocycle}\n c(x,y)+c(x+y,z)=c(y,z)+c(x,y+z).\n\\end{equation}\nThere is no function \\(a:G\\to\\mathbf F_2\\) such that\n\\begin{equation}\\label{eq:phase-flatten}\n c(x,y)=a(x)+a(y)-a(x+y)\\quad\\text{for every }x,y\\in G.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nNormalization follows from multiplication by zero. Expanding the two sides of \\eqref{eq:phase-cocycle} gives, respectively,\n\\[\n x_2y_1+(x_2+y_2)z_1,\n \\qquad y_2z_1+x_2(y_1+z_1),\n\\]\nwhich are equal by distributivity.","on}\\label{eq:phase-cocycle}\n c(x,y)+c(x+y,z)=c(y,z)+c(x,y+z).\n\\end{equation}\nThere is no function \\(a:G\\to\\mathbf F_2\\) such that\n\\begin{equation}\\label{eq:phase-flatten}\n c(x,y)=a(x)+a(y)-a(x+y)\\quad\\text{for every }x,y\\in G.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nNormalization follows from multiplication by zero. Expanding the two sides of \\eqref{eq:phase-cocycle} gives, respectively,\n\\[\n x_2y_1+(x_2+y_2)z_1,\n \\qquad y_2z_1+x_2(y_1+z_1),\n\\]\nwhich are equal by distributivity.","Every expression \\(a(x)+a(y)-a(x+y)\\) is symmetric in \\(x,y\\), because \\(G\\) is abelian. With \\(e_1=(1,0)\\) and \\(e_2=(0,1)\\), however,\n\\[\n c(e_1,e_2)=0,\\qquad c(e_2,e_1)=1.\n\\]\nThis contradicts the symmetry required by \\eqref{eq:phase-flatten}.\n\\end{proof}\n\\source{KS80}\nThe proof rules out all sixteen possible scalar functions \\(a\\), not just linear functions or selected trial gauges. The source \\code{ProjectiveExample.lean} proves the normalization, cocycle, asymmetry, and no-scalar-flattening statements. The matrix interpretation and group construction below are worked derivations of the finite model, not separate formal declarations.","\\plate{30_phase}{The entry in row $x$ and column $y$ is $c(x,y)=x_2y_1$. The two entries singled out in the text, $c(e_1,e_2)=0$ and $c(e_2,e_1)=1$, already obstruct every scalar flattening. Associativity still holds for every triple.}{fig:phase}","\\subsection{The phase is visible in actual matrices}\nOver \\(\\R\\), define\n\\[\n X=\\begin{pmatrix}0&1","\\subsection{The phase is visible in actual matrices}\nOver \\(\\R\\), define\n\\[\n X=\\begin{pmatrix}0&1\\\\1&0\\end{pmatrix},\\qquad\n Z=\\begin{pmatrix}1&0\\\\0&-1\\end{pmatrix},\\qquad\n U_{(p,q)}=X^pZ^q\\quad(p,q\\in\\{0,1\\}).\n\\]\nDirect multiplication gives \\(X^2=Z^2=I\\) and \\(ZX=-XZ\\). Moving \\(Z^{x_2}\\) past \\(X^{y_1}\\) therefore yields\n\\begin{equation}\\label{eq:phase-matrices}\n U_xU_y=(-1)^{c(x,y)}U_{x+y}.\n\\end{equation}\nFor example \\(U_{e_1}U_{e_2}=XZ\\), while \\(U_{e_2}U_{e_1}=-XZ\\). The two-bit labels add commutatively, but their chosen matrix lifts do not commute.","0&-1\\end{pmatrix},\\qquad\n U_{(p,q)}=X^pZ^q\\quad(p,q\\in\\{0,1\\}).\n\\]\nDirect multiplication gives \\(X^2=Z^2=I\\) and \\(ZX=-XZ\\). Moving \\(Z^{x_2}\\) past \\(X^{y_1}\\) therefore yields\n\\begin{equation}\\label{eq:phase-matrices}\n U_xU_y=(-1)^{c(x,y)}U_{x+y}.\n\\end{equation}\nFor example \\(U_{e_1}U_{e_2}=XZ\\), while \\(U_{e_2}U_{e_1}=-XZ\\). The two-bit labels add commutatively, but their chosen matrix lifts do not commute.","Associativity of three lifted matrices requires the same total sign on both parenthesizations. Equation~\\eqref{eq:phase-cocycle} is exactly that requirement:\n\\[\n (U_xU_y)U_z\n =(-1)^{c(x,y)+c(x+y,z)}U_{x+y+z},\n\\]\n\\[\n U_x(U_yU_z)\n =(-1)^{c(y,z)+c(x,y+z)}U_{x+y+z}.\n\\]\nThus a nonzero phase is not automatically a failed associativity law. It can be the data that make the composition law explicit.","Changing lifts to \\(U'_x=(-1)^{a(x)}U_x\\) changes the phase to\n\\[\n c'(x,y)=c(x,y)+a(x)+a(y)-a(x+y).\n\\]\nNo such change makes \\(c'\\) vanish identically. The antisymmetric part is unchanged by every scalar change of lift. In this model it is\n\\[\n \\kappa(x,y)=c(x,y)-c(y,x)\n            =x_2y_1-y_2x_1.\n\\]\nThis is a directly observable obstruction certificate for the model. Its nonzero value at \\((e_1,e_2)\\) is enough, without a search over gauges.","The same symmetry argument rules out removing this sign multiplier by arbitrary nonzero complex scalar lifts as well. A multiplicative scalar coboundary \\(b(x)b(y)/b(x+y)\\) on an abelian group is symmetric, whereas the multiplier in \\eqref{eq:phase-matrices} is not. This last sentence is an algebraic consequence written here. The retained Lean example uses the stated \\(\\mathbf F_2\\)-valued formulation.","\\subsection{Retaining the phase restores an ordinary multiplication law}\nInstead of trying to discard the phase, include it in the state. On the eight-element set \\(\\widetilde G=\\mathbf F_2\\times G\\), put\n\\begin{equation}\\label{eq:central-extension}\n (a,x)\\cdot(b,y)=(a+b+c(x,y),x+y).\n\\end{equation}\nThe first coordinate of the left-associated product of three elements is\n\\(a+b+d+c(x,y)+c(x+y,z)\\). The right-associated coordinate is\n\\(a+b+d+c(y,z)+c(x,y+z)\\). Equation~\\eqref{eq:phase-cocycle} proves equality. Normalization gives identity \\((0,0)\\), and\n\\[\n (a,x)^{-1}=(a+c(x,x),x).\n\\]\nBoth inverse products are the identity because \\(x+x=0\\) and arithmetic is modulo two. Hence \\eqref{eq:central-extension} defines a group.","res an ordinary multiplication law}\nInstead of trying to discard the phase, include it in the state. On the eight-element set \\(\\widetilde G=\\mathbf F_2\\times G\\), put\n\\begin{equation}\\label{eq:central-extension}\n (a,x)\\cdot(b,y)=(a+b+c(x,y),x+y).\n\\end{equation}\nThe first coordinate of the left-associated product of three elements is\n\\(a+b+d+c(x,y)+c(x+y,z)\\). The right-associated coordinate is\n\\(a+b+d+c(y,z)+c(x,y+z)\\). Equation~\\eqref{eq:phase-cocycle} proves equality. Normalization gives identity \\((0,0)\\), and\n\\[\n (a,x)^{-1}=(a+c(x,x),x).\n\\]\nBoth inverse products are the identity because \\(x+x=0\\) and arithmetic is modulo two. Hence \\eqref{eq:central-extension} defines a group.","eight-element set \\(\\widetilde G=\\mathbf F_2\\times G\\), put\n\\begin{equation}\\label{eq:central-extension}\n (a,x)\\cdot(b,y)=(a+b+c(x,y),x+y).\n\\end{equation}\nThe first coordinate of the left-associated product of three elements is\n\\(a+b+d+c(x,y)+c(x+y,z)\\). The right-associated coordinate is\n\\(a+b+d+c(y,z)+c(x,y+z)\\). Equation~\\eqref{eq:phase-cocycle} proves equality. Normalization gives identity \\((0,0)\\), and\n\\[\n (a,x)^{-1}=(a+c(x,x),x).\n\\]\nBoth inverse products are the identity because \\(x+x=0\\) and arithmetic is modulo two. Hence \\eqref{eq:central-extension} defines a group.","set \\(\\widetilde G=\\mathbf F_2\\times G\\), put\n\\begin{equation}\\label{eq:central-extension}\n (a,x)\\cdot(b,y)=(a+b+c(x,y),x+y).\n\\end{equation}\nThe first coordinate of the left-associated product of three elements is\n\\(a+b+d+c(x,y)+c(x+y,z)\\). The right-associated coordinate is\n\\(a+b+d+c(y,z)+c(x,y+z)\\). Equation~\\eqref{eq:phase-cocycle} proves equality. Normalization gives identity \\((0,0)\\), and\n\\[\n (a,x)^{-1}=(a+c(x,x),x).\n\\]\nBoth inverse products are the identity because \\(x+x=0\\) and arithmetic is modulo two. Hence \\eqref{eq:central-extension} defines a group.","-extension}\n (a,x)\\cdot(b,y)=(a+b+c(x,y),x+y).\n\\end{equation}\nThe first coordinate of the left-associated product of three elements is\n\\(a+b+d+c(x,y)+c(x+y,z)\\). The right-associated coordinate is\n\\(a+b+d+c(y,z)+c(x,y+z)\\). Equation~\\eqref{eq:phase-cocycle} proves equality. Normalization gives identity \\((0,0)\\), and\n\\[\n (a,x)^{-1}=(a+c(x,x),x).\n\\]\nBoth inverse products are the identity because \\(x+x=0\\) and arithmetic is modulo two. Hence \\eqref{eq:central-extension} defines a group.","Projection \\((a,x)\\mapsto x\\) is a homomorphism onto \\(G\\). Its kernel is \\(\\{(0,0),(1,0)\\}\\), which commutes with every element. The map\n\\[\n (a,x)\\longmapsto(-1)^aU_x\n\\]\nis a homomorphism by \\eqref{eq:phase-matrices}. Its eight matrices are distinct: they are \\(\\pm I,\\pm X,\\pm Z,\\pm XZ\\). It is therefore faithful. The phase extension is not an assertion that \\(G\\) itself has acquired an ordinary representation with these same lifts. The represented group has changed.","tached to the allowed changes and to a specified presentation problem. Enlarging the presentation can be mathematically productive, but it changes what is being realized and what information is carried. No general theorem that every obstruction can be removed by a finite enlargement is claimed. The finite checker enumerates all group triples, all scalar gauges, and all multiplication triples in \\(\\widetilde G\\), independently of the symmetry proof above.","not come with a numerical distance by default. In a declared linear model, the compatibility equations become a bounded operator. Closed range, holonomy, and conditioning then determine what small residuals actually guarantee.}\n\\section{Graph views, holonomy, and reconstruction}\\label{sec:graph}\n\\subsection{Actual transports on a finite graph}\nLet a finite connected simple graph have vertex set \\(V\\), root \\(o\\), and a chosen route \\(p_i:o\\to i\\) for every vertex, with the route at \\(o\\) empty. Use a common real normed fiber \\(H\\). An inverse consistent connection assigns continuous linear equivalences \\(\\tau_{ij}\\), with \\(\\tau_{ji}=\\tau_{ij}^{-1}\\). Our convention is that \\(\\tau_{ij}\\) carries the value at \\(j\\) into the coordinates at \\(i\\).","h a numerical distance by default. In a declared linear model, the compatibility equations become a bounded operator. Closed range, holonomy, and conditioning then determine what small residuals actually guarantee.}\n\\section{Graph views, holonomy, and reconstruction}\\label{sec:graph}\n\\subsection{Actual transports on a finite graph}\nLet a finite connected simple graph have vertex set \\(V\\), root \\(o\\), and a chosen route \\(p_i:o\\to i\\) for every vertex, with the route at \\(o\\) empty. Use a common real normed fiber \\(H\\). An inverse consistent connection assigns continuous linear equivalences \\(\\tau_{ij}\\), with \\(\\tau_{ji}=\\tau_{ij}^{-1}\\). Our convention is that \\(\\tau_{ij}\\) carries the value at \\(j\\) into the coordinates at \\(i\\).","default. In a declared linear model, the compatibility equations become a bounded operator. Closed range, holonomy, and conditioning then determine what small residuals actually guarantee.}\n\\section{Graph views, holonomy, and reconstruction}\\label{sec:graph}\n\\subsection{Actual transports on a finite graph}\nLet a finite connected simple graph have vertex set \\(V\\), root \\(o\\), and a chosen route \\(p_i:o\\to i\\) for every vertex, with the route at \\(o\\) empty. Use a common real normed fiber \\(H\\). An inverse consistent connection assigns continuous linear equivalences \\(\\tau_{ij}\\), with \\(\\tau_{ji}=\\tau_{ij}^{-1}\\). Our convention is that \\(\\tau_{ij}\\) carries the value at \\(j\\) into the coordinates at \\(i\\).","ity equations become a bounded operator. Closed range, holonomy, and conditioning then determine what small residuals actually guarantee.}\n\\section{Graph views, holonomy, and reconstruction}\\label{sec:graph}\n\\subsection{Actual transports on a finite graph}\nLet a finite connected simple graph have vertex set \\(V\\), root \\(o\\), and a chosen route \\(p_i:o\\to i\\) for every vertex, with the route at \\(o\\) empty. Use a common real normed fiber \\(H\\). An inverse consistent connection assigns continuous linear equivalences \\(\\tau_{ij}\\), with \\(\\tau_{ji}=\\tau_{ij}^{-1}\\). Our convention is that \\(\\tau_{ij}\\) carries the value at \\(j\\) into the coordinates at \\(i\\).","sed range, holonomy, and conditioning then determine what small residuals actually guarantee.}\n\\section{Graph views, holonomy, and reconstruction}\\label{sec:graph}\n\\subsection{Actual transports on a finite graph}\nLet a finite connected simple graph have vertex set \\(V\\), root \\(o\\), and a chosen route \\(p_i:o\\to i\\) for every vertex, with the route at \\(o\\) empty. Use a common real normed fiber \\(H\\). An inverse consistent connection assigns continuous linear equivalences \\(\\tau_{ij}\\), with \\(\\tau_{ji}=\\tau_{ij}^{-1}\\). Our convention is that \\(\\tau_{ij}\\) carries the value at \\(j\\) into the coordinates at \\(i\\).","lly guarantee.}\n\\section{Graph views, holonomy, and reconstruction}\\label{sec:graph}\n\\subsection{Actual transports on a finite graph}\nLet a finite connected simple graph have vertex set \\(V\\), root \\(o\\), and a chosen route \\(p_i:o\\to i\\) for every vertex, with the route at \\(o\\) empty. Use a common real normed fiber \\(H\\). An inverse consistent connection assigns continuous linear equivalences \\(\\tau_{ij}\\), with \\(\\tau_{ji}=\\tau_{ij}^{-1}\\). Our convention is that \\(\\tau_{ij}\\) carries the value at \\(j\\) into the coordinates at \\(i\\).","tion{Graph views, holonomy, and reconstruction}\\label{sec:graph}\n\\subsection{Actual transports on a finite graph}\nLet a finite connected simple graph have vertex set \\(V\\), root \\(o\\), and a chosen route \\(p_i:o\\to i\\) for every vertex, with the route at \\(o\\) empty. Use a common real normed fiber \\(H\\). An inverse consistent connection assigns continuous linear equivalences \\(\\tau_{ij}\\), with \\(\\tau_{ji}=\\tau_{ij}^{-1}\\). Our convention is that \\(\\tau_{ij}\\) carries the value at \\(j\\) into the coordinates at \\(i\\).","{sec:graph}\n\\subsection{Actual transports on a finite graph}\nLet a finite connected simple graph have vertex set \\(V\\), root \\(o\\), and a chosen route \\(p_i:o\\to i\\) for every vertex, with the route at \\(o\\) empty. Use a common real normed fiber \\(H\\). An inverse consistent connection assigns continuous linear equivalences \\(\\tau_{ij}\\), with \\(\\tau_{ji}=\\tau_{ij}^{-1}\\). Our convention is that \\(\\tau_{ij}\\) carries the value at \\(j\\) into the coordinates at \\(i\\).","sports on a finite graph}\nLet a finite connected simple graph have vertex set \\(V\\), root \\(o\\), and a chosen route \\(p_i:o\\to i\\) for every vertex, with the route at \\(o\\) empty. Use a common real normed fiber \\(H\\). An inverse consistent connection assigns continuous linear equivalences \\(\\tau_{ij}\\), with \\(\\tau_{ji}=\\tau_{ij}^{-1}\\). Our convention is that \\(\\tau_{ij}\\) carries the value at \\(j\\) into the coordinates at \\(i\\).","a finite connected simple graph have vertex set \\(V\\), root \\(o\\), and a chosen route \\(p_i:o\\to i\\) for every vertex, with the route at \\(o\\) empty. Use a common real normed fiber \\(H\\). An inverse consistent connection assigns continuous linear equivalences \\(\\tau_{ij}\\), with \\(\\tau_{ji}=\\tau_{ij}^{-1}\\). Our convention is that \\(\\tau_{ij}\\) carries the value at \\(j\\) into the coordinates at \\(i\\).","The candidate and residual spaces are finite products with maximum norms:\n\\[\n E=\\prod_{i\\in V}H,\\qquad\n R=\\prod_{(i,j)\\in\\vec E}H,\\qquad\n (dx)_{ij}=\\tau_{ij}x_j-x_i.\n\\]\nBoth orientations of an edge are retained. If \\(M\\) bounds the norms of all measured edge maps, then \\(\\norm d\\le M+1\\). The empty edge case has the usual zero residual.","Distinct walks are retained as arrows of the view category. The product along a walk \\(q=(i_0,\\ldots,i_m)\\) is\n\\[\n T_q=\\tau_{i_0i_1}\\cdots\\tau_{i_{m-1}i_m}.\n\\]\nConcatenation gives the restriction composition law. An exact vertex family defines a coherent walk-view section, and a coherent walk-view section satisfies every edge equation. Thus this graph construction returns to the original ambient realization type rather than creating a disconnected numerical model.","\\subsection{Transported residuals and path gains}\n\\begin{proposition}[The path telescope]\\label{KS40}\nFor a walk \\(q=(i_0,\\ldots,i_m)\\),\n\\[\n T_qx_{i_m}-x_{i_0}\n =\\sum_{k=0}^{m-1}\n (\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k})(dx)_{i_ki_{k+1}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with ea","\\subsection{Transported residuals and path gains}\n\\begin{proposition}[The path telescope]\\label{KS40}\nFor a walk \\(q=(i_0,\\ldots,i_m)\\),\n\\[\n T_qx_{i_m}-x_{i_0}\n =\\sum_{k=0}^{m-1}\n (\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k})(dx)_{i_ki_{k+1}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with","\\subsection{Transported residuals and path gains}\n\\begin{proposition}[The path telescope]\\label{KS40}\nFor a walk \\(q=(i_0,\\ldots,i_m)\\),\n\\[\n T_qx_{i_m}-x_{i_0}\n =\\sum_{k=0}^{m-1}\n (\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k})(dx)_{i_ki_{k+1}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk bot","\\subsection{Transported residuals and path gains}\n\\begin{proposition}[The path telescope]\\label{KS40}\nFor a walk \\(q=(i_0,\\ldots,i_m)\\),\n\\[\n T_qx_{i_m}-x_{i_0}\n =\\sum_{k=0}^{m-1}\n (\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k})(dx)_{i_ki_{k+1}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","\\begin{proposition}[The path telescope]\\label{KS40}\nFor a walk \\(q=(i_0,\\ldots,i_m)\\),\n\\[\n T_qx_{i_m}-x_{i_0}\n =\\sum_{k=0}^{m-1}\n (\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k})(dx)_{i_ki_{k+1}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","tion}[The path telescope]\\label{KS40}\nFor a walk \\(q=(i_0,\\ldots,i_m)\\),\n\\[\n T_qx_{i_m}-x_{i_0}\n =\\sum_{k=0}^{m-1}\n (\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k})(dx)_{i_ki_{k+1}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","\\(q=(i_0,\\ldots,i_m)\\),\n\\[\n T_qx_{i_m}-x_{i_0}\n =\\sum_{k=0}^{m-1}\n (\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k})(dx)_{i_ki_{k+1}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","{k=0}^{m-1}\n (\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k})(dx)_{i_ki_{k+1}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","s\\tau_{i_{k-1}i_k})(dx)_{i_ki_{k+1}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","(dx)_{i_ki_{k+1}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","}}.\n\\]\nThe empty prefix is the identity. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","y. If \\(\\gamma(q)\\) is the sum of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","of the products of preceding edge norms, then\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","}\\le\\gamma(q)\\norm{dx}_\\infty.\n\\]\n\\end{proposition}\n\\begin{proof}\nWrite \\(U_0=I\\) and \\(U_k=\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","\\tau_{i_0i_1}\\cdots\\tau_{i_{k-1}i_k}\\) for \\(1\\le k\\le m\\). The transport convention makes every \\(U_k\\) a map from the fiber at \\(i_k\\) to that at \\(i_0\\). Since \\(U_{k+1}=U_k\\tau_{i_ki_{k+1}}\\), the \\(k\\)-th transported defect is\n\\[\n U_k(dx)_{i_ki_{k+1}}\n =U_{k+1}x_{i_{k+1}}-U_kx_{i_k}.\n\\]\nSumming over \\(k=0,\\ldots,m-1\\) cancels each interior value once with each sign. The remaining terms are \\(U_mx_{i_m}-U_0x_{i_0}=T_qx_{i_m}-x_{i_0}\\), proving the identity with its actual ordered products. For an empty walk both sides are zero.","The triangle inequality and the operator norm bound now give\n\\[\n \\norm{T_qx_{i_m}-x_{i_0}}\n \\le\\sum_{k=0}^{m-1}\\norm{U_k}\\norm{(dx)_{i_ki_{k+1}}}\n \\le\\left(\\sum_{k=0}^{m-1}\\prod_{j=0}^{k-1}\\norm{\\tau_{i_ji_{j+1}}}\\right)\\norm{dx}_\\infty.\n\\]\nThe bracket is \\(\\gamma(q)\\), with the empty product equal to one. Repeated vertices or edges do not affect the cancellation. Reordering the edge maps would, in general, invalidate it.\n\\end{proof}\n\\source{KS40}\nFor isometric maps, \\(\\gamma(q)\\le|q|\\). For arbitrary bounded equivalences, route length cannot replace the amplification gain. Reversing a route can have a different gain, even though its transport is the inverse map.","x_{i_m}-x_{i_0}}\n \\le\\sum_{k=0}^{m-1}\\norm{U_k}\\norm{(dx)_{i_ki_{k+1}}}\n \\le\\left(\\sum_{k=0}^{m-1}\\prod_{j=0}^{k-1}\\norm{\\tau_{i_ji_{j+1}}}\\right)\\norm{dx}_\\infty.\n\\]\nThe bracket is \\(\\gamma(q)\\), with the empty product equal to one. Repeated vertices or edges do not affect the cancellation. Reordering the edge maps would, in general, invalidate it.\n\\end{proof}\n\\source{KS40}\nFor isometric maps, \\(\\gamma(q)\\le|q|\\). For arbitrary bounded equivalences, route length cannot replace the amplification gain. Reversing a route can have a different gain, even though its transport is the inverse map.","\\subsection{Holonomy is the remaining exactness condition}\nWrite \\(T_i=T_{p_i}\\). Reconstruction from a proposed root value is\n\\[\n (R_pz)_i=T_i^{-1}z.\n\\]\nFor each directed edge define the root loop operator\n\\[\n K_{ij}=T_i\\tau_{ij}T_j^{-1},\\qquad\n K=\\bigcap_{(i,j)\\in\\vec E}\\ker(K_{ij}-I).\n\\]\nEach \\(K_{ij}\\) is the transport around an actual rooted loop. No flatness assumption has been made.","\\begin{theorem}[Holonomy classification]\\label{KS41}\nEvaluation at the root and reconstruction give an equivalence\n\\[\n K\\simeq_L\\ker d.\n\\]\nA root value extends to an exact vertex family precisely when it lies in \\(K\\), and that extension is unique. The space \\(K\\) is independent of the chosen routes. Reconstruction is route independent on \\(K\\).\n\\end{theorem}\n\\begin{proof}\nEvery \\(T_i\\) is a continuous linear equivalence, being a finite ordered composition of such maps. Reconstruction is therefore well defined. For an edge \\((i,j)\\), apply \\(T_i\\) to its defect:\n\\[\n T_i\\bigl(\\tau_{ij}T_j^{-1}z-T_i^{-1}z\\bigr)\n =T_i\\tau_{ij}T_j^{-1}z-z=K_{ij}z-z.\n\\]\nBecause \\(T_i\\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \\(z\\). Thus \\(dR_pz=0\\) holds exactly for \\(z\\in K\\). The root route is emp","\\begin{theorem}[Holonomy classification]\\label{KS41}\nEvaluation at the root and reconstruction give an equivalence\n\\[\n K\\simeq_L\\ker d.\n\\]\nA root value extends to an exact vertex family precisely when it lies in \\(K\\), and that extension is unique. The space \\(K\\) is independent of the chosen routes. Reconstruction is route independent on \\(K\\).\n\\end{theorem}\n\\begin{proof}\nEvery \\(T_i\\) is a continuous linear equivalence, being a finite ordered composition of such maps. Reconstruction is therefore well defined. For an edge \\((i,j)\\), apply \\(T_i\\) to its defect:\n\\[\n T_i\\bigl(\\tau_{ij}T_j^{-1}z-T_i^{-1}z\\bigr)\n =T_i\\tau_{ij}T_j^{-1}z-z=K_{ij}z-z.\n\\]\nBecause \\(T_i\\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \\(z\\). Thus \\(dR_pz=0\\) holds exactly for \\(z\\in K\\). The root route is empty, so \\((R_pz)_o=z\\).","r d.\n\\]\nA root value extends to an exact vertex family precisely when it lies in \\(K\\), and that extension is unique. The space \\(K\\) is independent of the chosen routes. Reconstruction is route independent on \\(K\\).\n\\end{theorem}\n\\begin{proof}\nEvery \\(T_i\\) is a continuous linear equivalence, being a finite ordered composition of such maps. Reconstruction is therefore well defined. For an edge \\((i,j)\\), apply \\(T_i\\) to its defect:\n\\[\n T_i\\bigl(\\tau_{ij}T_j^{-1}z-T_i^{-1}z\\bigr)\n =T_i\\tau_{ij}T_j^{-1}z-z=K_{ij}z-z.\n\\]\nBecause \\(T_i\\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \\(z\\). Thus \\(dR_pz=0\\) holds exactly for \\(z\\in K\\). The root route is empty, so \\((R_pz)_o=z\\).","lue extends to an exact vertex family precisely when it lies in \\(K\\), and that extension is unique. The space \\(K\\) is independent of the chosen routes. Reconstruction is route independent on \\(K\\).\n\\end{theorem}\n\\begin{proof}\nEvery \\(T_i\\) is a continuous linear equivalence, being a finite ordered composition of such maps. Reconstruction is therefore well defined. For an edge \\((i,j)\\), apply \\(T_i\\) to its defect:\n\\[\n T_i\\bigl(\\tau_{ij}T_j^{-1}z-T_i^{-1}z\\bigr)\n =T_i\\tau_{ij}T_j^{-1}z-z=K_{ij}z-z.\n\\]\nBecause \\(T_i\\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \\(z\\). Thus \\(dR_pz=0\\) holds exactly for \\(z\\in K\\). The root route is empty, so \\((R_pz)_o=z\\).","vertex family precisely when it lies in \\(K\\), and that extension is unique. The space \\(K\\) is independent of the chosen routes. Reconstruction is route independent on \\(K\\).\n\\end{theorem}\n\\begin{proof}\nEvery \\(T_i\\) is a continuous linear equivalence, being a finite ordered composition of such maps. Reconstruction is therefore well defined. For an edge \\((i,j)\\), apply \\(T_i\\) to its defect:\n\\[\n T_i\\bigl(\\tau_{ij}T_j^{-1}z-T_i^{-1}z\\bigr)\n =T_i\\tau_{ij}T_j^{-1}z-z=K_{ij}z-z.\n\\]\nBecause \\(T_i\\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \\(z\\). Thus \\(dR_pz=0\\) holds exactly for \\(z\\in K\\). The root route is empty, so \\((R_pz)_o=z\\).","independent of the chosen routes. Reconstruction is route independent on \\(K\\).\n\\end{theorem}\n\\begin{proof}\nEvery \\(T_i\\) is a continuous linear equivalence, being a finite ordered composition of such maps. Reconstruction is therefore well defined. For an edge \\((i,j)\\), apply \\(T_i\\) to its defect:\n\\[\n T_i\\bigl(\\tau_{ij}T_j^{-1}z-T_i^{-1}z\\bigr)\n =T_i\\tau_{ij}T_j^{-1}z-z=K_{ij}z-z.\n\\]\nBecause \\(T_i\\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \\(z\\). Thus \\(dR_pz=0\\) holds exactly for \\(z\\in K\\). The root route is empty, so \\((R_pz)_o=z\\).","begin{proof}\nEvery \\(T_i\\) is a continuous linear equivalence, being a finite ordered composition of such maps. Reconstruction is therefore well defined. For an edge \\((i,j)\\), apply \\(T_i\\) to its defect:\n\\[\n T_i\\bigl(\\tau_{ij}T_j^{-1}z-T_i^{-1}z\\bigr)\n =T_i\\tau_{ij}T_j^{-1}z-z=K_{ij}z-z.\n\\]\nBecause \\(T_i\\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \\(z\\). Thus \\(dR_pz=0\\) holds exactly for \\(z\\in K\\). The root route is empty, so \\((R_pz)_o=z\\).","Every \\(T_i\\) is a continuous linear equivalence, being a finite ordered composition of such maps. Reconstruction is therefore well defined. For an edge \\((i,j)\\), apply \\(T_i\\) to its defect:\n\\[\n T_i\\bigl(\\tau_{ij}T_j^{-1}z-T_i^{-1}z\\bigr)\n =T_i\\tau_{ij}T_j^{-1}z-z=K_{ij}z-z.\n\\]\nBecause \\(T_i\\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \\(z\\). Thus \\(dR_pz=0\\) holds exactly for \\(z\\in K\\). The root route is empty, so \\((R_pz)_o=z\\).","linear equivalence, being a finite ordered composition of such maps. Reconstruction is therefore well defined. For an edge \\((i,j)\\), apply \\(T_i\\) to its defect:\n\\[\n T_i\\bigl(\\tau_{ij}T_j^{-1}z-T_i^{-1}z\\bigr)\n =T_i\\tau_{ij}T_j^{-1}z-z=K_{ij}z-z.\n\\]\nBecause \\(T_i\\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \\(z\\). Thus \\(dR_pz=0\\) holds exactly for \\(z\\in K\\). The root route is empty, so \\((R_pz)_o=z\\).","red composition of such maps. Reconstruction is therefore well defined. For an edge \\((i,j)\\), apply \\(T_i\\) to its defect:\n\\[\n T_i\\bigl(\\tau_{ij}T_j^{-1}z-T_i^{-1}z\\bigr)\n =T_i\\tau_{ij}T_j^{-1}z-z=K_{ij}z-z.\n\\]\nBecause \\(T_i\\) is injective, that edge defect is zero if and only if the corresponding root loop fixes \\(z\\). Thus \\(dR_pz=0\\) holds exactly for \\(z\\in K\\). The root route is empty, so \\((R_pz)_o=z\\).","Conversely, let \\(x\\in\\ker d\\). The path telescope along \\(p_i\\) has all zero summands, hence \\(T_ix_i=x_o\\) for every vertex. Applying \\(T_i^{-1}\\) gives \\(x_i=T_i^{-1}x_o\\), so \\(x=R_p(x_o)\\). The first part then places \\(x_o\\) in \\(K\\). We have proved both inverse equations and uniqueness of an exact extension of a root value.","The maps are linear. On the finite vertex product,\n\\[\n \\norm{R_pz}_\\infty\\le\\left(\\max_i\\norm{T_i^{-1}}\\right)\\norm z,\n \\qquad \\norm{x_o}\\le\\norm x_\\infty.\n\\]\nThese bounds also hold on the subspaces \\(K\\) and \\(\\ker d\\), proving continuity of both directions. No completeness is needed for these finite bounds.","For route independence, the proved equivalence identifies \\(K\\) with the image of root evaluation from the intrinsically defined space \\(\\ker d\\). That image does not refer to any chosen routes, so every route system yields the same \\(K\\). For \\(z\\in K\\), reconstruction using either system is an exact family with root \\(z\\), and uniqueness makes the families equal. For \\(z\\notin K\\), this uniqueness argument does not assert equality of the two generally inconsistent reconstructed families.\n\\end{proof}\n\\source{KS41}\nWith isometric transports the equivalence is an isometry for the vertex maximum norm, because every reconstructed coordinate has the root norm. With general bounded equivalences, it need not be an isometry.","\\subsection{Prescribed observations remain visible}\nIf a prescribed vertex set \\(U\\) contains \\(o\\), a candidate \\(x\\) has an exact repair preserving \\(x|_U\\) precisely when\n\\[\n x_o\\in K\\quad\\text{and}\\quad (R_px_o)|_U=x|_U.\n\\]\nNecessity follows from the classification. Sufficiency is the reconstructed family itself. The source identifies this criterion with the earlier obstruction class in \\(R/d(\\ker p_U)\\). A graph seed test is therefore an instance of the same protected correction problem, not a replacement for it.","\\chap{Closed range is the stability boundary}\\label{sec:closed}\n\\subsection{Projection exists before a residual estimate}\nLet \\(E\\) be a real Hilbert space and \\(R\\) a real Banach space. Let \\(D:E\\to R\\) be continuous linear. The kernel \\(K=\\ker D\\) is closed, so its orthogonal projection \\(q=P_K\\) exists~\\cite{mathlib-projection}.","\\begin{proposition}[Exact orthogonal repair]\\label{KS42}\nFor every \\(x\\in E\\),\n\\[\n D(qx)=0,\\quad q^2=q,\\quad\\norm{qx}\\le\\norm x,\n\\]\nand \\(qx\\) is the unique closest point to \\(x\\) in \\(K\\).\n\\end{proposition}\n\\begin{proof}\nBoundedness of \\(D\\) makes \\(K=\\ker D\\) closed. Lemma~\\ref{lem:projection-tool} constructs \\(q=P_K\\) and gives the orthogonal decomposition \\(x=qx+(x-qx)\\). The first term belongs to \\(K\\), so \\(Dqx=0\\). Projection fixes every point of \\(K\\), giving \\(q(qx)=qx\\).","proposition}[Exact orthogonal repair]\\label{KS42}\nFor every \\(x\\in E\\),\n\\[\n D(qx)=0,\\quad q^2=q,\\quad\\norm{qx}\\le\\norm x,\n\\]\nand \\(qx\\) is the unique closest point to \\(x\\) in \\(K\\).\n\\end{proposition}\n\\begin{proof}\nBoundedness of \\(D\\) makes \\(K=\\ker D\\) closed. Lemma~\\ref{lem:projection-tool} constructs \\(q=P_K\\) and gives the orthogonal decomposition \\(x=qx+(x-qx)\\). The first term belongs to \\(K\\), so \\(Dqx=0\\). Projection fixes every point of \\(K\\), giving \\(q(qx)=qx\\).","n}[Exact orthogonal repair]\\label{KS42}\nFor every \\(x\\in E\\),\n\\[\n D(qx)=0,\\quad q^2=q,\\quad\\norm{qx}\\le\\norm x,\n\\]\nand \\(qx\\) is the unique closest point to \\(x\\) in \\(K\\).\n\\end{proposition}\n\\begin{proof}\nBoundedness of \\(D\\) makes \\(K=\\ker D\\) closed. Lemma~\\ref{lem:projection-tool} constructs \\(q=P_K\\) and gives the orthogonal decomposition \\(x=qx+(x-qx)\\). The first term belongs to \\(K\\), so \\(Dqx=0\\). Projection fixes every point of \\(K\\), giving \\(q(qx)=qx\\).","every \\(x\\in E\\),\n\\[\n D(qx)=0,\\quad q^2=q,\\quad\\norm{qx}\\le\\norm x,\n\\]\nand \\(qx\\) is the unique closest point to \\(x\\) in \\(K\\).\n\\end{proposition}\n\\begin{proof}\nBoundedness of \\(D\\) makes \\(K=\\ker D\\) closed. Lemma~\\ref{lem:projection-tool} constructs \\(q=P_K\\) and gives the orthogonal decomposition \\(x=qx+(x-qx)\\). The first term belongs to \\(K\\), so \\(Dqx=0\\). Projection fixes every point of \\(K\\), giving \\(q(qx)=qx\\).","E\\),\n\\[\n D(qx)=0,\\quad q^2=q,\\quad\\norm{qx}\\le\\norm x,\n\\]\nand \\(qx\\) is the unique closest point to \\(x\\) in \\(K\\).\n\\end{proposition}\n\\begin{proof}\nBoundedness of \\(D\\) makes \\(K=\\ker D\\) closed. Lemma~\\ref{lem:projection-tool} constructs \\(q=P_K\\) and gives the orthogonal decomposition \\(x=qx+(x-qx)\\). The first term belongs to \\(K\\), so \\(Dqx=0\\). Projection fixes every point of \\(K\\), giving \\(q(qx)=qx\\).","Pythagoras applied to that decomposition gives \\(\\norm x^2=\\norm{qx}^2+\\norm{x-qx}^2\\), hence \\(\\norm{qx}\\le\\norm x\\). For any \\(y\\in K\\), the vector \\(qx-y\\) lies in \\(K\\), whereas \\(x-qx\\) lies in its orthogonal complement. Consequently\n\\[\n \\norm{x-y}^2=\\norm{x-qx}^2+\\norm{qx-y}^2\\ge\\norm{x-qx}^2.\n\\]\nEquality holds only when \\(\\norm{qx-y}=0\\), or \\(y=qx\\). This proves nearest-point uniqueness without any closedness assumption on \\(\\ran D\\). Closedness of the kernel and of the range serve different purposes.\n\\end{proof}\n\\source{KS42}\nNo closed range assumption on \\(D\\) is needed for this projection. Such an assumption enters when the size of the residual is asked to control the distance to the exact space.","ion gives \\(\\norm x^2=\\norm{qx}^2+\\norm{x-qx}^2\\), hence \\(\\norm{qx}\\le\\norm x\\). For any \\(y\\in K\\), the vector \\(qx-y\\) lies in \\(K\\), whereas \\(x-qx\\) lies in its orthogonal complement. Consequently\n\\[\n \\norm{x-y}^2=\\norm{x-qx}^2+\\norm{qx-y}^2\\ge\\norm{x-qx}^2.\n\\]\nEquality holds only when \\(\\norm{qx-y}=0\\), or \\(y=qx\\). This proves nearest-point uniqueness without any closedness assumption on \\(\\ran D\\). Closedness of the kernel and of the range serve different purposes.\n\\end{proof}\n\\source{KS42}\nNo closed range assumption on \\(D\\) is needed for this projection. Such an assumption enters when the size of the residual is asked to control the distance to the exact space.","whereas \\(x-qx\\) lies in its orthogonal complement. Consequently\n\\[\n \\norm{x-y}^2=\\norm{x-qx}^2+\\norm{qx-y}^2\\ge\\norm{x-qx}^2.\n\\]\nEquality holds only when \\(\\norm{qx-y}=0\\), or \\(y=qx\\). This proves nearest-point uniqueness without any closedness assumption on \\(\\ran D\\). Closedness of the kernel and of the range serve different purposes.\n\\end{proof}\n\\source{KS42}\nNo closed range assumption on \\(D\\) is needed for this projection. Such an assumption enters when the size of the residual is asked to control the distance to the exact space.","\\begin{theorem}[Closed range and quantitative exactness]\\label{KS43}\nThe following are equivalent:\n\\begin{enumerate}\n\\item \\(\\ran D\\) is closed.\n\\item There is \\(C\\ge0\\) with \\(\\norm{x-qx}\\le C\\norm{Dx}\\) for all \\(x\\).\n\\item There is \\(C\\ge0\\) such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target:","\\begin{theorem}[Closed range and quantitative exactness]\\label{KS43}\nThe following are equivalent:\n\\begin{enumerate}\n\\item \\(\\ran D\\) is closed.\n\\item There is \\(C\\ge0\\) with \\(\\norm{x-qx}\\le C\\norm{Dx}\\) for all \\(x\\).\n\\item There is \\(C\\ge0\\) such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$","\\begin{theorem}[Closed range and quantitative exactness]\\label{KS43}\nThe following are equivalent:\n\\begin{enumerate}\n\\item \\(\\ran D\\) is closed.\n\\item There is \\(C\\ge0\\) with \\(\\norm{x-qx}\\le C\\norm{Dx}\\) for all \\(x\\).\n\\item There is \\(C\\ge0\\) such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounde","\\begin{theorem}[Closed range and quantitative exactness]\\label{KS43}\nThe following are equivalent:\n\\begin{enumerate}\n\\item \\(\\ran D\\) is closed.\n\\item There is \\(C\\ge0\\) with \\(\\norm{x-qx}\\le C\\norm{Dx}\\) for all \\(x\\).\n\\item There is \\(C\\ge0\\) such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijec","\\begin{theorem}[Closed range and quantitative exactness]\\label{KS43}\nThe following are equivalent:\n\\begin{enumerate}\n\\item \\(\\ran D\\) is closed.\n\\item There is \\(C\\ge0\\) with \\(\\norm{x-qx}\\le C\\norm{Dx}\\) for all \\(x\\).\n\\item There is \\(C\\ge0\\) such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity","\\begin{theorem}[Closed range and quantitative exactness]\\label{KS43}\nThe following are equivalent:\n\\begin{enumerate}\n\\item \\(\\ran D\\) is closed.\n\\item There is \\(C\\ge0\\) with \\(\\norm{x-qx}\\le C\\norm{Dx}\\) for all \\(x\\).\n\\item There is \\(C\\ge0\\) such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is cla","\\begin{theorem}[Closed range and quantitative exactness]\\label{KS43}\nThe following are equivalent:\n\\begin{enumerate}\n\\item \\(\\ran D\\) is closed.\n\\item There is \\(C\\ge0\\) with \\(\\norm{x-qx}\\le C\\norm{Dx}\\) for all \\(x\\).\n\\item There is \\(C\\ge0\\) such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","ntitative exactness]\\label{KS43}\nThe following are equivalent:\n\\begin{enumerate}\n\\item \\(\\ran D\\) is closed.\n\\item There is \\(C\\ge0\\) with \\(\\norm{x-qx}\\le C\\norm{Dx}\\) for all \\(x\\).\n\\item There is \\(C\\ge0\\) such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","rm{x-qx}\\le C\\norm{Dx}\\) for all \\(x\\).\n\\item There is \\(C\\ge0\\) such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","orm{Dx}\\) for all \\(x\\).\n\\item There is \\(C\\ge0\\) such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","such that every \\(x\\) has some exact \\(y\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","\\) with \\(\\norm{x-y}\\le C\\norm{Dx}\\).\n\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","\\end{enumerate}\nWhen the range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","he range is closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","closed, the least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","least admissible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","missible constant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","ant is the operator norm of the inverse of \\(D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","D:K^\\perp\\to\\ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","ran D\\). In the zero operator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","rator case that inverse has zero dimensional domain and its norm is zero.\n\\end{theorem}\n\\begin{proof}\nLet $K=\\ker D$ and $K^\\perp$ carry the induced Hilbert norm. Define $D_\\perp u=Du$, but regard its target as $\\ran D$ with the norm inherited from $R$. It is injective: $Du=0$ puts $u$ in both $K$ and $K^\\perp$, so $u=0$. It is surjective onto that target: if $r=Dx$, then $u=x-qx\\in K^\\perp$ and $Du=Dx=r$. Thus it is a bounded linear bijection before any continuity of its inverse is claimed.","Suppose first that the range is closed. Because $R$ is Banach, $\\ran D$ is then Banach. The domain $K^\\perp$ is also complete. The bounded inverse theorem, proved in Appendix~\\ref{app:proof-tools}, gives a bounded inverse $A=D_\\perp^{-1}$. Applying it to $Dx=D(x-qx)$ yields\n\\[\n x-qx=A(Dx),\\qquad \\norm{x-qx}\\le\\norm A\\norm{Dx}.\n\\]\nHence condition 2 holds. If any $C\\ge0$ satisfies condition 2, then for every $r\\in\\ran D$ apply it to $u=Ar\\in K^\\perp$. Since $qu=0$, it gives $\\norm{Ar}\\le C\\norm r$. Taking the operator norm proves $\\norm A\\le C$. This establishes the least constant, not just a sufficient one. If $D=0$, both $K^\\perp$ and the range are zero spaces and $\\norm A=0$.","st that the range is closed. Because $R$ is Banach, $\\ran D$ is then Banach. The domain $K^\\perp$ is also complete. The bounded inverse theorem, proved in Appendix~\\ref{app:proof-tools}, gives a bounded inverse $A=D_\\perp^{-1}$. Applying it to $Dx=D(x-qx)$ yields\n\\[\n x-qx=A(Dx),\\qquad \\norm{x-qx}\\le\\norm A\\norm{Dx}.\n\\]\nHence condition 2 holds. If any $C\\ge0$ satisfies condition 2, then for every $r\\in\\ran D$ apply it to $u=Ar\\in K^\\perp$. Since $qu=0$, it gives $\\norm{Ar}\\le C\\norm r$. Taking the operator norm proves $\\norm A\\le C$. This establishes the least constant, not just a sufficient one. If $D=0$, both $K^\\perp$ and the range are zero spaces and $\\norm A=0$.","ed. Because $R$ is Banach, $\\ran D$ is then Banach. The domain $K^\\perp$ is also complete. The bounded inverse theorem, proved in Appendix~\\ref{app:proof-tools}, gives a bounded inverse $A=D_\\perp^{-1}$. Applying it to $Dx=D(x-qx)$ yields\n\\[\n x-qx=A(Dx),\\qquad \\norm{x-qx}\\le\\norm A\\norm{Dx}.\n\\]\nHence condition 2 holds. If any $C\\ge0$ satisfies condition 2, then for every $r\\in\\ran D$ apply it to $u=Ar\\in K^\\perp$. Since $qu=0$, it gives $\\norm{Ar}\\le C\\norm r$. Taking the operator norm proves $\\norm A\\le C$. This establishes the least constant, not just a sufficient one. If $D=0$, both $K^\\perp$ and the range are zero spaces and $\\norm A=0$.",", $\\ran D$ is then Banach. The domain $K^\\perp$ is also complete. The bounded inverse theorem, proved in Appendix~\\ref{app:proof-tools}, gives a bounded inverse $A=D_\\perp^{-1}$. Applying it to $Dx=D(x-qx)$ yields\n\\[\n x-qx=A(Dx),\\qquad \\norm{x-qx}\\le\\norm A\\norm{Dx}.\n\\]\nHence condition 2 holds. If any $C\\ge0$ satisfies condition 2, then for every $r\\in\\ran D$ apply it to $u=Ar\\in K^\\perp$. Since $qu=0$, it gives $\\norm{Ar}\\le C\\norm r$. Taking the operator norm proves $\\norm A\\le C$. This establishes the least constant, not just a sufficient one. If $D=0$, both $K^\\perp$ and the range are zero spaces and $\\norm A=0$.","n Banach. The domain $K^\\perp$ is also complete. The bounded inverse theorem, proved in Appendix~\\ref{app:proof-tools}, gives a bounded inverse $A=D_\\perp^{-1}$. Applying it to $Dx=D(x-qx)$ yields\n\\[\n x-qx=A(Dx),\\qquad \\norm{x-qx}\\le\\norm A\\norm{Dx}.\n\\]\nHence condition 2 holds. If any $C\\ge0$ satisfies condition 2, then for every $r\\in\\ran D$ apply it to $u=Ar\\in K^\\perp$. Since $qu=0$, it gives $\\norm{Ar}\\le C\\norm r$. Taking the operator norm proves $\\norm A\\le C$. This establishes the least constant, not just a sufficient one. If $D=0$, both $K^\\perp$ and the range are zero spaces and $\\norm A=0$.","ded inverse theorem, proved in Appendix~\\ref{app:proof-tools}, gives a bounded inverse $A=D_\\perp^{-1}$. Applying it to $Dx=D(x-qx)$ yields\n\\[\n x-qx=A(Dx),\\qquad \\norm{x-qx}\\le\\norm A\\norm{Dx}.\n\\]\nHence condition 2 holds. If any $C\\ge0$ satisfies condition 2, then for every $r\\in\\ran D$ apply it to $u=Ar\\in K^\\perp$. Since $qu=0$, it gives $\\norm{Ar}\\le C\\norm r$. Taking the operator norm proves $\\norm A\\le C$. This establishes the least constant, not just a sufficient one. If $D=0$, both $K^\\perp$ and the range are zero spaces and $\\norm A=0$.","rse $A=D_\\perp^{-1}$. Applying it to $Dx=D(x-qx)$ yields\n\\[\n x-qx=A(Dx),\\qquad \\norm{x-qx}\\le\\norm A\\norm{Dx}.\n\\]\nHence condition 2 holds. If any $C\\ge0$ satisfies condition 2, then for every $r\\in\\ran D$ apply it to $u=Ar\\in K^\\perp$. Since $qu=0$, it gives $\\norm{Ar}\\le C\\norm r$. Taking the operator norm proves $\\norm A\\le C$. This establishes the least constant, not just a sufficient one. If $D=0$, both $K^\\perp$ and the range are zero spaces and $\\norm A=0$.","rp^{-1}$. Applying it to $Dx=D(x-qx)$ yields\n\\[\n x-qx=A(Dx),\\qquad \\norm{x-qx}\\le\\norm A\\norm{Dx}.\n\\]\nHence condition 2 holds. If any $C\\ge0$ satisfies condition 2, then for every $r\\in\\ran D$ apply it to $u=Ar\\in K^\\perp$. Since $qu=0$, it gives $\\norm{Ar}\\le C\\norm r$. Taking the operator norm proves $\\norm A\\le C$. This establishes the least constant, not just a sufficient one. If $D=0$, both $K^\\perp$ and the range are zero spaces and $\\norm A=0$.","\\[\n x-qx=A(Dx),\\qquad \\norm{x-qx}\\le\\norm A\\norm{Dx}.\n\\]\nHence condition 2 holds. If any $C\\ge0$ satisfies condition 2, then for every $r\\in\\ran D$ apply it to $u=Ar\\in K^\\perp$. Since $qu=0$, it gives $\\norm{Ar}\\le C\\norm r$. Taking the operator norm proves $\\norm A\\le C$. This establishes the least constant, not just a sufficient one. If $D=0$, both $K^\\perp$ and the range are zero spaces and $\\norm A=0$.","Now assume condition 2. To prove closedness, take an arbitrary sequence $r_n\\in\\ran D$ with $r_n\\to r$ in $R$. Choose $x_n$ with $Dx_n=r_n$ and put $u_n=x_n-qx_n$. Then $u_n\\in K^\\perp$ and $Du_n=r_n$. Applying the estimate to $u_n-u_m$ gives\n\\[\n \\norm{u_n-u_m}\\le C\\norm{r_n-r_m}.\n\\]\nThus $(u_n)$ is Cauchy in the complete space $K^\\perp$. Let its limit be $u$. Continuity gives $Du=\\lim_n r_n=r$, so $r\\in\\ran D$. In a normed space, a set containing the limits of all its convergent sequences is closed. This proves condition 1 without presupposing that the initially unbounded inverse is continuous.","assume condition 2. To prove closedness, take an arbitrary sequence $r_n\\in\\ran D$ with $r_n\\to r$ in $R$. Choose $x_n$ with $Dx_n=r_n$ and put $u_n=x_n-qx_n$. Then $u_n\\in K^\\perp$ and $Du_n=r_n$. Applying the estimate to $u_n-u_m$ gives\n\\[\n \\norm{u_n-u_m}\\le C\\norm{r_n-r_m}.\n\\]\nThus $(u_n)$ is Cauchy in the complete space $K^\\perp$. Let its limit be $u$. Continuity gives $Du=\\lim_n r_n=r$, so $r\\in\\ran D$. In a normed space, a set containing the limits of all its convergent sequences is closed. This proves condition 1 without presupposing that the initially unbounded inverse is continuous.","Condition 2 implies condition 3 by choosing $y=qx$. Conversely, if condition 3 supplies an exact $y$ for a given $x$, the nearest-point property of $qx$ gives\n\\[\n \\norm{x-qx}\\le\\norm{x-y}\\le C\\norm{Dx}.\n\\]\nThe same $C$ works for all $x$, proving condition 2. This last implication does not require the selected answers $y$ to depend linearly or continuously on $x$.\n\\end{proof}\n\\source{KS43}\nThe last equivalence is important. A nonlinear choice of exact answers cannot bypass the failure of a uniform residual bound. Failure of closed range is not merely a defect of one selected projection algorithm.","\\subsection{A small residual far from exactness}\nOn \\(\\ell^2(\\N)\\), index the standard basis from one and define \\(De_n=n^{-1}e_n\\). The map is bounded and injective, so its exact space is zero. Yet\n\\[\n \\norm{e_n}=1,\\qquad\\norm{De_n}=1/n\\longrightarrow0.\n\\]\nNo finite \\(C\\) can bound \\(\\dist(e_n,\\ker D)\\) by \\(C\\norm{De_n}\\). The range is not closed. For example finite truncations of \\((1/n)_{n\\ge1}\\) belong to the range and converge in \\(\\ell^2\\), but a preimage of the limit would be the constant sequence, which does not belong to \\(\\ell^2\\).","tness}\nOn \\(\\ell^2(\\N)\\), index the standard basis from one and define \\(De_n=n^{-1}e_n\\). The map is bounded and injective, so its exact space is zero. Yet\n\\[\n \\norm{e_n}=1,\\qquad\\norm{De_n}=1/n\\longrightarrow0.\n\\]\nNo finite \\(C\\) can bound \\(\\dist(e_n,\\ker D)\\) by \\(C\\norm{De_n}\\). The range is not closed. For example finite truncations of \\((1/n)_{n\\ge1}\\) belong to the range and converge in \\(\\ell^2\\), but a preimage of the limit would be the constant sequence, which does not belong to \\(\\ell^2\\).","p is bounded and injective, so its exact space is zero. Yet\n\\[\n \\norm{e_n}=1,\\qquad\\norm{De_n}=1/n\\longrightarrow0.\n\\]\nNo finite \\(C\\) can bound \\(\\dist(e_n,\\ker D)\\) by \\(C\\norm{De_n}\\). The range is not closed. For example finite truncations of \\((1/n)_{n\\ge1}\\) belong to the range and converge in \\(\\ell^2\\), but a preimage of the limit would be the constant sequence, which does not belong to \\(\\ell^2\\).","Hilbert space used in Section~\\ref{sec:diagonal-regularization}. In the namespace \\path{KomorebiSpace.DiagonalRegularization}, \\path{T_not_closed_range} applies the general obstruction theorem to actual unit coordinate vectors. The declaration \\path{w_mem_closure_not_range} proves that the displayed limiting datum belongs to the closure of the range but not to the range itself. Formal coordinate $j$ represents the printed coordinate $n=j+1$.",":diagonal-regularization}. In the namespace \\path{KomorebiSpace.DiagonalRegularization}, \\path{T_not_closed_range} applies the general obstruction theorem to actual unit coordinate vectors. The declaration \\path{w_mem_closure_not_range} proves that the displayed limiting datum belongs to the closure of the range but not to the range itself. Formal coordinate $j$ represents the printed coordinate $n=j+1$.","\\subsection{Stable repair with protected data}\nThe preceding orthogonal projection does not generally preserve a separately declared observable. To join the analytic theory to Section~\\ref{sec:protected}, let \\(p:E\\to S\\) be continuous linear into a normed space. Define\n\\[\n M=\\ker p,\\qquad B=D|_M:M\\to R,\\qquad H=M\\cap\\ker D.\n\\]\nThe closed subspace \\(M\\) inherits its Hilbert structure from \\(E\\). A candidate \\(x\\) is \\emph{feasible} for protected repair when an exact \\(y\\) with \\(p(y)=p(x)\\) exists. The algebraic criterion is still \\(Dx\\in\\ran B\\).","table repair with protected data}\nThe preceding orthogonal projection does not generally preserve a separately declared observable. To join the analytic theory to Section~\\ref{sec:protected}, let \\(p:E\\to S\\) be continuous linear into a normed space. Define\n\\[\n M=\\ker p,\\qquad B=D|_M:M\\to R,\\qquad H=M\\cap\\ker D.\n\\]\nThe closed subspace \\(M\\) inherits its Hilbert structure from \\(E\\). A candidate \\(x\\) is \\emph{feasible} for protected repair when an exact \\(y\\) with \\(p(y)=p(x)\\) exists. The algebraic criterion is still \\(Dx\\in\\ran B\\).","rojection does not generally preserve a separately declared observable. To join the analytic theory to Section~\\ref{sec:protected}, let \\(p:E\\to S\\) be continuous linear into a normed space. Define\n\\[\n M=\\ker p,\\qquad B=D|_M:M\\to R,\\qquad H=M\\cap\\ker D.\n\\]\nThe closed subspace \\(M\\) inherits its Hilbert structure from \\(E\\). A candidate \\(x\\) is \\emph{feasible} for protected repair when an exact \\(y\\) with \\(p(y)=p(x)\\) exists. The algebraic criterion is still \\(Dx\\in\\ran B\\).","oes not generally preserve a separately declared observable. To join the analytic theory to Section~\\ref{sec:protected}, let \\(p:E\\to S\\) be continuous linear into a normed space. Define\n\\[\n M=\\ker p,\\qquad B=D|_M:M\\to R,\\qquad H=M\\cap\\ker D.\n\\]\nThe closed subspace \\(M\\) inherits its Hilbert structure from \\(E\\). A candidate \\(x\\) is \\emph{feasible} for protected repair when an exact \\(y\\) with \\(p(y)=p(x)\\) exists. The algebraic criterion is still \\(Dx\\in\\ran B\\).","able. To join the analytic theory to Section~\\ref{sec:protected}, let \\(p:E\\to S\\) be continuous linear into a normed space. Define\n\\[\n M=\\ker p,\\qquad B=D|_M:M\\to R,\\qquad H=M\\cap\\ker D.\n\\]\nThe closed subspace \\(M\\) inherits its Hilbert structure from \\(E\\). A candidate \\(x\\) is \\emph{feasible} for protected repair when an exact \\(y\\) with \\(p(y)=p(x)\\) exists. The algebraic criterion is still \\(Dx\\in\\ran B\\).","\\begin{theorem}[Nearest feasible protected repair]\\label{KS56}\nEvery feasible \\(x\\) has a unique closest exact answer \\(y_*\\) preserving \\(p(x)\\). Its correction \\(h_*=x-y_*\\) is the unique vector in the orthogonal complement of \\(\\ker B\\) inside \\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, ev","\\begin{theorem}[Nearest feasible protected repair]\\label{KS56}\nEvery feasible \\(x\\) has a unique closest exact answer \\(y_*\\) preserving \\(p(x)\\). Its correction \\(h_*=x-y_*\\) is the unique vector in the orthogonal complement of \\(\\ker B\\) inside \\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces a","\\begin{theorem}[Nearest feasible protected repair]\\label{KS56}\nEvery feasible \\(x\\) has a unique closest exact answer \\(y_*\\) preserving \\(p(x)\\). Its correction \\(h_*=x-y_*\\) is the unique vector in the orthogonal complement of \\(\\ker B\\) inside \\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected an","\\begin{theorem}[Nearest feasible protected repair]\\label{KS56}\nEvery feasible \\(x\\) has a unique closest exact answer \\(y_*\\) preserving \\(p(x)\\). Its correction \\(h_*=x-y_*\\) is the unique vector in the orthogonal complement of \\(\\ker B\\) inside \\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer","\\begin{theorem}[Nearest feasible protected repair]\\label{KS56}\nEvery feasible \\(x\\) has a unique closest exact answer \\(y_*\\) preserving \\(p(x)\\). Its correction \\(h_*=x-y_*\\) is the unique vector in the orthogonal complement of \\(\\ker B\\) inside \\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correcti","\\begin{theorem}[Nearest feasible protected repair]\\label{KS56}\nEvery feasible \\(x\\) has a unique closest exact answer \\(y_*\\) preserving \\(p(x)\\). Its correction \\(h_*=x-y_*\\) is the unique vector in the orthogonal complement of \\(\\ker B\\) inside \\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{","\\begin{theorem}[Nearest feasible protected repair]\\label{KS56}\nEvery feasible \\(x\\) has a unique closest exact answer \\(y_*\\) preserving \\(p(x)\\). Its correction \\(h_*=x-y_*\\) is the unique vector in the orthogonal complement of \\(\\ker B\\) inside \\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker","\\begin{theorem}[Nearest feasible protected repair]\\label{KS56}\nEvery feasible \\(x\\) has a unique closest exact answer \\(y_*\\) preserving \\(p(x)\\). Its correction \\(h_*=x-y_*\\) is the unique vector in the orthogonal complement of \\(\\ker B\\) inside \\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine","\\begin{theorem}[Nearest feasible protected repair]\\label{KS56}\nEvery feasible \\(x\\) has a unique closest exact answer \\(y_*\\) preserving \\(p(x)\\). Its correction \\(h_*=x-y_*\\) is the unique vector in the orthogonal complement of \\(\\ker B\\) inside \\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ke","est feasible protected repair]\\label{KS56}\nEvery feasible \\(x\\) has a unique closest exact answer \\(y_*\\) preserving \\(p(x)\\). Its correction \\(h_*=x-y_*\\) is the unique vector in the orthogonal complement of \\(\\ker B\\) inside \\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of wh…","\\(M\\) satisfying \\(Bh_*=Dx\\). For every other feasible answer \\(y\\),\n\\[\n \\norm{x-y}^2=\\norm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of whether its image is closed.","orm{x-y_*}^2+\\norm{y_*-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of whether its image is closed.","-y}^2.\n\\]\nMoreover, the following conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of whether its image is closed.","wing conditions are equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of whether its image is closed.","equivalent:\n\\[\n \\ran B\\text{ is closed}\n \\quad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of whether its image is closed.","uad\\Longleftrightarrow\\quad\n \\exists C\\ge0\\ \\forall\\text{ feasible }x:\n \\norm{x-y_*}\\le C\\norm{Dx}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of whether its image is closed.","}.\n\\]\nWhen these conditions hold, the least nonnegative constant is the transverse inverse norm of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of whether its image is closed.","of \\(B\\), including the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of whether its image is closed.","the zero-range case.\n\\end{theorem}\n\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of whether its image is closed.","\\begin{proof}\nFeasibility gives some $h_0=x-y_0\\in M$ with $Bh_0=Dx$. Conversely, every $h\\in M$ satisfying $Bh=Dx$ produces an exact protected answer $y=x-h$. Thus the answer problem and the correction problem are equivalent, with correction set\n\\[\n \\{h\\in M:Bh=Dx\\}=h_0+\\ker B.\n\\]\nThis is a nonempty closed affine subset of the Hilbert space $M$. Closedness of $\\ker B$ follows from continuity of $B$, independently of whether its image is closed.","Let $k_0=P^M_{\\ker B}h_0$ and $h_*=h_0-k_0$. Then $Bh_*=Bh_0=Dx$ and $h_*\\perp\\ker B$ in $M$. If a different initial correction is chosen, it differs from $h_0$ by an element of $\\ker B$. Its orthogonal-complement component is therefore again $h_*$. This proves that the construction is independent of the feasibility witness.","Every other correction can be written uniquely as $h_*+k$ with $k\\in\\ker B$. Orthogonality gives\n\\[\n \\norm{h_*+k}^2=\\norm{h_*}^2+\\norm k^2.\n\\]\nWriting $y_*=x-h_*$ and $y=x-h_*-k$, we have $x-y=h_*+k$ and $y_*-y=k$. Substitution proves the stated Pythagorean identity. Its last term vanishes only when $y=y_*$, establishing the unique closest answer.","If $\\ran B$ is closed, Theorem~\\ref{KS43} applied on $M$ gives a bounded inverse of $B$ on $(\\ker B)^{\\perp_M}$. The vector $h_*$ is exactly that inverse applied to $Dx$, so its norm obeys the required bound.","For the converse, suppose the bound holds for every feasible raw candidate. Every $x\\in M$ is feasible, because $y=0$ has $Dy=0$ and $py=px=0$. Its closest protected exact answer is $P^M_{\\ker B}x$. Thus the assumed estimate reduces on $M$ to\n\\[\n \\norm{x-P^M_{\\ker B}x}\\le C\\norm{Bx}\\qquad(x\\in M).\n\\]\nTheorem~\\ref{KS43} forces $\\ran B$ to be closed. The same restriction also proves that no smaller nonnegative constant than the transverse inverse norm can work. If the range is zero, feasible candidates are already exact under permitted correction and the least correction is zero, so the optimal gain is zero.\n\\end{proof}\n\\source{KS56}\nThe relevant stability boundary is therefore \\(D(\\ker p)\\), not automatically \\(D(E)\\). Restricting permitted changes can turn a well-conditioned unc","For the converse, suppose the bound holds for every feasible raw candidate. Every $x\\in M$ is feasible, because $y=0$ has $Dy=0$ and $py=px=0$. Its closest protected exact answer is $P^M_{\\ker B}x$. Thus the assumed estimate reduces on $M$ to\n\\[\n \\norm{x-P^M_{\\ker B}x}\\le C\\norm{Bx}\\qquad(x\\in M).\n\\]\nTheorem~\\ref{KS43} forces $\\ran B$ to be closed. The same restriction also proves that no smaller nonnegative constant than the transverse inverse norm can work. If the range is zero, feasible candidates are already exact under permitted correction and the least correction is zero, so the optimal gain is zero.\n\\end{proof}\n\\source{KS56}\nThe relevant stability boundary is therefore \\(D(\\ker p)\\), not automatically \\(D(E)\\). Restricting permitted changes can turn a well-conditioned unconstrained problem into an","For the converse, suppose the bound holds for every feasible raw candidate. Every $x\\in M$ is feasible, because $y=0$ has $Dy=0$ and $py=px=0$. Its closest protected exact answer is $P^M_{\\ker B}x$. Thus the assumed estimate reduces on $M$ to\n\\[\n \\norm{x-P^M_{\\ker B}x}\\le C\\norm{Bx}\\qquad(x\\in M).\n\\]\nTheorem~\\ref{KS43} forces $\\ran B$ to be closed. The same restriction also proves that no smaller nonnegative constant than the transverse inverse norm can work. If the range is zero, feasible candidates are already exact under permitted correction and the least correction is zero, so the optimal gain is zero.\n\\end{proof}\n\\source{KS56}\nThe relevant stability boundary is therefore \\(D(\\ker p)\\), not automatically \\(D(E)\\). Restricting permitted changes can turn a well-conditioned unconstrained problem into an obstructed","For the converse, suppose the bound holds for every feasible raw candidate. Every $x\\in M$ is feasible, because $y=0$ has $Dy=0$ and $py=px=0$. Its closest protected exact answer is $P^M_{\\ker B}x$. Thus the assumed estimate reduces on $M$ to\n\\[\n \\norm{x-P^M_{\\ker B}x}\\le C\\norm{Bx}\\qquad(x\\in M).\n\\]\nTheorem~\\ref{KS43} forces $\\ran B$ to be closed. The same restriction also proves that no smaller nonnegative constant than the transverse inverse norm can work. If the range is zero, feasible candidates are already exact under permitted correction and the least correction is zero, so the optimal gain is zero.\n\\end{proof}\n\\source{KS56}\nThe relevant stability boundary is therefore \\(D(\\ker p)\\), not automatically \\(D(E)\\). Restricting permitted changes can turn a well-conditioned unconstrained problem into an obstructed or poorly cond","For the converse, suppose the bound holds for every feasible raw candidate. Every $x\\in M$ is feasible, because $y=0$ has $Dy=0$ and $py=px=0$. Its closest protected exact answer is $P^M_{\\ker B}x$. Thus the assumed estimate reduces on $M$ to\n\\[\n \\norm{x-P^M_{\\ker B}x}\\le C\\norm{Bx}\\qquad(x\\in M).\n\\]\nTheorem~\\ref{KS43} forces $\\ran B$ to be closed. The same restriction also proves that no smaller nonnegative constant than the transverse inverse norm can work. If the range is zero, feasible candidates are already exact under permitted correction and the least correction is zero, so the optimal gain is zero.\n\\end{proof}\n\\source{KS56}\nThe relevant stability boundary is therefore \\(D(\\ker p)\\), not automatically \\(D(E)\\). Restricting permitted changes can turn a well-conditioned unconstrained problem into an obstructed or poorly conditioned protected one. The nearest protected answer is","For the converse, suppose the bound holds for every feasible raw candidate. Every $x\\in M$ is feasible, because $y=0$ has $Dy=0$ and $py=px=0$. Its closest protected exact answer is $P^M_{\\ker B}x$. Thus the assumed estimate reduces on $M$ to\n\\[\n \\norm{x-P^M_{\\ker B}x}\\le C\\norm{Bx}\\qquad(x\\in M).\n\\]\nTheorem~\\ref{KS43} forces $\\ran B$ to be closed. The same restriction also proves that no smaller nonnegative constant than the transverse inverse norm can work. If the range is zero, feasible candidates are already exact under permitted correction and the least correction is zero, so the optimal gain is zero.\n\\end{proof}\n\\source{KS56}\nThe relevant stability boundary is therefore \\(D(\\ker p)\\), not automatically \\(D(E)\\). Restricting permitted changes can turn a well-conditioned unconstrained problem into an obstructed or poorly conditioned protected one. The nearest protected answer is generally not \\(P_{\\ker D}x\\).","ad(x\\in M).\n\\]\nTheorem~\\ref{KS43} forces $\\ran B$ to be closed. The same restriction also proves that no smaller nonnegative constant than the transverse inverse norm can work. If the range is zero, feasible candidates are already exact under permitted correction and the least correction is zero, so the optimal gain is zero.\n\\end{proof}\n\\source{KS56}\nThe relevant stability boundary is therefore \\(D(\\ker p)\\), not automatically \\(D(E)\\). Restricting permitted changes can turn a well-conditioned unconstrained problem into an obstructed or poorly conditioned protected one. The nearest protected answer is generally not \\(P_{\\ker D}x\\).","forces $\\ran B$ to be closed. The same restriction also proves that no smaller nonnegative constant than the transverse inverse norm can work. If the range is zero, feasible candidates are already exact under permitted correction and the least correction is zero, so the optimal gain is zero.\n\\end{proof}\n\\source{KS56}\nThe relevant stability boundary is therefore \\(D(\\ker p)\\), not automatically \\(D(E)\\). Restricting permitted changes can turn a well-conditioned unconstrained problem into an obstructed or poorly conditioned protected one. The nearest protected answer is generally not \\(P_{\\ker D}x\\).","ready exact under permitted correction and the least correction is zero, so the optimal gain is zero.\n\\end{proof}\n\\source{KS56}\nThe relevant stability boundary is therefore \\(D(\\ker p)\\), not automatically \\(D(E)\\). Restricting permitted changes can turn a well-conditioned unconstrained problem into an obstructed or poorly conditioned protected one. The nearest protected answer is generally not \\(P_{\\ker D}x\\).","For \\(D(x_1,x_2,x_3)=x_1-x_2\\), \\(p(x)=(x_1,x_3)\\), and \\(x=(2,0,5)\\), the ordinary orthogonal answer is \\((1,1,5)\\). The protected answer is \\((2,2,5)\\). The respective squared errors are two and four. Both have zero residual, but only the latter preserves the declared data. The theorem is formalized in \\code{ProtectedHilbertRepair.lean}, using the actual operator on \\(\\ker p\\). The displayed three-coordinate values are an illustrative calculation, not a separate universal algorithm.","\\chap{Holonomy repair and transport amplification}\\label{sec:holonomyrepair}\n\\subsection{An orthogonal anchor, not a global least squares answer}\nReturn to the graph model with a real Hilbert fiber. The root residual is\n\\[\n \\mathcal H z=(K_{ij}z-z)_{ij},\\qquad K=\\ker\\mathcal H.\n\\]\nLet \\(q=P_K\\), and define\n\\[\n Px=R_p(qx_o).\n\\]\nThis is always exact, fixes exact families, and is idempotent. It is the unique exact family preserving the feasible component \\(qx_o\\) of the anchor. Its orthogonality is on the anchor fiber. The vertex and edge norms are still maximum norms.","ec:holonomyrepair}\n\\subsection{An orthogonal anchor, not a global least squares answer}\nReturn to the graph model with a real Hilbert fiber. The root residual is\n\\[\n \\mathcal H z=(K_{ij}z-z)_{ij},\\qquad K=\\ker\\mathcal H.\n\\]\nLet \\(q=P_K\\), and define\n\\[\n Px=R_p(qx_o).\n\\]\nThis is always exact, fixes exact families, and is idempotent. It is the unique exact family preserving the feasible component \\(qx_o\\) of the anchor. Its orthogonality is on the anchor fiber. The vertex and edge norms are still maximum norms.","For example on a two-vertex identity connection, \\((0,2)\\) with root value zero is repaired to \\((0,0)\\). The constant family \\((1,1)\\) is closer in the maximum norm, but it does not preserve the anchor. The operator \\(P\\) is not being claimed to be a global nearest-point projection in that norm.","\\subsection{Every factor in the bound has a meaning}\nDefine the route gain\n\\[\n A_p=\\max_i\\gamma(\\overline{p_i}),\n\\]\nusing reverse routes so that the error is measured in each destination view. The path telescope gives\n\\[\n \\norm{x-R_p(x_o)}_\\infty\\le A_p\\norm{dx}_\\infty.\n\\]\nLet \\(N_p\\) be the residual normalization \\((N_pr)_{ij}=T_ir_{ij}\\). The exact identity \\(\\mathcal H=N_pdR_p\\) implies\n\\[\n \\norm{\\mathcal H(x_o)}_\\infty\n \\le\\norm{N_p}(\\norm d A_p+1)\\norm{dx}_\\infty.\n\\]","\\begin{theorem}[Amplified holonomy repair]\\label{KS44}\nIf \\(\\ran\\mathcal H\\) is closed and \\(\\kappa\\) is its transverse inverse gain, then\n\\begin{equation}\\label{eq:graphgain}\n \\norm{x-Px}_\\infty\\le\n \\left[A_p+\\norm{R_p}\\kappa\\norm{N_p}(\\norm d A_p+1)\\right]\n \\norm{dx}_\\infty.\n\\end{equation}\nThe range condition is necessary for the existence of any uniform bound of distance to an exact family by \\(\\norm{dx}_\\infty\\). It is automatic for a fixed finite dimensional fiber, but not uniformly quantitative across all connections.\n\\end{theorem}\n\\begin{proof}\nFirst quantify the residual seen at the root. The identity \\(\\mathcal H=N_pdR_p\\), applied to \\(x_o\\), gives\n\\[\n \\begin{aligned}\n \\norm{\\mathcal H(x_o)}_\\infty\n &=\\norm{N_p\\bigl(d(R_px_o-x)+dx\\bigr)}_\\infty","begin{theorem}[Amplified holonomy repair]\\label{KS44}\nIf \\(\\ran\\mathcal H\\) is closed and \\(\\kappa\\) is its transverse inverse gain, then\n\\begin{equation}\\label{eq:graphgain}\n \\norm{x-Px}_\\infty\\le\n \\left[A_p+\\norm{R_p}\\kappa\\norm{N_p}(\\norm d A_p+1)\\right]\n \\norm{dx}_\\infty.\n\\end{equation}\nThe range condition is necessary for the existence of any uniform bound of distance to an exact family by \\(\\norm{dx}_\\infty\\). It is automatic for a fixed finite dimensional fiber, but not uniformly quantitative across all connections.\n\\end{theorem}\n\\begin{proof}\nFirst quantify the residual seen at the root. The identity \\(\\mathcal H=N_pdR_p\\), applied to \\(x_o\\), gives\n\\[\n \\begin{aligned}\n \\norm{\\mathcal H(x_o)}_\\infty\n &=\\norm{N_p\\bigl(d(R_px_o-x)+dx\\bigr)}_\\infty","}\\kappa\\norm{N_p}(\\norm d A_p+1)\\right]\n \\norm{dx}_\\infty.\n\\end{equation}\nThe range condition is necessary for the existence of any uniform bound of distance to an exact family by \\(\\norm{dx}_\\infty\\). It is automatic for a fixed finite dimensional fiber, but not uniformly quantitative across all connections.\n\\end{theorem}\n\\begin{proof}\nFirst quantify the residual seen at the root. The identity \\(\\mathcal H=N_pdR_p\\), applied to \\(x_o\\), gives\n\\[\n \\begin{aligned}\n \\norm{\\mathcal H(x_o)}_\\infty\n &=\\norm{N_p\\bigl(d(R_px_o-x)+dx\\bigr)}_\\infty","ght]\n \\norm{dx}_\\infty.\n\\end{equation}\nThe range condition is necessary for the existence of any uniform bound of distance to an exact family by \\(\\norm{dx}_\\infty\\). It is automatic for a fixed finite dimensional fiber, but not uniformly quantitative across all connections.\n\\end{theorem}\n\\begin{proof}\nFirst quantify the residual seen at the root. The identity \\(\\mathcal H=N_pdR_p\\), applied to \\(x_o\\), gives\n\\[\n \\begin{aligned}\n \\norm{\\mathcal H(x_o)}_\\infty\n &=\\norm{N_p\\bigl(d(R_px_o-x)+dx\\bigr)}_\\infty","_\\infty.\n\\end{equation}\nThe range condition is necessary for the existence of any uniform bound of distance to an exact family by \\(\\norm{dx}_\\infty\\). It is automatic for a fixed finite dimensional fiber, but not uniformly quantitative across all connections.\n\\end{theorem}\n\\begin{proof}\nFirst quantify the residual seen at the root. The identity \\(\\mathcal H=N_pdR_p\\), applied to \\(x_o\\), gives\n\\[\n \\begin{aligned}\n \\norm{\\mathcal H(x_o)}_\\infty\n &=\\norm{N_p\\bigl(d(R_px_o-x)+dx\\bigr)}_\\infty\\\\\n &\\le\\norm{N_p}\\bigl(\\norm d\\norm{R_px_o-x}_\\infty+\\norm{dx}_\\infty\\bigr)\\\\\n &\\le\\norm{N_p}(\\norm d A_p+1)\\norm{dx}_\\infty.\n \\end{aligned}\n\\]\nClosed range of \\(\\mathcal H\\) lets Theorem~\\ref{KS43} control the anchor correction by\n\\(\\norm{x_o-qx_o}\\le\\kappa\\norm{\\mathcal H(x_o)}_\\infty\\). Insert the uncorrected reconstruction between \\(x\\) and \\(Px=R_pqx_o\\):\n\\[\n \\norm{x-Px}_\\infty\n \\le\\norm{x-R_px_o}_\\infty+\\norm{R_p}\\norm{x_o-qx_o}.\n\\]\nSubstituting the two preceding bounds yields \\eqref{eq:graphgain}. The finite maximum-norm residual product is a Banach space because its fiber is Hilbert, so th…","&\\le\\norm{N_p}(\\norm d A_p+1)\\norm{dx}_\\infty.\n \\end{aligned}\n\\]\nClosed range of \\(\\mathcal H\\) lets Theorem~\\ref{KS43} control the anchor correction by\n\\(\\norm{x_o-qx_o}\\le\\kappa\\norm{\\mathcal H(x_o)}_\\infty\\). Insert the uncorrected reconstruction between \\(x\\) and \\(Px=R_pqx_o\\):\n\\[\n \\norm{x-Px}_\\infty\n \\le\\norm{x-R_px_o}_\\infty+\\norm{R_p}\\norm{x_o-qx_o}.\n\\]\nSubstituting the two preceding bounds yields \\eqref{eq:graphgain}. The finite maximum-norm residual product is a Banach space because its fiber is Hilbert, so the target completeness needed in Theorem~\\ref{KS43} is available.","For necessity, suppose a constant \\(C\\) satisfies\n\\(\\dist(w,\\ker d)\\le C\\norm{dw}_\\infty\\) for every vertex family \\(w\\). Fix \\(z\\in K^\\perp\\) and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mat","For necessity, suppose a constant \\(C\\) satisfies\n\\(\\dist(w,\\ker d)\\le C\\norm{dw}_\\infty\\) for every vertex family \\(w\\). Fix \\(z\\in K^\\perp\\) and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the","For necessity, suppose a constant \\(C\\) satisfies\n\\(\\dist(w,\\ker d)\\le C\\norm{dw}_\\infty\\) for every vertex family \\(w\\). Fix \\(z\\in K^\\perp\\) and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a unifo","For necessity, suppose a constant \\(C\\) satisfies\n\\(\\dist(w,\\ker d)\\le C\\norm{dw}_\\infty\\) for every vertex family \\(w\\). Fix \\(z\\in K^\\perp\\) and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of t","For necessity, suppose a constant \\(C\\) satisfies\n\\(\\dist(w,\\ker d)\\le C\\norm{dw}_\\infty\\) for every vertex family \\(w\\). Fix \\(z\\in K^\\perp\\) and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\","For necessity, suppose a constant \\(C\\) satisfies\n\\(\\dist(w,\\ker d)\\le C\\norm{dw}_\\infty\\) for every vertex family \\(w\\). Fix \\(z\\in K^\\perp\\) and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(","For necessity, suppose a constant \\(C\\) satisfies\n\\(\\dist(w,\\ker d)\\le C\\norm{dw}_\\infty\\) for every vertex family \\(w\\). Fix \\(z\\in K^\\perp\\) and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\)","For necessity, suppose a constant \\(C\\) satisfies\n\\(\\dist(w,\\ker d)\\le C\\norm{dw}_\\infty\\) for every vertex family \\(w\\). Fix \\(z\\in K^\\perp\\) and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a unif","For necessity, suppose a constant \\(C\\) satisfies\n\\(\\dist(w,\\ker d)\\le C\\norm{dw}_\\infty\\) for every vertex family \\(w\\). Fix \\(z\\in K^\\perp\\) and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain","\\(\\dist(w,\\ker d)\\le C\\norm{dw}_\\infty\\) for every vertex family \\(w\\). Fix \\(z\\in K^\\perp\\) and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can…",") and take \\(w=R_pz\\). Every exact family \\(e\\) has \\(e_o\\in K\\), and root evaluation has norm at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\)…","at most one. Therefore\n\\[\n \\norm z=\\dist(z,K)\\le\\norm{z-e_o}\\le\\norm{R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\…","R_pz-e}_\\infty.\n\\]\nTaking the infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\infty.\n\\]\nThe exact repair is route independent for a fixed root and …","e infimum over \\(e\\in\\ker d\\) is legitimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\infty.\n\\]\nThe exact repair is route independent for a fixed root and inner product. A chosen uppe…","itimate without assuming that this global nearest point is attained. It gives\n\\[\n \\norm z\\le C\\norm{dR_pz}_\\infty\n \\le C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\infty.\n\\]\nThe exact repair is route independent for a fixed root and inner product. A chosen upper bound involving route gains need n…","e C\\norm{N_p^{-1}}\\norm{\\mathcal Hz}_\\infty.\n\\]\nHere \\(N_p\\) is a continuous linear equivalence on the finite edge product because every component \\(T_i\\) is one. For a general root vector, remove its projection onto \\(K\\); this leaves the same \\(\\mathcal H\\)-image and puts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\infty.\n\\]\nThe exact repair is route independent for a fixed root and inner product. A chosen upper bound involving route gains need not have the same numerical value for different routes.","uts it in \\(K^\\perp\\). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\infty.\n\\]\nThe exact repair is route independent for a fixed root and inner product. A chosen upper bound involving route gains need not have the same numerical value for different routes.","). Thus the displayed estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\infty.\n\\]\nThe exact repair is route independent for a fixed root and inner product. A chosen upper bound involving route gains need not have the same numerical value for different routes.","d estimate is a uniform residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\infty.\n\\]\nThe exact repair is route independent for a fixed root and inner product. A chosen upper bound involving route gains need not have the same numerical value for different routes.","rm residual bound of the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\infty.\n\\]\nThe exact repair is route independent for a fixed root and inner product. A chosen upper bound involving route gains need not have the same numerical value for different routes.","the kind in Theorem~\\ref{KS43}, which forces \\(\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\infty.\n\\]\nThe exact repair is route independent for a fixed root and inner product. A chosen upper bound involving route gains need not have the same numerical value for different routes.","\\ran\\mathcal H\\) to be closed. A finite-dimensional fiber makes that image finite dimensional and hence closed. It does not give a uniform numerical value of the inverse gain across different connections.\n\\end{proof}\n\\source{KS44}\nFor isometric transports, one can use \\(\\norm{R_p}\\le1\\), \\(\\norm{N_p}\\le1\\), \\(\\norm d\\le2\\), and \\(A_p\\le r\\), where \\(r\\) is the route radius. The bound reduces to\n\\[\n \\norm{x-Px}_\\infty\\le[r+\\kappa(2r+1)]\\norm{dx}_\\infty.\n\\]\nThe exact repair is route independent for a fixed root and inner product. A chosen upper bound involving route gains need not have the same numerical value for different routes.","\\subsection{Flatness does not remove amplification}\nConsider a scalar chain with \\(n+1\\) vertices, where \\(n\\ge1\\), and coordinates \\(e_i(z)=2^iz\\). Set \\(\\tau_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Bo","\\subsection{Flatness does not remove amplification}\nConsider a scalar chain with \\(n+1\\) vertices, where \\(n\\ge1\\), and coordinates \\(e_i(z)=2^iz\\). Set \\(\\tau_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every e","\\subsection{Flatness does not remove amplification}\nConsider a scalar chain with \\(n+1\\) vertices, where \\(n\\ge1\\), and coordinates \\(e_i(z)=2^iz\\). Set \\(\\tau_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual","\\subsection{Flatness does not remove amplification}\nConsider a scalar chain with \\(n+1\\) vertices, where \\(n\\ge1\\), and coordinates \\(e_i(z)=2^iz\\). Set \\(\\tau_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual norm one. Root-preserving repa","\\subsection{Flatness does not remove amplification}\nConsider a scalar chain with \\(n+1\\) vertices, where \\(n\\ge1\\), and coordinates \\(e_i(z)=2^iz\\). Set \\(\\tau_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual norm one. Root-preserving repair has error \\(2^n-1\\), which exceeds the route-length estimate once \\(n\\ge2\\). The vertical scale is logarithmic.}{fig:amplification}","bsection{Flatness does not remove amplification}\nConsider a scalar chain with \\(n+1\\) vertices, where \\(n\\ge1\\), and coordinates \\(e_i(z)=2^iz\\). Set \\(\\tau_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual norm one. Root-preserving repair has error \\(2^n-1\\), which exceeds the route-length estimate once \\(n\\ge2\\). The vertical scale is logarithmic.}{fig:amplification}","Flatness does not remove amplification}\nConsider a scalar chain with \\(n+1\\) vertices, where \\(n\\ge1\\), and coordinates \\(e_i(z)=2^iz\\). Set \\(\\tau_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual norm one. Root-preserving repair has error \\(2^n-1\\), which exceeds the route-length estimate once \\(n\\ge2\\). The vertical scale is logarithmic.}{fig:amplification}","tion}\nConsider a scalar chain with \\(n+1\\) vertices, where \\(n\\ge1\\), and coordinates \\(e_i(z)=2^iz\\). Set \\(\\tau_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual norm one. Root-preserving repair has error \\(2^n-1\\), which exceeds the route-length estimate once \\(n\\ge2\\). The vertical scale is logarithmic.}{fig:amplification}","r a scalar chain with \\(n+1\\) vertices, where \\(n\\ge1\\), and coordinates \\(e_i(z)=2^iz\\). Set \\(\\tau_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual norm one. Root-preserving repair has error \\(2^n-1\\), which exceeds the route-length estimate once \\(n\\ge2\\). The vertical scale is logarithmic.}{fig:amplification}","rdinates \\(e_i(z)=2^iz\\). Set \\(\\tau_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual norm one. Root-preserving repair has error \\(2^n-1\\), which exceeds the route-length estimate once \\(n\\ge2\\). The vertical scale is logarithmic.}{fig:amplification}","u_{ij}=e_ie_j^{-1}\\). This connection is flat, but reverse transport along one edge multiplies norms by two. Let\n\\[\n x_i=8\\cdot2^i-1,\\qquad y_i=7\\cdot2^i.\n\\]\nThe family \\(y\\) is exact and preserves \\(x_o=7\\). On a forward edge the residual of \\(x\\) is \\(1/2\\), and on its reverse it is \\(-1\\). Thus\n\\[\n \\norm{dx}_\\infty=1,\\qquad\n \\norm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual norm one. Root-preserving repair has error \\(2^n-1\\), which exceeds the route-length estimate once \\(n\\ge2\\). The vertical scale is logarithmic.}{fig:amplification}","rm{x-Px}_\\infty=2^n-1.\n\\]\nFor \\(n\\ge2\\), a route length bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual norm one. Root-preserving repair has error \\(2^n-1\\), which exceeds the route-length estimate once \\(n\\ge2\\). The vertical scale is logarithmic.}{fig:amplification}","bound of \\(n\\) fails. This is amplification with trivial holonomy, distinct from poor conditioning of the fixed space of a nearly trivial loop.\n\\source{KS45}\n\\plate{17_amplification}{A flat connection can still amplify a fixed residual. Both orientations of every edge are measured, giving residual norm one. Root-preserving repair has error \\(2^n-1\\), which exceeds the route-length estimate once \\(n\\ge2\\). The vertical scale is logarithmic.}{fig:amplification}","\\subsection{Small holonomy can still be poorly conditioned}\nIn \\(\\R^2\\), the rational rotation\n\\[\n U_t=\\frac1{1+t^2}\n \\begin{pmatrix}1-t^2&-2t","\\subsection{Small holonomy can still be poorly conditioned}\nIn \\(\\R^2\\), the rational rotation\n\\[\n U_t=\\frac1{1+t^2}\n \\begin{pmatrix}1-t^2&-2t\\\\2t&1-t^2\\end{pmatrix}\n\\]\nsatisfies\n\\[\n \\norm{(U_t-I)z}^2=\\frac{4t^2}{1+t^2}\\norm z^2.\n\\]\nFor \\(t\\ne0\\) its fixed space is zero and its residual gain is \\(\\sqrt{1+t^2}/(2|t|)\\), which diverges as \\(t\\to0\\). At \\(t=0\\), the entire fiber is fixed. Therefore small loop discrepancy does not supply a uniform stable realization estimate. The fixed directions and their conditioning must be analyzed together.","2t&1-t^2\\end{pmatrix}\n\\]\nsatisfies\n\\[\n \\norm{(U_t-I)z}^2=\\frac{4t^2}{1+t^2}\\norm z^2.\n\\]\nFor \\(t\\ne0\\) its fixed space is zero and its residual gain is \\(\\sqrt{1+t^2}/(2|t|)\\), which diverges as \\(t\\to0\\). At \\(t=0\\), the entire fiber is fixed. Therefore small loop discrepancy does not supply a uniform stable realization estimate. The fixed directions and their conditioning must be analyzed together.","\\chap{Quantitative descent in a common Hilbert model}\\label{sec:quantitative}\n\\subsection{Local exactness and matching are separate operators}\nLet \\(E\\) be a real Hilbert space of raw local candidates. Let \\(D:E\\to R\\) and \\(C:E\\to Y\\) be continuous linear maps into Banach spaces, measuring local equation defects and overlap mismatch, respectively. Put\n\\[\n L=\\ker D,\\qquad B=C|_L,\n \\qquad M=\\ker D\\cap\\ker C.\n\\]\nThe exact matched families form \\(M\\). First project to \\(L\\), then project inside \\(L\\) to \\(\\ker B\\). Denote the resulting map by \\(P\\).","odel}\\label{sec:quantitative}\n\\subsection{Local exactness and matching are separate operators}\nLet \\(E\\) be a real Hilbert space of raw local candidates. Let \\(D:E\\to R\\) and \\(C:E\\to Y\\) be continuous linear maps into Banach spaces, measuring local equation defects and overlap mismatch, respectively. Put\n\\[\n L=\\ker D,\\qquad B=C|_L,\n \\qquad M=\\ker D\\cap\\ker C.\n\\]\nThe exact matched families form \\(M\\). First project to \\(L\\), then project inside \\(L\\) to \\(\\ker B\\). Denote the resulting map by \\(P\\).","tive}\n\\subsection{Local exactness and matching are separate operators}\nLet \\(E\\) be a real Hilbert space of raw local candidates. Let \\(D:E\\to R\\) and \\(C:E\\to Y\\) be continuous linear maps into Banach spaces, measuring local equation defects and overlap mismatch, respectively. Put\n\\[\n L=\\ker D,\\qquad B=C|_L,\n \\qquad M=\\ker D\\cap\\ker C.\n\\]\nThe exact matched families form \\(M\\). First project to \\(L\\), then project inside \\(L\\) to \\(\\ker B\\). Denote the resulting map by \\(P\\).","ctness and matching are separate operators}\nLet \\(E\\) be a real Hilbert space of raw local candidates. Let \\(D:E\\to R\\) and \\(C:E\\to Y\\) be continuous linear maps into Banach spaces, measuring local equation defects and overlap mismatch, respectively. Put\n\\[\n L=\\ker D,\\qquad B=C|_L,\n \\qquad M=\\ker D\\cap\\ker C.\n\\]\nThe exact matched families form \\(M\\). First project to \\(L\\), then project inside \\(L\\) to \\(\\ker B\\). Denote the resulting map by \\(P\\).","d matching are separate operators}\nLet \\(E\\) be a real Hilbert space of raw local candidates. Let \\(D:E\\to R\\) and \\(C:E\\to Y\\) be continuous linear maps into Banach spaces, measuring local equation defects and overlap mismatch, respectively. Put\n\\[\n L=\\ker D,\\qquad B=C|_L,\n \\qquad M=\\ker D\\cap\\ker C.\n\\]\nThe exact matched families form \\(M\\). First project to \\(L\\), then project inside \\(L\\) to \\(\\ker B\\). Denote the resulting map by \\(P\\).","(E\\) be a real Hilbert space of raw local candidates. Let \\(D:E\\to R\\) and \\(C:E\\to Y\\) be continuous linear maps into Banach spaces, measuring local equation defects and overlap mismatch, respectively. Put\n\\[\n L=\\ker D,\\qquad B=C|_L,\n \\qquad M=\\ker D\\cap\\ker C.\n\\]\nThe exact matched families form \\(M\\). First project to \\(L\\), then project inside \\(L\\) to \\(\\ker B\\). Denote the resulting map by \\(P\\).","\\begin{theorem}[The two-stage repair is the joint projection]\\label{KS46}\nThe operator \\(P\\) is the orthogonal projection onto \\(M\\), equivalently onto the kernel of \\(\\Delta=(D,C)\\). If \\(u=x-P_Lx\\) and \\(v=P_Lx-Px\\), then\n\\[\n u\\perp v,\\qquad\\norm{x-Px}^2=\\norm u^2+\\norm v^2.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe subspace \\(L=\\ker D\\) is closed, hence Hilbert with the inherited inner product. The restriction \\(B=C|_L\\) is continuous, so its kernel is a closed subspace of \\(L\\). Under the inclusion into \\(E\\), this kernel is exactly \\(M=L\\cap\\ker C\\). Thus both projections used to define \\(P\\) exist.","em}[The two-stage repair is the joint projection]\\label{KS46}\nThe operator \\(P\\) is the orthogonal projection onto \\(M\\), equivalently onto the kernel of \\(\\Delta=(D,C)\\). If \\(u=x-P_Lx\\) and \\(v=P_Lx-Px\\), then\n\\[\n u\\perp v,\\qquad\\norm{x-Px}^2=\\norm u^2+\\norm v^2.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe subspace \\(L=\\ker D\\) is closed, hence Hilbert with the inherited inner product. The restriction \\(B=C|_L\\) is continuous, so its kernel is a closed subspace of \\(L\\). Under the inclusion into \\(E\\), this kernel is exactly \\(M=L\\cap\\ker C\\). Thus both projections used to define \\(P\\) exist.","e operator \\(P\\) is the orthogonal projection onto \\(M\\), equivalently onto the kernel of \\(\\Delta=(D,C)\\). If \\(u=x-P_Lx\\) and \\(v=P_Lx-Px\\), then\n\\[\n u\\perp v,\\qquad\\norm{x-Px}^2=\\norm u^2+\\norm v^2.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe subspace \\(L=\\ker D\\) is closed, hence Hilbert with the inherited inner product. The restriction \\(B=C|_L\\) is continuous, so its kernel is a closed subspace of \\(L\\). Under the inclusion into \\(E\\), this kernel is exactly \\(M=L\\cap\\ker C\\). Thus both projections used to define \\(P\\) exist.","al projection onto \\(M\\), equivalently onto the kernel of \\(\\Delta=(D,C)\\). If \\(u=x-P_Lx\\) and \\(v=P_Lx-Px\\), then\n\\[\n u\\perp v,\\qquad\\norm{x-Px}^2=\\norm u^2+\\norm v^2.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe subspace \\(L=\\ker D\\) is closed, hence Hilbert with the inherited inner product. The restriction \\(B=C|_L\\) is continuous, so its kernel is a closed subspace of \\(L\\). Under the inclusion into \\(E\\), this kernel is exactly \\(M=L\\cap\\ker C\\). Thus both projections used to define \\(P\\) exist.","valently onto the kernel of \\(\\Delta=(D,C)\\). If \\(u=x-P_Lx\\) and \\(v=P_Lx-Px\\), then\n\\[\n u\\perp v,\\qquad\\norm{x-Px}^2=\\norm u^2+\\norm v^2.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe subspace \\(L=\\ker D\\) is closed, hence Hilbert with the inherited inner product. The restriction \\(B=C|_L\\) is continuous, so its kernel is a closed subspace of \\(L\\). Under the inclusion into \\(E\\), this kernel is exactly \\(M=L\\cap\\ker C\\). Thus both projections used to define \\(P\\) exist.","_Lx\\) and \\(v=P_Lx-Px\\), then\n\\[\n u\\perp v,\\qquad\\norm{x-Px}^2=\\norm u^2+\\norm v^2.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe subspace \\(L=\\ker D\\) is closed, hence Hilbert with the inherited inner product. The restriction \\(B=C|_L\\) is continuous, so its kernel is a closed subspace of \\(L\\). Under the inclusion into \\(E\\), this kernel is exactly \\(M=L\\cap\\ker C\\). Thus both projections used to define \\(P\\) exist.","For an input \\(x\\), write \\(\\ell=P_Lx\\) and let \\(Px\\) be the projection of \\(\\ell\\) onto \\(\\ker B\\) inside \\(L\\). Then \\(Px\\in M\\), \\(u=x-\\ell\\in L^\\perp\\), and \\(v=\\ell-Px\\in L\\) is orthogonal to \\(M\\). Since \\(M\\subseteq L\\), the vector \\(u\\) is also orthogonal to \\(M\\). Their sum \\(x-Px=u+v\\) is therefore orthogonal to \\(M\\). A vector in \\(M\\) with that error is the unique orthogonal projection of \\(x\\), so \\(P=P_M\\).","input \\(x\\), write \\(\\ell=P_Lx\\) and let \\(Px\\) be the projection of \\(\\ell\\) onto \\(\\ker B\\) inside \\(L\\). Then \\(Px\\in M\\), \\(u=x-\\ell\\in L^\\perp\\), and \\(v=\\ell-Px\\in L\\) is orthogonal to \\(M\\). Since \\(M\\subseteq L\\), the vector \\(u\\) is also orthogonal to \\(M\\). Their sum \\(x-Px=u+v\\) is therefore orthogonal to \\(M\\). A vector in \\(M\\) with that error is the unique orthogonal projection of \\(x\\), so \\(P=P_M\\).","), write \\(\\ell=P_Lx\\) and let \\(Px\\) be the projection of \\(\\ell\\) onto \\(\\ker B\\) inside \\(L\\). Then \\(Px\\in M\\), \\(u=x-\\ell\\in L^\\perp\\), and \\(v=\\ell-Px\\in L\\) is orthogonal to \\(M\\). Since \\(M\\subseteq L\\), the vector \\(u\\) is also orthogonal to \\(M\\). Their sum \\(x-Px=u+v\\) is therefore orthogonal to \\(M\\). A vector in \\(M\\) with that error is the unique orthogonal projection of \\(x\\), so \\(P=P_M\\).","The two errors themselves are orthogonal because \\(u\\in L^\\perp\\) and \\(v\\in L\\). Expanding the squared norm of their sum gives \\(\\norm{x-Px}^2=\\norm u^2+\\norm v^2\\). Finally \\(\\ker(D,C)=\\ker D\\cap\\ker C=M\\), so this is also the projection onto the kernel of the joint defect. The proof uses a second projection \\emph{inside} \\(L\\). Replacing it by the ambient projection onto \\(\\ker C\\) would invalidate the membership step and need not preserve local exactness.\n\\end{proof}\n\\source{KS46}\nThis identity does not require the local and overlap operators to commute. The second projection is taken in the already exact subspace, not in the whole raw space without restriction.","For a concrete failure of the unrestricted procedure, take \\(L=\\{(a,0)\\}\\) and \\(\\ker C=\\{(a,a)\\}\\) in \\(\\R^2\\). Their intersection is zero. Starting from \\((1,0)\\), projection to \\(L\\) does nothing, while subsequent projection to \\(\\ker C\\) gives \\((1/2,1/2)\\), which is no longer in \\(L\\). The restricted second stage instead gives the correct joint answer zero.","\\subsection{A stable repair requires a stable matching problem}\nSuppose \\(\\ran D\\) and \\(\\ran B\\) are closed, with residual gains \\(a,b\\). Then\n\\[\n \\norm u\\le a\\norm{Dx},\\qquad\n \\norm v\\le b\\norm{CP_Lx}\n \\le b(\\norm{Cx}+\\norm C\\,a\\norm{Dx}).\n\\]\nTherefore\n\\begin{equation}\\label{eq:jointgain}\n \\norm{x-Px}\\le\n a(1+b\\norm C)\\norm{Dx}+b\\norm{Cx}.\n\\end{equation}\nThe Pythagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while i","\\subsection{A stable repair requires a stable matching problem}\nSuppose \\(\\ran D\\) and \\(\\ran B\\) are closed, with residual gains \\(a,b\\). Then\n\\[\n \\norm u\\le a\\norm{Dx},\\qquad\n \\norm v\\le b\\norm{CP_Lx}\n \\le b(\\norm{Cx}+\\norm C\\,a\\norm{Dx}).\n\\]\nTherefore\n\\begin{equation}\\label{eq:jointgain}\n \\norm{x-Px}\\le\n a(1+b\\norm C)\\norm{Dx}+b\\norm{Cx}.\n\\end{equation}\nThe Pythagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-erro","\\subsection{A stable repair requires a stable matching problem}\nSuppose \\(\\ran D\\) and \\(\\ran B\\) are closed, with residual gains \\(a,b\\). Then\n\\[\n \\norm u\\le a\\norm{Dx},\\qquad\n \\norm v\\le b\\norm{CP_Lx}\n \\le b(\\norm{Cx}+\\norm C\\,a\\norm{Dx}).\n\\]\nTherefore\n\\begin{equation}\\label{eq:jointgain}\n \\norm{x-Px}\\le\n a(1+b\\norm C)\\norm{Dx}+b\\norm{Cx}.\n\\end{equation}\nThe Pythagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither i","\\subsection{A stable repair requires a stable matching problem}\nSuppose \\(\\ran D\\) and \\(\\ran B\\) are closed, with residual gains \\(a,b\\). Then\n\\[\n \\norm u\\le a\\norm{Dx},\\qquad\n \\norm v\\le b\\norm{CP_Lx}\n \\le b(\\norm{Cx}+\\norm C\\,a\\norm{Dx}).\n\\]\nTherefore\n\\begin{equation}\\label{eq:jointgain}\n \\norm{x-Px}\\le\n a(1+b\\norm C)\\norm{Dx}+b\\norm{Cx}.\n\\end{equation}\nThe Pythagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound o","\\subsection{A stable repair requires a stable matching problem}\nSuppose \\(\\ran D\\) and \\(\\ran B\\) are closed, with residual gains \\(a,b\\). Then\n\\[\n \\norm u\\le a\\norm{Dx},\\qquad\n \\norm v\\le b\\norm{CP_Lx}\n \\le b(\\norm{Cx}+\\norm C\\,a\\norm{Dx}).\n\\]\nTherefore\n\\begin{equation}\\label{eq:jointgain}\n \\norm{x-Px}\\le\n a(1+b\\norm C)\\norm{Dx}+b\\norm{Cx}.\n\\end{equation}\nThe Pythagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).","stable repair requires a stable matching problem}\nSuppose \\(\\ran D\\) and \\(\\ran B\\) are closed, with residual gains \\(a,b\\). Then\n\\[\n \\norm u\\le a\\norm{Dx},\\qquad\n \\norm v\\le b\\norm{CP_Lx}\n \\le b(\\norm{Cx}+\\norm C\\,a\\norm{Dx}).\n\\]\nTherefore\n\\begin{equation}\\label{eq:jointgain}\n \\norm{x-Px}\\le\n a(1+b\\norm C)\\norm{Dx}+b\\norm{Cx}.\n\\end{equation}\nThe Pythagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).","gains \\(a,b\\). Then\n\\[\n \\norm u\\le a\\norm{Dx},\\qquad\n \\norm v\\le b\\norm{CP_Lx}\n \\le b(\\norm{Cx}+\\norm C\\,a\\norm{Dx}).\n\\]\nTherefore\n\\begin{equation}\\label{eq:jointgain}\n \\norm{x-Px}\\le\n a(1+b\\norm C)\\norm{Dx}+b\\norm{Cx}.\n\\end{equation}\nThe Pythagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).","uad\n \\norm v\\le b\\norm{CP_Lx}\n \\le b(\\norm{Cx}+\\norm C\\,a\\norm{Dx}).\n\\]\nTherefore\n\\begin{equation}\\label{eq:jointgain}\n \\norm{x-Px}\\le\n a(1+b\\norm C)\\norm{Dx}+b\\norm{Cx}.\n\\end{equation}\nThe Pythagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).","ion}\\label{eq:jointgain}\n \\norm{x-Px}\\le\n a(1+b\\norm C)\\norm{Dx}+b\\norm{Cx}.\n\\end{equation}\nThe Pythagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).","a(1+b\\norm C)\\norm{Dx}+b\\norm{Cx}.\n\\end{equation}\nThe Pythagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).","hagorean identity gives the sharper estimate\n\\[\n \\norm{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).","m{x-Px}^2\\le\n a^2\\norm{Dx}^2+b^2\\bigl(\\norm{Cx}+a\\norm C\\norm{Dx}\\bigr)^2.\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).",".\n\\]\nFor the maximum norm on \\(R\\times Y\\), two sufficient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).","ient gains are therefore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).","fore\n\\[\n G_2=\\sqrt{a^2+b^2(1+a\\norm C)^2}\n \\le G_1=a(1+b\\norm C)+b.\n\\]\nThe inequality follows from \\(\\sqrt{u^2+v^2}\\le u+v\\) for nonnegative \\(u,v\\). The formal interface supplies the sum-type gain \\(G_1\\), while its squared-error theorem also yields \\(G_2\\). Neither is claimed to be the least joint gain. In either bound, \\(\\norm C\\) may be replaced by any certified bound on \\(C\\) restricted to \\((\\ker D)^\\perp\\).","\\begin{theorem}[Closed range for the joint defect]\\label{KS47}\nAssume \\(\\ran D\\) is closed. Then\n\\[\n \\ran(D,C)\\text{ is closed}\\quad\\Longleftrightarrow\\quad\n \\ran(C|_{\\ker D})\\text{ is closed}.\n\\]\nUnder either condition, a finite joint residual-to-repair bound exists.\n\\end{theorem}\n\\begin{proof}\nSuppose the range of the joint map is closed. To test closedness of $\\ran B$, take a sequence $Bk_n\\to y$ with $k_n\\in\\ker D$. The joint images are\n\\[\n (Dk_n,Ck_n)=(0,Bk_n)\\longrightarrow(0,y).\n\\]\nClosedness supplies $x\\in E$ with $(Dx,Cx)=(0,y)$. The first coordinate gives $x\\in\\ker D$, and the second gives $Bx=y$. Hence the matching range is closed. This direction did not require closedness of $\\ran D$.","eorem}[Closed range for the joint defect]\\label{KS47}\nAssume \\(\\ran D\\) is closed. Then\n\\[\n \\ran(D,C)\\text{ is closed}\\quad\\Longleftrightarrow\\quad\n \\ran(C|_{\\ker D})\\text{ is closed}.\n\\]\nUnder either condition, a finite joint residual-to-repair bound exists.\n\\end{theorem}\n\\begin{proof}\nSuppose the range of the joint map is closed. To test closedness of $\\ran B$, take a sequence $Bk_n\\to y$ with $k_n\\in\\ker D$. The joint images are\n\\[\n (Dk_n,Ck_n)=(0,Bk_n)\\longrightarrow(0,y).\n\\]\nClosedness supplies $x\\in E$ with $(Dx,Cx)=(0,y)$. The first coordinate gives $x\\in\\ker D$, and the second gives $Bx=y$. Hence the matching range is closed. This direction did not require closedness of $\\ran D$.","ct]\\label{KS47}\nAssume \\(\\ran D\\) is closed. Then\n\\[\n \\ran(D,C)\\text{ is closed}\\quad\\Longleftrightarrow\\quad\n \\ran(C|_{\\ker D})\\text{ is closed}.\n\\]\nUnder either condition, a finite joint residual-to-repair bound exists.\n\\end{theorem}\n\\begin{proof}\nSuppose the range of the joint map is closed. To test closedness of $\\ran B$, take a sequence $Bk_n\\to y$ with $k_n\\in\\ker D$. The joint images are\n\\[\n (Dk_n,Ck_n)=(0,Bk_n)\\longrightarrow(0,y).\n\\]\nClosedness supplies $x\\in E$ with $(Dx,Cx)=(0,y)$. The first coordinate gives $x\\in\\ker D$, and the second gives $Bx=y$. Hence the matching range is closed. This direction did not require closedness of $\\ran D$.","{ is closed}\\quad\\Longleftrightarrow\\quad\n \\ran(C|_{\\ker D})\\text{ is closed}.\n\\]\nUnder either condition, a finite joint residual-to-repair bound exists.\n\\end{theorem}\n\\begin{proof}\nSuppose the range of the joint map is closed. To test closedness of $\\ran B$, take a sequence $Bk_n\\to y$ with $k_n\\in\\ker D$. The joint images are\n\\[\n (Dk_n,Ck_n)=(0,Bk_n)\\longrightarrow(0,y).\n\\]\nClosedness supplies $x\\in E$ with $(Dx,Cx)=(0,y)$. The first coordinate gives $x\\in\\ker D$, and the second gives $Bx=y$. Hence the matching range is closed. This direction did not require closedness of $\\ran D$.","uad\\Longleftrightarrow\\quad\n \\ran(C|_{\\ker D})\\text{ is closed}.\n\\]\nUnder either condition, a finite joint residual-to-repair bound exists.\n\\end{theorem}\n\\begin{proof}\nSuppose the range of the joint map is closed. To test closedness of $\\ran B$, take a sequence $Bk_n\\to y$ with $k_n\\in\\ker D$. The joint images are\n\\[\n (Dk_n,Ck_n)=(0,Bk_n)\\longrightarrow(0,y).\n\\]\nClosedness supplies $x\\in E$ with $(Dx,Cx)=(0,y)$. The first coordinate gives $x\\in\\ker D$, and the second gives $Bx=y$. Hence the matching range is closed. This direction did not require closedness of $\\ran D$.","})\\text{ is closed}.\n\\]\nUnder either condition, a finite joint residual-to-repair bound exists.\n\\end{theorem}\n\\begin{proof}\nSuppose the range of the joint map is closed. To test closedness of $\\ran B$, take a sequence $Bk_n\\to y$ with $k_n\\in\\ker D$. The joint images are\n\\[\n (Dk_n,Ck_n)=(0,Bk_n)\\longrightarrow(0,y).\n\\]\nClosedness supplies $x\\in E$ with $(Dx,Cx)=(0,y)$. The first coordinate gives $x\\in\\ker D$, and the second gives $Bx=y$. Hence the matching range is closed. This direction did not require closedness of $\\ran D$.","r condition, a finite joint residual-to-repair bound exists.\n\\end{theorem}\n\\begin{proof}\nSuppose the range of the joint map is closed. To test closedness of $\\ran B$, take a sequence $Bk_n\\to y$ with $k_n\\in\\ker D$. The joint images are\n\\[\n (Dk_n,Ck_n)=(0,Bk_n)\\longrightarrow(0,y).\n\\]\nClosedness supplies $x\\in E$ with $(Dx,Cx)=(0,y)$. The first coordinate gives $x\\in\\ker D$, and the second gives $Bx=y$. Hence the matching range is closed. This direction did not require closedness of $\\ran D$.","Suppose the range of the joint map is closed. To test closedness of $\\ran B$, take a sequence $Bk_n\\to y$ with $k_n\\in\\ker D$. The joint images are\n\\[\n (Dk_n,Ck_n)=(0,Bk_n)\\longrightarrow(0,y).\n\\]\nClosedness supplies $x\\in E$ with $(Dx,Cx)=(0,y)$. The first coordinate gives $x\\in\\ker D$, and the second gives $Bx=y$. Hence the matching range is closed. This direction did not require closedness of $\\ran D$.","For the reverse direction, assume both $\\ran D$ and $\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint","For the reverse direction, assume both $\\ran D$ and $\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, provi","For the reverse direction, assume both $\\ran D$ and $\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its ma","For the reverse direction, assume both $\\ran D$ and $\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so The","For the reverse direction, assume both $\\ran D$ and $\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43}","For the reverse direction, assume both $\\ran D$ and $\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an expl","For the reverse direction, assume both $\\ran D$ and $\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not neces","For the reverse direction, assume both $\\ran D$ and $\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof","For the reverse direction, assume both $\\ran D$ and $\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof}\n\\sour","For the reverse direction, assume both $\\ran D$ and $\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof}\n\\source{KS47}","$\\ran B$ are closed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof}\n\\source{KS47}","losed, and take joint images converging to $(r,y)$. Thus $Dx_n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof}\n\\source{KS47}","n\\to r$ and $Cx_n\\to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof}\n\\source{KS47}","to y$. Decompose each input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof}\n\\source{KS47}","ach input orthogonally as\n\\[\n x_n=u_n+k_n,\\quad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof}\n\\source{KS47}","ad\n u_n\\in(\\ker D)^\\perp,\\quad k_n\\in\\ker D.\n\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof}\n\\source{KS47}","\\]\nBecause $r\\in\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof}\n\\source{KS47}","\\ran D$ and the transverse inverse $A$ of $D$ is bounded by Theorem~\\ref{KS43}, the first components determine\n\\[\n u_n=A(Dx_n)\\longrightarrow A(r)=u,\n \\qquad Du=r.\n\\]\nContinuity of $C$ now gives\n\\[\n Bk_n=Ck_n=Cx_n-Cu_n\\longrightarrow y-Cu.\n\\]\nClosedness of $\\ran B$ supplies $k\\in\\ker D$ with $Bk=y-Cu$. There is no need for the sequence $k_n$ itself to converge. The single vector $u+k$ satisfies\n\\[\n D(u+k)=r,\n \\qquad C(u+k)=Cu+Bk=y.\n\\]\nThus the joint limit lies in the joint range, proving closedness. The target product is Banach under its maximum norm, so Theorem~\\ref{KS43} applies to the joint map and gives a finite repair gain. The earlier two-stage estimate supplies an explicit, though not necessarily least, value.\n\\end{proof}\n\\source{KS47}","\\chap{Controlled motion of exact spaces}\\label{sec:motion}\n\\subsection{A two-sided condition number theorem}\nLet \\(D,D':E\\to R\\) be continuous linear maps from a real Hilbert space to the same normed target. Write \\(q=P_{\\ker D}\\), \\(q'=P_{\\ker D'}\\). Suppose\n\\[\n \\norm{(I-q)x}\\le C\\norm{Dx},\\qquad\n \\norm{(I-q')x}\\le C'\\norm{D'x},\\qquad C,C'\\ge0,\n\\]\nand \\(\\norm{D-D'}\\le\\varepsilon\\).","\\begin{theorem}[Motion of the exact projector]\\label{KS48}\nWithout either kernel inclusion as a hypothesis,\n\\begin{equation}\\label{eq:motion}\n \\norm{q-q'}\\le\\max(C,C')\\varepsilon.\n\\end{equation}\nIn particular a uniform residual gain \\(M\\) makes \\(D\\mapsto P_{\\ker D}\\) \\(M\\)-Lipschitz on the class of operators carrying that gain.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\nor","\\begin{theorem}[Motion of the exact projector]\\label{KS48}\nWithout either kernel inclusion as a hypothesis,\n\\begin{equation}\\label{eq:motion}\n \\norm{q-q'}\\le\\max(C,C')\\varepsilon.\n\\end{equation}\nIn particular a uniform residual gain \\(M\\) makes \\(D\\mapsto P_{\\ker D}\\) \\(M\\)-Lipschitz on the class of operators carrying that gain.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of t","\\begin{theorem}[Motion of the exact projector]\\label{KS48}\nWithout either kernel inclusion as a hypothesis,\n\\begin{equation}\\label{eq:motion}\n \\norm{q-q'}\\le\\max(C,C')\\varepsilon.\n\\end{equation}\nIn particular a uniform residual gain \\(M\\) makes \\(D\\mapsto P_{\\ker D}\\) \\(M\\)-Lipschitz on the class of operators carrying that gain.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball","\\begin{theorem}[Motion of the exact projector]\\label{KS48}\nWithout either kernel inclusion as a hypothesis,\n\\begin{equation}\\label{eq:motion}\n \\norm{q-q'}\\le\\max(C,C')\\varepsilon.\n\\end{equation}\nIn particular a uniform residual gain \\(M\\) makes \\(D\\mapsto P_{\\ker D}\\) \\(M\\)-Lipschitz on the class of operators carrying that gain.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$","\\begin{theorem}[Motion of the exact projector]\\label{KS48}\nWithout either kernel inclusion as a hypothesis,\n\\begin{equation}\\label{eq:motion}\n \\norm{q-q'}\\le\\max(C,C')\\varepsilon.\n\\end{equation}\nIn particular a uniform residual gain \\(M\\) makes \\(D\\mapsto P_{\\ker D}\\) \\(M\\)-Lipschitz on the class of operators carrying that gain.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, n","eorem}[Motion of the exact projector]\\label{KS48}\nWithout either kernel inclusion as a hypothesis,\n\\begin{equation}\\label{eq:motion}\n \\norm{q-q'}\\le\\max(C,C')\\varepsilon.\n\\end{equation}\nIn particular a uniform residual gain \\(M\\) makes \\(D\\mapsto P_{\\ker D}\\) \\(M\\)-Lipschitz on the class of operators carrying that gain.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, not just a coarser bound in terms o…","quation}\\label{eq:motion}\n \\norm{q-q'}\\le\\max(C,C')\\varepsilon.\n\\end{equation}\nIn particular a uniform residual gain \\(M\\) makes \\(D\\mapsto P_{\\ker D}\\) \\(M\\)-Lipschitz on the class of operators carrying that gain.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, not just a coarser bound in terms of $\\norm x$.","\\ker D}\\) \\(M\\)-Lipschitz on the class of operators carrying that gain.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, not just a coarser bound in terms of $\\norm x$.","he class of operators carrying that gain.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, not just a coarser bound in terms of $\\norm x$.","that gain.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, not just a coarser bound in terms of $\\norm x$.","in.\n\\end{theorem}\n\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, not just a coarser bound in terms of $\\norm x$.","\\begin{proof}\nFor $z\\in\\ker D'$, the old residual bound gives\n\\[\n \\norm{(I-q)z}\\le C\\norm{Dz}\n =C\\norm{(D-D')z}\\le C\\varepsilon\\norm z.\n\\]\nInterchanging the roles gives\n\\[\n \\norm{(I-q')w}\\le C'\\varepsilon\\norm w\n \\qquad(w\\in\\ker D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, not just a coarser bound in terms of $\\norm x$.","r D).                                      \\tag{*}\n\\]\nWe need a dual form of (*). For $v\\in(\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, not just a coarser bound in terms of $\\norm x$.","\\ker D')^\\perp$ and any $y\\in E$, self-adjointness of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, not just a coarser bound in terms of $\\norm x$.","ss of $q$ and $v=(I-q')v$ give\n\\[\n |\\langle qv,y\\rangle|\n =|\\langle v,qy\\rangle|\n =|\\langle v,(I-q')qy\\rangle|\n \\le C'\\varepsilon\\norm v\\norm{qy}\n \\le C'\\varepsilon\\norm v\\norm y.\n\\]\nThe Hilbert norm equals the supremum of these pairings on the unit ball, so\n$\\norm{qv}\\le C'\\varepsilon\\norm v$. This proves the precise restriction estimate needed on $(I-q')x$, not just a coarser bound in terms of $\\norm x$.","Now write\n\\[\n (q-q')x=q(I-q')x-(I-q)q'x.\n\\]\nThe two displayed terms lie in $\\ker D$ and $(\\ker D)^\\perp$, respectively, so they are orthogonal. The input components $(I-q')x$ and $q'x$ are also orthogonal. Applying the dual estimate to the first and the original cross estimate to the second gives\n\\[\n\\begin{aligned}\n \\norm{(q-q')x}^2\n &=\\norm{q(I-q')x}^2+\\norm{(I-q)q'x}^2","Now write\n\\[\n (q-q')x=q(I-q')x-(I-q)q'x.\n\\]\nThe two displayed terms lie in $\\ker D$ and $(\\ker D)^\\perp$, respectively, so they are orthogonal. The input components $(I-q')x$ and $q'x$ are also orthogonal. Applying the dual estimate to the first and the original cross estimate to the second gives\n\\[\n\\begin{aligned}\n \\norm{(q-q')x}^2\n &=\\norm{q(I-q')x}^2+\\norm{(I-q)q'x}^2\\\\\n &\\le (C'\\varepsilon)^2\\norm{(I-q')x}^2\n      +(C\\varepsilon)^2\\norm{q'x}^2\\\\\n &\\le\\varepsilon^2\\max(C,C')^2\n       \\bigl(\\norm{(I-q')x}^2+\\norm{q'x}^2\\bigr)\\\\\n &=\\varepsilon^2\\max(C,C')^2\\norm x^2.\n\\end{aligned}\n\\]\nThe bound $\\norm{D-D'}\\le\\varepsilon$ implies $\\varepsilon\\ge0$, so taking square roots yields the operator estimate. If the common gain is $M$, substitute $C=C'=M$ and $\\varepsilon=\\norm{D-D'}$. This is exactly the Lipschitz inequality on the class carrying that gain. The use of two orthogonal decompositions explains the maximum coefficient instead of a sum.\n\\end{proof}\n\\source{KS48}\nThere is no smallness threshold when both certificates are already supplied. The estimate can be uninformative when thei…","&=\\varepsilon^2\\max(C,C')^2\\norm x^2.\n\\end{aligned}\n\\]\nThe bound $\\norm{D-D'}\\le\\varepsilon$ implies $\\varepsilon\\ge0$, so taking square roots yields the operator estimate. If the common gain is $M$, substitute $C=C'=M$ and $\\varepsilon=\\norm{D-D'}$. This is exactly the Lipschitz inequality on the class carrying that gain. The use of two orthogonal decompositions explains the maximum coefficient instead of a sum.\n\\end{proof}\n\\source{KS48}\nThere is no smallness threshold when both certificates are already supplied. The estimate can be uninformative when their gains are large. Knowing only the old certificate is a different situation.","For changing input as well as model, the same theorem gives\n\\[\n \\norm{qx-q'y}\\le\\norm{x-y}+\\max(C,C')\\varepsilon\\norm y.\n\\]\nIt also bounds the change in distance to the exact space by \\(\\max(C,C')\\varepsilon\\norm x\\).","\\subsection{Constructing the new certificate}\nAssume the target \\(R\\) is Banach and \\(A:R\\to E\\) is a continuous linear right inverse of \\(D\\), with \\(DA=I\\), \\(\\norm A\\le a\\), and \\(a\\varepsilon<1\\). Set\n\\[\n T=(D-D')A,\\qquad Z=(I-T)^{-1},\\qquad A'=AZ.\n\\]\nThe inverse is supplied by the convergent Neumann series because \\(\\norm T\\le a\\varepsilon<1\\).","\\begin{theorem}[Stable surjectivity and derived repair]\\label{KS49}\nThe perturbed map is surjective, with\n\\[\n D'A'=I,\\qquad \\norm{A'}\\le\\frac{a}{1-a\\varepsilon}.\n\\]\nIts exact projection satisfies\n\\[\n \\norm{x-q'x}\\le\\frac{a}{1-a\\varepsilon}\\norm{D'x},\\qquad\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe map $T=(D-D')A$ acts on the Banach target $R$ and satisfies $\\norm T\\le a\\varepsilon<1$. The operator series\n\\[\n Z=\\sum_{n=0}^\\infty T^n\n\\]\nconverges in operator norm and obeys $\\norm Z\\le(1-a\\varepsilon)^{-1}$. Multiplying its finite partial sums on either side by $I-T$ gives $I-T^{N+1}$. Passing to the operator-norm limit proves $(I-T)Z=Z(I-T)=I$. Appendix~\\ref{app:proof-tools} supplies the full series argument, including completeness of the operator space.","n{theorem}[Stable surjectivity and derived repair]\\label{KS49}\nThe perturbed map is surjective, with\n\\[\n D'A'=I,\\qquad \\norm{A'}\\le\\frac{a}{1-a\\varepsilon}.\n\\]\nIts exact projection satisfies\n\\[\n \\norm{x-q'x}\\le\\frac{a}{1-a\\varepsilon}\\norm{D'x},\\qquad\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe map $T=(D-D')A$ acts on the Banach target $R$ and satisfies $\\norm T\\le a\\varepsilon<1$. The operator series\n\\[\n Z=\\sum_{n=0}^\\infty T^n\n\\]\nconverges in operator norm and obeys $\\norm Z\\le(1-a\\varepsilon)^{-1}$. Multiplying its finite partial sums on either side by $I-T$ gives $I-T^{N+1}$. Passing to the operator-norm limit proves $(I-T)Z=Z(I-T)=I$. Appendix~\\ref{app:proof-tools} supplies the full series argument, including completeness of the operator space.","\\label{KS49}\nThe perturbed map is surjective, with\n\\[\n D'A'=I,\\qquad \\norm{A'}\\le\\frac{a}{1-a\\varepsilon}.\n\\]\nIts exact projection satisfies\n\\[\n \\norm{x-q'x}\\le\\frac{a}{1-a\\varepsilon}\\norm{D'x},\\qquad\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe map $T=(D-D')A$ acts on the Banach target $R$ and satisfies $\\norm T\\le a\\varepsilon<1$. The operator series\n\\[\n Z=\\sum_{n=0}^\\infty T^n\n\\]\nconverges in operator norm and obeys $\\norm Z\\le(1-a\\varepsilon)^{-1}$. Multiplying its finite partial sums on either side by $I-T$ gives $I-T^{N+1}$. Passing to the operator-norm limit proves $(I-T)Z=Z(I-T)=I$. Appendix~\\ref{app:proof-tools} supplies the full series argument, including completeness of the operator space.","d \\norm{A'}\\le\\frac{a}{1-a\\varepsilon}.\n\\]\nIts exact projection satisfies\n\\[\n \\norm{x-q'x}\\le\\frac{a}{1-a\\varepsilon}\\norm{D'x},\\qquad\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe map $T=(D-D')A$ acts on the Banach target $R$ and satisfies $\\norm T\\le a\\varepsilon<1$. The operator series\n\\[\n Z=\\sum_{n=0}^\\infty T^n\n\\]\nconverges in operator norm and obeys $\\norm Z\\le(1-a\\varepsilon)^{-1}$. Multiplying its finite partial sums on either side by $I-T$ gives $I-T^{N+1}$. Passing to the operator-norm limit proves $(I-T)Z=Z(I-T)=I$. Appendix~\\ref{app:proof-tools} supplies the full series argument, including completeness of the operator space.","}\\le\\frac{a}{1-a\\varepsilon}\\norm{D'x},\\qquad\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe map $T=(D-D')A$ acts on the Banach target $R$ and satisfies $\\norm T\\le a\\varepsilon<1$. The operator series\n\\[\n Z=\\sum_{n=0}^\\infty T^n\n\\]\nconverges in operator norm and obeys $\\norm Z\\le(1-a\\varepsilon)^{-1}$. Multiplying its finite partial sums on either side by $I-T$ gives $I-T^{N+1}$. Passing to the operator-norm limit proves $(I-T)Z=Z(I-T)=I$. Appendix~\\ref{app:proof-tools} supplies the full series argument, including completeness of the operator space.","{1-a\\varepsilon}\\norm{D'x},\\qquad\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe map $T=(D-D')A$ acts on the Banach target $R$ and satisfies $\\norm T\\le a\\varepsilon<1$. The operator series\n\\[\n Z=\\sum_{n=0}^\\infty T^n\n\\]\nconverges in operator norm and obeys $\\norm Z\\le(1-a\\varepsilon)^{-1}$. Multiplying its finite partial sums on either side by $I-T$ gives $I-T^{N+1}$. Passing to the operator-norm limit proves $(I-T)Z=Z(I-T)=I$. Appendix~\\ref{app:proof-tools} supplies the full series argument, including completeness of the operator space.","a\\varepsilon}{1-a\\varepsilon}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe map $T=(D-D')A$ acts on the Banach target $R$ and satisfies $\\norm T\\le a\\varepsilon<1$. The operator series\n\\[\n Z=\\sum_{n=0}^\\infty T^n\n\\]\nconverges in operator norm and obeys $\\norm Z\\le(1-a\\varepsilon)^{-1}$. Multiplying its finite partial sums on either side by $I-T$ gives $I-T^{N+1}$. Passing to the operator-norm limit proves $(I-T)Z=Z(I-T)=I$. Appendix~\\ref{app:proof-tools} supplies the full series argument, including completeness of the operator space.","Since $DA=I$, one has $D'A=DA-(D-D')A=I-T$. Therefore $A'=AZ$ satisfies\n\\[\n D'A'=(I-T)Z=I,\\qquad\n \\norm{A'}\\le\\norm A\\norm Z\\le\\frac a{1-a\\varepsilon}.\n\\]\nThe first equation explicitly supplies a preimage of every target vector, proving surjectivity.","For any $x$, put $y=x-A'D'x$. Then $D'y=D'x-D'A'D'x=0$, so $y$ is a valid exact competitor for orthogonal projection. Its error gives\n\\[\n \\norm{x-q'x}\\le\\norm{x-y}=\\norm{A'D'x}\n \\le\\frac a{1-a\\varepsilon}\\norm{D'x}.\n\\]\nThe same argument using $A$ gives $\\norm{x-qx}\\le a\\norm{Dx}$. Both coefficients are nonnegative and $a/(1-a\\varepsilon)\\ge a$. Substitution into Theorem~\\ref{KS48} yields\n\\[\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\nNo assertion that $I-A'D'$ is orthogonal was needed. It supplies an exact comparison vector, while best approximation supplies the orthogonal bound.\n\\end{proof}\n\\source{KS49}\nIf a measured residual \\(y\\) satisfies \\(\\norm{D'x-y}\\le\\eta\\), the distance to the ne","For any $x$, put $y=x-A'D'x$. Then $D'y=D'x-D'A'D'x=0$, so $y$ is a valid exact competitor for orthogonal projection. Its error gives\n\\[\n \\norm{x-q'x}\\le\\norm{x-y}=\\norm{A'D'x}\n \\le\\frac a{1-a\\varepsilon}\\norm{D'x}.\n\\]\nThe same argument using $A$ gives $\\norm{x-qx}\\le a\\norm{Dx}$. Both coefficients are nonnegative and $a/(1-a\\varepsilon)\\ge a$. Substitution into Theorem~\\ref{KS48} yields\n\\[\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\nNo assertion that $I-A'D'$ is orthogonal was needed. It supplies an exact comparison vector, while best approximation supplies the orthogonal bound.\n\\end{proof}\n\\source{KS49}\nIf a measured residual \\(y\\) satisfies \\(\\norm{D'x-y}\\le\\eta\\), the distance to the new exact space is","For any $x$, put $y=x-A'D'x$. Then $D'y=D'x-D'A'D'x=0$, so $y$ is a valid exact competitor for orthogonal projection. Its error gives\n\\[\n \\norm{x-q'x}\\le\\norm{x-y}=\\norm{A'D'x}\n \\le\\frac a{1-a\\varepsilon}\\norm{D'x}.\n\\]\nThe same argument using $A$ gives $\\norm{x-qx}\\le a\\norm{Dx}$. Both coefficients are nonnegative and $a/(1-a\\varepsilon)\\ge a$. Substitution into Theorem~\\ref{KS48} yields\n\\[\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\nNo assertion that $I-A'D'$ is orthogonal was needed. It supplies an exact comparison vector, while best approximation supplies the orthogonal bound.\n\\end{proof}\n\\source{KS49}\nIf a measured residual \\(y\\) satisfies \\(\\norm{D'x-y}\\le\\eta\\), the distance to the new exact space is at most \\(a(\\norm y+\\eta)/","For any $x$, put $y=x-A'D'x$. Then $D'y=D'x-D'A'D'x=0$, so $y$ is a valid exact competitor for orthogonal projection. Its error gives\n\\[\n \\norm{x-q'x}\\le\\norm{x-y}=\\norm{A'D'x}\n \\le\\frac a{1-a\\varepsilon}\\norm{D'x}.\n\\]\nThe same argument using $A$ gives $\\norm{x-qx}\\le a\\norm{Dx}$. Both coefficients are nonnegative and $a/(1-a\\varepsilon)\\ge a$. Substitution into Theorem~\\ref{KS48} yields\n\\[\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\nNo assertion that $I-A'D'$ is orthogonal was needed. It supplies an exact comparison vector, while best approximation supplies the orthogonal bound.\n\\end{proof}\n\\source{KS49}\nIf a measured residual \\(y\\) satisfies \\(\\norm{D'x-y}\\le\\eta\\), the distance to the new exact space is at most \\(a(\\norm y+\\eta)/(1-a\\var","For any $x$, put $y=x-A'D'x$. Then $D'y=D'x-D'A'D'x=0$, so $y$ is a valid exact competitor for orthogonal projection. Its error gives\n\\[\n \\norm{x-q'x}\\le\\norm{x-y}=\\norm{A'D'x}\n \\le\\frac a{1-a\\varepsilon}\\norm{D'x}.\n\\]\nThe same argument using $A$ gives $\\norm{x-qx}\\le a\\norm{Dx}$. Both coefficients are nonnegative and $a/(1-a\\varepsilon)\\ge a$. Substitution into Theorem~\\ref{KS48} yields\n\\[\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\nNo assertion that $I-A'D'$ is orthogonal was needed. It supplies an exact comparison vector, while best approximation supplies the orthogonal bound.\n\\end{proof}\n\\source{KS49}\nIf a measured residual \\(y\\) satisfies \\(\\norm{D'x-y}\\le\\eta\\), the distance to the new exact space is at most \\(a(\\norm y+\\eta)/(1-a\\varepsilon)\\). The uncertainty radius remains a quantity to justify in the application. The theorem does not estimate it statistically.","valid exact competitor for orthogonal projection. Its error gives\n\\[\n \\norm{x-q'x}\\le\\norm{x-y}=\\norm{A'D'x}\n \\le\\frac a{1-a\\varepsilon}\\norm{D'x}.\n\\]\nThe same argument using $A$ gives $\\norm{x-qx}\\le a\\norm{Dx}$. Both coefficients are nonnegative and $a/(1-a\\varepsilon)\\ge a$. Substitution into Theorem~\\ref{KS48} yields\n\\[\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\nNo assertion that $I-A'D'$ is orthogonal was needed. It supplies an exact comparison vector, while best approximation supplies the orthogonal bound.\n\\end{proof}\n\\source{KS49}\nIf a measured residual \\(y\\) satisfies \\(\\norm{D'x-y}\\le\\eta\\), the distance to the new exact space is at most \\(a(\\norm y+\\eta)/(1-a\\varepsilon)\\). The uncertainty radius remains a quantity to justify in the application. The theorem does not estimate it statistically.","ves $\\norm{x-qx}\\le a\\norm{Dx}$. Both coefficients are nonnegative and $a/(1-a\\varepsilon)\\ge a$. Substitution into Theorem~\\ref{KS48} yields\n\\[\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\nNo assertion that $I-A'D'$ is orthogonal was needed. It supplies an exact comparison vector, while best approximation supplies the orthogonal bound.\n\\end{proof}\n\\source{KS49}\nIf a measured residual \\(y\\) satisfies \\(\\norm{D'x-y}\\le\\eta\\), the distance to the new exact space is at most \\(a(\\norm y+\\eta)/(1-a\\varepsilon)\\). The uncertainty radius remains a quantity to justify in the application. The theorem does not estimate it statistically.","\\le a\\norm{Dx}$. Both coefficients are nonnegative and $a/(1-a\\varepsilon)\\ge a$. Substitution into Theorem~\\ref{KS48} yields\n\\[\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\nNo assertion that $I-A'D'$ is orthogonal was needed. It supplies an exact comparison vector, while best approximation supplies the orthogonal bound.\n\\end{proof}\n\\source{KS49}\nIf a measured residual \\(y\\) satisfies \\(\\norm{D'x-y}\\le\\eta\\), the distance to the new exact space is at most \\(a(\\norm y+\\eta)/(1-a\\varepsilon)\\). The uncertainty radius remains a quantity to justify in the application. The theorem does not estimate it statistically.","n)\\ge a$. Substitution into Theorem~\\ref{KS48} yields\n\\[\n \\norm{q-q'}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}.\n\\]\nNo assertion that $I-A'D'$ is orthogonal was needed. It supplies an exact comparison vector, while best approximation supplies the orthogonal bound.\n\\end{proof}\n\\source{KS49}\nIf a measured residual \\(y\\) satisfies \\(\\norm{D'x-y}\\le\\eta\\), the distance to the new exact space is at most \\(a(\\norm y+\\eta)/(1-a\\varepsilon)\\). The uncertainty radius remains a quantity to justify in the application. The theorem does not estimate it statistically.","For a surjective Hilbert-source operator, the right inverse need not be a separate unexplained input. The inverse of \\(D:(\\ker D)^\\perp\\to R\\), followed by inclusion, constructs it. A known residual certificate bounds its norm. Surjectivity is still substantive, and cannot be assumed for a redundant residual map whose target contains unattainable directions.","\\subsection{An equivalence between the moving kernels}\n\\begin{theorem}[Explicit kernel transport]\\label{KS50}\nUnder the hypotheses of Theorem~\\ref{KS49},\n\\[\n J:\\ker D\\simeq_L\\ker D',\\qquad Jx=x-A'D'x,\n\\]\nwith inverse \\(y\\mapsto y-ADy\\). Moreover\n\\[\n \\norm{Jx-x}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}\\norm x.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $R_0=I-AD$ and $R_1=I-A'D'$. From $DA=D'A'=I$ we obtain $DR_0=0$ and $D'R_1=0$. Each $R_i$ fixes its corresponding kernel. Thus $R_1$ restricts to a map from $\\ker D$ to $\\ker D'$, while $R_0$ restricts in the other direction.","n equivalence between the moving kernels}\n\\begin{theorem}[Explicit kernel transport]\\label{KS50}\nUnder the hypotheses of Theorem~\\ref{KS49},\n\\[\n J:\\ker D\\simeq_L\\ker D',\\qquad Jx=x-A'D'x,\n\\]\nwith inverse \\(y\\mapsto y-ADy\\). Moreover\n\\[\n \\norm{Jx-x}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}\\norm x.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $R_0=I-AD$ and $R_1=I-A'D'$. From $DA=D'A'=I$ we obtain $DR_0=0$ and $D'R_1=0$. Each $R_i$ fixes its corresponding kernel. Thus $R_1$ restricts to a map from $\\ker D$ to $\\ker D'$, while $R_0$ restricts in the other direction.","tween the moving kernels}\n\\begin{theorem}[Explicit kernel transport]\\label{KS50}\nUnder the hypotheses of Theorem~\\ref{KS49},\n\\[\n J:\\ker D\\simeq_L\\ker D',\\qquad Jx=x-A'D'x,\n\\]\nwith inverse \\(y\\mapsto y-ADy\\). Moreover\n\\[\n \\norm{Jx-x}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}\\norm x.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $R_0=I-AD$ and $R_1=I-A'D'$. From $DA=D'A'=I$ we obtain $DR_0=0$ and $D'R_1=0$. Each $R_i$ fixes its corresponding kernel. Thus $R_1$ restricts to a map from $\\ker D$ to $\\ker D'$, while $R_0$ restricts in the other direction.","xplicit kernel transport]\\label{KS50}\nUnder the hypotheses of Theorem~\\ref{KS49},\n\\[\n J:\\ker D\\simeq_L\\ker D',\\qquad Jx=x-A'D'x,\n\\]\nwith inverse \\(y\\mapsto y-ADy\\). Moreover\n\\[\n \\norm{Jx-x}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}\\norm x.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $R_0=I-AD$ and $R_1=I-A'D'$. From $DA=D'A'=I$ we obtain $DR_0=0$ and $D'R_1=0$. Each $R_i$ fixes its corresponding kernel. Thus $R_1$ restricts to a map from $\\ker D$ to $\\ker D'$, while $R_0$ restricts in the other direction.","l{KS50}\nUnder the hypotheses of Theorem~\\ref{KS49},\n\\[\n J:\\ker D\\simeq_L\\ker D',\\qquad Jx=x-A'D'x,\n\\]\nwith inverse \\(y\\mapsto y-ADy\\). Moreover\n\\[\n \\norm{Jx-x}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}\\norm x.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $R_0=I-AD$ and $R_1=I-A'D'$. From $DA=D'A'=I$ we obtain $DR_0=0$ and $D'R_1=0$. Each $R_i$ fixes its corresponding kernel. Thus $R_1$ restricts to a map from $\\ker D$ to $\\ker D'$, while $R_0$ restricts in the other direction.","r the hypotheses of Theorem~\\ref{KS49},\n\\[\n J:\\ker D\\simeq_L\\ker D',\\qquad Jx=x-A'D'x,\n\\]\nwith inverse \\(y\\mapsto y-ADy\\). Moreover\n\\[\n \\norm{Jx-x}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}\\norm x.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $R_0=I-AD$ and $R_1=I-A'D'$. From $DA=D'A'=I$ we obtain $DR_0=0$ and $D'R_1=0$. Each $R_i$ fixes its corresponding kernel. Thus $R_1$ restricts to a map from $\\ker D$ to $\\ker D'$, while $R_0$ restricts in the other direction.","of Theorem~\\ref{KS49},\n\\[\n J:\\ker D\\simeq_L\\ker D',\\qquad Jx=x-A'D'x,\n\\]\nwith inverse \\(y\\mapsto y-ADy\\). Moreover\n\\[\n \\norm{Jx-x}\\le\\frac{a\\varepsilon}{1-a\\varepsilon}\\norm x.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $R_0=I-AD$ and $R_1=I-A'D'$. From $DA=D'A'=I$ we obtain $DR_0=0$ and $D'R_1=0$. Each $R_i$ fixes its corresponding kernel. Thus $R_1$ restricts to a map from $\\ker D$ to $\\ker D'$, while $R_0$ restricts in the other direction.","The inverse identities require two additional cancellations. First,\n\\[\n R_0A'=(I-AD)AZ=(A-ADA)Z=0.\n\\]\nSecond, using $D'A=I-T$ and $Z(I-T)=I$,\n\\[\n R_1A=A-AZD'A=A-AZ(I-T)=0.\n\\]\nFor $x\\in\\ker D$, these identities give\n\\[\n R_0R_1x=R_0(x-A'D'x)=R_0x=x.\n\\]\nFor $y\\in\\ker D'$, they give\n\\[\n R_1R_0y=R_1(y-ADy)=R_1y=y.\n\\]\nBoth restricted maps are bounded and linear as restrictions of bounded linear operators. They are therefore inverse continuous linear maps, establishing the asserted equivalence.","Finally, if $Dx=0$, then $D'x=(D'-D)x$. Hence\n\\[\n \\norm{Jx-x}=\\norm{A'D'x}\n \\le\\norm{A'}\\varepsilon\\norm x\n \\le\\frac{a\\varepsilon}{1-a\\varepsilon}\\norm x.\n\\]\nThis derivation uses the selected right inverse $A$ throughout. It proves a controlled transport of exact vectors, not a claim that the selected map is the nearest-point projection onto the new kernel.\n\\end{proof}\n\\source{KS50}\nThe map \\(J\\) depends on the selected splitting. It is not generally the restriction of the orthogonal projection \\(q'\\). Equivalence of exact spaces and nearest-point repair answer different questions.","rm x\n \\le\\frac{a\\varepsilon}{1-a\\varepsilon}\\norm x.\n\\]\nThis derivation uses the selected right inverse $A$ throughout. It proves a controlled transport of exact vectors, not a claim that the selected map is the nearest-point projection onto the new kernel.\n\\end{proof}\n\\source{KS50}\nThe map \\(J\\) depends on the selected splitting. It is not generally the restriction of the orthogonal projection \\(q'\\). Equivalence of exact spaces and nearest-point repair answer different questions.","\\subsection{An all-real worked model}\nLet \\(D_t(x,y)=x+ty\\). Its kernel is the line \\(\\{(-tb,b):b\\in\\R\\}\\), and its orthogonal projector is\n\\begin{equation}\\label{eq:lineprojector}\n q_t=\\frac1{1+t^2}\n \\begin{pmatrix}t^2&-t","\\subsection{An all-real worked model}\nLet \\(D_t(x,y)=x+ty\\). Its kernel is the line \\(\\{(-tb,b):b\\in\\R\\}\\), and its orthogonal projector is\n\\begin{equation}\\label{eq:lineprojector}\n q_t=\\frac1{1+t^2}\n \\begin{pmatrix}t^2&-t\\\\-t&1\\end{pmatrix}.\n\\end{equation}\nThe common right inverse \\(r\\mapsto(r,0)\\) supplies gain at most one for every \\(t\\), and \\(\\norm{D_s-D_t}=|s-t|\\). The general theorem gives \\(\\norm{q_s-q_t}\\le|s-t|\\).","-t&1\\end{pmatrix}.\n\\end{equation}\nThe common right inverse \\(r\\mapsto(r,0)\\) supplies gain at most one for every \\(t\\), and \\(\\norm{D_s-D_t}=|s-t|\\). The general theorem gives \\(\\norm{q_s-q_t}\\le|s-t|\\).","\\begin{proposition}[Exact motion of the line]\\label{KS51}\nFor every real \\(t\\),\n\\[\n \\norm{(q_t-q_0)x}^2=\\frac{t^2}{1+t^2}\\norm x^2,\n \\qquad \\norm{q_t-q_0}=\\frac{|t|}{\\sqrt{1+t^2}}.\n\\]\nIn the small perturbation interface, kernel transport sends \\((0,b)\\) to \\((-tb,b)\\). Orthogonal projection instead sends it to \\((-tb/(1+t^2),b/(1+t^2))\\).\n\\end{proposition}\n\\begin{proof}\nThe vector \\(v_t=(-t,1)\\) spans \\(\\ker D_t\\) and has squared norm \\(1+t^2>0\\). Projection onto this line is\n\\[\n x\\longmapsto\\frac{\\langle x,v_t\\rangle}{1+t^2}v_t,\n\\]\nwhich gives the matrix in \\eqref{eq:lineprojector}. In particular it is the orthogonal projector, not merely a map whose image is contained in the kernel.","proposition}[Exact motion of the line]\\label{KS51}\nFor every real \\(t\\),\n\\[\n \\norm{(q_t-q_0)x}^2=\\frac{t^2}{1+t^2}\\norm x^2,\n \\qquad \\norm{q_t-q_0}=\\frac{|t|}{\\sqrt{1+t^2}}.\n\\]\nIn the small perturbation interface, kernel transport sends \\((0,b)\\) to \\((-tb,b)\\). Orthogonal projection instead sends it to \\((-tb/(1+t^2),b/(1+t^2))\\).\n\\end{proposition}\n\\begin{proof}\nThe vector \\(v_t=(-t,1)\\) spans \\(\\ker D_t\\) and has squared norm \\(1+t^2>0\\). Projection onto this line is\n\\[\n x\\longmapsto\\frac{\\langle x,v_t\\rangle}{1+t^2}v_t,\n\\]\nwhich gives the matrix in \\eqref{eq:lineprojector}. In particular it is the orthogonal projector, not merely a map whose image is contained in the kernel.","ine]\\label{KS51}\nFor every real \\(t\\),\n\\[\n \\norm{(q_t-q_0)x}^2=\\frac{t^2}{1+t^2}\\norm x^2,\n \\qquad \\norm{q_t-q_0}=\\frac{|t|}{\\sqrt{1+t^2}}.\n\\]\nIn the small perturbation interface, kernel transport sends \\((0,b)\\) to \\((-tb,b)\\). Orthogonal projection instead sends it to \\((-tb/(1+t^2),b/(1+t^2))\\).\n\\end{proposition}\n\\begin{proof}\nThe vector \\(v_t=(-t,1)\\) spans \\(\\ker D_t\\) and has squared norm \\(1+t^2>0\\). Projection onto this line is\n\\[\n x\\longmapsto\\frac{\\langle x,v_t\\rangle}{1+t^2}v_t,\n\\]\nwhich gives the matrix in \\eqref{eq:lineprojector}. In particular it is the orthogonal projector, not merely a map whose image is contained in the kernel.","\\norm{(q_t-q_0)x}^2=\\frac{t^2}{1+t^2}\\norm x^2,\n \\qquad \\norm{q_t-q_0}=\\frac{|t|}{\\sqrt{1+t^2}}.\n\\]\nIn the small perturbation interface, kernel transport sends \\((0,b)\\) to \\((-tb,b)\\). Orthogonal projection instead sends it to \\((-tb/(1+t^2),b/(1+t^2))\\).\n\\end{proposition}\n\\begin{proof}\nThe vector \\(v_t=(-t,1)\\) spans \\(\\ker D_t\\) and has squared norm \\(1+t^2>0\\). Projection onto this line is\n\\[\n x\\longmapsto\\frac{\\langle x,v_t\\rangle}{1+t^2}v_t,\n\\]\nwhich gives the matrix in \\eqref{eq:lineprojector}. In particular it is the orthogonal projector, not merely a map whose image is contained in the kernel.","Subtract \\(q_0=\\operatorname{diag}(0,1)\\), obtaining the symmetric matrix\n\\[\n B_t=q_t-q_0=\\frac1{1+t^2}\n \\begin{pmatrix}t^2&-t","Subtract \\(q_0=\\operatorname{diag}(0,1)\\), obtaining the symmetric matrix\n\\[\n B_t=q_t-q_0=\\frac1{1+t^2}\n \\begin{pmatrix}t^2&-t\\\\-t&-t^2\\end{pmatrix}.\n\\]\nThe two off-diagonal entries of its square are zero, since \\(-t^3+t^3=0\\). Both diagonal entries are \\((t^4+t^2)/(1+t^2)^2=t^2/(1+t^2)\\). Thus\n\\[\n B_t^{\\mathsf T}B_t=B_t^2=\\frac{t^2}{1+t^2}I.\n\\]\nPairing this identity with an arbitrary vector proves the all-vector norm formula. It bounds the operator norm above by \\(|t|/\\sqrt{1+t^2}\\), and evaluating any unit vector gives equality. The argument includes \\(t=0\\), where both sides vanish.","-t&-t^2\\end{pmatrix}.\n\\]\nThe two off-diagonal entries of its square are zero, since \\(-t^3+t^3=0\\). Both diagonal entries are \\((t^4+t^2)/(1+t^2)^2=t^2/(1+t^2)\\). Thus\n\\[\n B_t^{\\mathsf T}B_t=B_t^2=\\frac{t^2}{1+t^2}I.\n\\]\nPairing this identity with an arbitrary vector proves the all-vector norm formula. It bounds the operator norm above by \\(|t|/\\sqrt{1+t^2}\\), and evaluating any unit vector gives equality. The argument includes \\(t=0\\), where both sides vanish.","off-diagonal entries of its square are zero, since \\(-t^3+t^3=0\\). Both diagonal entries are \\((t^4+t^2)/(1+t^2)^2=t^2/(1+t^2)\\). Thus\n\\[\n B_t^{\\mathsf T}B_t=B_t^2=\\frac{t^2}{1+t^2}I.\n\\]\nPairing this identity with an arbitrary vector proves the all-vector norm formula. It bounds the operator norm above by \\(|t|/\\sqrt{1+t^2}\\), and evaluating any unit vector gives equality. The argument includes \\(t=0\\), where both sides vanish.","For the common section \\(A(r)=(r,0)\\), the perturbation inverse leaves \\(A\\) unchanged because \\(D_tA=I\\). The constructed transport is therefore \\(Jx=x-AD_tx\\) on the old kernel. At \\(x=(0,b)\\), it gives \\((-tb,b)\\). Substitution into the projection formula gives instead \\((-tb/(1+t^2),b/(1+t^2))\\). The strict-radius theorem applies when \\(|t|<1\\), but these explicit line and projection calculations are valid for every real \\(t\\).\n\\end{proof}\n\\source{KS51}","r,0)\\), the perturbation inverse leaves \\(A\\) unchanged because \\(D_tA=I\\). The constructed transport is therefore \\(Jx=x-AD_tx\\) on the old kernel. At \\(x=(0,b)\\), it gives \\((-tb,b)\\). Substitution into the projection formula gives instead \\((-tb/(1+t^2),b/(1+t^2))\\). The strict-radius theorem applies when \\(|t|<1\\), but these explicit line and projection calculations are valid for every real \\(t\\).\n\\end{proof}\n\\source{KS51}","\\subsection{What fails without the hypotheses}\nTake \\(D_0=\\operatorname{diag}(1,0)\\) and \\(D_t=\\operatorname{diag}(1,t)\\). The operator difference has norm \\(|t|\\). For every nonzero \\(t\\), however, the new kernel is zero, while the old kernel is the vertical axis. Their projections differ in norm by one. Operator closeness alone does not imply kernel continuity.","More generally, suppose a unit vector \\(x\\) is exact for the original operator but orthogonal to the new kernel. Every new residual gain \\(C'\\) obeys\n\\[\n 1=\\norm{x-q'x}\\le C'\\norm{D'x}\\le C'\\varepsilon.\n\\]\nLosing an exact direction under a small operator change forces the new gain to grow. At the strict radius boundary, \\(D=I\\), \\(A=I\\), and \\(D'=0\\) have \\(a=\\varepsilon=1\\), yet \\(D'\\) is not surjective. Thus the strict condition in Theorem~\\ref{KS49} cannot be replaced universally by a non-strict one.\n\\source{KS52}\n\\plate{18_rank_loss}{Two families can have the same small operator-change norm and very different exact-space motion. A moving line retains a uniform residual gain. The rank-changing diagonal family loses an exact dire","More generally, suppose a unit vector \\(x\\) is exact for the original operator but orthogonal to the new kernel. Every new residual gain \\(C'\\) obeys\n\\[\n 1=\\norm{x-q'x}\\le C'\\norm{D'x}\\le C'\\varepsilon.\n\\]\nLosing an exact direction under a small operator change forces the new gain to grow. At the strict radius boundary, \\(D=I\\), \\(A=I\\), and \\(D'=0\\) have \\(a=\\varepsilon=1\\), yet \\(D'\\) is not surjective. Thus the strict condition in Theorem~\\ref{KS49} cannot be replaced universally by a non-strict one.\n\\source{KS52}\n\\plate{18_rank_loss}{Two families can have the same small operator-change norm and very different exact-space motion. A moving line retains a uniform residual gain. The rank-changing diagonal family loses an exact direction for every nonzero parameter, so its projector gap remains one.}{fig:rankloss}","\\subsection{Motion of the actual descent repair}\nCompare pairs \\((D,C)\\) and \\((D',C')\\) on the same raw Hilbert space. Suppose each pair has closed local range and closed overlap range on its own locally exact subspace. Let \\(G,G'\\) be the resulting joint residual gains. If\n\\[\n \\norm{D-D'}\\le\\varepsilon_D,\\qquad\n \\norm{C-C'}\\le\\varepsilon_C,\n\\]\nthen the maximum norm on the target product gives\n\\[\n \\norm{(D,C)-(D',C')}\\le\\max(\\varepsilon_D,\\varepsilon_C).\n\\]\nApplying Theorem~\\ref{KS48} to the joint operators, and using the actual two-stage identity of Theorem~\\ref","\\subsection{Motion of the actual descent repair}\nCompare pairs \\((D,C)\\) and \\((D',C')\\) on the same raw Hilbert space. Suppose each pair has closed local range and closed overlap range on its own locally exact subspace. Let \\(G,G'\\) be the resulting joint residual gains. If\n\\[\n \\norm{D-D'}\\le\\varepsilon_D,\\qquad\n \\norm{C-C'}\\le\\varepsilon_C,\n\\]\nthen the maximum norm on the target product gives\n\\[\n \\norm{(D,C)-(D',C')}\\le\\max(\\varepsilon_D,\\varepsilon_C).\n\\]\nApplying Theorem~\\ref{KS48} to the joint operators, and using the actual two-stage identity of Theorem~\\ref{KS46}, gives\n\\b","\\subsection{Motion of the actual descent repair}\nCompare pairs \\((D,C)\\) and \\((D',C')\\) on the same raw Hilbert space. Suppose each pair has closed local range and closed overlap range on its own locally exact subspace. Let \\(G,G'\\) be the resulting joint residual gains. If\n\\[\n \\norm{D-D'}\\le\\varepsilon_D,\\qquad\n \\norm{C-C'}\\le\\varepsilon_C,\n\\]\nthen the maximum norm on the target product gives\n\\[\n \\norm{(D,C)-(D',C')}\\le\\max(\\varepsilon_D,\\varepsilon_C).\n\\]\nApplying Theorem~\\ref{KS48} to the joint operators, and using the actual two-stage identity of Theorem~\\ref{KS46}, gives\n\\begin{equation}\\label{eq:descentmotion}\n \\norm{P_{D,C}-P_{D',C'}}\n \\le\\max(G,G')\\max(\\varepsilon_D,\\varepsilon_C).\n\\end{equation}\nThis is a compar","\\subsection{Motion of the actual descent repair}\nCompare pairs \\((D,C)\\) and \\((D',C')\\) on the same raw Hilbert space. Suppose each pair has closed local range and closed overlap range on its own locally exact subspace. Let \\(G,G'\\) be the resulting joint residual gains. If\n\\[\n \\norm{D-D'}\\le\\varepsilon_D,\\qquad\n \\norm{C-C'}\\le\\varepsilon_C,\n\\]\nthen the maximum norm on the target product gives\n\\[\n \\norm{(D,C)-(D',C')}\\le\\max(\\varepsilon_D,\\varepsilon_C).\n\\]\nApplying Theorem~\\ref{KS48} to the joint operators, and using the actual two-stage identity of Theorem~\\ref{KS46}, gives\n\\begin{equation}\\label{eq:descentmotion}\n \\norm{P_{D,C}-P_{D',C'}}\n \\le\\max(G,G')\\max(\\varepsilon_D,\\varepsilon_C).\n\\end{equation}\nThis is a comparison of the original repair maps, not a replacement definition that happens to have a similar kernel. Both the loc","\\subsection{Motion of the actual descent repair}\nCompare pairs \\((D,C)\\) and \\((D',C')\\) on the same raw Hilbert space. Suppose each pair has closed local range and closed overlap range on its own locally exact subspace. Let \\(G,G'\\) be the resulting joint residual gains. If\n\\[\n \\norm{D-D'}\\le\\varepsilon_D,\\qquad\n \\norm{C-C'}\\le\\varepsilon_C,\n\\]\nthen the maximum norm on the target product gives\n\\[\n \\norm{(D,C)-(D',C')}\\le\\max(\\varepsilon_D,\\varepsilon_C).\n\\]\nApplying Theorem~\\ref{KS48} to the joint operators, and using the actual two-stage identity of Theorem~\\ref{KS46}, gives\n\\begin{equation}\\label{eq:descentmotion}\n \\norm{P_{D,C}-P_{D',C'}}\n \\le\\max(G,G')\\max(\\varepsilon_D,\\varepsilon_C).\n\\end{equation}\nThis is a comparison of the original repair maps, not a replacement definition that happens to have a similar kernel. Both the local equations and overlap conditions can change. The raw space and its inner product remain fixed.\n\\source{KS53}","ally exact subspace. Let \\(G,G'\\) be the resulting joint residual gains. If\n\\[\n \\norm{D-D'}\\le\\varepsilon_D,\\qquad\n \\norm{C-C'}\\le\\varepsilon_C,\n\\]\nthen the maximum norm on the target product gives\n\\[\n \\norm{(D,C)-(D',C')}\\le\\max(\\varepsilon_D,\\varepsilon_C).\n\\]\nApplying Theorem~\\ref{KS48} to the joint operators, and using the actual two-stage identity of Theorem~\\ref{KS46}, gives\n\\begin{equation}\\label{eq:descentmotion}\n \\norm{P_{D,C}-P_{D',C'}}\n \\le\\max(G,G')\\max(\\varepsilon_D,\\varepsilon_C).\n\\end{equation}\nThis is a comparison of the original repair maps, not a replacement definition that happens to have a similar kernel. Both the local equations and overlap conditions can change. The raw space and its inner product remain fixed.\n\\source{KS53}","\\subsection{The statement carried by a certificate}\nLet \\(D:E\\to R\\) be bounded linear from a real Hilbert space into a real normed space. A residual certificate with gain \\(C\\ge0\\) is the universally quantified estimate\n\\[\n \\norm{x-qx}\\le C\\norm{Dx}\\quad\\text{for every }x\\in E,\n \\qquad q=P_{\\ker D}.\n\\]\nA reported number \\(C\\) is not a certificate unless this estimate has been justified in the stated model. A successful test at one input is likewise not the universally quantified estimate.","Suppose the measured vector is \\(y\\), with a certified error radius \\(\\eta\\) such that \\(\\norm{Dx-y}\\le\\eta\\). The triangle inequality converts the measured ball into a bound for the true residual.","\\begin{proposition}[Measurement and operator uncertainty]\\label{KS69}\nWith the reference certificate above,\n\\[\n \\norm{x-qx}\\le C(\\norm y+\\eta)\n \\quad\\text{if }\\norm{Dx-y}\\le\\eta.\n\\]\nMore generally, let \\(\\widetilde D:E\\to R\\) be the bounded linear operator actually used in the measurement. If\n\\[\n \\norm{D-\\widetilde D}\\le\\varepsilon,\n \\qquad\\norm{\\widetilde D x-y}\\le\\eta,\n\\]\nthen\n\\begin{equation}\\label{eq:measured-reference}\n \\norm{x-qx}\\le C(\\norm y+\\eta+\\varepsilon\\norm x).\n\\end{equation}\nThe projection in this conclusion is still the projection for the reference operator \\(D\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion follows from \\(\\norm{Dx}\\le\\norm y+\\eta\\). For the second, decompose\n\\[\n Dx=(D-\\widetilde D)x+(\\widetilde D x-y)+y.\n\\]\nIts norm is at most \\(\\varepsilon\\norm x+\\eta+\\norm y\\). Apply the reference residual certificate.\n\\end{proof}\n\\source{KS","\\begin{proposition}[Measurement and operator uncertainty]\\label{KS69}\nWith the reference certificate above,\n\\[\n \\norm{x-qx}\\le C(\\norm y+\\eta)\n \\quad\\text{if }\\norm{Dx-y}\\le\\eta.\n\\]\nMore generally, let \\(\\widetilde D:E\\to R\\) be the bounded linear operator actually used in the measurement. If\n\\[\n \\norm{D-\\widetilde D}\\le\\varepsilon,\n \\qquad\\norm{\\widetilde D x-y}\\le\\eta,\n\\]\nthen\n\\begin{equation}\\label{eq:measured-reference}\n \\norm{x-qx}\\le C(\\norm y+\\eta+\\varepsilon\\norm x).\n\\end{equation}\nThe projection in this conclusion is still the projection for the reference operator \\(D\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion follows from \\(\\norm{Dx}\\le\\norm y+\\eta\\). For the second, decompose\n\\[\n Dx=(D-\\widetilde D)x+(\\widetilde D x-y)+y.\n\\]\nIts norm is at most \\(\\varepsilon\\norm x+\\eta+\\norm y\\). Apply the reference residual certificate.\n\\end{proof}\n\\source{KS69}\nThe input-size term has a mathematical p","\\begin{proposition}[Measurement and operator uncertainty]\\label{KS69}\nWith the reference certificate above,\n\\[\n \\norm{x-qx}\\le C(\\norm y+\\eta)\n \\quad\\text{if }\\norm{Dx-y}\\le\\eta.\n\\]\nMore generally, let \\(\\widetilde D:E\\to R\\) be the bounded linear operator actually used in the measurement. If\n\\[\n \\norm{D-\\widetilde D}\\le\\varepsilon,\n \\qquad\\norm{\\widetilde D x-y}\\le\\eta,\n\\]\nthen\n\\begin{equation}\\label{eq:measured-reference}\n \\norm{x-qx}\\le C(\\norm y+\\eta+\\varepsilon\\norm x).\n\\end{equation}\nThe projection in this conclusion is still the projection for the reference operator \\(D\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion follows from \\(\\norm{Dx}\\le\\norm y+\\eta\\). For the second, decompose\n\\[\n Dx=(D-\\widetilde D)x+(\\widetilde D x-y)+y.\n\\]\nIts norm is at most \\(\\varepsilon\\norm x+\\eta+\\norm y\\). Apply the reference residual certificate.\n\\end{proof}\n\\source{KS69}\nThe input-size term has a mathematical purpose. A small error in an operator can produce a large error on a large input. It cannot generally be replaced by a number depending only on the meas","\\begin{proposition}[Measurement and operator uncertainty]\\label{KS69}\nWith the reference certificate above,\n\\[\n \\norm{x-qx}\\le C(\\norm y+\\eta)\n \\quad\\text{if }\\norm{Dx-y}\\le\\eta.\n\\]\nMore generally, let \\(\\widetilde D:E\\to R\\) be the bounded linear operator actually used in the measurement. If\n\\[\n \\norm{D-\\widetilde D}\\le\\varepsilon,\n \\qquad\\norm{\\widetilde D x-y}\\le\\eta,\n\\]\nthen\n\\begin{equation}\\label{eq:measured-reference}\n \\norm{x-qx}\\le C(\\norm y+\\eta+\\varepsilon\\norm x).\n\\end{equation}\nThe projection in this conclusion is still the projection for the reference operator \\(D\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion follows from \\(\\norm{Dx}\\le\\norm y+\\eta\\). For the second, decompose\n\\[\n Dx=(D-\\widetilde D)x+(\\widetilde D x-y)+y.\n\\]\nIts norm is at most \\(\\varepsilon\\norm x+\\eta+\\norm y\\). Apply the reference residual certificate.\n\\end{proof}\n\\source{KS69}\nThe input-size term has a mathematical purpose. A small error in an operator can produce a large error on a large input. It cannot generally be replaced by a number depending only on the measured residual. Even a meas…","More generally, let \\(\\widetilde D:E\\to R\\) be the bounded linear operator actually used in the measurement. If\n\\[\n \\norm{D-\\widetilde D}\\le\\varepsilon,\n \\qquad\\norm{\\widetilde D x-y}\\le\\eta,\n\\]\nthen\n\\begin{equation}\\label{eq:measured-reference}\n \\norm{x-qx}\\le C(\\norm y+\\eta+\\varepsilon\\norm x).\n\\end{equation}\nThe projection in this conclusion is still the projection for the reference operator \\(D\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion follows from \\(\\norm{Dx}\\le\\norm y+\\eta\\). For the second, decompose\n\\[\n Dx=(D-\\widetilde D)x+(\\widetilde D x-y)+y.\n\\]\nIts norm is at most \\(\\varepsilon\\norm x+\\eta+\\norm y\\). Apply the reference residual certificate.\n\\end{proof}\n\\source{KS69}\nThe input-size term has a mathematical purpose. A small error in an operator can produce a large error on a large input. It cannot generally be replaced by a number depending only on the measured residual. Even a measurement of zero may reflect a difference between the measured kernel and the reference kernel.","or actually used in the measurement. If\n\\[\n \\norm{D-\\widetilde D}\\le\\varepsilon,\n \\qquad\\norm{\\widetilde D x-y}\\le\\eta,\n\\]\nthen\n\\begin{equation}\\label{eq:measured-reference}\n \\norm{x-qx}\\le C(\\norm y+\\eta+\\varepsilon\\norm x).\n\\end{equation}\nThe projection in this conclusion is still the projection for the reference operator \\(D\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion follows from \\(\\norm{Dx}\\le\\norm y+\\eta\\). For the second, decompose\n\\[\n Dx=(D-\\widetilde D)x+(\\widetilde D x-y)+y.\n\\]\nIts norm is at most \\(\\varepsilon\\norm x+\\eta+\\norm y\\). Apply the reference residual certificate.\n\\end{proof}\n\\source{KS69}\nThe input-size term has a mathematical purpose. A small error in an operator can produce a large error on a large input. It cannot generally be replaced by a number depending only on the measured residual. Even a measurement of zero may reflect a difference between the measured kernel and the reference kernel.","etilde D}\\le\\varepsilon,\n \\qquad\\norm{\\widetilde D x-y}\\le\\eta,\n\\]\nthen\n\\begin{equation}\\label{eq:measured-reference}\n \\norm{x-qx}\\le C(\\norm y+\\eta+\\varepsilon\\norm x).\n\\end{equation}\nThe projection in this conclusion is still the projection for the reference operator \\(D\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion follows from \\(\\norm{Dx}\\le\\norm y+\\eta\\). For the second, decompose\n\\[\n Dx=(D-\\widetilde D)x+(\\widetilde D x-y)+y.\n\\]\nIts norm is at most \\(\\varepsilon\\norm x+\\eta+\\norm y\\). Apply the reference residual certificate.\n\\end{proof}\n\\source{KS69}\nThe input-size term has a mathematical purpose. A small error in an operator can produce a large error on a large input. It cannot generally be replaced by a number depending only on the measured residual. Even a measurement of zero may reflect a difference between the measured kernel and the reference kernel.","n\n\\begin{equation}\\label{eq:measured-reference}\n \\norm{x-qx}\\le C(\\norm y+\\eta+\\varepsilon\\norm x).\n\\end{equation}\nThe projection in this conclusion is still the projection for the reference operator \\(D\\).\n\\end{proposition}\n\\begin{proof}\nThe first assertion follows from \\(\\norm{Dx}\\le\\norm y+\\eta\\). For the second, decompose\n\\[\n Dx=(D-\\widetilde D)x+(\\widetilde D x-y)+y.\n\\]\nIts norm is at most \\(\\varepsilon\\norm x+\\eta+\\norm y\\). Apply the reference residual certificate.\n\\end{proof}\n\\source{KS69}\nThe input-size term has a mathematical purpose. A small error in an operator can produce a large error on a large input. It cannot generally be replaced by a number depending only on the measured residual. Even a measurement of zero may reflect a difference between the measured kernel and the reference kernel.","\\subsection{A vanishing measured residual can still miss the reference space}\nFor a concrete family, take \\(E=\\R^2\\) with the Euclidean norm and \\(R=\\R\\), and define\n\\[\n D(a,b)=a,\\qquad \\widetilde D(a,b)=a+tb.\n\\]\nThe reference projection is \\(q(a,b)=(0,b)\\), its least residual gain is one, and \\(\\norm{D-\\widetilde D}\\le|t|\\). On the input \\(x=(-tB,B)\\), the measured operator returns exactly zero. Its distance to the reference kernel is nevertheless \\(|tB|\\).","s the reference space}\nFor a concrete family, take \\(E=\\R^2\\) with the Euclidean norm and \\(R=\\R\\), and define\n\\[\n D(a,b)=a,\\qquad \\widetilde D(a,b)=a+tb.\n\\]\nThe reference projection is \\(q(a,b)=(0,b)\\), its least residual gain is one, and \\(\\norm{D-\\widetilde D}\\le|t|\\). On the input \\(x=(-tB,B)\\), the measured operator returns exactly zero. Its distance to the reference kernel is nevertheless \\(|tB|\\).","Setting \\(\\eta=0\\) and \\(y=0\\) in \\eqref{eq:measured-reference} gives the valid upper bound \\(|t|\\sqrt{1+t^2}|B|\\). Deleting the input-size term would give zero and would be false whenever \\(tB\\ne0\\). The input is exact for the changed model and not exact for the reference model. This is precisely the distinction that the estimate is designed to retain.","\\subsection{A preserved kernel gives a stronger conclusion}\nA useful regime occurs when the changed operator annihilates every old exact vector:\n\\[\n \\widetilde D q=0.\n\\]\nThis says that the old kernel is contained in the new kernel. Equality is not assumed. Let \\(\\norm{D-\\widetilde D}\\le\\varepsilon\\) and suppose the strict budget \\(C\\varepsilon<1\\) holds.","\\begin{theorem}[Kernel preservation and a strict error budget]\\label{KS70}\nUnder these hypotheses,\n\\[\n \\ker\\widetilde D=\\ker D,\n \\qquad\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}\\norm{\\widetilde D x}.\n\\]\nFor a measurement with \\(\\norm{\\widetilde D x-y}\\le\\eta\\), the bound becomes\n\\[\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}(\\norm y+\\eta).\n\\]\nThus the exact space is unchanged and the residual gain degrades by an explicit factor.\n\\end{theorem}\n\\begin{proof}\nLet \\(q=P_{\\ker D}\\), and write \\(u=x-qx\\). The stated kernel inclusion makes both \\(Dqx\\) and \\(\\widetilde Dqx\\) zero. Hence \\(Du=Dx\\) and \\(\\widetilde Du=\\widetilde Dx\\). Applying the reference residual bound to \\(u\\), whose projection onto \\(\\ker D\\) is zero, gives\n\\[\n \\begin{aligned}\n \\norm u&\\le C\\norm{Du}\n \\le C\\norm{\\widetilde Du}+C\\norm{(D-\\widetilde D)u}","]\\label{KS70}\nUnder these hypotheses,\n\\[\n \\ker\\widetilde D=\\ker D,\n \\qquad\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}\\norm{\\widetilde D x}.\n\\]\nFor a measurement with \\(\\norm{\\widetilde D x-y}\\le\\eta\\), the bound becomes\n\\[\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}(\\norm y+\\eta).\n\\]\nThus the exact space is unchanged and the residual gain degrades by an explicit factor.\n\\end{theorem}\n\\begin{proof}\nLet \\(q=P_{\\ker D}\\), and write \\(u=x-qx\\). The stated kernel inclusion makes both \\(Dqx\\) and \\(\\widetilde Dqx\\) zero. Hence \\(Du=Dx\\) and \\(\\widetilde Du=\\widetilde Dx\\). Applying the reference residual bound to \\(u\\), whose projection onto \\(\\ker D\\) is zero, gives\n\\[\n \\begin{aligned}\n \\norm u&\\le C\\norm{Du}\n \\le C\\norm{\\widetilde Du}+C\\norm{(D-\\widetilde D)u}","potheses,\n\\[\n \\ker\\widetilde D=\\ker D,\n \\qquad\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}\\norm{\\widetilde D x}.\n\\]\nFor a measurement with \\(\\norm{\\widetilde D x-y}\\le\\eta\\), the bound becomes\n\\[\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}(\\norm y+\\eta).\n\\]\nThus the exact space is unchanged and the residual gain degrades by an explicit factor.\n\\end{theorem}\n\\begin{proof}\nLet \\(q=P_{\\ker D}\\), and write \\(u=x-qx\\). The stated kernel inclusion makes both \\(Dqx\\) and \\(\\widetilde Dqx\\) zero. Hence \\(Du=Dx\\) and \\(\\widetilde Du=\\widetilde Dx\\). Applying the reference residual bound to \\(u\\), whose projection onto \\(\\ker D\\) is zero, gives\n\\[\n \\begin{aligned}\n \\norm u&\\le C\\norm{Du}\n \\le C\\norm{\\widetilde Du}+C\\norm{(D-\\widetilde D)u}","rm{x-qx}\\le\\frac{C}{1-C\\varepsilon}\\norm{\\widetilde D x}.\n\\]\nFor a measurement with \\(\\norm{\\widetilde D x-y}\\le\\eta\\), the bound becomes\n\\[\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}(\\norm y+\\eta).\n\\]\nThus the exact space is unchanged and the residual gain degrades by an explicit factor.\n\\end{theorem}\n\\begin{proof}\nLet \\(q=P_{\\ker D}\\), and write \\(u=x-qx\\). The stated kernel inclusion makes both \\(Dqx\\) and \\(\\widetilde Dqx\\) zero. Hence \\(Du=Dx\\) and \\(\\widetilde Du=\\widetilde Dx\\). Applying the reference residual bound to \\(u\\), whose projection onto \\(\\ker D\\) is zero, gives\n\\[\n \\begin{aligned}\n \\norm u&\\le C\\norm{Du}\n \\le C\\norm{\\widetilde Du}+C\\norm{(D-\\widetilde D)u}","frac{C}{1-C\\varepsilon}\\norm{\\widetilde D x}.\n\\]\nFor a measurement with \\(\\norm{\\widetilde D x-y}\\le\\eta\\), the bound becomes\n\\[\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}(\\norm y+\\eta).\n\\]\nThus the exact space is unchanged and the residual gain degrades by an explicit factor.\n\\end{theorem}\n\\begin{proof}\nLet \\(q=P_{\\ker D}\\), and write \\(u=x-qx\\). The stated kernel inclusion makes both \\(Dqx\\) and \\(\\widetilde Dqx\\) zero. Hence \\(Du=Dx\\) and \\(\\widetilde Du=\\widetilde Dx\\). Applying the reference residual bound to \\(u\\), whose projection onto \\(\\ker D\\) is zero, gives\n\\[\n \\begin{aligned}\n \\norm u&\\le C\\norm{Du}\n \\le C\\norm{\\widetilde Du}+C\\norm{(D-\\widetilde D)u}","idetilde D x}.\n\\]\nFor a measurement with \\(\\norm{\\widetilde D x-y}\\le\\eta\\), the bound becomes\n\\[\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}(\\norm y+\\eta).\n\\]\nThus the exact space is unchanged and the residual gain degrades by an explicit factor.\n\\end{theorem}\n\\begin{proof}\nLet \\(q=P_{\\ker D}\\), and write \\(u=x-qx\\). The stated kernel inclusion makes both \\(Dqx\\) and \\(\\widetilde Dqx\\) zero. Hence \\(Du=Dx\\) and \\(\\widetilde Du=\\widetilde Dx\\). Applying the reference residual bound to \\(u\\), whose projection onto \\(\\ker D\\) is zero, gives\n\\[\n \\begin{aligned}\n \\norm u&\\le C\\norm{Du}\n \\le C\\norm{\\widetilde Du}+C\\norm{(D-\\widetilde D)u}","\\]\nFor a measurement with \\(\\norm{\\widetilde D x-y}\\le\\eta\\), the bound becomes\n\\[\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}(\\norm y+\\eta).\n\\]\nThus the exact space is unchanged and the residual gain degrades by an explicit factor.\n\\end{theorem}\n\\begin{proof}\nLet \\(q=P_{\\ker D}\\), and write \\(u=x-qx\\). The stated kernel inclusion makes both \\(Dqx\\) and \\(\\widetilde Dqx\\) zero. Hence \\(Du=Dx\\) and \\(\\widetilde Du=\\widetilde Dx\\). Applying the reference residual bound to \\(u\\), whose projection onto \\(\\ker D\\) is zero, gives\n\\[\n \\begin{aligned}\n \\norm u&\\le C\\norm{Du}\n \\le C\\norm{\\widetilde Du}+C\\norm{(D-\\widetilde D)u}","ecomes\n\\[\n \\norm{x-qx}\\le\\frac{C}{1-C\\varepsilon}(\\norm y+\\eta).\n\\]\nThus the exact space is unchanged and the residual gain degrades by an explicit factor.\n\\end{theorem}\n\\begin{proof}\nLet \\(q=P_{\\ker D}\\), and write \\(u=x-qx\\). The stated kernel inclusion makes both \\(Dqx\\) and \\(\\widetilde Dqx\\) zero. Hence \\(Du=Dx\\) and \\(\\widetilde Du=\\widetilde Dx\\). Applying the reference residual bound to \\(u\\), whose projection onto \\(\\ker D\\) is zero, gives\n\\[\n \\begin{aligned}\n \\norm u&\\le C\\norm{Du}\n \\le C\\norm{\\widetilde Du}+C\\norm{(D-\\widetilde D)u}","c{C}{1-C\\varepsilon}(\\norm y+\\eta).\n\\]\nThus the exact space is unchanged and the residual gain degrades by an explicit factor.\n\\end{theorem}\n\\begin{proof}\nLet \\(q=P_{\\ker D}\\), and write \\(u=x-qx\\). The stated kernel inclusion makes both \\(Dqx\\) and \\(\\widetilde Dqx\\) zero. Hence \\(Du=Dx\\) and \\(\\widetilde Du=\\widetilde Dx\\). Applying the reference residual bound to \\(u\\), whose projection onto \\(\\ker D\\) is zero, gives\n\\[\n \\begin{aligned}\n \\norm u&\\le C\\norm{Du}\n \\le C\\norm{\\widetilde Du}+C\\norm{(D-\\widetilde D)u}","y+\\eta).\n\\]\nThus the exact space is unchanged and the residual gain degrades by an explicit factor.\n\\end{theorem}\n\\begin{proof}\nLet \\(q=P_{\\ker D}\\), and write \\(u=x-qx\\). The stated kernel inclusion makes both \\(Dqx\\) and \\(\\widetilde Dqx\\) zero. Hence \\(Du=Dx\\) and \\(\\widetilde Du=\\widetilde Dx\\). Applying the reference residual bound to \\(u\\), whose projection onto \\(\\ker D\\) is zero, gives\n\\[\n \\begin{aligned}\n \\norm u&\\le C\\norm{Du}\n \\le C\\norm{\\widetilde Du}+C\\norm{(D-\\widetilde D)u}\\\\\n &\\le C\\norm{\\widetilde Dx}+C\\varepsilon\\norm u.\n \\end{aligned}\n\\]\nSubtract the final term. The strict inequality \\(C\\varepsilon<1\\) makes \\(1-C\\varepsilon\\) positive, so division preserves the inequality and gives the claimed residual estimate. This also covers \\(C=0\\) without division by \\(C\\).","&\\le C\\norm{\\widetilde Dx}+C\\varepsilon\\norm u.\n \\end{aligned}\n\\]\nSubtract the final term. The strict inequality \\(C\\varepsilon<1\\) makes \\(1-C\\varepsilon\\) positive, so division preserves the inequality and gives the claimed residual estimate. This also covers \\(C=0\\) without division by \\(C\\).","If \\(\\widetilde Dx=0\\), the same estimate forces \\(u=0\\), hence \\(x=qx\\in\\ker D\\). This is the inclusion not initially assumed. Combining it with the supplied inclusion proves \\(\\ker\\widetilde D=\\ker D\\), and their orthogonal projections coincide in the fixed Hilbert geometry. Finally \\(\\norm{\\widetilde Dx-y}\\le\\eta\\) gives \\(\\norm{\\widetilde Dx}\\le\\norm y+\\eta\\). Substitution proves the measured certificate. The kernel-preservation hypothesis is what removed \\(qx\\) before applying the perturbation estimate. Without it, one obtains a term proportional to \\(\\norm x\\), not the absorbable term proportional to \\(\\norm u\\).\n\\end{proof}\n\\source{KS70}\nThis result does not require the reference operator to be surjectiv","If \\(\\widetilde Dx=0\\), the same estimate forces \\(u=0\\), hence \\(x=qx\\in\\ker D\\). This is the inclusion not initially assumed. Combining it with the supplied inclusion proves \\(\\ker\\widetilde D=\\ker D\\), and their orthogonal projections coincide in the fixed Hilbert geometry. Finally \\(\\norm{\\widetilde Dx-y}\\le\\eta\\) gives \\(\\norm{\\widetilde Dx}\\le\\norm y+\\eta\\). Substitution proves the measured certificate. The kernel-preservation hypothesis is what removed \\(qx\\) before applying the perturbation estimate. Without it, one obtains a term proportional to \\(\\norm x\\), not the absorbable term proportional to \\(\\norm u\\).\n\\end{proof}\n\\source{KS70}\nThis result does not require the reference operator to be surjective. It is different from the split-","If \\(\\widetilde Dx=0\\), the same estimate forces \\(u=0\\), hence \\(x=qx\\in\\ker D\\). This is the inclusion not initially assumed. Combining it with the supplied inclusion proves \\(\\ker\\widetilde D=\\ker D\\), and their orthogonal projections coincide in the fixed Hilbert geometry. Finally \\(\\norm{\\widetilde Dx-y}\\le\\eta\\) gives \\(\\norm{\\widetilde Dx}\\le\\norm y+\\eta\\). Substitution proves the measured certificate. The kernel-preservation hypothesis is what removed \\(qx\\) before applying the perturbation estimate. Without it, one obtains a term proportional to \\(\\norm x\\), not the absorbable term proportional to \\(\\norm u\\).\n\\end{proof}\n\\source{KS70}\nThis result does not require the reference operator to be surjective. It is different from the split-surjective perturbation t","If \\(\\widetilde Dx=0\\), the same estimate forces \\(u=0\\), hence \\(x=qx\\in\\ker D\\). This is the inclusion not initially assumed. Combining it with the supplied inclusion proves \\(\\ker\\widetilde D=\\ker D\\), and their orthogonal projections coincide in the fixed Hilbert geometry. Finally \\(\\norm{\\widetilde Dx-y}\\le\\eta\\) gives \\(\\norm{\\widetilde Dx}\\le\\norm y+\\eta\\). Substitution proves the measured certificate. The kernel-preservation hypothesis is what removed \\(qx\\) before applying the perturbation estimate. Without it, one obtains a term proportional to \\(\\norm x\\), not the absorbable term proportional to \\(\\norm u\\).\n\\end{proof}\n\\source{KS70}\nThis result does not require the reference operator to be surjective. It is different from the split-surjective perturbation theorem, which allows the kernel to move and constructs an actual transport between the old and new kernels. Both theorems","If \\(\\widetilde Dx=0\\), the same estimate forces \\(u=0\\), hence \\(x=qx\\in\\ker D\\). This is the inclusion not initially assumed. Combining it with the supplied inclusion proves \\(\\ker\\widetilde D=\\ker D\\), and their orthogonal projections coincide in the fixed Hilbert geometry. Finally \\(\\norm{\\widetilde Dx-y}\\le\\eta\\) gives \\(\\norm{\\widetilde Dx}\\le\\norm y+\\eta\\). Substitution proves the measured certificate. The kernel-preservation hypothesis is what removed \\(qx\\) before applying the perturbation estimate. Without it, one obtains a term proportional to \\(\\norm x\\), not the absorbable term proportional to \\(\\norm u\\).\n\\end{proof}\n\\source{KS70}\nThis result does not require the reference operator to be surjective. It is different from the split-surjective perturbation theorem, which allows the kernel to move and constructs an actual transport between the old and new kernels. Both theorems have a strict quantitative radius, but they retain different structural data.","x=qx\\in\\ker D\\). This is the inclusion not initially assumed. Combining it with the supplied inclusion proves \\(\\ker\\widetilde D=\\ker D\\), and their orthogonal projections coincide in the fixed Hilbert geometry. Finally \\(\\norm{\\widetilde Dx-y}\\le\\eta\\) gives \\(\\norm{\\widetilde Dx}\\le\\norm y+\\eta\\). Substitution proves the measured certificate. The kernel-preservation hypothesis is what removed \\(qx\\) before applying the perturbation estimate. Without it, one obtains a term proportional to \\(\\norm x\\), not the absorbable term proportional to \\(\\norm u\\).\n\\end{proof}\n\\source{KS70}\nThis result does not require the reference operator to be surjective. It is different from the split-surjective perturbation theorem, which allows the kernel to move and constructs an actual transport between the old and new kernels. Both theorems have a strict quantitative radius, but they retain different structural data.","usion proves \\(\\ker\\widetilde D=\\ker D\\), and their orthogonal projections coincide in the fixed Hilbert geometry. Finally \\(\\norm{\\widetilde Dx-y}\\le\\eta\\) gives \\(\\norm{\\widetilde Dx}\\le\\norm y+\\eta\\). Substitution proves the measured certificate. The kernel-preservation hypothesis is what removed \\(qx\\) before applying the perturbation estimate. Without it, one obtains a term proportional to \\(\\norm x\\), not the absorbable term proportional to \\(\\norm u\\).\n\\end{proof}\n\\source{KS70}\nThis result does not require the reference operator to be surjective. It is different from the split-surjective perturbation theorem, which allows the kernel to move and constructs an actual transport between the old and new kernels. Both theorems have a strict quantitative radius, but they retain different structural data.","rthogonal projections coincide in the fixed Hilbert geometry. Finally \\(\\norm{\\widetilde Dx-y}\\le\\eta\\) gives \\(\\norm{\\widetilde Dx}\\le\\norm y+\\eta\\). Substitution proves the measured certificate. The kernel-preservation hypothesis is what removed \\(qx\\) before applying the perturbation estimate. Without it, one obtains a term proportional to \\(\\norm x\\), not the absorbable term proportional to \\(\\norm u\\).\n\\end{proof}\n\\source{KS70}\nThis result does not require the reference operator to be surjective. It is different from the split-surjective perturbation theorem, which allows the kernel to move and constructs an actual transport between the old and new kernels. Both theorems have a strict quantitative radius, but they retain different structural data.","For \\(D(a,b)=a\\) and \\(\\widetilde D(a,b)=(1-t)a\\), with \\(0\\le t<1\\), the kernel is preserved and the exact gain is \\(1/(1-t)\\). The bound is attained on transverse inputs. At \\(t=1\\), the changed operator is zero and its kernel becomes all of \\(\\R^2\\). This illustrates why the strict endpoint cannot simply be included.","\\plate{22_gain_budget}{For the kernel-preserving family \\(\\widetilde D_t(a,b)=(1-t)a\\), the exact residual gain is \\(1/(1-t)\\). It deteriorates as the strict budget is exhausted. At the excluded endpoint the kernel changes, so the finite-radius estimate does not extend by substituting \\(t=1\\).}{fig:gain-budget}","\\subsection{Norm conventions are part of the certificate}\nOn a finite family of real Hilbert fibers, the maximum norm and the Euclidean product norm satisfy\n\\[\n \\norm x_\\infty\\le\\norm x_2\\le\\sqrt{n}\\norm x_\\infty,\n\\]\nwhere \\(n\\) is the number of components. The first inequality follows because each squared component is one term in the sum. The second follows by bounding every component by the maximum. These inequalities transfer bounds with explicit dimension factors. They do not assert that the same numerical best constant remains optimal.","\\subsection{Extended directed distance}\nLet \\(\\mathcal C\\) be a category and let \\(W\\) assign each arrow a value in the extended nonnegative reals \\([0,\\infty]\\). Suppose\n\\[\n W(\\id_A)=0,\\qquad W(gf)\\le W(f)+W(g).\n\\]\nThe order of ordinary composition is first \\(f\\), then \\(g\\). Define\n\\[\n d_W(A,B)=\\inf_{f:A\\to B}W(f).\n\\]\nThe infimum of an empty family is \\(\\infty\\). No assumption of connectedness, finite cost, symmetry, or attainment is made.","\\begin{proposition}[Directed cost geometry]\\label{KS78}\nFor every arrow \\(f:A\\to B\\),\n\\[\n d_W(A,B)\\le W(f),\\qquad d_W(A,A)=0,\n\\]\nand for all objects,\n\\[\n d_W(A,C)\\le d_W(A,B)+d_W(B,C).\n\\]\nThe symmetric function\n\\[\n d_W^{\\mathrm{sym}}(A,B)=\\max\\{d_W(A,B),d_W(B,A)\\}\n\\]\nhas zero diagonal, is symmetric, and satisfies the extended triangle inequality. The relation \\(d_W^{\\mathrm{sym}}(A,B)=0\\) is an equivalence relation.\n\\end{proposition}\n\\begin{proof}\nThe infimum is bounded above by each of its entries. Apply this to the identity to obtain the zero diagonal. If either term on the right of the triangle inequality is infinite, there is nothing to prove. Otherwise, for every \\(\\varepsilon>0\\), choose arrows with costs less than the corresponding infima plus \\(\\varepsilon\\). Thei","\\begin{proposition}[Directed cost geometry]\\label{KS78}\nFor every arrow \\(f:A\\to B\\),\n\\[\n d_W(A,B)\\le W(f),\\qquad d_W(A,A)=0,\n\\]\nand for all objects,\n\\[\n d_W(A,C)\\le d_W(A,B)+d_W(B,C).\n\\]\nThe symmetric function\n\\[\n d_W^{\\mathrm{sym}}(A,B)=\\max\\{d_W(A,B),d_W(B,A)\\}\n\\]\nhas zero diagonal, is symmetric, and satisfies the extended triangle inequality. The relation \\(d_W^{\\mathrm{sym}}(A,B)=0\\) is an equivalence relation.\n\\end{proposition}\n\\begin{proof}\nThe infimum is bounded above by each of its entries. Apply this to the identity to obtain the zero diagonal. If either term on the right of the triangle inequality is infinite, there is nothing to prove. Otherwise, for every \\(\\varepsilon>0\\), choose arrows with costs less than the corresponding infima plus \\(\\varepsilon\\). Their composite has cost less than\n\\(d_W(A,B)+d_","\\begin{proposition}[Directed cost geometry]\\label{KS78}\nFor every arrow \\(f:A\\to B\\),\n\\[\n d_W(A,B)\\le W(f),\\qquad d_W(A,A)=0,\n\\]\nand for all objects,\n\\[\n d_W(A,C)\\le d_W(A,B)+d_W(B,C).\n\\]\nThe symmetric function\n\\[\n d_W^{\\mathrm{sym}}(A,B)=\\max\\{d_W(A,B),d_W(B,A)\\}\n\\]\nhas zero diagonal, is symmetric, and satisfies the extended triangle inequality. The relation \\(d_W^{\\mathrm{sym}}(A,B)=0\\) is an equivalence relation.\n\\end{proposition}\n\\begin{proof}\nThe infimum is bounded above by each of its entries. Apply this to the identity to obtain the zero diagonal. If either term on the right of the triangle inequality is infinite, there is nothing to prove. Otherwise, for every \\(\\varepsilon>0\\), choose arrows with costs less than the corresponding infima plus \\(\\varepsilon\\). Their composite has cost less than\n\\(d_W(A,B)+d_W(B,C)+2\\varepsilon\\). Letting \\(\\varepsilon\\) tend to zero","\\begin{proposition}[Directed cost geometry]\\label{KS78}\nFor every arrow \\(f:A\\to B\\),\n\\[\n d_W(A,B)\\le W(f),\\qquad d_W(A,A)=0,\n\\]\nand for all objects,\n\\[\n d_W(A,C)\\le d_W(A,B)+d_W(B,C).\n\\]\nThe symmetric function\n\\[\n d_W^{\\mathrm{sym}}(A,B)=\\max\\{d_W(A,B),d_W(B,A)\\}\n\\]\nhas zero diagonal, is symmetric, and satisfies the extended triangle inequality. The relation \\(d_W^{\\mathrm{sym}}(A,B)=0\\) is an equivalence relation.\n\\end{proposition}\n\\begin{proof}\nThe infimum is bounded above by each of its entries. Apply this to the identity to obtain the zero diagonal. If either term on the right of the triangle inequality is infinite, there is nothing to prove. Otherwise, for every \\(\\varepsilon>0\\), choose arrows with costs less than the corresponding infima plus \\(\\varepsilon\\). Their composite has cost less than\n\\(d_W(A,B)+d_W(B,C)+2\\varepsilon\\). Letting \\(\\varepsilon\\) tend to zero gives the result. This argument does not require an optimizing arrow.","agonal, is symmetric, and satisfies the extended triangle inequality. The relation \\(d_W^{\\mathrm{sym}}(A,B)=0\\) is an equivalence relation.\n\\end{proposition}\n\\begin{proof}\nThe infimum is bounded above by each of its entries. Apply this to the identity to obtain the zero diagonal. If either term on the right of the triangle inequality is infinite, there is nothing to prove. Otherwise, for every \\(\\varepsilon>0\\), choose arrows with costs less than the corresponding infima plus \\(\\varepsilon\\). Their composite has cost less than\n\\(d_W(A,B)+d_W(B,C)+2\\varepsilon\\). Letting \\(\\varepsilon\\) tend to zero gives the result. This argument does not require an optimizing arrow.","(d_W^{\\mathrm{sym}}(A,B)=0\\) is an equivalence relation.\n\\end{proposition}\n\\begin{proof}\nThe infimum is bounded above by each of its entries. Apply this to the identity to obtain the zero diagonal. If either term on the right of the triangle inequality is infinite, there is nothing to prove. Otherwise, for every \\(\\varepsilon>0\\), choose arrows with costs less than the corresponding infima plus \\(\\varepsilon\\). Their composite has cost less than\n\\(d_W(A,B)+d_W(B,C)+2\\varepsilon\\). Letting \\(\\varepsilon\\) tend to zero gives the result. This argument does not require an optimizing arrow.","ion.\n\\end{proposition}\n\\begin{proof}\nThe infimum is bounded above by each of its entries. Apply this to the identity to obtain the zero diagonal. If either term on the right of the triangle inequality is infinite, there is nothing to prove. Otherwise, for every \\(\\varepsilon>0\\), choose arrows with costs less than the corresponding infima plus \\(\\varepsilon\\). Their composite has cost less than\n\\(d_W(A,B)+d_W(B,C)+2\\varepsilon\\). Letting \\(\\varepsilon\\) tend to zero gives the result. This argument does not require an optimizing arrow.","imum is bounded above by each of its entries. Apply this to the identity to obtain the zero diagonal. If either term on the right of the triangle inequality is infinite, there is nothing to prove. Otherwise, for every \\(\\varepsilon>0\\), choose arrows with costs less than the corresponding infima plus \\(\\varepsilon\\). Their composite has cost less than\n\\(d_W(A,B)+d_W(B,C)+2\\varepsilon\\). Letting \\(\\varepsilon\\) tend to zero gives the result. This argument does not require an optimizing arrow.","Each of the two directed distances from \\(A\\) to \\(C\\) and from \\(C\\) to \\(A\\) is bounded by \\(d_W^{\\mathrm{sym}}(A,B)+d_W^{\\mathrm{sym}}(B,C)\\). Taking their maximum proves the symmetric triangle law. Zero diagonal and symmetry give reflexivity and symmetry of the zero-distance relation. The triangle inequality and nonnegativity give transitivity.\n\\end{proof}\n\\source{KS78}\nThe formal source proves these laws directly in \\(\\R_{\\ge0}^{\\infty}\\) and constructs the zero-distance setoid. It does not identify zero distance with categorical isomorphism. Passing to the quotient can separate zero-distance classes, but would still not prove that those classes are the intended semantic objects.","stances from \\(A\\) to \\(C\\) and from \\(C\\) to \\(A\\) is bounded by \\(d_W^{\\mathrm{sym}}(A,B)+d_W^{\\mathrm{sym}}(B,C)\\). Taking their maximum proves the symmetric triangle law. Zero diagonal and symmetry give reflexivity and symmetry of the zero-distance relation. The triangle inequality and nonnegativity give transitivity.\n\\end{proof}\n\\source{KS78}\nThe formal source proves these laws directly in \\(\\R_{\\ge0}^{\\infty}\\) and constructs the zero-distance setoid. It does not identify zero distance with categorical isomorphism. Passing to the quotient can separate zero-distance classes, but would still not prove that those classes are the intended semantic objects.","\\subsection{A concrete path category}\nConsider four vertices \\(A,B,C,D\\), and the free path category generated by\n\\[\n A\\xrightarrow{0}B,\\qquad B\\xrightarrow{0}A,\\qquad\n A\\xrightarrow{2}C,\\qquad C\\xrightarrow{5}A.\n\\]\nThere are no arrows incident to \\(D\\), apart from its identity. A path costs the sum of its edge labels. Concatenation is additive and empty paths cost zero, so all cost axioms hold. The directed distances are\n\\[\n \\begin{array}{c|rrrr}\n d_W&A&B&C&D","category}\nConsider four vertices \\(A,B,C,D\\), and the free path category generated by\n\\[\n A\\xrightarrow{0}B,\\qquad B\\xrightarrow{0}A,\\qquad\n A\\xrightarrow{2}C,\\qquad C\\xrightarrow{5}A.\n\\]\nThere are no arrows incident to \\(D\\), apart from its identity. A path costs the sum of its edge labels. Concatenation is additive and empty paths cost zero, so all cost axioms hold. The directed distances are\n\\[\n \\begin{array}{c|rrrr}\n d_W&A&B&C&D\\\\\\hline\n A&0&0&2&\\infty\\\\\n B&0&0&2&\\infty\\\\\n C&5&5&0&\\infty\\\\\n D&\\infty&\\infty&\\infty&0\n \\end{array}\n\\]\nFor example, the path from \\(B\\) through \\(A\\) to \\(C\\) costs two. Every path from \\(C\\) to \\(B\\) must first pay the edge cost five and can then move from \\(A\\) to \\(B\\) for free. Nonnegative loops cannot improve either cost.","D&\\infty&\\infty&\\infty&0\n \\end{array}\n\\]\nFor example, the path from \\(B\\) through \\(A\\) to \\(C\\) costs two. Every path from \\(C\\) to \\(B\\) must first pay the edge cost five and can then move from \\(A\\) to \\(B\\) for free. Nonnegative loops cannot improve either cost.","\\plate{29_transport_cost}{Directed path costs need not be symmetric or finite. Vertices $A$ and $B$ have zero cost in both directions, but they are not isomorphic in the free path category. The distinct zero-cost arrows have no imposed inverse relations.}{fig:transport-cost}","The symmetric zero-distance classes are \\(\\{A,B\\}\\), \\(\\{C\\}\\), and \\(\\{D\\}\\). The distance between the first two classes is five. The third class is infinitely far from each of them. Yet \\(A\\) and \\(B\\) are not isomorphic in this category. Any path from one to the other has positive length, so concatenating it with a return path cannot give the length-zero identity. This is a categorical distinction even though the numerical cost is zero.","\\subsection{Why compatibility alone can collapse the geometry}\n\\begin{proposition}[The zero-arrow collapse]\\label{KS79}\nSuppose \\(\\mathcal C\\) has zero morphisms and \\(W(0:A\\to B)=0\\) for every pair of objects. Then\n\\[\n d_W(A,B)=0\\quad\\text{for all }A,B.\n\\]\n\\end{proposition}\n\\begin{proof}\nThe zero arrow is one member of the family over which the infimum is taken. Thus \\(0\\le d_W(A,B)\\le W(0)=0\\).\n\\end{proof}\n\\source{KS79}\nFor the preadditive tower differential, the zero family of maps is exact. Consequently any proposed nonnegative arrow cost that vanishes on every exact map also vanishes on every zero map. If it satisfies the cost axioms at all, its object distance collapses by the proposition.","The qualification is important. A raw operator defect norm need not satisfy the unweighted composition inequality required of \\(W\\). The transport Leibniz identity gives coefficients involving the surrounding maps. It is not legitimate to first call an arbitrary defect norm a subadditive cost and then use the metric construction.","\\subsection{Preservation changes the admissible arrows}\nOne response to collapse is to restrict arrows to those satisfying a declared preservation contract. For objects equipped with maps \\(p_A:A\\to S\\), a morphism can be required to satisfy\n\\[\n p_Bf=p_A.\n\\]\nIdentities and composition preserve this equation. Unless \\(p_A=0\\), the zero arrow is not admissible. In other applications the arrows might be required to be isomorphisms or to retain a particular observable. These are different choices of category, not automatic consequences of small residuals.","\\subsection{The two limits that must be established}\nLet \\((M,\\dist)\\) be a complete metric space. Let \\(\\omega:M\\to[0,\\infty)\\) be a nonnegative residual function, and call a point exact when \\(\\omega(x)=0\\). We regard \\(\\omega\\) as a real-valued function when discussing continuity. Assume it is lower semicontinuous, so that every sublevel set \\(\\{x:\\omega(x)\\le a\\}\\) is closed.","For an approximation sequence \\(f_n\\), the statements\n\\[\n \\omega(f_n)\\longrightarrow0\n \\quad\\text{and}\\quad\n f_n\\longrightarrow x\n\\]\nare different. The first concerns the observed discrepancy. The second concerns the mathematical candidate in the chosen metric. Neither implies the other without additional hypotheses.","\\begin{theorem}[Summable movement gives exactification]\\label{KS71}\nSuppose \\(b_n\\ge0\\), \\(\\sum_{n=0}^{\\infty}b_n<\\infty\\), and\n\\[\n \\dist(f_n,f_{n+1})\\le b_n,\\qquad \\omega(f_n)\\longrightarrow0.\n\\]\nThen there is a limit \\(x\\in M\\) satisfying\n\\[\n f_n\\longrightarrow x,\\qquad\\omega(x)=0,\\qquad\n \\dist(f_n,x)\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary pro","\\begin{theorem}[Summable movement gives exactification]\\label{KS71}\nSuppose \\(b_n\\ge0\\), \\(\\sum_{n=0}^{\\infty}b_n<\\infty\\), and\n\\[\n \\dist(f_n,f_{n+1})\\le b_n,\\qquad \\omega(f_n)\\longrightarrow0.\n\\]\nThen there is a limit \\(x\\in M\\) satisfying\n\\[\n f_n\\longrightarrow x,\\qquad\\omega(x)=0,\\qquad\n \\dist(f_n,x)\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pa","\\begin{theorem}[Summable movement gives exactification]\\label{KS71}\nSuppose \\(b_n\\ge0\\), \\(\\sum_{n=0}^{\\infty}b_n<\\infty\\), and\n\\[\n \\dist(f_n,f_{n+1})\\le b_n,\\qquad \\omega(f_n)\\longrightarrow0.\n\\]\nThen there is a limit \\(x\\in M\\) satisfying\n\\[\n f_n\\longrightarrow x,\\qquad\\omega(x)=0,\\qquad\n \\dist(f_n,x)\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hy","\\begin{theorem}[Summable movement gives exactification]\\label{KS71}\nSuppose \\(b_n\\ge0\\), \\(\\sum_{n=0}^{\\infty}b_n<\\infty\\), and\n\\[\n \\dist(f_n,f_{n+1})\\le b_n,\\qquad \\omega(f_n)\\longrightarrow0.\n\\]\nThen there is a limit \\(x\\in M\\) satisfying\n\\[\n f_n\\longrightarrow x,\\qquad\\omega(x)=0,\\qquad\n \\dist(f_n,x)\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limi","\\begin{theorem}[Summable movement gives exactification]\\label{KS71}\nSuppose \\(b_n\\ge0\\), \\(\\sum_{n=0}^{\\infty}b_n<\\infty\\), and\n\\[\n \\dist(f_n,f_{n+1})\\le b_n,\\qquad \\omega(f_n)\\longrightarrow0.\n\\]\nThen there is a limit \\(x\\in M\\) satisfying\n\\[\n f_n\\longrightarrow x,\\qquad\\omega(x)=0,\\qquad\n \\dist(f_n,x)\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limit \\(x\\in M\\).","on]\\label{KS71}\nSuppose \\(b_n\\ge0\\), \\(\\sum_{n=0}^{\\infty}b_n<\\infty\\), and\n\\[\n \\dist(f_n,f_{n+1})\\le b_n,\\qquad \\omega(f_n)\\longrightarrow0.\n\\]\nThen there is a limit \\(x\\in M\\) satisfying\n\\[\n f_n\\longrightarrow x,\\qquad\\omega(x)=0,\\qquad\n \\dist(f_n,x)\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limit \\(x\\in M\\).","\\(\\sum_{n=0}^{\\infty}b_n<\\infty\\), and\n\\[\n \\dist(f_n,f_{n+1})\\le b_n,\\qquad \\omega(f_n)\\longrightarrow0.\n\\]\nThen there is a limit \\(x\\in M\\) satisfying\n\\[\n f_n\\longrightarrow x,\\qquad\\omega(x)=0,\\qquad\n \\dist(f_n,x)\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limit \\(x\\in M\\).","ere is a limit \\(x\\in M\\) satisfying\n\\[\n f_n\\longrightarrow x,\\qquad\\omega(x)=0,\\qquad\n \\dist(f_n,x)\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limit \\(x\\in M\\).","ad\\omega(x)=0,\\qquad\n \\dist(f_n,x)\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limit \\(x\\in M\\).","m_{k=n}^{\\infty}b_k.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limit \\(x\\in M\\).","end{theorem}\n\\begin{proof}\nFor integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limit \\(x\\in M\\).","For integers \\(m>n\\), apply the triangle inequality successively along the finite chain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limit \\(x\\in M\\).","ain \\(f_n,f_{n+1},\\ldots,f_m\\). It gives\n\\[\n \\dist(f_n,f_m)\\le\\sum_{k=n}^{m-1}\\dist(f_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limit \\(x\\in M\\).","_k,f_{k+1})\n \\le\\sum_{k=n}^{m-1}b_k\\le\\sum_{k=n}^{\\infty}b_k.\n\\]\nGiven \\(\\varepsilon>0\\), convergence of the nonnegative series supplies \\(N\\) for which the final tail is below \\(\\varepsilon\\) whenever \\(n\\ge N\\). Interchanging \\(m,n\\) when necessary proves the Cauchy condition for all pairs of indices at least \\(N\\). Completeness, not residual decay, is the hypothesis that now supplies a limit \\(x\\in M\\).","Keep \\(n\\) fixed. The reverse triangle inequality says\n\\[\n \\bigl|\\dist(f_n,f_m)-\\dist(f_n,x)\\bigr|\\le\\dist(f_m,x)\\longrightarrow0.\n\\]\nPassing to the limit in the uniform tail bound proves the claimed distance certificate at that \\(n\\).","It remains to prove exactness of the limit. Fix \\(\\varepsilon>0\\). Since \\(\\omega(f_m)\\to0\\), an entire tail of the sequence belongs to the sublevel set \\(C_\\varepsilon=\\{z:\\omega(z)\\le\\varepsilon\\}\\). Lower semicontinuity makes this set closed, so the limit of that tail belongs to it. Hence \\(\\omega(x)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\). If \\(\\omega(x)>0\\), choosing \\(\\varepsilon=\\omega(x)/2\\) contradicts this inequality. Since \\(\\omega\\) is nonnegative, \\(\\omega(x)=0\\). We have not used, or asserted, continuity of \\(\\omega\\) or the stronger equality \\(\\omega(f_m)\\to\\omega(x)\\) for arbitrary convergent inputs.\n\\end{proof}\n\\source{KS71}\nThe formal theorem is slightly more general: it is stated in a complete pseudometric space. In a metri","It remains to prove exactness of the limit. Fix \\(\\varepsilon>0\\). Since \\(\\omega(f_m)\\to0\\), an entire tail of the sequence belongs to the sublevel set \\(C_\\varepsilon=\\{z:\\omega(z)\\le\\varepsilon\\}\\). Lower semicontinuity makes this set closed, so the limit of that tail belongs to it. Hence \\(\\omega(x)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\). If \\(\\omega(x)>0\\), choosing \\(\\varepsilon=\\omega(x)/2\\) contradicts this inequality. Since \\(\\omega\\) is nonnegative, \\(\\omega(x)=0\\). We have not used, or asserted, continuity of \\(\\omega\\) or the stronger equality \\(\\omega(f_m)\\to\\omega(x)\\) for arbitrary convergent inputs.\n\\end{proof}\n\\source{KS71}\nThe formal theorem is slightly more general: it is stated in a complete pseudometric space. In a metric space the limit of this f","It remains to prove exactness of the limit. Fix \\(\\varepsilon>0\\). Since \\(\\omega(f_m)\\to0\\), an entire tail of the sequence belongs to the sublevel set \\(C_\\varepsilon=\\{z:\\omega(z)\\le\\varepsilon\\}\\). Lower semicontinuity makes this set closed, so the limit of that tail belongs to it. Hence \\(\\omega(x)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\). If \\(\\omega(x)>0\\), choosing \\(\\varepsilon=\\omega(x)/2\\) contradicts this inequality. Since \\(\\omega\\) is nonnegative, \\(\\omega(x)=0\\). We have not used, or asserted, continuity of \\(\\omega\\) or the stronger equality \\(\\omega(f_m)\\to\\omega(x)\\) for arbitrary convergent inputs.\n\\end{proof}\n\\source{KS71}\nThe formal theorem is slightly more general: it is stated in a complete pseudometric space. In a metric space the limit of this fixed sequence is unique. That does not mean different admissible approximation procedures select the same ex","It remains to prove exactness of the limit. Fix \\(\\varepsilon>0\\). Since \\(\\omega(f_m)\\to0\\), an entire tail of the sequence belongs to the sublevel set \\(C_\\varepsilon=\\{z:\\omega(z)\\le\\varepsilon\\}\\). Lower semicontinuity makes this set closed, so the limit of that tail belongs to it. Hence \\(\\omega(x)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\). If \\(\\omega(x)>0\\), choosing \\(\\varepsilon=\\omega(x)/2\\) contradicts this inequality. Since \\(\\omega\\) is nonnegative, \\(\\omega(x)=0\\). We have not used, or asserted, continuity of \\(\\omega\\) or the stronger equality \\(\\omega(f_m)\\to\\omega(x)\\) for arbitrary convergent inputs.\n\\end{proof}\n\\source{KS71}\nThe formal theorem is slightly more general: it is stated in a complete pseudometric space. In a metric space the limit of this fixed sequence is unique. That does not mean different admissible approximation procedures select the same exact point.","varepsilon>0\\). Since \\(\\omega(f_m)\\to0\\), an entire tail of the sequence belongs to the sublevel set \\(C_\\varepsilon=\\{z:\\omega(z)\\le\\varepsilon\\}\\). Lower semicontinuity makes this set closed, so the limit of that tail belongs to it. Hence \\(\\omega(x)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\). If \\(\\omega(x)>0\\), choosing \\(\\varepsilon=\\omega(x)/2\\) contradicts this inequality. Since \\(\\omega\\) is nonnegative, \\(\\omega(x)=0\\). We have not used, or asserted, continuity of \\(\\omega\\) or the stronger equality \\(\\omega(f_m)\\to\\omega(x)\\) for arbitrary convergent inputs.\n\\end{proof}\n\\source{KS71}\nThe formal theorem is slightly more general: it is stated in a complete pseudometric space. In a metric space the limit of this fixed sequence is unique. That does not mean different admissible approximation procedures select the same exact point.","ega(f_m)\\to0\\), an entire tail of the sequence belongs to the sublevel set \\(C_\\varepsilon=\\{z:\\omega(z)\\le\\varepsilon\\}\\). Lower semicontinuity makes this set closed, so the limit of that tail belongs to it. Hence \\(\\omega(x)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\). If \\(\\omega(x)>0\\), choosing \\(\\varepsilon=\\omega(x)/2\\) contradicts this inequality. Since \\(\\omega\\) is nonnegative, \\(\\omega(x)=0\\). We have not used, or asserted, continuity of \\(\\omega\\) or the stronger equality \\(\\omega(f_m)\\to\\omega(x)\\) for arbitrary convergent inputs.\n\\end{proof}\n\\source{KS71}\nThe formal theorem is slightly more general: it is stated in a complete pseudometric space. In a metric space the limit of this fixed sequence is unique. That does not mean different admissible approximation procedures select the same exact point.","sublevel set \\(C_\\varepsilon=\\{z:\\omega(z)\\le\\varepsilon\\}\\). Lower semicontinuity makes this set closed, so the limit of that tail belongs to it. Hence \\(\\omega(x)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\). If \\(\\omega(x)>0\\), choosing \\(\\varepsilon=\\omega(x)/2\\) contradicts this inequality. Since \\(\\omega\\) is nonnegative, \\(\\omega(x)=0\\). We have not used, or asserted, continuity of \\(\\omega\\) or the stronger equality \\(\\omega(f_m)\\to\\omega(x)\\) for arbitrary convergent inputs.\n\\end{proof}\n\\source{KS71}\nThe formal theorem is slightly more general: it is stated in a complete pseudometric space. In a metric space the limit of this fixed sequence is unique. That does not mean different admissible approximation procedures select the same exact point.","ega(z)\\le\\varepsilon\\}\\). Lower semicontinuity makes this set closed, so the limit of that tail belongs to it. Hence \\(\\omega(x)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\). If \\(\\omega(x)>0\\), choosing \\(\\varepsilon=\\omega(x)/2\\) contradicts this inequality. Since \\(\\omega\\) is nonnegative, \\(\\omega(x)=0\\). We have not used, or asserted, continuity of \\(\\omega\\) or the stronger equality \\(\\omega(f_m)\\to\\omega(x)\\) for arbitrary convergent inputs.\n\\end{proof}\n\\source{KS71}\nThe formal theorem is slightly more general: it is stated in a complete pseudometric space. In a metric space the limit of this fixed sequence is unique. That does not mean different admissible approximation procedures select the same exact point.","\\subsection{Variable rates and a movement budget}\nLet \\(q_n,C_n\\ge0\\), and define the prefix product\n\\[\n Q_0=1,\\qquad Q_{n+1}=q_nQ_n.\n\\]\nAssume\n\\begin{equation}\\label{eq:variable-rate}\n \\omega(f_{n+1})\\le q_n\\omega(f_n),\\qquad\n \\dist(f_n,f_{n+1})\\le C_n\\omega(f_n).\n\\end{equation}","\\begin{theorem}[Variable-rate exactification]\\label{KS72}\nIf \\(Q_n\\to0\\) and \\(\\sum_n C_nQ_n<\\infty\\), then \\((f_n)\\) has an exact limit and\n\\begin{equation}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero","\\begin{theorem}[Variable-rate exactification]\\label{KS72}\nIf \\(Q_n\\to0\\) and \\(\\sum_n C_nQ_n<\\infty\\), then \\((f_n)\\) has an exact limit and\n\\begin{equation}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption the","\\begin{theorem}[Variable-rate exactification]\\label{KS72}\nIf \\(Q_n\\to0\\) and \\(\\sum_n C_nQ_n<\\infty\\), then \\((f_n)\\) has an exact limit and\n\\begin{equation}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dis","\\begin{theorem}[Variable-rate exactification]\\label{KS72}\nIf \\(Q_n\\to0\\) and \\(\\sum_n C_nQ_n<\\infty\\), then \\((f_n)\\) has an exact limit and\n\\begin{equation}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All a","\\begin{theorem}[Variable-rate exactification]\\label{KS72}\nIf \\(Q_n\\to0\\) and \\(\\sum_n C_nQ_n<\\infty\\), then \\((f_n)\\) has an exact limit and\n\\begin{equation}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold,","\\begin{theorem}[Variable-rate exactification]\\label{KS72}\nIf \\(Q_n\\to0\\) and \\(\\sum_n C_nQ_n<\\infty\\), then \\((f_n)\\) has an exact limit and\n\\begin{equation}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificat","\\begin{theorem}[Variable-rate exactification]\\label{KS72}\nIf \\(Q_n\\to0\\) and \\(\\sum_n C_nQ_n<\\infty\\), then \\((f_n)\\) has an exact limit and\n\\begin{equation}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make","ctification]\\label{KS72}\nIf \\(Q_n\\to0\\) and \\(\\sum_n C_nQ_n<\\infty\\), then \\((f_n)\\) has an exact limit and\n\\begin{equation}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make every step zero and every residual zero. No ca…","<\\infty\\), then \\((f_n)\\) has an exact limit and\n\\begin{equation}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make every step zero and every residual zero. No cancellation of that value was used.","limit and\n\\begin{equation}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make every step zero and every residual zero. No cancellation of that value was used.","n}\\label{eq:variable-tail}\n \\dist(f_n,x)\\le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make every step zero and every residual zero. No cancellation of that value was used.","le\\omega(f_0)\\sum_{k=n}^{\\infty}C_kQ_k.\n\\end{equation}\nFor constant \\(q_n=q\\in[0,1)\\) and constant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make every step zero and every residual zero. No cancellation of that value was used.","nstant \\(C_n=C\\), the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make every step zero and every residual zero. No cancellation of that value was used.",", the sharper restarted estimate is\n\\[\n \\dist(f_n,x)\\le\\frac{C}{1-q}\\,\\omega(f_n).\n\\]\n\\end{theorem}\n\\begin{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make every step zero and every residual zero. No cancellation of that value was used.","n{proof}\nThe estimate \\(\\omega(f_n)\\le Q_n\\omega(f_0)\\) follows by induction. At \\(n=0\\) it is equality. If it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make every step zero and every residual zero. No cancellation of that value was used.","it holds at \\(n\\), nonnegativity of \\(q_n\\) permits multiplication and gives\n\\[\n \\omega(f_{n+1})\\le q_n\\omega(f_n)\n \\le q_nQ_n\\omega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make every step zero and every residual zero. No cancellation of that value was used.","ega(f_0)=Q_{n+1}\\omega(f_0).\n\\]\nBecause \\(Q_n\\to0\\) and \\(\\omega(f_0)\\) is a fixed finite number, the residual tends to zero. The movement assumption then gives\n\\[\n \\dist(f_n,f_{n+1})\\le C_nQ_n\\omega(f_0).\n\\]\nThis is a summable nonnegative budget by the hypothesis on \\(\\sum C_nQ_n\\). All assumptions of Theorem~\\ref{KS71} now hold, and its tail certificate is exactly \\eqref{eq:variable-tail}. If \\(\\omega(f_0)=0\\), the same estimates make every step zero and every residual zero. No cancellation of that value was used.","For constant rates, the global estimate can be sharpened by restarting at the current index. Fix \\(n\\). A new induction on \\(j\\) gives\n\\[\n \\omega(f_{n+j})\\le q^j\\omega(f_n),\\qquad\n \\dist(f_{n+j},f_{n+j+1})\\le Cq^j\\omega(f_n).\n\\]\nSum the finite chain from \\(n\\) to \\(n+m\\), and then let \\(m\\to\\infty\\) using the already established limit. The geometric series equals \\(1/(1-q)\\), including \\(q=0\\), where its first term is one and all later terms vanish. This proves the restarted bound. The variable-rate proof itself requires only nonnegative \\(q_n\\), decay of their cumulative products, and summability of the movement weights. It does not insert an unstated upper bound \\(q_n<1\\) at each step.\n\\end{proof}\n\\source{KS72}\nThe individual factors \\(q_n\\) need not all be below one. Temporary residual","For constant rates, the global estimate can be sharpened by restarting at the current index. Fix \\(n\\). A new induction on \\(j\\) gives\n\\[\n \\omega(f_{n+j})\\le q^j\\omega(f_n),\\qquad\n \\dist(f_{n+j},f_{n+j+1})\\le Cq^j\\omega(f_n).\n\\]\nSum the finite chain from \\(n\\) to \\(n+m\\), and then let \\(m\\to\\infty\\) using the already established limit. The geometric series equals \\(1/(1-q)\\), including \\(q=0\\), where its first term is one and all later terms vanish. This proves the restarted bound. The variable-rate proof itself requires only nonnegative \\(q_n\\), decay of their cumulative products, and summability of the movement weights. It does not insert an unstated upper bound \\(q_n<1\\) at each step.\n\\end{proof}\n\\source{KS72}\nThe individual factors \\(q_n\\) need not all be below one. Temporary residual increases are permitted if","For constant rates, the global estimate can be sharpened by restarting at the current index. Fix \\(n\\). A new induction on \\(j\\) gives\n\\[\n \\omega(f_{n+j})\\le q^j\\omega(f_n),\\qquad\n \\dist(f_{n+j},f_{n+j+1})\\le Cq^j\\omega(f_n).\n\\]\nSum the finite chain from \\(n\\) to \\(n+m\\), and then let \\(m\\to\\infty\\) using the already established limit. The geometric series equals \\(1/(1-q)\\), including \\(q=0\\), where its first term is one and all later terms vanish. This proves the restarted bound. The variable-rate proof itself requires only nonnegative \\(q_n\\), decay of their cumulative products, and summability of the movement weights. It does not insert an unstated upper bound \\(q_n<1\\) at each step.\n\\end{proof}\n\\source{KS72}\nThe individual factors \\(q_n\\) need not all be below one. Temporary residual increases are permitted if the prefix products decay and the weighted movements remain summable. Conversely, a decaying produc","For constant rates, the global estimate can be sharpened by restarting at the current index. Fix \\(n\\). A new induction on \\(j\\) gives\n\\[\n \\omega(f_{n+j})\\le q^j\\omega(f_n),\\qquad\n \\dist(f_{n+j},f_{n+j+1})\\le Cq^j\\omega(f_n).\n\\]\nSum the finite chain from \\(n\\) to \\(n+m\\), and then let \\(m\\to\\infty\\) using the already established limit. The geometric series equals \\(1/(1-q)\\), including \\(q=0\\), where its first term is one and all later terms vanish. This proves the restarted bound. The variable-rate proof itself requires only nonnegative \\(q_n\\), decay of their cumulative products, and summability of the movement weights. It does not insert an unstated upper bound \\(q_n<1\\) at each step.\n\\end{proof}\n\\source{KS72}\nThe individual factors \\(q_n\\) need not all be below one. Temporary residual increases are permitted if the prefix products decay and the weighted movements remain summable. Conversely, a decaying product is not enough when its movemen","For constant rates, the global estimate can be sharpened by restarting at the current index. Fix \\(n\\). A new induction on \\(j\\) gives\n\\[\n \\omega(f_{n+j})\\le q^j\\omega(f_n),\\qquad\n \\dist(f_{n+j},f_{n+j+1})\\le Cq^j\\omega(f_n).\n\\]\nSum the finite chain from \\(n\\) to \\(n+m\\), and then let \\(m\\to\\infty\\) using the already established limit. The geometric series equals \\(1/(1-q)\\), including \\(q=0\\), where its first term is one and all later terms vanish. This proves the restarted bound. The variable-rate proof itself requires only nonnegative \\(q_n\\), decay of their cumulative products, and summability of the movement weights. It does not insert an unstated upper bound \\(q_n<1\\) at each step.\n\\end{proof}\n\\source{KS72}\nThe individual factors \\(q_n\\) need not all be below one. Temporary residual increases are permitted if the prefix products decay and the weighted movements remain summable. Conversely, a decaying product is not enough when its movement budget di","For constant rates, the global estimate can be sharpened by restarting at the current index. Fix \\(n\\). A new induction on \\(j\\) gives\n\\[\n \\omega(f_{n+j})\\le q^j\\omega(f_n),\\qquad\n \\dist(f_{n+j},f_{n+j+1})\\le Cq^j\\omega(f_n).\n\\]\nSum the finite chain from \\(n\\) to \\(n+m\\), and then let \\(m\\to\\infty\\) using the already established limit. The geometric series equals \\(1/(1-q)\\), including \\(q=0\\), where its first term is one and all later terms vanish. This proves the restarted bound. The variable-rate proof itself requires only nonnegative \\(q_n\\), decay of their cumulative products, and summability of the movement weights. It does not insert an unstated upper bound \\(q_n<1\\) at each step.\n\\end{proof}\n\\source{KS72}\nThe individual factors \\(q_n\\) need not all be below one. Temporary residual increases are permitted if the prefix products decay and the weighted movements remain summable. Conversely, a decaying product is not enough when its movement budget diverges.","f_{n+j})\\le q^j\\omega(f_n),\\qquad\n \\dist(f_{n+j},f_{n+j+1})\\le Cq^j\\omega(f_n).\n\\]\nSum the finite chain from \\(n\\) to \\(n+m\\), and then let \\(m\\to\\infty\\) using the already established limit. The geometric series equals \\(1/(1-q)\\), including \\(q=0\\), where its first term is one and all later terms vanish. This proves the restarted bound. The variable-rate proof itself requires only nonnegative \\(q_n\\), decay of their cumulative products, and summability of the movement weights. It does not insert an unstated upper bound \\(q_n<1\\) at each step.\n\\end{proof}\n\\source{KS72}\nThe individual factors \\(q_n\\) need not all be below one. Temporary residual increases are permitted if the prefix products decay and the weighted movements remain summable. Conversely, a decaying product is not enough when its movement budget diverges.","d then let \\(m\\to\\infty\\) using the already established limit. The geometric series equals \\(1/(1-q)\\), including \\(q=0\\), where its first term is one and all later terms vanish. This proves the restarted bound. The variable-rate proof itself requires only nonnegative \\(q_n\\), decay of their cumulative products, and summability of the movement weights. It does not insert an unstated upper bound \\(q_n<1\\) at each step.\n\\end{proof}\n\\source{KS72}\nThe individual factors \\(q_n\\) need not all be below one. Temporary residual increases are permitted if the prefix products decay and the weighted movements remain summable. Conversely, a decaying product is not enough when its movement budget diverges.","ric series equals \\(1/(1-q)\\), including \\(q=0\\), where its first term is one and all later terms vanish. This proves the restarted bound. The variable-rate proof itself requires only nonnegative \\(q_n\\), decay of their cumulative products, and summability of the movement weights. It does not insert an unstated upper bound \\(q_n<1\\) at each step.\n\\end{proof}\n\\source{KS72}\nThe individual factors \\(q_n\\) need not all be below one. Temporary residual increases are permitted if the prefix products decay and the weighted movements remain summable. Conversely, a decaying product is not enough when its movement budget diverges.","\\subsection{A convergent path with no uniform geometric factor}\nIn \\(\\R^2\\) with the maximum norm, fix \\(s\\in\\R\\) and take\n\\[\n f_n=(s,(n+1)^{-1}),\\qquad \\omega(u,r)=|r|.\n\\]\nThe protected coordinate is the first one. Direct calculation gives\n\\[\n q_n=\\frac{n+1}{n+2},\\quad Q_n=\\frac1{n+1},\\quad\n C_n=\\frac1{n+2},\\quad\n \\dist(f_n,f_{n+1})=\\frac1{(n+1)(n+2)}.\n\\]\nNo constant \\(q<1\\) bounds all the ratios \\(q_n\\), since they tend to one. Nevertheless\n\\[\n \\sum_{k=n}^{\\infty}C_kQ_k\n =\\sum_{k=n}^{\\infty}\\left(\\frac1{k+1}-\\frac1{k+2}\\right)\n =\\frac1{n+1}.\n\\]\nThe limit is \\((s,0)\\), and the error bound is an equality. The model shows why the variable-rate theorem is not merely a rephrasing of a fixed geometric contraction.","bsection{A convergent path with no uniform geometric factor}\nIn \\(\\R^2\\) with the maximum norm, fix \\(s\\in\\R\\) and take\n\\[\n f_n=(s,(n+1)^{-1}),\\qquad \\omega(u,r)=|r|.\n\\]\nThe protected coordinate is the first one. Direct calculation gives\n\\[\n q_n=\\frac{n+1}{n+2},\\quad Q_n=\\frac1{n+1},\\quad\n C_n=\\frac1{n+2},\\quad\n \\dist(f_n,f_{n+1})=\\frac1{(n+1)(n+2)}.\n\\]\nNo constant \\(q<1\\) bounds all the ratios \\(q_n\\), since they tend to one. Nevertheless\n\\[\n \\sum_{k=n}^{\\infty}C_kQ_k\n =\\sum_{k=n}^{\\infty}\\left(\\frac1{k+1}-\\frac1{k+2}\\right)\n =\\frac1{n+1}.\n\\]\nThe limit is \\((s,0)\\), and the error bound is an equality. The model shows why the variable-rate theorem is not merely a rephrasing of a fixed geometric contraction.","metric factor}\nIn \\(\\R^2\\) with the maximum norm, fix \\(s\\in\\R\\) and take\n\\[\n f_n=(s,(n+1)^{-1}),\\qquad \\omega(u,r)=|r|.\n\\]\nThe protected coordinate is the first one. Direct calculation gives\n\\[\n q_n=\\frac{n+1}{n+2},\\quad Q_n=\\frac1{n+1},\\quad\n C_n=\\frac1{n+2},\\quad\n \\dist(f_n,f_{n+1})=\\frac1{(n+1)(n+2)}.\n\\]\nNo constant \\(q<1\\) bounds all the ratios \\(q_n\\), since they tend to one. Nevertheless\n\\[\n \\sum_{k=n}^{\\infty}C_kQ_k\n =\\sum_{k=n}^{\\infty}\\left(\\frac1{k+1}-\\frac1{k+2}\\right)\n =\\frac1{n+1}.\n\\]\nThe limit is \\((s,0)\\), and the error bound is an equality. The model shows why the variable-rate theorem is not merely a rephrasing of a fixed geometric contraction.",".\n\\]\nThe protected coordinate is the first one. Direct calculation gives\n\\[\n q_n=\\frac{n+1}{n+2},\\quad Q_n=\\frac1{n+1},\\quad\n C_n=\\frac1{n+2},\\quad\n \\dist(f_n,f_{n+1})=\\frac1{(n+1)(n+2)}.\n\\]\nNo constant \\(q<1\\) bounds all the ratios \\(q_n\\), since they tend to one. Nevertheless\n\\[\n \\sum_{k=n}^{\\infty}C_kQ_k\n =\\sum_{k=n}^{\\infty}\\left(\\frac1{k+1}-\\frac1{k+2}\\right)\n =\\frac1{n+1}.\n\\]\nThe limit is \\((s,0)\\), and the error bound is an equality. The model shows why the variable-rate theorem is not merely a rephrasing of a fixed geometric contraction.","\\plate{25_summable_path}{Polynomial residual decay can still give an exact limit with a sharp error certificate. The residual $1/(n+1)$ is not summable, but the actual movement $1/((n+1)(n+2))$ is. The theorem uses the movement series.}{fig:summable-path}","\\subsection{Residual bursts can be compatible with convergence}\nSet \\(q_{2k}=2\\), \\(q_{2k+1}=1/8\\). Then\n\\[\n Q_{2k}=4^{-k},\\qquad Q_{2k+1}=2\\cdot4^{-k}.\n\\]\nFor \\(f_n=(s,Q_n)\\), the residual doubles at the start of every pair of steps. Choose \\(C_{2k}=1\\) and \\(C_{2k+1}=7/8\\). These give equality in the movement bound, and\n\\[\n \\sum_{n=0}^{\\infty}C_nQ_n\n =\\sum_{k=0}^{\\infty}\\left(1+\\frac74\\right)4^{-k}\n =\\frac{11}{3}.\n\\]\nThus \\(f_n\\to(s,0)\\) despite infinitely many increases. Starting at an even index \\(2k\\), the remaining movement budget is \\(\\frac{11}{3}4^{-k}\\). At an odd index \\(2k+1\\), it is \\(\\frac83 4^{-k}\\). These bounds are larger than the actual distance because the path repeatedly moves away and returns.","n be compatible with convergence}\nSet \\(q_{2k}=2\\), \\(q_{2k+1}=1/8\\). Then\n\\[\n Q_{2k}=4^{-k},\\qquad Q_{2k+1}=2\\cdot4^{-k}.\n\\]\nFor \\(f_n=(s,Q_n)\\), the residual doubles at the start of every pair of steps. Choose \\(C_{2k}=1\\) and \\(C_{2k+1}=7/8\\). These give equality in the movement bound, and\n\\[\n \\sum_{n=0}^{\\infty}C_nQ_n\n =\\sum_{k=0}^{\\infty}\\left(1+\\frac74\\right)4^{-k}\n =\\frac{11}{3}.\n\\]\nThus \\(f_n\\to(s,0)\\) despite infinitely many increases. Starting at an even index \\(2k\\), the remaining movement budget is \\(\\frac{11}{3}4^{-k}\\). At an odd index \\(2k+1\\), it is \\(\\frac83 4^{-k}\\). These bounds are larger than the actual distance because the path repeatedly moves away and returns.","Q_{2k}=4^{-k},\\qquad Q_{2k+1}=2\\cdot4^{-k}.\n\\]\nFor \\(f_n=(s,Q_n)\\), the residual doubles at the start of every pair of steps. Choose \\(C_{2k}=1\\) and \\(C_{2k+1}=7/8\\). These give equality in the movement bound, and\n\\[\n \\sum_{n=0}^{\\infty}C_nQ_n\n =\\sum_{k=0}^{\\infty}\\left(1+\\frac74\\right)4^{-k}\n =\\frac{11}{3}.\n\\]\nThus \\(f_n\\to(s,0)\\) despite infinitely many increases. Starting at an even index \\(2k\\), the remaining movement budget is \\(\\frac{11}{3}4^{-k}\\). At an odd index \\(2k+1\\), it is \\(\\frac83 4^{-k}\\). These bounds are larger than the actual distance because the path repeatedly moves away and returns.","-k}.\n\\]\nFor \\(f_n=(s,Q_n)\\), the residual doubles at the start of every pair of steps. Choose \\(C_{2k}=1\\) and \\(C_{2k+1}=7/8\\). These give equality in the movement bound, and\n\\[\n \\sum_{n=0}^{\\infty}C_nQ_n\n =\\sum_{k=0}^{\\infty}\\left(1+\\frac74\\right)4^{-k}\n =\\frac{11}{3}.\n\\]\nThus \\(f_n\\to(s,0)\\) despite infinitely many increases. Starting at an even index \\(2k\\), the remaining movement budget is \\(\\frac{11}{3}4^{-k}\\). At an odd index \\(2k+1\\), it is \\(\\frac83 4^{-k}\\). These bounds are larger than the actual distance because the path repeatedly moves away and returns.","l doubles at the start of every pair of steps. Choose \\(C_{2k}=1\\) and \\(C_{2k+1}=7/8\\). These give equality in the movement bound, and\n\\[\n \\sum_{n=0}^{\\infty}C_nQ_n\n =\\sum_{k=0}^{\\infty}\\left(1+\\frac74\\right)4^{-k}\n =\\frac{11}{3}.\n\\]\nThus \\(f_n\\to(s,0)\\) despite infinitely many increases. Starting at an even index \\(2k\\), the remaining movement budget is \\(\\frac{11}{3}4^{-k}\\). At an odd index \\(2k+1\\), it is \\(\\frac83 4^{-k}\\). These bounds are larger than the actual distance because the path repeatedly moves away and returns.","start of every pair of steps. Choose \\(C_{2k}=1\\) and \\(C_{2k+1}=7/8\\). These give equality in the movement bound, and\n\\[\n \\sum_{n=0}^{\\infty}C_nQ_n\n =\\sum_{k=0}^{\\infty}\\left(1+\\frac74\\right)4^{-k}\n =\\frac{11}{3}.\n\\]\nThus \\(f_n\\to(s,0)\\) despite infinitely many increases. Starting at an even index \\(2k\\), the remaining movement budget is \\(\\frac{11}{3}4^{-k}\\). At an odd index \\(2k+1\\), it is \\(\\frac83 4^{-k}\\). These bounds are larger than the actual distance because the path repeatedly moves away and returns.","\\plate{26_residual_bursts}{The residual increases at every even-to-odd step, but each two-step product is $1/4$. A convergent product and a summable movement budget control the whole path. Monotonicity of individual residual values is not assumed by the variable-rate theorem.}{fig:residual-bursts}","\\subsection{Protected meaning survives a controlled limit}\nReturn to Banach and normed linear spaces, with continuous linear maps \\(d:E\\to R\\) and \\(p:E\\to S\\). Put \\(\\omega(f)=\\norm{df}\\). The residual is continuous, hence lower semicontinuous.","\\begin{proposition}[Controlled paths kill the actual obstruction]\\label{KS73}\nSuppose \\(E\\) is complete, \\(\\norm{df_n}\\to0\\), \\(p(f_n)=p(f_0)\\), and \\(\\dist(f_n,f_{n+1})\\le b_n\\) for a summable nonnegative budget. Then there is \\(x\\) with\n\\[\n f_n\\to x,\\qquad dx=0,\\qquad p(x)=p(f_0),\n\\]\nand the constrained obstruction \\([df_0]\\in R/d(\\ker p)\\) vanishes.\n\\end{proposition}\n\\begin{proof}\nThe function \\(f\\mapsto\\norm{df}\\) is continuous because \\(d\\) is bounded linear and the norm is continuous. It is therefore an admissible lower semicontinuous residual for Theorem~\\ref{KS71}. That theorem supplies a limit \\(x\\in E\\), together with \\(\\norm{dx}=0\\), which means \\(dx=0\\).","egin{proposition}[Controlled paths kill the actual obstruction]\\label{KS73}\nSuppose \\(E\\) is complete, \\(\\norm{df_n}\\to0\\), \\(p(f_n)=p(f_0)\\), and \\(\\dist(f_n,f_{n+1})\\le b_n\\) for a summable nonnegative budget. Then there is \\(x\\) with\n\\[\n f_n\\to x,\\qquad dx=0,\\qquad p(x)=p(f_0),\n\\]\nand the constrained obstruction \\([df_0]\\in R/d(\\ker p)\\) vanishes.\n\\end{proposition}\n\\begin{proof}\nThe function \\(f\\mapsto\\norm{df}\\) is continuous because \\(d\\) is bounded linear and the norm is continuous. It is therefore an admissible lower semicontinuous residual for Theorem~\\ref{KS71}. That theorem supplies a limit \\(x\\in E\\), together with \\(\\norm{dx}=0\\), which means \\(dx=0\\).","actual obstruction]\\label{KS73}\nSuppose \\(E\\) is complete, \\(\\norm{df_n}\\to0\\), \\(p(f_n)=p(f_0)\\), and \\(\\dist(f_n,f_{n+1})\\le b_n\\) for a summable nonnegative budget. Then there is \\(x\\) with\n\\[\n f_n\\to x,\\qquad dx=0,\\qquad p(x)=p(f_0),\n\\]\nand the constrained obstruction \\([df_0]\\in R/d(\\ker p)\\) vanishes.\n\\end{proposition}\n\\begin{proof}\nThe function \\(f\\mapsto\\norm{df}\\) is continuous because \\(d\\) is bounded linear and the norm is continuous. It is therefore an admissible lower semicontinuous residual for Theorem~\\ref{KS71}. That theorem supplies a limit \\(x\\in E\\), together with \\(\\norm{dx}=0\\), which means \\(dx=0\\).",". Then there is \\(x\\) with\n\\[\n f_n\\to x,\\qquad dx=0,\\qquad p(x)=p(f_0),\n\\]\nand the constrained obstruction \\([df_0]\\in R/d(\\ker p)\\) vanishes.\n\\end{proposition}\n\\begin{proof}\nThe function \\(f\\mapsto\\norm{df}\\) is continuous because \\(d\\) is bounded linear and the norm is continuous. It is therefore an admissible lower semicontinuous residual for Theorem~\\ref{KS71}. That theorem supplies a limit \\(x\\in E\\), together with \\(\\norm{dx}=0\\), which means \\(dx=0\\).","\\[\n f_n\\to x,\\qquad dx=0,\\qquad p(x)=p(f_0),\n\\]\nand the constrained obstruction \\([df_0]\\in R/d(\\ker p)\\) vanishes.\n\\end{proposition}\n\\begin{proof}\nThe function \\(f\\mapsto\\norm{df}\\) is continuous because \\(d\\) is bounded linear and the norm is continuous. It is therefore an admissible lower semicontinuous residual for Theorem~\\ref{KS71}. That theorem supplies a limit \\(x\\in E\\), together with \\(\\norm{dx}=0\\), which means \\(dx=0\\).","uad p(x)=p(f_0),\n\\]\nand the constrained obstruction \\([df_0]\\in R/d(\\ker p)\\) vanishes.\n\\end{proposition}\n\\begin{proof}\nThe function \\(f\\mapsto\\norm{df}\\) is continuous because \\(d\\) is bounded linear and the norm is continuous. It is therefore an admissible lower semicontinuous residual for Theorem~\\ref{KS71}. That theorem supplies a limit \\(x\\in E\\), together with \\(\\norm{dx}=0\\), which means \\(dx=0\\).","Continuity of \\(p\\) gives \\(p(f_n)\\to p(x)\\). The left side is the constant sequence with value \\(p(f_0)\\). Uniqueness of limits in the normed target therefore gives \\(p(x)=p(f_0)\\). Define the actual vector \\(a=f_0-x\\). Linearity, rather than a closure argument, now yields\n\\[\n p(a)=p(f_0)-p(x)=0,\\qquad da=df_0-dx=df_0.\n\\]\nThus \\(a\\in\\ker p\\) and \\(df_0\\in d(\\ker p)\\). Membership in this algebraic image is precisely what makes the class \\([df_0]\\) zero. Merely knowing membership in the closure of that image would not justify the same conclusion.\n\\end{proof}\n\\source{KS73}\nThis proposition converts a controlled approximation argument into a statement about the algebraic quotient defined earlier. Merely showing","Continuity of \\(p\\) gives \\(p(f_n)\\to p(x)\\). The left side is the constant sequence with value \\(p(f_0)\\). Uniqueness of limits in the normed target therefore gives \\(p(x)=p(f_0)\\). Define the actual vector \\(a=f_0-x\\). Linearity, rather than a closure argument, now yields\n\\[\n p(a)=p(f_0)-p(x)=0,\\qquad da=df_0-dx=df_0.\n\\]\nThus \\(a\\in\\ker p\\) and \\(df_0\\in d(\\ker p)\\). Membership in this algebraic image is precisely what makes the class \\([df_0]\\) zero. Merely knowing membership in the closure of that image would not justify the same conclusion.\n\\end{proof}\n\\source{KS73}\nThis proposition converts a controlled approximation argument into a statement about the algebraic quotient defined earlier. Merely showing that \\(df_n\\to0\\) would","Continuity of \\(p\\) gives \\(p(f_n)\\to p(x)\\). The left side is the constant sequence with value \\(p(f_0)\\). Uniqueness of limits in the normed target therefore gives \\(p(x)=p(f_0)\\). Define the actual vector \\(a=f_0-x\\). Linearity, rather than a closure argument, now yields\n\\[\n p(a)=p(f_0)-p(x)=0,\\qquad da=df_0-dx=df_0.\n\\]\nThus \\(a\\in\\ker p\\) and \\(df_0\\in d(\\ker p)\\). Membership in this algebraic image is precisely what makes the class \\([df_0]\\) zero. Merely knowing membership in the closure of that image would not justify the same conclusion.\n\\end{proof}\n\\source{KS73}\nThis proposition converts a controlled approximation argument into a statement about the algebraic quotient defined earlier. Merely showing that \\(df_n\\to0\\) would not produce the admissible vector \\(f_0-x\\), because there mig","Continuity of \\(p\\) gives \\(p(f_n)\\to p(x)\\). The left side is the constant sequence with value \\(p(f_0)\\). Uniqueness of limits in the normed target therefore gives \\(p(x)=p(f_0)\\). Define the actual vector \\(a=f_0-x\\). Linearity, rather than a closure argument, now yields\n\\[\n p(a)=p(f_0)-p(x)=0,\\qquad da=df_0-dx=df_0.\n\\]\nThus \\(a\\in\\ker p\\) and \\(df_0\\in d(\\ker p)\\). Membership in this algebraic image is precisely what makes the class \\([df_0]\\) zero. Merely knowing membership in the closure of that image would not justify the same conclusion.\n\\end{proof}\n\\source{KS73}\nThis proposition converts a controlled approximation argument into a statement about the algebraic quotient defined earlier. Merely showing that \\(df_n\\to0\\) would not produce the admissible vector \\(f_0-x\\), because there might be no candidate limit \\(x\\).","). The left side is the constant sequence with value \\(p(f_0)\\). Uniqueness of limits in the normed target therefore gives \\(p(x)=p(f_0)\\). Define the actual vector \\(a=f_0-x\\). Linearity, rather than a closure argument, now yields\n\\[\n p(a)=p(f_0)-p(x)=0,\\qquad da=df_0-dx=df_0.\n\\]\nThus \\(a\\in\\ker p\\) and \\(df_0\\in d(\\ker p)\\). Membership in this algebraic image is precisely what makes the class \\([df_0]\\) zero. Merely knowing membership in the closure of that image would not justify the same conclusion.\n\\end{proof}\n\\source{KS73}\nThis proposition converts a controlled approximation argument into a statement about the algebraic quotient defined earlier. Merely showing that \\(df_n\\to0\\) would not produce the admissible vector \\(f_0-x\\), because there might be no candidate limit \\(x\\).","Thus \\(a\\in\\ker p\\) and \\(df_0\\in d(\\ker p)\\). Membership in this algebraic image is precisely what makes the class \\([df_0]\\) zero. Merely knowing membership in the closure of that image would not justify the same conclusion.\n\\end{proof}\n\\source{KS73}\nThis proposition converts a controlled approximation argument into a statement about the algebraic quotient defined earlier. Merely showing that \\(df_n\\to0\\) would not produce the admissible vector \\(f_0-x\\), because there might be no candidate limit \\(x\\).","ship in this algebraic image is precisely what makes the class \\([df_0]\\) zero. Merely knowing membership in the closure of that image would not justify the same conclusion.\n\\end{proof}\n\\source{KS73}\nThis proposition converts a controlled approximation argument into a statement about the algebraic quotient defined earlier. Merely showing that \\(df_n\\to0\\) would not produce the admissible vector \\(f_0-x\\), because there might be no candidate limit \\(x\\).","the class \\([df_0]\\) zero. Merely knowing membership in the closure of that image would not justify the same conclusion.\n\\end{proof}\n\\source{KS73}\nThis proposition converts a controlled approximation argument into a statement about the algebraic quotient defined earlier. Merely showing that \\(df_n\\to0\\) would not produce the admissible vector \\(f_0-x\\), because there might be no candidate limit \\(x\\).","\\subsection{When two repair paths select the same point}\n\\begin{proposition}[Path independence requires a comparison]\\label{KS74}\nLet \\(f_n\\to x\\) and \\(g_n\\to y\\) in a metric space. If\n\\[\n \\dist(f_n,g_n)\\le\\beta_n,\\qquad\\beta_n\\to0,\n\\]\nthen \\(x=y\\). In contrast, exactness and preservation of a nonzero observable alone do not force a unique answer.\n\\end{proposition}\n\\begin{proof}\nFor each \\(n\\), the triangle inequality gives\n\\[\n \\dist(x,y)\\le\\dist(x,f_n)+\\dist(f_n,g_n)+\\dist(g_n,y)\n \\le\\dist(x,f_n)+\\beta_n+\\dist(g_n,y).\n\\]\nGiven any \\(\\varepsilon>0\\), choose \\(n\\) large enough that the three terms on the right are each below \\(\\varepsilon/3\\). Thus \\(\\dist(x,y)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\), so the distance is zero. The metric separation axiom gives \\(x=y\\). In a pseudometric space only zero distance would follo","\\subsection{When two repair paths select the same point}\n\\begin{proposition}[Path independence requires a comparison]\\label{KS74}\nLet \\(f_n\\to x\\) and \\(g_n\\to y\\) in a metric space. If\n\\[\n \\dist(f_n,g_n)\\le\\beta_n,\\qquad\\beta_n\\to0,\n\\]\nthen \\(x=y\\). In contrast, exactness and preservation of a nonzero observable alone do not force a unique answer.\n\\end{proposition}\n\\begin{proof}\nFor each \\(n\\), the triangle inequality gives\n\\[\n \\dist(x,y)\\le\\dist(x,f_n)+\\dist(f_n,g_n)+\\dist(g_n,y)\n \\le\\dist(x,f_n)+\\beta_n+\\dist(g_n,y).\n\\]\nGiven any \\(\\varepsilon>0\\), choose \\(n\\) large enough that the three terms on the right are each below \\(\\varepsilon/3\\). Thus \\(\\dist(x,y)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\), so the distance is zero. The metric separation axiom gives \\(x=y\\). In a pseudometric space only zero distance would follow without an addi","\\subsection{When two repair paths select the same point}\n\\begin{proposition}[Path independence requires a comparison]\\label{KS74}\nLet \\(f_n\\to x\\) and \\(g_n\\to y\\) in a metric space. If\n\\[\n \\dist(f_n,g_n)\\le\\beta_n,\\qquad\\beta_n\\to0,\n\\]\nthen \\(x=y\\). In contrast, exactness and preservation of a nonzero observable alone do not force a unique answer.\n\\end{proposition}\n\\begin{proof}\nFor each \\(n\\), the triangle inequality gives\n\\[\n \\dist(x,y)\\le\\dist(x,f_n)+\\dist(f_n,g_n)+\\dist(g_n,y)\n \\le\\dist(x,f_n)+\\beta_n+\\dist(g_n,y).\n\\]\nGiven any \\(\\varepsilon>0\\), choose \\(n\\) large enough that the three terms on the right are each below \\(\\varepsilon/3\\). Thus \\(\\dist(x,y)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\), so the distance is zero. The metric separation axiom gives \\(x=y\\). In a pseudometric space only zero distance would follow without an additional separation hypothesis.","hs select the same point}\n\\begin{proposition}[Path independence requires a comparison]\\label{KS74}\nLet \\(f_n\\to x\\) and \\(g_n\\to y\\) in a metric space. If\n\\[\n \\dist(f_n,g_n)\\le\\beta_n,\\qquad\\beta_n\\to0,\n\\]\nthen \\(x=y\\). In contrast, exactness and preservation of a nonzero observable alone do not force a unique answer.\n\\end{proposition}\n\\begin{proof}\nFor each \\(n\\), the triangle inequality gives\n\\[\n \\dist(x,y)\\le\\dist(x,f_n)+\\dist(f_n,g_n)+\\dist(g_n,y)\n \\le\\dist(x,f_n)+\\beta_n+\\dist(g_n,y).\n\\]\nGiven any \\(\\varepsilon>0\\), choose \\(n\\) large enough that the three terms on the right are each below \\(\\varepsilon/3\\). Thus \\(\\dist(x,y)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\), so the distance is zero. The metric separation axiom gives \\(x=y\\). In a pseudometric space only zero distance would follow without an additional separation hypothesis.","o x\\) and \\(g_n\\to y\\) in a metric space. If\n\\[\n \\dist(f_n,g_n)\\le\\beta_n,\\qquad\\beta_n\\to0,\n\\]\nthen \\(x=y\\). In contrast, exactness and preservation of a nonzero observable alone do not force a unique answer.\n\\end{proposition}\n\\begin{proof}\nFor each \\(n\\), the triangle inequality gives\n\\[\n \\dist(x,y)\\le\\dist(x,f_n)+\\dist(f_n,g_n)+\\dist(g_n,y)\n \\le\\dist(x,f_n)+\\beta_n+\\dist(g_n,y).\n\\]\nGiven any \\(\\varepsilon>0\\), choose \\(n\\) large enough that the three terms on the right are each below \\(\\varepsilon/3\\). Thus \\(\\dist(x,y)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\), so the distance is zero. The metric separation axiom gives \\(x=y\\). In a pseudometric space only zero distance would follow without an additional separation hypothesis.","a metric space. If\n\\[\n \\dist(f_n,g_n)\\le\\beta_n,\\qquad\\beta_n\\to0,\n\\]\nthen \\(x=y\\). In contrast, exactness and preservation of a nonzero observable alone do not force a unique answer.\n\\end{proposition}\n\\begin{proof}\nFor each \\(n\\), the triangle inequality gives\n\\[\n \\dist(x,y)\\le\\dist(x,f_n)+\\dist(f_n,g_n)+\\dist(g_n,y)\n \\le\\dist(x,f_n)+\\beta_n+\\dist(g_n,y).\n\\]\nGiven any \\(\\varepsilon>0\\), choose \\(n\\) large enough that the three terms on the right are each below \\(\\varepsilon/3\\). Thus \\(\\dist(x,y)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\), so the distance is zero. The metric separation axiom gives \\(x=y\\). In a pseudometric space only zero distance would follow without an additional separation hypothesis.","]\nthen \\(x=y\\). In contrast, exactness and preservation of a nonzero observable alone do not force a unique answer.\n\\end{proposition}\n\\begin{proof}\nFor each \\(n\\), the triangle inequality gives\n\\[\n \\dist(x,y)\\le\\dist(x,f_n)+\\dist(f_n,g_n)+\\dist(g_n,y)\n \\le\\dist(x,f_n)+\\beta_n+\\dist(g_n,y).\n\\]\nGiven any \\(\\varepsilon>0\\), choose \\(n\\) large enough that the three terms on the right are each below \\(\\varepsilon/3\\). Thus \\(\\dist(x,y)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\), so the distance is zero. The metric separation axiom gives \\(x=y\\). In a pseudometric space only zero distance would follow without an additional separation hypothesis.","ast, exactness and preservation of a nonzero observable alone do not force a unique answer.\n\\end{proposition}\n\\begin{proof}\nFor each \\(n\\), the triangle inequality gives\n\\[\n \\dist(x,y)\\le\\dist(x,f_n)+\\dist(f_n,g_n)+\\dist(g_n,y)\n \\le\\dist(x,f_n)+\\beta_n+\\dist(g_n,y).\n\\]\nGiven any \\(\\varepsilon>0\\), choose \\(n\\) large enough that the three terms on the right are each below \\(\\varepsilon/3\\). Thus \\(\\dist(x,y)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\), so the distance is zero. The metric separation axiom gives \\(x=y\\). In a pseudometric space only zero distance would follow without an additional separation hypothesis.","ervable alone do not force a unique answer.\n\\end{proposition}\n\\begin{proof}\nFor each \\(n\\), the triangle inequality gives\n\\[\n \\dist(x,y)\\le\\dist(x,f_n)+\\dist(f_n,g_n)+\\dist(g_n,y)\n \\le\\dist(x,f_n)+\\beta_n+\\dist(g_n,y).\n\\]\nGiven any \\(\\varepsilon>0\\), choose \\(n\\) large enough that the three terms on the right are each below \\(\\varepsilon/3\\). Thus \\(\\dist(x,y)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\), so the distance is zero. The metric separation axiom gives \\(x=y\\). In a pseudometric space only zero distance would follow without an additional separation hypothesis.","For each \\(n\\), the triangle inequality gives\n\\[\n \\dist(x,y)\\le\\dist(x,f_n)+\\dist(f_n,g_n)+\\dist(g_n,y)\n \\le\\dist(x,f_n)+\\beta_n+\\dist(g_n,y).\n\\]\nGiven any \\(\\varepsilon>0\\), choose \\(n\\) large enough that the three terms on the right are each below \\(\\varepsilon/3\\). Thus \\(\\dist(x,y)\\le\\varepsilon\\) for every positive \\(\\varepsilon\\), so the distance is zero. The metric separation axiom gives \\(x=y\\). In a pseudometric space only zero distance would follow without an additional separation hypothesis.","For the negative assertion let \\(E=\\R\\times(\\R\\times\\R)\\), \\(d(s,u,r)=r\\), and \\(p(s,u,r)=s\\). Both \\(x=(1,0,0)\\) and \\(y=(1,1,0)\\) have zero defect and protected value one, but their middle coordinates differ. They are distinct points of the same feasible fiber. Constant sequences at those points meet exactness and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path s","For the negative assertion let \\(E=\\R\\times(\\R\\times\\R)\\), \\(d(s,u,r)=r\\), and \\(p(s,u,r)=s\\). Both \\(x=(1,0,0)\\) and \\(y=(1,1,0)\\) have zero defect and protected value one, but their middle coordinates differ. They are distinct points of the same feasible fiber. Constant sequences at those points meet exactness and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometr","For the negative assertion let \\(E=\\R\\times(\\R\\times\\R)\\), \\(d(s,u,r)=r\\), and \\(p(s,u,r)=s\\). Both \\(x=(1,0,0)\\) and \\(y=(1,1,0)\\) have zero defect and protected value one, but their middle coordinates differ. They are distinct points of the same feasible fiber. Constant sequences at those points meet exactness and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. T","For the negative assertion let \\(E=\\R\\times(\\R\\times\\R)\\), \\(d(s,u,r)=r\\), and \\(p(s,u,r)=s\\). Both \\(x=(1,0,0)\\) and \\(y=(1,1,0)\\) have zero defect and protected value one, but their middle coordinates differ. They are distinct points of the same feasible fiber. Constant sequences at those points meet exactness and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. Their limits are \\((s","For the negative assertion let \\(E=\\R\\times(\\R\\times\\R)\\), \\(d(s,u,r)=r\\), and \\(p(s,u,r)=s\\). Both \\(x=(1,0,0)\\) and \\(y=(1,1,0)\\) have zero defect and protected value one, but their middle coordinates differ. They are distinct points of the same feasible fiber. Constant sequences at those points meet exactness and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. Their limits are \\((s,a,0)\\). For diffe","ct and protected value one, but their middle coordinates differ. They are distinct points of the same feasible fiber. Constant sequences at those points meet exactness and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. Their limits are \\((s,a,0)\\). For different \\(a,b\\),\n\\[\n \\dist(f_n^{(a)},f_n^{(b)})=|a-b|(1-2^{-n})\\longrightarrow|a-b|,\n\\]\nnot zero. The missing uniqueness hypothesis is visible in the exact direction along which the paths separate.","distinct points of the same feasible fiber. Constant sequences at those points meet exactness and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. Their limits are \\((s,a,0)\\). For different \\(a,b\\),\n\\[\n \\dist(f_n^{(a)},f_n^{(b)})=|a-b|(1-2^{-n})\\longrightarrow|a-b|,\n\\]\nnot zero. The missing uniqueness hypothesis is visible in the exact direction along which the paths separate.","the same feasible fiber. Constant sequences at those points meet exactness and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. Their limits are \\((s,a,0)\\). For different \\(a,b\\),\n\\[\n \\dist(f_n^{(a)},f_n^{(b)})=|a-b|(1-2^{-n})\\longrightarrow|a-b|,\n\\]\nnot zero. The missing uniqueness hypothesis is visible in the exact direction along which the paths separate.","onstant sequences at those points meet exactness and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. Their limits are \\((s,a,0)\\). For different \\(a,b\\),\n\\[\n \\dist(f_n^{(a)},f_n^{(b)})=|a-b|(1-2^{-n})\\longrightarrow|a-b|,\n\\]\nnot zero. The missing uniqueness hypothesis is visible in the exact direction along which the paths separate.","ess and preservation but not a vanishing comparison bound, so these conditions alone cannot imply path independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. Their limits are \\((s,a,0)\\). For different \\(a,b\\),\n\\[\n \\dist(f_n^{(a)},f_n^{(b)})=|a-b|(1-2^{-n})\\longrightarrow|a-b|,\n\\]\nnot zero. The missing uniqueness hypothesis is visible in the exact direction along which the paths separate.","ath independence.\n\\end{proof}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. Their limits are \\((s,a,0)\\). For different \\(a,b\\),\n\\[\n \\dist(f_n^{(a)},f_n^{(b)})=|a-b|(1-2^{-n})\\longrightarrow|a-b|,\n\\]\nnot zero. The missing uniqueness hypothesis is visible in the exact direction along which the paths separate.","f}\n\\source{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. Their limits are \\((s,a,0)\\). For different \\(a,b\\),\n\\[\n \\dist(f_n^{(a)},f_n^{(b)})=|a-b|(1-2^{-n})\\longrightarrow|a-b|,\n\\]\nnot zero. The missing uniqueness hypothesis is visible in the exact direction along which the paths separate.","e{KS74}\nA stronger worked example uses actual convergent paths from the \\emph{same} initial input:\n\\[\n f_n^{(a)}=(s,a(1-2^{-n}),2^{-n})\\qquad(a\\in\\R).\n\\]\nAll paths begin at \\((s,0,1)\\), keep \\(p=s\\), and have residual \\(2^{-n}\\). Their maximum-norm movements are \\(\\frac12\\max\\{|a|,1\\}2^{-n}\\), so every path satisfies geometric exactification. Their limits are \\((s,a,0)\\). For different \\(a,b\\),\n\\[\n \\dist(f_n^{(a)},f_n^{(b)})=|a-b|(1-2^{-n})\\longrightarrow|a-b|,\n\\]\nnot zero. The missing uniqueness hypothesis is visible in the exact direction along which the paths separate.","\\plate{27_path_limits}{All paths start at the same candidate, have the same residual at each step, and preserve the same coordinate $s$. They converge to different exact points because the coordinate $u$ is unprotected. A vanishing residual is not a path-comparison budget.}{fig:path-limits}","\\subsection{Changing the sampling schedule does not preserve a rate}\nA residual profile \\(\\eta_n\\) indexed by resolution is not an iteration trajectory. Nevertheless a useful convergence property is invariant under a cofinal schedule when the profile is monotone.","\\begin{proposition}[Cofinal sampling of a decreasing profile]\\label{KS75}\nSuppose \\(\\eta_n\\ge0\\) is nonincreasing and \\(\\phi:\\N\\to\\N\\) is monotone and unbounded. Then\n\\[\n \\eta_n\\to0\\quad\\Longleftrightarrow\\quad\\eta_{\\phi(n)}\\to0.\n\\]\nWithout the monotonicity of \\(\\eta\\), the reverse implication can fail even for a cofinal schedule.\n\\end{proposition}\n\\begin{proof}\nThe schedule tends to infinity, so convergence of the full profile implies convergence of the sampled one. Conversely, choose a sampled value below any prescribed \\(\\varepsilon>0\\). Every later index of the original profile has no larger value, by monotonicity. This proves convergence of the full profile.","\\N\\) is monotone and unbounded. Then\n\\[\n \\eta_n\\to0\\quad\\Longleftrightarrow\\quad\\eta_{\\phi(n)}\\to0.\n\\]\nWithout the monotonicity of \\(\\eta\\), the reverse implication can fail even for a cofinal schedule.\n\\end{proposition}\n\\begin{proof}\nThe schedule tends to infinity, so convergence of the full profile implies convergence of the sampled one. Conversely, choose a sampled value below any prescribed \\(\\varepsilon>0\\). Every later index of the original profile has no larger value, by monotonicity. This proves convergence of the full profile.","For failure without monotonicity, take \\(\\eta_{2n}=0\\), \\(\\eta_{2n+1}=1\\), and \\(\\phi(n)=2n\\). All sampled values vanish, but infinitely many original values equal one.\n\\end{proof}\n\\source{KS75}\nEven in the decreasing case, the numerical rate changes. The profile \\(\\eta_n=1/(n+1)\\), sampled at \\(\\phi(n)=2^n-1\\), becomes \\(2^{-n}\\). This is exponential decay in the \\emph{number of observations}, not a proof of a faster correction process at the original resolution scale.","\\subsection{Three hypotheses that cannot simply be dropped}\n\\textbf{Summable movement.} In maximum-norm \\(\\R^3\\), let\n\\[\n f_n=(s,H_n,(n+1)^{-1}),\\qquad\n H_n=\\sum_{j=1}^{n}\\frac1j.\n\\]\nThe first coordinate is protected and the residual is the last coordinate. It tends to zero with the same ratios as the polynomial example. The movement equals \\(1/(n+1)\\), however, and the unprotected coordinate diverges. Here \\(C_n=1\\), so \\(\\sum C_nQ_n\\) is the divergent harmonic series. Residual decay and semantic preservation hold, but exactification of this path fails.","hree hypotheses that cannot simply be dropped}\n\\textbf{Summable movement.} In maximum-norm \\(\\R^3\\), let\n\\[\n f_n=(s,H_n,(n+1)^{-1}),\\qquad\n H_n=\\sum_{j=1}^{n}\\frac1j.\n\\]\nThe first coordinate is protected and the residual is the last coordinate. It tends to zero with the same ratios as the polynomial example. The movement equals \\(1/(n+1)\\), however, and the unprotected coordinate diverges. Here \\(C_n=1\\), so \\(\\sum C_nQ_n\\) is the divergent harmonic series. Residual decay and semantic preservation hold, but exactification of this path fails.","s that cannot simply be dropped}\n\\textbf{Summable movement.} In maximum-norm \\(\\R^3\\), let\n\\[\n f_n=(s,H_n,(n+1)^{-1}),\\qquad\n H_n=\\sum_{j=1}^{n}\\frac1j.\n\\]\nThe first coordinate is protected and the residual is the last coordinate. It tends to zero with the same ratios as the polynomial example. The movement equals \\(1/(n+1)\\), however, and the unprotected coordinate diverges. Here \\(C_n=1\\), so \\(\\sum C_nQ_n\\) is the divergent harmonic series. Residual decay and semantic preservation hold, but exactification of this path fails.","\\textbf{Completeness.} In the metric space \\((0,\\infty)\\), take \\(f_n=1/(n+1)\\) and \\(\\omega(x)=x\\). The steps are summable and the residual tends to zero, but there is no limit in this space. Completing the space adds the missing zero. It is essential to say whether that completion belongs to the intended model.","\\textbf{Lower semicontinuity.} In \\(\\R\\), define \\(\\omega(0)=1\\) and \\(\\omega(x)=0\\) for \\(x\\ne0\\). The same positive sequence converges with summable movement and residual zero at every step, yet its limit is not exact. The zero sublevel set is not closed. This example isolates why convergence of measurements along a sequence must interact correctly with the ambient topology.","\\subsection{Distance to the actual correction image}\nLet \\(R\\) be a normed additive group and \\(B\\le R\\) an additive subgroup. In the protected linear model, \\(B=d(\\ker p)\\). For a residual \\(\\Omega\\in R\\), define\n\\[\n \\nu_B(\\Omega)=\\inf_{b\\in B}\\norm{\\Omega-b}.\n\\]\nThe subgroup contains zero, so this is a finite nonnegative real number. It is an infimum, not an assertion that a nearest admissible residual exists.","\\begin{proposition}[The floor detects closure]\\label{KS76}\nFor every \\(b\\in B\\),\n\\[\n 0\\le\\nu_B(\\Omega)\\le\\norm{\\Omega-b}.\n\\]\nMoreover,\n\\[\n \\nu_B(\\Omega)=0\\quad\\Longleftrightarrow\\quad\\Omega\\in\\overline B.\n\\]\nIf \\(B\\) is closed, this is equivalent to \\(\\Omega\\in B\\).\n\\end{proposition}\n\\begin{proof}\nThe set of candidate distances is nonempty because \\(0\\in B\\), bounded below by zero, and contains the finite value \\(\\norm\\Omega\\). Its infimum is therefore a finite nonnegative real, and the defining lower-bound property gives \\(\\nu_B(\\Omega)\\le\\norm{\\Omega-b}\\) for every \\(b\\in B\\).","n}[The floor detects closure]\\label{KS76}\nFor every \\(b\\in B\\),\n\\[\n 0\\le\\nu_B(\\Omega)\\le\\norm{\\Omega-b}.\n\\]\nMoreover,\n\\[\n \\nu_B(\\Omega)=0\\quad\\Longleftrightarrow\\quad\\Omega\\in\\overline B.\n\\]\nIf \\(B\\) is closed, this is equivalent to \\(\\Omega\\in B\\).\n\\end{proposition}\n\\begin{proof}\nThe set of candidate distances is nonempty because \\(0\\in B\\), bounded below by zero, and contains the finite value \\(\\norm\\Omega\\). Its infimum is therefore a finite nonnegative real, and the defining lower-bound property gives \\(\\nu_B(\\Omega)\\le\\norm{\\Omega-b}\\) for every \\(b\\in B\\).","ver,\n\\[\n \\nu_B(\\Omega)=0\\quad\\Longleftrightarrow\\quad\\Omega\\in\\overline B.\n\\]\nIf \\(B\\) is closed, this is equivalent to \\(\\Omega\\in B\\).\n\\end{proposition}\n\\begin{proof}\nThe set of candidate distances is nonempty because \\(0\\in B\\), bounded below by zero, and contains the finite value \\(\\norm\\Omega\\). Its infimum is therefore a finite nonnegative real, and the defining lower-bound property gives \\(\\nu_B(\\Omega)\\le\\norm{\\Omega-b}\\) for every \\(b\\in B\\).","ow\\quad\\Omega\\in\\overline B.\n\\]\nIf \\(B\\) is closed, this is equivalent to \\(\\Omega\\in B\\).\n\\end{proposition}\n\\begin{proof}\nThe set of candidate distances is nonempty because \\(0\\in B\\), bounded below by zero, and contains the finite value \\(\\norm\\Omega\\). Its infimum is therefore a finite nonnegative real, and the defining lower-bound property gives \\(\\nu_B(\\Omega)\\le\\norm{\\Omega-b}\\) for every \\(b\\in B\\).","Suppose \\(\\nu_B(\\Omega)=0\\) and fix \\(\\varepsilon>0\\). There must be \\(b\\in B\\) with \\(\\norm{\\Omega-b}<\\varepsilon\\). Otherwise every candidate distance would be at least \\(\\varepsilon\\), making \\(\\varepsilon\\) a lower bound and forcing the infimum to be at least \\(\\varepsilon\\), a contradiction. Thus every open ball about \\(\\Omega\\) meets \\(B\\), which is the closure criterion.","Conversely, if \\(\\Omega\\in\\overline B\\), every such ball supplies \\(b\\) with \\(\\norm{\\Omega-b}<\\varepsilon\\). The infimum then lies between zero and \\(\\varepsilon\\) for every positive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quo","Conversely, if \\(\\Omega\\in\\overline B\\), every such ball supplies \\(b\\) with \\(\\norm{\\Omega-b}<\\varepsilon\\). The infimum then lies between zero and \\(\\varepsilon\\) for every positive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebr","Conversely, if \\(\\Omega\\in\\overline B\\), every such ball supplies \\(b\\) with \\(\\norm{\\Omega-b}<\\varepsilon\\). The infimum then lies between zero and \\(\\varepsilon\\) for every positive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when","Conversely, if \\(\\Omega\\in\\overline B\\), every such ball supplies \\(b\\) with \\(\\norm{\\Omega-b}<\\varepsilon\\). The infimum then lies between zero and \\(\\varepsilon\\) for every positive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the","Conversely, if \\(\\Omega\\in\\overline B\\), every such ball supplies \\(b\\) with \\(\\norm{\\Omega-b}<\\varepsilon\\). The infimum then lies between zero and \\(\\varepsilon\\) for every positive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the floor on residual representatives","Conversely, if \\(\\Omega\\in\\overline B\\), every such ball supplies \\(b\\) with \\(\\norm{\\Omega-b}<\\varepsilon\\). The infimum then lies between zero and \\(\\varepsilon\\) for every positive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the floor on residual representatives and proves the closure cr","Conversely, if \\(\\Omega\\in\\overline B\\), every such ball supplies \\(b\\) with \\(\\norm{\\Omega-b}<\\varepsilon\\). The infimum then lies between zero and \\(\\varepsilon\\) for every positive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the floor on residual representatives and proves the closure criterion. This quotient interpretation is its elementary translation-","Conversely, if \\(\\Omega\\in\\overline B\\), every such ball supplies \\(b\\) with \\(\\norm{\\Omega-b}<\\varepsilon\\). The infimum then lies between zero and \\(\\varepsilon\\) for every positive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the floor on residual representatives and proves the closure criterion. This quotient interpretation is its elementary translation-invariance calculation, not a new claim that every obstruction quotien","imum then lies between zero and \\(\\varepsilon\\) for every positive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the floor on residual representatives and proves the closure criterion. This quotient interpretation is its elementary translation-invariance calculation, not a new claim that every obstruction quotient carries a separated norm.","r every positive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the floor on residual representatives and proves the closure criterion. This quotient interpretation is its elementary translation-invariance calculation, not a new claim that every obstruction quotient carries a separated norm.","ositive \\(\\varepsilon\\), so it is zero. If \\(B\\) is closed, membership in its closure is exactly membership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the floor on residual representatives and proves the closure criterion. This quotient interpretation is its elementary translation-invariance calculation, not a new claim that every obstruction quotient carries a separated norm.","bership in \\(B\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the floor on residual representatives and proves the closure criterion. This quotient interpretation is its elementary translation-invariance calculation, not a new claim that every obstruction quotient carries a separated norm.","\\). No step asserts that the infimum has a minimizing element when \\(B\\) is not closed, and closedness alone is being used here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the floor on residual representatives and proves the closure criterion. This quotient interpretation is its elementary translation-invariance calculation, not a new claim that every obstruction quotient carries a separated norm.","sed here only for the zero-floor membership conclusion.\n\\end{proof}\n\\source{KS76}\nTranslation by a member \\(b_0\\in B\\) does not change the floor: replacing \\(b\\) by \\(b-b_0\\) gives the same set of distances. Thus the floor is well defined on the algebraic quotient \\(R/B\\). When \\(B\\) is a linear subspace, this is the usual quotient seminorm. A nonzero algebraic class can have seminorm zero when \\(B\\) is not closed. The retained source implements the floor on residual representatives and proves the closure criterion. This quotient interpretation is its elementary translation-invariance calculation, not a new claim that every obstruction quotient carries a separated norm.","midrule\n\\(\\Omega\\in B\\) & Zero & Exact cancellation exists, and the floor is zero.","\\(\\Omega\\in\\overline B\\setminus B\\) & Nonzero & Arbitrarily small corrected residuals exist, but no exact cancellation exists.","\\(\\Omega\\notin\\overline B\\) & Nonzero & The floor is strictly positive and blocks every allowed correction.","In finite-dimensional real or complex linear models, every linear subspace is closed. The middle row is therefore absent. It becomes possible in infinite dimensions, or for additive subgroups that are not closed real-linear subspaces. For example the rational subgroup of \\(\\R\\) is dense, so the distance from any irrational real to that subgroup is zero even though its coset is nonzero.","Closedness also should not be confused with a numerical error bound back in the candidate space. The statement \\(\\Omega\\in B\\) supplies an admissible correction, but not a bound on its size. The closed-range estimates of Sections~\\ref{sec:closed} and~\\ref{sec:motion}, or their protected counterparts, are the additional analytic tools that control that size in their declared settings.","\\subsection{A persistent floor blocks every correction sequence}\nThe correction image can vary with the resolution or the admissibility rule. Let \\(B_n\\le R\\) and let \\(\\Omega_n\\in R\\) be given.","\\begin{proposition}[Uniformly positive floors obstruct asymptotic repair]\\label{KS77}\nSuppose \\(c>0\\) and eventually\n\\[\n c\\le\\nu_{B_n}(\\Omega_n).\n\\]\nFor any choices \\(b_n\\in B_n\\), it is impossible that\n\\[\n \\norm{\\Omega_n-b_n}\\longrightarrow0.\n\\]\n\\end{proposition}\n\\begin{proof}\nEvery allowed correction satisfies\n\\(\\norm{\\Omega_n-b_n}\\ge\\nu_{B_n}(\\Omega_n)\\). The corrected norms are therefore eventually at least \\(c\\). They cannot converge to zero.\n\\end{proof}\n\\source{KS77}\nThe quantifier covers all admissible sequences, not just the output of one repair routine. On the other hand, the hypothesis is an eventually \\emph{uniform} positive lower bound. Having a nonzero obstruction class at each stage is not the same statement. The classes might have zero floors, or their positive floors might","\\begin{proposition}[Uniformly positive floors obstruct asymptotic repair]\\label{KS77}\nSuppose \\(c>0\\) and eventually\n\\[\n c\\le\\nu_{B_n}(\\Omega_n).\n\\]\nFor any choices \\(b_n\\in B_n\\), it is impossible that\n\\[\n \\norm{\\Omega_n-b_n}\\longrightarrow0.\n\\]\n\\end{proposition}\n\\begin{proof}\nEvery allowed correction satisfies\n\\(\\norm{\\Omega_n-b_n}\\ge\\nu_{B_n}(\\Omega_n)\\). The corrected norms are therefore eventually at least \\(c\\). They cannot converge to zero.\n\\end{proof}\n\\source{KS77}\nThe quantifier covers all admissible sequences, not just the output of one repair routine. On the other hand, the hypothesis is an eventually \\emph{uniform} positive lower bound. Having a nonzero obstruction class at each stage is not the same statement. The classes might have zero floors, or their positive floors might tend to zero.","n}[Uniformly positive floors obstruct asymptotic repair]\\label{KS77}\nSuppose \\(c>0\\) and eventually\n\\[\n c\\le\\nu_{B_n}(\\Omega_n).\n\\]\nFor any choices \\(b_n\\in B_n\\), it is impossible that\n\\[\n \\norm{\\Omega_n-b_n}\\longrightarrow0.\n\\]\n\\end{proposition}\n\\begin{proof}\nEvery allowed correction satisfies\n\\(\\norm{\\Omega_n-b_n}\\ge\\nu_{B_n}(\\Omega_n)\\). The corrected norms are therefore eventually at least \\(c\\). They cannot converge to zero.\n\\end{proof}\n\\source{KS77}\nThe quantifier covers all admissible sequences, not just the output of one repair routine. On the other hand, the hypothesis is an eventually \\emph{uniform} positive lower bound. Having a nonzero obstruction class at each stage is not the same statement. The classes might have zero floors, or their positive floors might tend to zero.","A finite-dimensional illustration uses \\(R=\\R^2\\), \\(B=\\R\\times\\{0\\}\\), and \\(\\Omega_n=(n,1)\\). Every correction leaves the second coordinate equal to one, so the floor is one independently of \\(n\\). Replacing the residuals by \\((n,1/(n+1))\\) gives nonzero algebraic classes at every stage, but their floors tend to zero. The persistent-floor theorem applies to the first sequence and not the second.","\\subsection{The bridge back to exactification}\nProposition~\\ref{KS73} and Proposition~\\ref{KS76} fit together closely. If corrections produce a norm-convergent, protected path with vanishing residual, the actual quotient class vanishes. If only the corrected residuals tend to zero, all that follows is membership in \\(\\overline{d(\\ker p)}\\).","\\subsection{The complete spaces and the preservation rule}\nIndex sequences by the positive integers. Set\n\\[\n E=\\ell^2(\\N_{>0};\\R)\\oplus\\R,\\qquad\n R=\\ell^2(\\N_{>0};\\R),\\qquad\n \\norm{(u,t)}_E^2=\\sum_{n\\ge1}|u_n|^2+|t|^2.\n\\]\nLet \\(w=(1/n)_{n\\ge1}\\). It belongs to \\(\\ell^2\\), because\n\\[\n \\sum_{n\\ge1}\\frac1{n^2}\n \\le1+\\sum_{n\\ge2}\\frac1{n(n-1)}=2.\n\\]\nDefine the defect and protected observable by\n\\[\n d(u,t)_n=\\frac{u_n+t}{n},\\qquad p(u,t)=t.\n\\]\nThe diagonal map \\(T:u\\mapsto(u_n/n)_n\\) has operator norm one. Since \\(d(u,t)=Tu+tw\\),\n\\[\n \\norm{d(u,t)}\\le\\norm u+|t|\\norm w\n \\le\\sqrt{1+\\norm w^2}\\,\\norm{(u,t)}_E.\n\\]\nThus both \\(d\\) and \\(p\\) are continuous linear maps between the declared complete spaces. No failure of completeness or continuity is built into the model.","ex sequences by the positive integers. Set\n\\[\n E=\\ell^2(\\N_{>0};\\R)\\oplus\\R,\\qquad\n R=\\ell^2(\\N_{>0};\\R),\\qquad\n \\norm{(u,t)}_E^2=\\sum_{n\\ge1}|u_n|^2+|t|^2.\n\\]\nLet \\(w=(1/n)_{n\\ge1}\\). It belongs to \\(\\ell^2\\), because\n\\[\n \\sum_{n\\ge1}\\frac1{n^2}\n \\le1+\\sum_{n\\ge2}\\frac1{n(n-1)}=2.\n\\]\nDefine the defect and protected observable by\n\\[\n d(u,t)_n=\\frac{u_n+t}{n},\\qquad p(u,t)=t.\n\\]\nThe diagonal map \\(T:u\\mapsto(u_n/n)_n\\) has operator norm one. Since \\(d(u,t)=Tu+tw\\),\n\\[\n \\norm{d(u,t)}\\le\\norm u+|t|\\norm w\n \\le\\sqrt{1+\\norm w^2}\\,\\norm{(u,t)}_E.\n\\]\nThus both \\(d\\) and \\(p\\) are continuous linear maps between the declared complete spaces. No failure of completeness or continuity is built into the model.","^2(\\N_{>0};\\R)\\oplus\\R,\\qquad\n R=\\ell^2(\\N_{>0};\\R),\\qquad\n \\norm{(u,t)}_E^2=\\sum_{n\\ge1}|u_n|^2+|t|^2.\n\\]\nLet \\(w=(1/n)_{n\\ge1}\\). It belongs to \\(\\ell^2\\), because\n\\[\n \\sum_{n\\ge1}\\frac1{n^2}\n \\le1+\\sum_{n\\ge2}\\frac1{n(n-1)}=2.\n\\]\nDefine the defect and protected observable by\n\\[\n d(u,t)_n=\\frac{u_n+t}{n},\\qquad p(u,t)=t.\n\\]\nThe diagonal map \\(T:u\\mapsto(u_n/n)_n\\) has operator norm one. Since \\(d(u,t)=Tu+tw\\),\n\\[\n \\norm{d(u,t)}\\le\\norm u+|t|\\norm w\n \\le\\sqrt{1+\\norm w^2}\\,\\norm{(u,t)}_E.\n\\]\nThus both \\(d\\) and \\(p\\) are continuous linear maps between the declared complete spaces. No failure of completeness or continuity is built into the model.","us\\R,\\qquad\n R=\\ell^2(\\N_{>0};\\R),\\qquad\n \\norm{(u,t)}_E^2=\\sum_{n\\ge1}|u_n|^2+|t|^2.\n\\]\nLet \\(w=(1/n)_{n\\ge1}\\). It belongs to \\(\\ell^2\\), because\n\\[\n \\sum_{n\\ge1}\\frac1{n^2}\n \\le1+\\sum_{n\\ge2}\\frac1{n(n-1)}=2.\n\\]\nDefine the defect and protected observable by\n\\[\n d(u,t)_n=\\frac{u_n+t}{n},\\qquad p(u,t)=t.\n\\]\nThe diagonal map \\(T:u\\mapsto(u_n/n)_n\\) has operator norm one. Since \\(d(u,t)=Tu+tw\\),\n\\[\n \\norm{d(u,t)}\\le\\norm u+|t|\\norm w\n \\le\\sqrt{1+\\norm w^2}\\,\\norm{(u,t)}_E.\n\\]\nThus both \\(d\\) and \\(p\\) are continuous linear maps between the declared complete spaces. No failure of completeness or continuity is built into the model.","\\(w=(1/n)_{n\\ge1}\\). It belongs to \\(\\ell^2\\), because\n\\[\n \\sum_{n\\ge1}\\frac1{n^2}\n \\le1+\\sum_{n\\ge2}\\frac1{n(n-1)}=2.\n\\]\nDefine the defect and protected observable by\n\\[\n d(u,t)_n=\\frac{u_n+t}{n},\\qquad p(u,t)=t.\n\\]\nThe diagonal map \\(T:u\\mapsto(u_n/n)_n\\) has operator norm one. Since \\(d(u,t)=Tu+tw\\),\n\\[\n \\norm{d(u,t)}\\le\\norm u+|t|\\norm w\n \\le\\sqrt{1+\\norm w^2}\\,\\norm{(u,t)}_E.\n\\]\nThus both \\(d\\) and \\(p\\) are continuous linear maps between the declared complete spaces. No failure of completeness or continuity is built into the model.","_{n\\ge1}\\). It belongs to \\(\\ell^2\\), because\n\\[\n \\sum_{n\\ge1}\\frac1{n^2}\n \\le1+\\sum_{n\\ge2}\\frac1{n(n-1)}=2.\n\\]\nDefine the defect and protected observable by\n\\[\n d(u,t)_n=\\frac{u_n+t}{n},\\qquad p(u,t)=t.\n\\]\nThe diagonal map \\(T:u\\mapsto(u_n/n)_n\\) has operator norm one. Since \\(d(u,t)=Tu+tw\\),\n\\[\n \\norm{d(u,t)}\\le\\norm u+|t|\\norm w\n \\le\\sqrt{1+\\norm w^2}\\,\\norm{(u,t)}_E.\n\\]\nThus both \\(d\\) and \\(p\\) are continuous linear maps between the declared complete spaces. No failure of completeness or continuity is built into the model.","Take the raw candidate \\(f_0=(0,1)\\). Its residual is \\(w\\), and the coordinate that must be preserved is \\(t=1\\). An exact protected answer would require\n\\[\n u_n=-1\\quad\\text{for every }n.\n\\]\nThat sequence does not lie in \\(\\ell^2\\). Consequently no answer in \\(E\\) is both exact and protected.","In fact \\(\\ker d=\\{0\\}\\). If \\(d(u,t)=0\\), then \\(u_n=-t\\) for every \\(n\\). A constant square-summable sequence must be zero, so \\(t=0\\) and \\(u=0\\). There is an exact answer after abandoning the preservation rule, namely zero, but it changes \\(p(f_0)=1\\) to zero. Even a trivial ambiguity space does not imply feasibility.","\\subsection{The obstruction class is nonzero while its floor is zero}\nThe admissible correction space is \\(\\ker p=\\ell^2\\oplus\\{0\\}\\), with correction image\n\\[\n B=d(\\ker p)=\\ran T\n  =\\{r\\in\\ell^2:(nr_n)_{n\\ge1}\\in\\ell^2\\}.\n\\]\nIt contains every finitely supported sequence, so it is dense in \\(R\\). But \\(w\\notin B\\), because \\((nw_n)_n=(1,1,\\ldots)\\) is not square summable. Therefore\n\\[\n [w]\\ne0\\text{ in }R/B,\\qquad \\nu_B(w)=0.\n\\]\nThis is the middle row of the preceding classification. The example is a worked Hilbert-space derivation without a separate Lean instance. Its finite-coordinate calculations are independently tested. The general floor and exactification results in Sections~\\ref{sec:variable-exactification} and~\\ref{sec:floor} have formal counterparts.","\\subsection{A complete calculation of the finite repairs}\nLet \\(\\mathbf 1_N\\) be one in coordinates \\(1,\\ldots,N\\) and zero elsewhere. Put\n\\[\n a_N=(\\mathbf 1_N,0),\\qquad f_N=f_0-a_N=(-\\mathbf 1_N,1).\n\\]\nThen \\(p(f_N)=1\\), and the first \\(N\\) residual coordinates vanish exactly. The remaining residual and movement satisfy\n\\[\n d(f_N)_n=\\begin{cases}0&n\\le N,","\\subsection{A complete calculation of the finite repairs}\nLet \\(\\mathbf 1_N\\) be one in coordinates \\(1,\\ldots,N\\) and zero elsewhere. Put\n\\[\n a_N=(\\mathbf 1_N,0),\\qquad f_N=f_0-a_N=(-\\mathbf 1_N,1).\n\\]\nThen \\(p(f_N)=1\\), and the first \\(N\\) residual coordinates vanish exactly. The remaining residual and movement satisfy\n\\[\n d(f_N)_n=\\begin{cases}0&n\\le N,\\\\1/n&n>N,\\end{cases}\n \\qquad \\norm{a_N}^2=N,\\qquad \\norm{f_N}^2=N+1.\n\\]\nFor every \\(N\\ge1\\), telescoping comparisons give\n\\begin{equation}\\label{eq:escape-tail}\n \\frac1{N+1}\\le\\norm{d(f_N)}^2\n =\\sum_{n>N}\\frac1{n^2}\\le\\frac1N.\n\\end{equation}\nIndeed \\(1/[n(n+1)]\\le1/n^2\\le1/[n(n-1)]\\) for \\(n\\ge2\\). Summing the two telescoping bounds proves \\eqref{eq:escape-tail} without numerically evaluating an infinite series.","1/n&n>N,\\end{cases}\n \\qquad \\norm{a_N}^2=N,\\qquad \\norm{f_N}^2=N+1.\n\\]\nFor every \\(N\\ge1\\), telescoping comparisons give\n\\begin{equation}\\label{eq:escape-tail}\n \\frac1{N+1}\\le\\norm{d(f_N)}^2\n =\\sum_{n>N}\\frac1{n^2}\\le\\frac1N.\n\\end{equation}\nIndeed \\(1/[n(n+1)]\\le1/n^2\\le1/[n(n-1)]\\) for \\(n\\ge2\\). Summing the two telescoping bounds proves \\eqref{eq:escape-tail} without numerically evaluating an infinite series.","The residual tends to zero, but the corrections grow like \\(\\sqrt N\\). Successive candidates differ by a unit coordinate, so\n\\[\n \\norm{f_{N+1}-f_N}=1.\n\\]\nThe movement series diverges and the sequence is not Cauchy. This is exactly the hypothesis missing from the summable exactification theorem.","\\begin{center}\n\\begin{tabular}{@{}rrrr@{}}\n\\toprule Repaired coordinates \\(N\\) & \\(\\norm{a_N}\\) & Residual square interval & Finite inverse gain","midrule\n1 & 1 & \\([1/2,1]\\) & 1","4 & 2 & \\([1/5,1/4]\\) & 4","16 & 4 & \\([1/17,1/16]\\) & 16","64 & 8 & \\([1/65,1/64]\\) & 64","\\fixedplate{28_escaping_repair}{The squared correction size is $N$, while the infinite residual square lies between $1/(N+1)$ and $1/N$. The plotted band consists of proved bounds, not a numerical truncation of the tail. Smaller residuals require larger changes and do not produce a convergent candidate.}{fig:escaping-repair}","\\subsection{Escape is not an accident of this particular algorithm}\nSuppose an arbitrary protected candidate \\((u,1)\\) satisfies \\(\\norm{d(u,1)}\\le\\varepsilon\\). On its first \\(N\\) coordinates,\n\\[\n \\sum_{n=1}^{N}|u_n+1|^2\n \\le N^2\\sum_{n=1}^{N}\\frac{|u_n+1|^2}{n^2}\n \\le N^2\\varepsilon^2.\n\\]\nThe reverse triangle inequality in \\(\\R^N\\) gives\n\\begin{equation}\\label{eq:escape-necessity}\n \\norm u\\ge\\sqrt N-N\\varepsilon\\qquad(N\\ge1).\n\\end{equation}\nIf a sequence of protected candidates has residual tending to zero, then for each fixed \\(N\\), its eventual correction norms are at least, for example, \\(\\sqrt N/2\\). Taking \\(N\\) arbitrarily large shows that these norms must tend to infinity. There is no uniformly bounded family of protected corrections with arbitrarily small residual.","d(u,1)}\\le\\varepsilon\\). On its first \\(N\\) coordinates,\n\\[\n \\sum_{n=1}^{N}|u_n+1|^2\n \\le N^2\\sum_{n=1}^{N}\\frac{|u_n+1|^2}{n^2}\n \\le N^2\\varepsilon^2.\n\\]\nThe reverse triangle inequality in \\(\\R^N\\) gives\n\\begin{equation}\\label{eq:escape-necessity}\n \\norm u\\ge\\sqrt N-N\\varepsilon\\qquad(N\\ge1).\n\\end{equation}\nIf a sequence of protected candidates has residual tending to zero, then for each fixed \\(N\\), its eventual correction norms are at least, for example, \\(\\sqrt N/2\\). Taking \\(N\\) arbitrarily large shows that these norms must tend to infinity. There is no uniformly bounded family of protected corrections with arbitrarily small residual.","m_{n=1}^{N}|u_n+1|^2\n \\le N^2\\sum_{n=1}^{N}\\frac{|u_n+1|^2}{n^2}\n \\le N^2\\varepsilon^2.\n\\]\nThe reverse triangle inequality in \\(\\R^N\\) gives\n\\begin{equation}\\label{eq:escape-necessity}\n \\norm u\\ge\\sqrt N-N\\varepsilon\\qquad(N\\ge1).\n\\end{equation}\nIf a sequence of protected candidates has residual tending to zero, then for each fixed \\(N\\), its eventual correction norms are at least, for example, \\(\\sqrt N/2\\). Taking \\(N\\) arbitrarily large shows that these norms must tend to infinity. There is no uniformly bounded family of protected corrections with arbitrarily small residual.","\\le N^2\\sum_{n=1}^{N}\\frac{|u_n+1|^2}{n^2}\n \\le N^2\\varepsilon^2.\n\\]\nThe reverse triangle inequality in \\(\\R^N\\) gives\n\\begin{equation}\\label{eq:escape-necessity}\n \\norm u\\ge\\sqrt N-N\\varepsilon\\qquad(N\\ge1).\n\\end{equation}\nIf a sequence of protected candidates has residual tending to zero, then for each fixed \\(N\\), its eventual correction norms are at least, for example, \\(\\sqrt N/2\\). Taking \\(N\\) arbitrarily large shows that these norms must tend to infinity. There is no uniformly bounded family of protected corrections with arbitrarily small residual.","\\subsection{Every finite operator is invertible, but the gains diverge}\nOn \\(\\R^N\\), let\n\\[\n T_N=\\operatorname{diag}(1,1/2,\\ldots,1/N).\n\\]\nIt is invertible and \\(\\norm{T_N^{-1}}=N\\) in the Euclidean norm. The upper bound follows by multiplying each coordinate by an integer at most \\(N\\), and equality is attained at the last coordinate vector. Full finite rank therefore supplies no uniform gain as \\(N\\to\\infty\\).","ubsection{Every finite operator is invertible, but the gains diverge}\nOn \\(\\R^N\\), let\n\\[\n T_N=\\operatorname{diag}(1,1/2,\\ldots,1/N).\n\\]\nIt is invertible and \\(\\norm{T_N^{-1}}=N\\) in the Euclidean norm. The upper bound follows by multiplying each coordinate by an integer at most \\(N\\), and equality is attained at the last coordinate vector. Full finite rank therefore supplies no uniform gain as \\(N\\to\\infty\\).","In the infinite space, \\(\\norm{e_n}=1\\) but \\(\\norm{Te_n}=1/n\\), while \\(\\ker T=\\{0\\}\\). A uniform estimate\n\\(\\dist(u,\\ker T)\\le C\\norm{Tu}\\) would force \\(1\\le C/n\\) for every \\(n\\), which is impossible. This agrees with the nonclosed range already established directly. The low residual directions do not reflect new exact vectors. They reflect increasingly ill-conditioned approximate directions.","\\subsection{A coherent inverse-limit point still exists in a larger space}\nAt each finite level, the protected exact equations have the unique solution\n\\[\n (-1,\\ldots,-1;1)\\in\\R^N\\oplus\\R.\n\\]\nTruncation maps the solution at level \\(N+1\\) to the one at level \\(N\\). These finite solution sets are singletons, so their inverse limit is also a singleton. Its coordinate description is\n\\[\n (-1,-1,\\ldots;1)\\in\\R^{\\N_{>0}}\\times\\R.\n\\]\nThe coherent point is not missing. It lies outside the intended \\(\\ell^2\\)-based semantic space.","the protected exact equations have the unique solution\n\\[\n (-1,\\ldots,-1;1)\\in\\R^N\\oplus\\R.\n\\]\nTruncation maps the solution at level \\(N+1\\) to the one at level \\(N\\). These finite solution sets are singletons, so their inverse limit is also a singleton. Its coordinate description is\n\\[\n (-1,-1,\\ldots;1)\\in\\R^{\\N_{>0}}\\times\\R.\n\\]\nThe coherent point is not missing. It lies outside the intended \\(\\ell^2\\)-based semantic space.","This example does not contradict finite-fiber existence. It distinguishes the ordinary inverse limit of finite views from the image of a separate comparison map that sends square-summable semantic objects to their finite prefixes. The prescribed coherent point is not in that image. Replacing \\(\\ell^2\\) by all sequences makes the equation solvable, but changes the topology and the class of allowed objects.","\\subsection{The correction space is the protected kernel}\nLet $E,Y,Z$ be real Hilbert spaces, let $d:E\\to Y$ and $p:E\\to Z$ be bounded linear maps, and fix a raw candidate $f\\in E$. Write\n\\[\n H=\\ker p,\\qquad T=d|_H:H\\to Y,\\qquad r=df.\n\\]\nThe space $H$ is a closed Hilbert subspace with the norm inherited from $E$. An allowed correction is $a\\in H$, and its repaired candidate is $f-a$. Thus\n\\[\n p(f-a)=pf,\\qquad d(f-a)=r-Ta.\n\\]\nThe protected exact problem is precisely $Ta=r$. Restricting the domain before optimizing is important. A penalty on $\\norm{pa}$ would only discourage a change in the protected data, not prohibit it.","e real Hilbert spaces, let $d:E\\to Y$ and $p:E\\to Z$ be bounded linear maps, and fix a raw candidate $f\\in E$. Write\n\\[\n H=\\ker p,\\qquad T=d|_H:H\\to Y,\\qquad r=df.\n\\]\nThe space $H$ is a closed Hilbert subspace with the norm inherited from $E$. An allowed correction is $a\\in H$, and its repaired candidate is $f-a$. Thus\n\\[\n p(f-a)=pf,\\qquad d(f-a)=r-Ta.\n\\]\nThe protected exact problem is precisely $Ta=r$. Restricting the domain before optimizing is important. A penalty on $\\norm{pa}$ would only discourage a change in the protected data, not prohibit it.","and fix a raw candidate $f\\in E$. Write\n\\[\n H=\\ker p,\\qquad T=d|_H:H\\to Y,\\qquad r=df.\n\\]\nThe space $H$ is a closed Hilbert subspace with the norm inherited from $E$. An allowed correction is $a\\in H$, and its repaired candidate is $f-a$. Thus\n\\[\n p(f-a)=pf,\\qquad d(f-a)=r-Ta.\n\\]\nThe protected exact problem is precisely $Ta=r$. Restricting the domain before optimizing is important. A penalty on $\\norm{pa}$ would only discourage a change in the protected data, not prohibit it.","The adjoint below always means $T^*:Y\\to H$, taken with the inherited Hilbert metric on $H$. In ambient coordinates this is the projection of $d^*$ onto $H$. A coordinate basis of $\\ker p$ need not be orthonormal. Its Gram matrix must then be included in any calculation of correction energy.","\\subsection{The compactness needed for the existence argument}\n\\begin{lemma}[Bounded sequences have weakly convergent subsequences]\\label{lem:weak-subsequence}\nEvery bounded sequence $(a_n)$ in a real Hilbert space has a subsequence converging weakly to some $a$. Moreover,\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{n_k}}.\n\\]\n\\end{lemma}\n\\begin{proof}\nChoose $M\\ge0$ with $\\norm{a_n}\\le M$. If $M=0$, every term is zero and the conclusion is immediate. Otherwise let $H_0$ be the closed linear span of the countably many vectors $a_n$. Finite rational linear combinations of those vectors form a countable dense set, so $H_0$ is separable. Apply Gram--Schmidt to a countable dense sequence, omitting zero remainders. This gives either a finite orthonormal basis or a countable orthonormal sequence $(e_j)$ whose closed span is $H_0$. The ambient Hilbert space itself need not be separable.","ent}\n\\begin{lemma}[Bounded sequences have weakly convergent subsequences]\\label{lem:weak-subsequence}\nEvery bounded sequence $(a_n)$ in a real Hilbert space has a subsequence converging weakly to some $a$. Moreover,\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{n_k}}.\n\\]\n\\end{lemma}\n\\begin{proof}\nChoose $M\\ge0$ with $\\norm{a_n}\\le M$. If $M=0$, every term is zero and the conclusion is immediate. Otherwise let $H_0$ be the closed linear span of the countably many vectors $a_n$. Finite rational linear combinations of those vectors form a countable dense set, so $H_0$ is separable. Apply Gram--Schmidt to a countable dense sequence, omitting zero remainders. This gives either a finite orthonormal basis or a countable orthonormal sequence $(e_j)$ whose closed span is $H_0$. The ambient Hilbert space itself need not be separable.","bsequences]\\label{lem:weak-subsequence}\nEvery bounded sequence $(a_n)$ in a real Hilbert space has a subsequence converging weakly to some $a$. Moreover,\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{n_k}}.\n\\]\n\\end{lemma}\n\\begin{proof}\nChoose $M\\ge0$ with $\\norm{a_n}\\le M$. If $M=0$, every term is zero and the conclusion is immediate. Otherwise let $H_0$ be the closed linear span of the countably many vectors $a_n$. Finite rational linear combinations of those vectors form a countable dense set, so $H_0$ is separable. Apply Gram--Schmidt to a countable dense sequence, omitting zero remainders. This gives either a finite orthonormal basis or a countable orthonormal sequence $(e_j)$ whose closed span is $H_0$. The ambient Hilbert space itself need not be separable.","ace has a subsequence converging weakly to some $a$. Moreover,\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{n_k}}.\n\\]\n\\end{lemma}\n\\begin{proof}\nChoose $M\\ge0$ with $\\norm{a_n}\\le M$. If $M=0$, every term is zero and the conclusion is immediate. Otherwise let $H_0$ be the closed linear span of the countably many vectors $a_n$. Finite rational linear combinations of those vectors form a countable dense set, so $H_0$ is separable. Apply Gram--Schmidt to a countable dense sequence, omitting zero remainders. This gives either a finite orthonormal basis or a countable orthonormal sequence $(e_j)$ whose closed span is $H_0$. The ambient Hilbert space itself need not be separable.",", every term is zero and the conclusion is immediate. Otherwise let $H_0$ be the closed linear span of the countably many vectors $a_n$. Finite rational linear combinations of those vectors form a countable dense set, so $H_0$ is separable. Apply Gram--Schmidt to a countable dense sequence, omitting zero remainders. This gives either a finite orthonormal basis or a countable orthonormal sequence $(e_j)$ whose closed span is $H_0$. The ambient Hilbert space itself need not be separable.","conclusion is immediate. Otherwise let $H_0$ be the closed linear span of the countably many vectors $a_n$. Finite rational linear combinations of those vectors form a countable dense set, so $H_0$ is separable. Apply Gram--Schmidt to a countable dense sequence, omitting zero remainders. This gives either a finite orthonormal basis or a countable orthonormal sequence $(e_j)$ whose closed span is $H_0$. The ambient Hilbert space itself need not be separable.","Each scalar sequence $\\langle a_n,e_j\\rangle$ is bounded by $M$. For the first coordinate choose a convergent subsequence, for the second choose a subsequence of that subsequence, and continue. Taking the $k$th term from the $k$th nested subsequence, with increasing original indices, yields $a_{n_k}$ for which every fixed coordinate converges. Call the coordinate limits $c_j$. In the finite-dimensional case only finitely many successive selections are required.","For each finite $J$, Bessel's inequality and passage to the limit give\n\\[\n \\sum_{j=1}^Jc_j^2\n =\\lim_k\\sum_{j=1}^J\\langle a_{n_k},e_j\\rangle^2\\le M^2.\n\\]\nThe nondecreasing partial sums are therefore bounded. Orthogonality makes the partial vector sums $\\sum_{j\\le J}c_je_j$ Cauchy, so completeness defines $a\\in H_0$ with $\\norm a\\le M$.","To verify weak convergence against $h\\in H_0$, choose a finite basis combination $h_J$ close to $h$. Then\n\\[\n |\\langle a_{n_k}-a,h\\rangle|\n \\le |\\langle a_{n_k}-a,h_J\\rangle|\n      +2M\\norm{h-h_J}.\n\\]\nFor a prescribed tolerance, choose $h_J$ first to control the second term and then $k$ to control the first by finitely many coordinate limits. For a general $h$ in the ambient space, write $h=P_{H_0}h+h_\\perp$. The orthogonal component pairs to zero with every $a_{n_k}$ and with $a$, reducing to the preceding case. This proves weak convergence against every test vector.","asis combination $h_J$ close to $h$. Then\n\\[\n |\\langle a_{n_k}-a,h\\rangle|\n \\le |\\langle a_{n_k}-a,h_J\\rangle|\n      +2M\\norm{h-h_J}.\n\\]\nFor a prescribed tolerance, choose $h_J$ first to control the second term and then $k$ to control the first by finitely many coordinate limits. For a general $h$ in the ambient space, write $h=P_{H_0}h+h_\\perp$. The orthogonal component pairs to zero with every $a_{n_k}$ and with $a$, reducing to the preceding case. This proves weak convergence against every test vector.","_J$ close to $h$. Then\n\\[\n |\\langle a_{n_k}-a,h\\rangle|\n \\le |\\langle a_{n_k}-a,h_J\\rangle|\n      +2M\\norm{h-h_J}.\n\\]\nFor a prescribed tolerance, choose $h_J$ first to control the second term and then $k$ to control the first by finitely many coordinate limits. For a general $h$ in the ambient space, write $h=P_{H_0}h+h_\\perp$. The orthogonal component pairs to zero with every $a_{n_k}$ and with $a$, reducing to the preceding case. This proves weak convergence against every test vector.","If $a\\ne0$, testing against $a/\\norm a$ gives\n\\[\n \\norm a=\\lim_k\\langle a_{n_k},a/\\norm a\\rangle\n \\le\\liminf_k\\norm{a_{n_k}}.\n\\]\nThe zero limit case follows from nonnegativity. Thus the lower-semicontinuity conclusion concerns the actual norms on the selected subsequence, not only the original bound $M$.\n\\end{proof}","\\begin{theorem}[Bounded approximate feasibility]\\label{WR01}\nFor a bounded linear map $T:H\\to Y$ between real Hilbert spaces and $r\\in Y$, the following are equivalent:\n\\[\n \\begin{split}\n &r\\in\\ran T,","\\begin{theorem}[Bounded approximate feasibility]\\label{WR01}\nFor a bounded linear map $T:H\\to Y$ between real Hilbert spaces and $r\\in Y$, the following are equivalent:\n\\[\n \\begin{split}\n &r\\in\\ran T,\\\\\n &\\text{there is a bounded sequence }(a_n)\\subset H\n       \\text{ with }\\norm{Ta_n-r}\\longrightarrow0.\n \\end{split}\n\\]\nIf all $\\norm{a_n}\\le M$, an exact solution exists with norm at most $M$.\n\\end{theorem}\n\\begin{proof}\nIf $Ta=r$, the constant sequence $a_n=a$ is bounded and has zero residual. For the converse, let $\\norm{a_n}\\le M$ and $Ta_n\\to r$ in norm. By the preceding lemma there is a subsequence $a_{n_k}$ converging weakly to $a\\in H$ with $\\norm a\\le M$.","&\\text{there is a bounded sequence }(a_n)\\subset H\n       \\text{ with }\\norm{Ta_n-r}\\longrightarrow0.\n \\end{split}\n\\]\nIf all $\\norm{a_n}\\le M$, an exact solution exists with norm at most $M$.\n\\end{theorem}\n\\begin{proof}\nIf $Ta=r$, the constant sequence $a_n=a$ is bounded and has zero residual. For the converse, let $\\norm{a_n}\\le M$ and $Ta_n\\to r$ in norm. By the preceding lemma there is a subsequence $a_{n_k}$ converging weakly to $a\\in H$ with $\\norm a\\le M$.","there is a bounded sequence }(a_n)\\subset H\n       \\text{ with }\\norm{Ta_n-r}\\longrightarrow0.\n \\end{split}\n\\]\nIf all $\\norm{a_n}\\le M$, an exact solution exists with norm at most $M$.\n\\end{theorem}\n\\begin{proof}\nIf $Ta=r$, the constant sequence $a_n=a$ is bounded and has zero residual. For the converse, let $\\norm{a_n}\\le M$ and $Ta_n\\to r$ in norm. By the preceding lemma there is a subsequence $a_{n_k}$ converging weakly to $a\\in H$ with $\\norm a\\le M$.","_n)\\subset H\n       \\text{ with }\\norm{Ta_n-r}\\longrightarrow0.\n \\end{split}\n\\]\nIf all $\\norm{a_n}\\le M$, an exact solution exists with norm at most $M$.\n\\end{theorem}\n\\begin{proof}\nIf $Ta=r$, the constant sequence $a_n=a$ is bounded and has zero residual. For the converse, let $\\norm{a_n}\\le M$ and $Ta_n\\to r$ in norm. By the preceding lemma there is a subsequence $a_{n_k}$ converging weakly to $a\\in H$ with $\\norm a\\le M$.","\\text{ with }\\norm{Ta_n-r}\\longrightarrow0.\n \\end{split}\n\\]\nIf all $\\norm{a_n}\\le M$, an exact solution exists with norm at most $M$.\n\\end{theorem}\n\\begin{proof}\nIf $Ta=r$, the constant sequence $a_n=a$ is bounded and has zero residual. For the converse, let $\\norm{a_n}\\le M$ and $Ta_n\\to r$ in norm. By the preceding lemma there is a subsequence $a_{n_k}$ converging weakly to $a\\in H$ with $\\norm a\\le M$.","For every fixed $y\\in Y$, the adjoint identity and weak convergence give\n\\[\n \\langle Ta,y\\rangle\n =\\langle a,T^*y\\rangle\n =\\lim_k\\langle a_{n_k},T^*y\\rangle\n =\\lim_k\\langle Ta_{n_k},y\\rangle.\n\\]\nThe last limit equals $\\langle r,y\\rangle$ because norm convergence implies convergence of each continuous pairing. Thus $\\langle Ta-r,y\\rangle=0$ for every $y$. Taking $y=Ta-r$ gives $\\norm{Ta-r}^2=0$, hence $Ta=r$.","very fixed $y\\in Y$, the adjoint identity and weak convergence give\n\\[\n \\langle Ta,y\\rangle\n =\\langle a,T^*y\\rangle\n =\\lim_k\\langle a_{n_k},T^*y\\rangle\n =\\lim_k\\langle Ta_{n_k},y\\rangle.\n\\]\nThe last limit equals $\\langle r,y\\rangle$ because norm convergence implies convergence of each continuous pairing. Thus $\\langle Ta-r,y\\rangle=0$ for every $y$. Taking $y=Ta-r$ gives $\\norm{Ta-r}^2=0$, hence $Ta=r$.","This proof does not claim that $Ta_{n_k}$ converges in norm merely because $a_{n_k}$ converges weakly. Its norm convergence to $r$ was a separate hypothesis. The boundedness of the candidate corrections is used to obtain a weakly convergent subsequence, and their accurate residuals are used to identify its image exactly. The bound $\\norm a\\le M$ comes from weak lower semicontinuity, so the exact answer stays inside the same closed budget ball.\n\\end{proof}\n\\written{WR01}","An immediate contrapositive is useful. If $r\\in\\overline{\\ran T}\\setminus\\ran T$, then every sequence with $Ta_n\\to r$ satisfies $\\norm{a_n}\\to\\infty$. Otherwise infinitely many terms would lie in one bounded ball, contradicting the theorem along that subsequence. This recovers the escape mechanism of Section~\\ref{sec:escape} without using the diagonal formula.","\\subsection{Existence does not make an arbitrary sequence converge strongly}\nTake $H=Y=\\ell^2$ and $Te_n=e_n/n$. The vectors $a_n=e_n$ have norm one and satisfy $Ta_n\\to0$. An exact answer exists, namely zero. Yet $\\norm{e_n-e_m}=\\sqrt2$ for $n\\ne m$, so no subsequence converges in norm.","Suppose now that $A_1\\supseteq A_2\\supseteq\\cdots$ are nonempty closed affine subsets of a real Hilbert correction space $H$. A typical $A_n$ consists of corrections satisfying the first $n$ bounded linear observation equations. The sets are affine because the equations are linear, and closed because the observations are continuous.","Let $a_n$ be the unique point of $A_n$ nearest to zero, and put $m_n=\\norm{a_n}$. The Hilbert projection theorem supplies these points. It can also be proved directly: a sequence approaching the minimum norm is Cauchy by the parallelogram identity, its limit belongs to the closed convex set, and the same identity gives uniqueness.","\\begin{theorem}[A uniform finite budget yields a strong realization]\\label{WR02}\nFor the nested affine sets above, $m_n$ is nondecreasing, and for $m\\ge n$,\n\\begin{equation}\\label{eq:nested-energy}\n \\norm{a_m-a_n}^2=m_m^2-m_n^2.\n\\end{equation}\nTheir intersection is nonempty if and only if $\\sup_n m_n<\\infty$. In that case $a_n$ converges in norm to the unique least norm point $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n","\\begin{theorem}[A uniform finite budget yields a strong realization]\\label{WR02}\nFor the nested affine sets above, $m_n$ is nondecreasing, and for $m\\ge n$,\n\\begin{equation}\\label{eq:nested-energy}\n \\norm{a_m-a_n}^2=m_m^2-m_n^2.\n\\end{equation}\nTheir intersection is nonempty if and only if $\\sup_n m_n<\\infty$. In that case $a_n$ converges in norm to the unique least norm point $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm","\\begin{theorem}[A uniform finite budget yields a strong realization]\\label{WR02}\nFor the nested affine sets above, $m_n$ is nondecreasing, and for $m\\ge n$,\n\\begin{equation}\\label{eq:nested-energy}\n \\norm{a_m-a_n}^2=m_m^2-m_n^2.\n\\end{equation}\nTheir intersection is nonempty if and only if $\\sup_n m_n<\\infty$. In that case $a_n$ converges in norm to the unique least norm point $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nA","\\begin{theorem}[A uniform finite budget yields a strong realization]\\label{WR02}\nFor the nested affine sets above, $m_n$ is nondecreasing, and for $m\\ge n$,\n\\begin{equation}\\label{eq:nested-energy}\n \\norm{a_m-a_n}^2=m_m^2-m_n^2.\n\\end{equation}\nTheir intersection is nonempty if and only if $\\sup_n m_n<\\infty$. In that case $a_n$ converges in norm to the unique least norm point $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","eorem}[A uniform finite budget yields a strong realization]\\label{WR02}\nFor the nested affine sets above, $m_n$ is nondecreasing, and for $m\\ge n$,\n\\begin{equation}\\label{eq:nested-energy}\n \\norm{a_m-a_n}^2=m_m^2-m_n^2.\n\\end{equation}\nTheir intersection is nonempty if and only if $\\sup_n m_n<\\infty$. In that case $a_n$ converges in norm to the unique least norm point $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","d affine sets above, $m_n$ is nondecreasing, and for $m\\ge n$,\n\\begin{equation}\\label{eq:nested-energy}\n \\norm{a_m-a_n}^2=m_m^2-m_n^2.\n\\end{equation}\nTheir intersection is nonempty if and only if $\\sup_n m_n<\\infty$. In that case $a_n$ converges in norm to the unique least norm point $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","mpty if and only if $\\sup_n m_n<\\infty$. In that case $a_n$ converges in norm to the unique least norm point $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","nly if $\\sup_n m_n<\\infty$. In that case $a_n$ converges in norm to the unique least norm point $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","nfty$. In that case $a_n$ converges in norm to the unique least norm point $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","to the unique least norm point $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","t $a_*\\in\\bigcap_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","_n A_n$, with\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","orm{a_n}^2.\n\\]\nIf the intersection is empty, then $m_n\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\nThe nearest-point theorem for nonempty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","empty closed convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","d convex subsets of a Hilbert space, proved in Appendix~\\ref{app:proof-tools}, supplies $a_n$. Write $A_n=a_n+L_n$, where $L_n$ is its closed direction subspace. For $\\ell\\in L_n$ and every real $t$, the vector $a_n+t\\ell$ belongs to $A_n$. Minimality implies\n\\[\n 0\\le\\norm{a_n+t\\ell}^2-\\norm{a_n}^2\n =2t\\langle a_n,\\ell\\rangle+t^2\\norm\\ell^2.\n\\]\nAllowing both signs of arbitrarily small $t$ forces $\\langle a_n,\\ell\\rangle=0$.","If $m\\ge n$, nesting gives $a_m\\in A_n$, so $a_m-a_n\\in L_n$. Expanding its norm therefore gives\n\\[\n \\norm{a_m}^2\n =\\norm{a_n+(a_m-a_n)}^2\n =\\norm{a_n}^2+\\norm{a_m-a_n}^2.\n\\]\nThis is the exact energy identity. It also makes $m_n^2$ nondecreasing.","Suppose these energies are bounded, and let $L=\\lim_n m_n^2$. Given $\\epsilon>0$, choose $N$ with $L-m_N^2<\\epsilon^2$. For $m\\ge n\\ge N$,\n\\[\n \\norm{a_m-a_n}^2=m_m^2-m_n^2\\le L-m_N^2<\\epsilon^2.\n\\]\nThus $(a_n)$ is Cauchy and has a norm limit $a_*$. For each fixed $n$, all terms from that index onward belong to the closed set $A_n$, so $a_*\\in A_n$. This places $a_*$ in their entire intersection.","If $a$ lies in the intersection, then $\\norm a\\ge\\norm{a_n}$ for every $n$. Norm continuity gives $\\norm a\\ge\\norm{a_*}$. Uniqueness follows from the nearest-point theorem applied to the intersection, which is closed and convex and now known to be nonempty. Alternatively, the midpoint of two different minimizers would have strictly smaller norm by the parallelogram identity. Taking $m\\to\\infty$ in the finite energy identity yields\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nConversely, any element of the intersection bounds every minimum norm from above. Therefore an empty intersection forces the nondecreasing sequence $m_n$ to be unbounded. A nondecreasing unbounded nonnegative sequence t","If $a$ lies in the intersection, then $\\norm a\\ge\\norm{a_n}$ for every $n$. Norm continuity gives $\\norm a\\ge\\norm{a_*}$. Uniqueness follows from the nearest-point theorem applied to the intersection, which is closed and convex and now known to be nonempty. Alternatively, the midpoint of two different minimizers would have strictly smaller norm by the parallelogram identity. Taking $m\\to\\infty$ in the finite energy identity yields\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nConversely, any element of the intersection bounds every minimum norm from above. Therefore an empty intersection forces the nondecreasing sequence $m_n$ to be unbounded. A nondecreasing unbounded nonnegative sequence tends to infinity, giving the last conclusion.\n\\end{pro","If $a$ lies in the intersection, then $\\norm a\\ge\\norm{a_n}$ for every $n$. Norm continuity gives $\\norm a\\ge\\norm{a_*}$. Uniqueness follows from the nearest-point theorem applied to the intersection, which is closed and convex and now known to be nonempty. Alternatively, the midpoint of two different minimizers would have strictly smaller norm by the parallelogram identity. Taking $m\\to\\infty$ in the finite energy identity yields\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nConversely, any element of the intersection bounds every minimum norm from above. Therefore an empty intersection forces the nondecreasing sequence $m_n$ to be unbounded. A nondecreasing unbounded nonnegative sequence tends to infinity, giving the last conclusion.\n\\end{proof}\n\\written{W","If $a$ lies in the intersection, then $\\norm a\\ge\\norm{a_n}$ for every $n$. Norm continuity gives $\\norm a\\ge\\norm{a_*}$. Uniqueness follows from the nearest-point theorem applied to the intersection, which is closed and convex and now known to be nonempty. Alternatively, the midpoint of two different minimizers would have strictly smaller norm by the parallelogram identity. Taking $m\\to\\infty$ in the finite energy identity yields\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nConversely, any element of the intersection bounds every minimum norm from above. Therefore an empty intersection forces the nondecreasing sequence $m_n$ to be unbounded. A nondecreasing unbounded nonnegative sequence tends to infinity, giving the last conclusion.\n\\end{proof}\n\\written{WR02}","$\\norm a\\ge\\norm{a_n}$ for every $n$. Norm continuity gives $\\norm a\\ge\\norm{a_*}$. Uniqueness follows from the nearest-point theorem applied to the intersection, which is closed and convex and now known to be nonempty. Alternatively, the midpoint of two different minimizers would have strictly smaller norm by the parallelogram identity. Taking $m\\to\\infty$ in the finite energy identity yields\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nConversely, any element of the intersection bounds every minimum norm from above. Therefore an empty intersection forces the nondecreasing sequence $m_n$ to be unbounded. A nondecreasing unbounded nonnegative sequence tends to infinity, giving the last conclusion.\n\\end{proof}\n\\written{WR02}","known to be nonempty. Alternatively, the midpoint of two different minimizers would have strictly smaller norm by the parallelogram identity. Taking $m\\to\\infty$ in the finite energy identity yields\n\\[\n \\norm{a_*-a_n}^2=\\norm{a_*}^2-\\norm{a_n}^2.\n\\]\nConversely, any element of the intersection bounds every minimum norm from above. Therefore an empty intersection forces the nondecreasing sequence $m_n$ to be unbounded. A nondecreasing unbounded nonnegative sequence tends to infinity, giving the last conclusion.\n\\end{proof}\n\\written{WR02}","For the attainable diagonal datum $r_n=1/n^2$ under $Tu=(u_n/n)$, with $n\\ge1$, the first $N$ equations force correction coordinate $n$ to equal $1/n$ for $n\\le N$. Their minimum energy completion is zero beyond $N$, and\n\\[\n m_N^2=\\sum_{n=1}^{N}\\frac1{n^2}\\le2.\n\\]\nThese corrections converge strongly to $(1/n)$. For the escaping datum $r_n=1/n$, the corresponding energy is $m_N^2=N$. Every finite equation is again solved, but the budget test now rules out an ambient solution.","\\chap{The attainable residual at a fixed correction budget}\\label{sec:budget-frontier}\nThe unrestricted floor $\\dist(r,\\ran T)$ forgets how large a correction must be. To retain that information, fix a nonnegative budget $M$ and define\n\\[\n \\beta_{T,r}(M)=\\inf_{\\norm a\\le M}\\norm{r-Ta}.\n\\]\nThe norm is the actual correction norm in $H=\\ker p$. It is not the norm of arbitrary coordinates used to represent that kernel. The function $\\beta$ describes the smallest residual compatible with a declared amount of allowed change.","a fixed correction budget}\\label{sec:budget-frontier}\nThe unrestricted floor $\\dist(r,\\ran T)$ forgets how large a correction must be. To retain that information, fix a nonnegative budget $M$ and define\n\\[\n \\beta_{T,r}(M)=\\inf_{\\norm a\\le M}\\norm{r-Ta}.\n\\]\nThe norm is the actual correction norm in $H=\\ker p$. It is not the norm of arbitrary coordinates used to represent that kernel. The function $\\beta$ describes the smallest residual compatible with a declared amount of allowed change.","\\subsection{At every finite budget the residual minimum exists}\n\\begin{theorem}[Attainment and shape of the budget profile]\\label{WR03}\nFor a bounded linear map between real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has l","\\subsection{At every finite budget the residual minimum exists}\n\\begin{theorem}[Attainment and shape of the budget profile]\\label{WR03}\nFor a bounded linear map between real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$.","\\subsection{At every finite budget the residual minimum exists}\n\\begin{theorem}[Attainment and shape of the budget profile]\\label{WR03}\nFor a bounded linear map between real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$.","\\subsection{At every finite budget the residual minimum exists}\n\\begin{theorem}[Attainment and shape of the budget profile]\\label{WR03}\nFor a bounded linear map between real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target give","\\subsection{At every finite budget the residual minimum exists}\n\\begin{theorem}[Attainment and shape of the budget profile]\\label{WR03}\nFor a bounded linear map between real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\n","ubsection{At every finite budget the residual minimum exists}\n\\begin{theorem}[Attainment and shape of the budget profile]\\label{WR03}\nFor a bounded linear map between real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{…","very finite budget the residual minimum exists}\n\\begin{theorem}[Attainment and shape of the budget profile]\\label{WR03}\nFor a bounded linear map between real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\b…","sidual minimum exists}\n\\begin{theorem}[Attainment and shape of the budget profile]\\label{WR03}\nFor a bounded linear map between real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse in…","t and shape of the budget profile]\\label{WR03}\nFor a bounded linear map between real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum,…","r map between real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zer…","ween real Hilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and th…","ilbert spaces, the infimum defining $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conc…","ing $\\beta(M)$ is attained for every $M\\ge0$. The profile is nonincreasing and convex. For $0\\le M_1\\le M_2$,\n\\[\n 0\\le\\beta(M_1)-\\beta(M_2)\\le\\norm T(M_2-M_1).\n\\]\nWriting $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","$q=\\norm{r-b}$, one has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","ne has\n\\[\n \\lim_{M\\to\\infty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","fty}\\beta(M)=q.\n\\]\nThe equality $\\beta(M)=q$ at a finite budget is equivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","ivalent to the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","o the existence of $a$ with $Ta=b$ and $\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","\\norm a\\le M$. In particular,\n\\[\n \\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","\\beta(M)=0\\quad\\Longleftrightarrow\\quad\n \\exists a\\in H:\\ Ta=r,\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.",",\\ \\norm a\\le M.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe feasible ball is nonempty because it contains zero, and the infimum lies between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","s between zero and $\\norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","norm r$. Choose $a_n$ in that ball with $\\norm{r-Ta_n}\\to\\beta(M)$. A weakly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","ly convergent subsequence has limit $a$ with $\\norm a\\le M$. Bounded linearity makes $r-Ta_{n_k}$ converge weakly to $r-Ta$. Applying weak lower semicontinuity in the target gives\n\\[\n \\norm{r-Ta}\\le\\liminf_k\\norm{r-Ta_{n_k}}=\\beta(M).\n\\]\nThe reverse inequality holds by the definition of the infimum, so $a$ attains it. When $M=0$ the only feasible correction is zero, and the same conclusion reads $\\beta(0)=\\norm r$.","For convexity, take minimizers $a_1,a_2$ at budgets $M_1,M_2$ and $t\\in[0,1]$. The candidate $a_t=ta_1+(1-t)a_2$ satisfies\n\\[\n \\norm{a_t}\\le tM_1+(1-t)M_2,\n\\]\nand its residual is $t(r-Ta_1)+(1-t)(r-Ta_2)$. The triangle inequality therefore gives\n\\[\n \\beta(tM_1+(1-t)M_2)\\le t\\beta(M_1)+(1-t)\\beta(M_2).\n\\]\nMonotonicity follows from inclusion of balls. If $0\\le M_1\\le M_2$ and $M_2>0$, choose a minimizer $a_2$ at $M_2$ and put $a_1'=(M_1/M_2)a_2$. Then $\\norm{a_1'}\\le M_1$ and $\\norm{a_2-a_1'}\\le M_2-M_1$, so\n\\[\n \\beta(M_1)\\le\\norm{r-Ta_1'}\n \\le\\beta(M_2)+\\norm T(M_2-M_1).\n\\]\nTogether with monotonicity this proves the Lipschitz estimate. The omitted case $M_2=0$ has $M_1=0$ and is equality.","nvexity, take minimizers $a_1,a_2$ at budgets $M_1,M_2$ and $t\\in[0,1]$. The candidate $a_t=ta_1+(1-t)a_2$ satisfies\n\\[\n \\norm{a_t}\\le tM_1+(1-t)M_2,\n\\]\nand its residual is $t(r-Ta_1)+(1-t)(r-Ta_2)$. The triangle inequality therefore gives\n\\[\n \\beta(tM_1+(1-t)M_2)\\le t\\beta(M_1)+(1-t)\\beta(M_2).\n\\]\nMonotonicity follows from inclusion of balls. If $0\\le M_1\\le M_2$ and $M_2>0$, choose a minimizer $a_2$ at $M_2$ and put $a_1'=(M_1/M_2)a_2$. Then $\\norm{a_1'}\\le M_1$ and $\\norm{a_2-a_1'}\\le M_2-M_1$, so\n\\[\n \\beta(M_1)\\le\\norm{r-Ta_1'}\n \\le\\beta(M_2)+\\norm T(M_2-M_1).\n\\]\nTogether with monotonicity this proves the Lipschitz estimate. The omitted case $M_2=0$ has $M_1=0$ and is equality.","take minimizers $a_1,a_2$ at budgets $M_1,M_2$ and $t\\in[0,1]$. The candidate $a_t=ta_1+(1-t)a_2$ satisfies\n\\[\n \\norm{a_t}\\le tM_1+(1-t)M_2,\n\\]\nand its residual is $t(r-Ta_1)+(1-t)(r-Ta_2)$. The triangle inequality therefore gives\n\\[\n \\beta(tM_1+(1-t)M_2)\\le t\\beta(M_1)+(1-t)\\beta(M_2).\n\\]\nMonotonicity follows from inclusion of balls. If $0\\le M_1\\le M_2$ and $M_2>0$, choose a minimizer $a_2$ at $M_2$ and put $a_1'=(M_1/M_2)a_2$. Then $\\norm{a_1'}\\le M_1$ and $\\norm{a_2-a_1'}\\le M_2-M_1$, so\n\\[\n \\beta(M_1)\\le\\norm{r-Ta_1'}\n \\le\\beta(M_2)+\\norm T(M_2-M_1).\n\\]\nTogether with monotonicity this proves the Lipschitz estimate. The omitted case $M_2=0$ has $M_1=0$ and is equality.","s $a_1,a_2$ at budgets $M_1,M_2$ and $t\\in[0,1]$. The candidate $a_t=ta_1+(1-t)a_2$ satisfies\n\\[\n \\norm{a_t}\\le tM_1+(1-t)M_2,\n\\]\nand its residual is $t(r-Ta_1)+(1-t)(r-Ta_2)$. The triangle inequality therefore gives\n\\[\n \\beta(tM_1+(1-t)M_2)\\le t\\beta(M_1)+(1-t)\\beta(M_2).\n\\]\nMonotonicity follows from inclusion of balls. If $0\\le M_1\\le M_2$ and $M_2>0$, choose a minimizer $a_2$ at $M_2$ and put $a_1'=(M_1/M_2)a_2$. Then $\\norm{a_1'}\\le M_1$ and $\\norm{a_2-a_1'}\\le M_2-M_1$, so\n\\[\n \\beta(M_1)\\le\\norm{r-Ta_1'}\n \\le\\beta(M_2)+\\norm T(M_2-M_1).\n\\]\nTogether with monotonicity this proves the Lipschitz estimate. The omitted case $M_2=0$ has $M_1=0$ and is equality.","1,M_2$ and $t\\in[0,1]$. The candidate $a_t=ta_1+(1-t)a_2$ satisfies\n\\[\n \\norm{a_t}\\le tM_1+(1-t)M_2,\n\\]\nand its residual is $t(r-Ta_1)+(1-t)(r-Ta_2)$. The triangle inequality therefore gives\n\\[\n \\beta(tM_1+(1-t)M_2)\\le t\\beta(M_1)+(1-t)\\beta(M_2).\n\\]\nMonotonicity follows from inclusion of balls. If $0\\le M_1\\le M_2$ and $M_2>0$, choose a minimizer $a_2$ at $M_2$ and put $a_1'=(M_1/M_2)a_2$. Then $\\norm{a_1'}\\le M_1$ and $\\norm{a_2-a_1'}\\le M_2-M_1$, so\n\\[\n \\beta(M_1)\\le\\norm{r-Ta_1'}\n \\le\\beta(M_2)+\\norm T(M_2-M_1).\n\\]\nTogether with monotonicity this proves the Lipschitz estimate. The omitted case $M_2=0$ has $M_1=0$ and is equality.","he candidate $a_t=ta_1+(1-t)a_2$ satisfies\n\\[\n \\norm{a_t}\\le tM_1+(1-t)M_2,\n\\]\nand its residual is $t(r-Ta_1)+(1-t)(r-Ta_2)$. The triangle inequality therefore gives\n\\[\n \\beta(tM_1+(1-t)M_2)\\le t\\beta(M_1)+(1-t)\\beta(M_2).\n\\]\nMonotonicity follows from inclusion of balls. If $0\\le M_1\\le M_2$ and $M_2>0$, choose a minimizer $a_2$ at $M_2$ and put $a_1'=(M_1/M_2)a_2$. Then $\\norm{a_1'}\\le M_1$ and $\\norm{a_2-a_1'}\\le M_2-M_1$, so\n\\[\n \\beta(M_1)\\le\\norm{r-Ta_1'}\n \\le\\beta(M_2)+\\norm T(M_2-M_1).\n\\]\nTogether with monotonicity this proves the Lipschitz estimate. The omitted case $M_2=0$ has $M_1=0$ and is equality.","atisfies\n\\[\n \\norm{a_t}\\le tM_1+(1-t)M_2,\n\\]\nand its residual is $t(r-Ta_1)+(1-t)(r-Ta_2)$. The triangle inequality therefore gives\n\\[\n \\beta(tM_1+(1-t)M_2)\\le t\\beta(M_1)+(1-t)\\beta(M_2).\n\\]\nMonotonicity follows from inclusion of balls. If $0\\le M_1\\le M_2$ and $M_2>0$, choose a minimizer $a_2$ at $M_2$ and put $a_1'=(M_1/M_2)a_2$. Then $\\norm{a_1'}\\le M_1$ and $\\norm{a_2-a_1'}\\le M_2-M_1$, so\n\\[\n \\beta(M_1)\\le\\norm{r-Ta_1'}\n \\le\\beta(M_2)+\\norm T(M_2-M_1).\n\\]\nTogether with monotonicity this proves the Lipschitz estimate. The omitted case $M_2=0$ has $M_1=0$ and is equality.","(1-t)M_2)\\le t\\beta(M_1)+(1-t)\\beta(M_2).\n\\]\nMonotonicity follows from inclusion of balls. If $0\\le M_1\\le M_2$ and $M_2>0$, choose a minimizer $a_2$ at $M_2$ and put $a_1'=(M_1/M_2)a_2$. Then $\\norm{a_1'}\\le M_1$ and $\\norm{a_2-a_1'}\\le M_2-M_1$, so\n\\[\n \\beta(M_1)\\le\\norm{r-Ta_1'}\n \\le\\beta(M_2)+\\norm T(M_2-M_1).\n\\]\nTogether with monotonicity this proves the Lipschitz estimate. The omitted case $M_2=0$ has $M_1=0$ and is equality.","t\\beta(M_1)+(1-t)\\beta(M_2).\n\\]\nMonotonicity follows from inclusion of balls. If $0\\le M_1\\le M_2$ and $M_2>0$, choose a minimizer $a_2$ at $M_2$ and put $a_1'=(M_1/M_2)a_2$. Then $\\norm{a_1'}\\le M_1$ and $\\norm{a_2-a_1'}\\le M_2-M_1$, so\n\\[\n \\beta(M_1)\\le\\norm{r-Ta_1'}\n \\le\\beta(M_2)+\\norm T(M_2-M_1).\n\\]\nTogether with monotonicity this proves the Lipschitz estimate. The omitted case $M_2=0$ has $M_1=0$ and is equality.","Since $r-b$ is orthogonal to $C$ and $b-Ta\\in C$, every correction satisfies\n\\begin{equation}\\label{eq:range-decomposition}\n \\norm{r-Ta}^2=q^2+\\norm{b-Ta}^2.\n\\end{equation}\nThus $\\beta(M)\\ge q$. Given $\\epsilon>0$, range density in $C$ supplies one correction $a$ with $\\norm{b-Ta}<\\epsilon$. For every $M\\ge\\norm a$ it follows that\n\\[\n q\\le\\beta(M)\\le\\sqrt{q^2+\\epsilon^2}.\n\\]\nFirst letting the budget increase and then letting $\\epsilon\\downarrow0$ proves the limit.","C$ and $b-Ta\\in C$, every correction satisfies\n\\begin{equation}\\label{eq:range-decomposition}\n \\norm{r-Ta}^2=q^2+\\norm{b-Ta}^2.\n\\end{equation}\nThus $\\beta(M)\\ge q$. Given $\\epsilon>0$, range density in $C$ supplies one correction $a$ with $\\norm{b-Ta}<\\epsilon$. For every $M\\ge\\norm a$ it follows that\n\\[\n q\\le\\beta(M)\\le\\sqrt{q^2+\\epsilon^2}.\n\\]\nFirst letting the budget increase and then letting $\\epsilon\\downarrow0$ proves the limit.","Finally, if $\\beta(M)=q$, choose an attaining correction in that ball. The orthogonal identity forces $\\norm{b-Ta}^2=0$, hence $Ta=b$. Conversely such a correction attains residual $q$. Setting $q=0$ proves the exact-feasibility criterion. Attainment at fixed budget is essential to the forward implication: a zero unrestricted infimum would not itself supply a preimage.\n\\end{proof}\n\\written{WR03}","Although $\\ran T$ may not be closed, the image $T(\\{a:\\norm a\\le M\\})$ of each closed budget ball is closed and convex. To see closedness, take a convergent sequence in that image, lift it to the bounded ball, and apply the weak limit argument. The full range is the increasing union of these closed images. Passing to an unlimited budget can therefore lose closedness even though every fixed-budget optimization is well posed.","The nearest reachable image $y_M=P_{T(\\{a:\\norm a\\le M\\})}r$ is unique, so the best residual vector $r-y_M$ is unique as well. Corrections mapping to $y_M$ can differ by vectors in $\\ker T$. Among those corrections there is a unique least norm correction. These two uniqueness statements should not be confused with injectivity of $T$.","\\subsection{A dual witness can rule out every correction in the budget}\n\\begin{theorem}[A sharp budget certificate]\\label{WR04}\nFor every $M\\ge0$,\n\\begin{equation}\\label{eq:budget-dual}\n \\beta_{T,r}(M)=\n \\max_{\\norm z\\le1}\\bigl(\\langle r,z\\rangle-M\\norm{T^*z}\\bigr).\n\\end{equation}\nThus any single $z$ with $\\norm z\\le1$ and\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}>\\varepsilon\n\\]\ncertifies that no correction of norm at most $M$ has residual at most $\\varepsilon$.\n\\end{theorem}\n\\begin{proof}\nFor every $\\norm a\\le M$ and $\\norm z\\le1$, the adjoint identity and Cauchy--Schwarz give\n\\[\n\\begin{aligned}\n \\langle r,z\\rangle-M\\norm{T^*z}\n &=\\langle r-Ta,z\\rangle+\\langle a,T^*z\\rangle-M\\norm{T^*z}","ness can rule out every correction in the budget}\n\\begin{theorem}[A sharp budget certificate]\\label{WR04}\nFor every $M\\ge0$,\n\\begin{equation}\\label{eq:budget-dual}\n \\beta_{T,r}(M)=\n \\max_{\\norm z\\le1}\\bigl(\\langle r,z\\rangle-M\\norm{T^*z}\\bigr).\n\\end{equation}\nThus any single $z$ with $\\norm z\\le1$ and\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}>\\varepsilon\n\\]\ncertifies that no correction of norm at most $M$ has residual at most $\\varepsilon$.\n\\end{theorem}\n\\begin{proof}\nFor every $\\norm a\\le M$ and $\\norm z\\le1$, the adjoint identity and Cauchy--Schwarz give\n\\[\n\\begin{aligned}\n \\langle r,z\\rangle-M\\norm{T^*z}\n &=\\langle r-Ta,z\\rangle+\\langle a,T^*z\\rangle-M\\norm{T^*z}","orrection in the budget}\n\\begin{theorem}[A sharp budget certificate]\\label{WR04}\nFor every $M\\ge0$,\n\\begin{equation}\\label{eq:budget-dual}\n \\beta_{T,r}(M)=\n \\max_{\\norm z\\le1}\\bigl(\\langle r,z\\rangle-M\\norm{T^*z}\\bigr).\n\\end{equation}\nThus any single $z$ with $\\norm z\\le1$ and\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}>\\varepsilon\n\\]\ncertifies that no correction of norm at most $M$ has residual at most $\\varepsilon$.\n\\end{theorem}\n\\begin{proof}\nFor every $\\norm a\\le M$ and $\\norm z\\le1$, the adjoint identity and Cauchy--Schwarz give\n\\[\n\\begin{aligned}\n \\langle r,z\\rangle-M\\norm{T^*z}\n &=\\langle r-Ta,z\\rangle+\\langle a,T^*z\\rangle-M\\norm{T^*z}","et certificate]\\label{WR04}\nFor every $M\\ge0$,\n\\begin{equation}\\label{eq:budget-dual}\n \\beta_{T,r}(M)=\n \\max_{\\norm z\\le1}\\bigl(\\langle r,z\\rangle-M\\norm{T^*z}\\bigr).\n\\end{equation}\nThus any single $z$ with $\\norm z\\le1$ and\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}>\\varepsilon\n\\]\ncertifies that no correction of norm at most $M$ has residual at most $\\varepsilon$.\n\\end{theorem}\n\\begin{proof}\nFor every $\\norm a\\le M$ and $\\norm z\\le1$, the adjoint identity and Cauchy--Schwarz give\n\\[\n\\begin{aligned}\n \\langle r,z\\rangle-M\\norm{T^*z}\n &=\\langle r-Ta,z\\rangle+\\langle a,T^*z\\rangle-M\\norm{T^*z}","el{WR04}\nFor every $M\\ge0$,\n\\begin{equation}\\label{eq:budget-dual}\n \\beta_{T,r}(M)=\n \\max_{\\norm z\\le1}\\bigl(\\langle r,z\\rangle-M\\norm{T^*z}\\bigr).\n\\end{equation}\nThus any single $z$ with $\\norm z\\le1$ and\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}>\\varepsilon\n\\]\ncertifies that no correction of norm at most $M$ has residual at most $\\varepsilon$.\n\\end{theorem}\n\\begin{proof}\nFor every $\\norm a\\le M$ and $\\norm z\\le1$, the adjoint identity and Cauchy--Schwarz give\n\\[\n\\begin{aligned}\n \\langle r,z\\rangle-M\\norm{T^*z}\n &=\\langle r-Ta,z\\rangle+\\langle a,T^*z\\rangle-M\\norm{T^*z}","ual}\n \\beta_{T,r}(M)=\n \\max_{\\norm z\\le1}\\bigl(\\langle r,z\\rangle-M\\norm{T^*z}\\bigr).\n\\end{equation}\nThus any single $z$ with $\\norm z\\le1$ and\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}>\\varepsilon\n\\]\ncertifies that no correction of norm at most $M$ has residual at most $\\varepsilon$.\n\\end{theorem}\n\\begin{proof}\nFor every $\\norm a\\le M$ and $\\norm z\\le1$, the adjoint identity and Cauchy--Schwarz give\n\\[\n\\begin{aligned}\n \\langle r,z\\rangle-M\\norm{T^*z}\n &=\\langle r-Ta,z\\rangle+\\langle a,T^*z\\rangle-M\\norm{T^*z}\\\\\n &\\le\\langle r-Ta,z\\rangle\n \\le\\norm{r-Ta}.\n\\end{aligned}\n\\]\nTaking the infimum over corrections proves that every proposed dual value is at most $\\beta(M)$. This part alone already proves the impossibility certificate.","&\\le\\langle r-Ta,z\\rangle\n \\le\\norm{r-Ta}.\n\\end{aligned}\n\\]\nTaking the infimum over corrections proves that every proposed dual value is at most $\\beta(M)$. This part alone already proves the impossibility certificate.","To prove sharpness, let $a_M$ be an attaining correction from Theorem~\\ref{WR03} and put $v=r-Ta_M$. If $v=0$, then $\\beta(M)=0$ and $z=0$ attains the dual bound. Assume henceforth $v\\ne0$ and set $z=v/\\norm v$.","Fix any feasible $a$ and put $h=a-a_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\no","Fix any feasible $a$ and put $h=a-a_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle","Fix any feasible $a$ and put $h=a-a_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm","Fix any feasible $a$ and put $h=a-a_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have const","Fix any feasible $a$ and put $h=a-a_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness a","Fix any feasible $a$ and put $h=a-a_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations","Fix any feasible $a$ and put $h=a-a_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations in a convex bal","Fix any feasible $a$ and put $h=a-a_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations in a convex ball, not an unproved exch","Fix any feasible $a$ and put $h=a-a_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations in a convex ball, not an unproved exchange of an infimum and a supremum.\n\\end{proof}\n\\w","Fix any feasible $a$ and put $h=a-a_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations in a convex ball, not an unproved exchange of an infimum and a supremum.\n\\end{proof}\n\\written{WR04}","_M$. The segment $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations in a convex ball, not an unproved exchange of an infimum and a supremum.\n\\end{proof}\n\\written{WR04}","t $a_M+th$ is feasible for $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations in a convex ball, not an unproved exchange of an infimum and a supremum.\n\\end{proof}\n\\written{WR04}","r $0\\le t\\le1$. Minimality gives, for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations in a convex ball, not an unproved exchange of an infimum and a supremum.\n\\end{proof}\n\\written{WR04}","for $t>0$,\n\\[\n 0\\le\\norm{v-tTh}^2-\\norm v^2\n   =-2t\\langle v,Th\\rangle+t^2\\norm{Th}^2.\n\\]\nDivide by $t$ and let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations in a convex ball, not an unproved exchange of an infimum and a supremum.\n\\end{proof}\n\\written{WR04}","let $t\\downarrow0$. It follows that $\\langle v,T(a-a_M)\\rangle\\le0$, hence\n\\[\n \\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations in a convex ball, not an unproved exchange of an infimum and a supremum.\n\\end{proof}\n\\written{WR04}","\\langle T^*z,a\\rangle\\le\\langle T^*z,a_M\\rangle\n \\quad(\\norm a\\le M).\n\\]\nThe support value of a Hilbert ball can be computed directly. If $T^*z\\ne0$ and $M>0$, it is attained at $a=M T^*z/\\norm{T^*z}$ and equals $M\\norm{T^*z}$. If either factor is zero, every feasible pairing is zero and the same formula holds. Thus\n\\[\n \\langle T^*z,a_M\\rangle=M\\norm{T^*z}.\n\\]\nSubstituting this equality gives\n\\[\n \\langle r,z\\rangle-M\\norm{T^*z}\n =\\langle r-Ta_M,z\\rangle=\\norm v=\\beta(M).\n\\]\nWe have constructed an attaining witness and thereby justified the word ``maximum.'' The proof uses one-sided variations in a convex ball, not an unproved exchange of an infimum and a supremum.\n\\end{proof}\n\\written{WR04}","In finite dimensions this is a checkable inequality, not merely an optimization prescription. A program may propose the correction or the witness. The displayed inequality can be verified independently of the proposal procedure. In a protected ambient calculation, use the restricted adjoint, not $d^*$ without projection onto $\\ker p$.","\\subsection{A two-dimensional calculation of the supporting witness}\nTake\n\\[\n T=\\begin{pmatrix}1&0\\\\0&1/2\\end{pmatrix},\\qquad r=(1,1),\\qquad\n a=(1/2,2/5),\\qquad v=r-Ta=(1/2,4/5).\n\\]\nThen $T^*v=a$. At the budget $M=\\norm a=\\sqrt{41}/10$, the normalized vector $z=v/\\norm v$ gives\n\\[\n \\langle r,v\\rangle-\\norm a\\norm{T^*v}\n =\\frac{13}{10}-\\frac{41}{100}=\\frac{89}{100}=\\norm v^2.\n\\]\nConsequently $a$ is a globally optimal budgeted correction and $\\beta(M)=\\sqrt{89}/10$. It is not necessary to compare it with a dense numerical grid of competing vectors. The supporting witness proves optimality over the entire ball.","0&1/2\\end{pmatrix},\\qquad r=(1,1),\\qquad\n a=(1/2,2/5),\\qquad v=r-Ta=(1/2,4/5).\n\\]\nThen $T^*v=a$. At the budget $M=\\norm a=\\sqrt{41}/10$, the normalized vector $z=v/\\norm v$ gives\n\\[\n \\langle r,v\\rangle-\\norm a\\norm{T^*v}\n =\\frac{13}{10}-\\frac{41}{100}=\\frac{89}{100}=\\norm v^2.\n\\]\nConsequently $a$ is a globally optimal budgeted correction and $\\beta(M)=\\sqrt{89}/10$. It is not necessary to compare it with a dense numerical grid of competing vectors. The supporting witness proves optimality over the entire ball.","\\subsection{How measurement and model uncertainty affect the budget}\n\\begin{theorem}[A robust budget profile]\\label{WR05}\nSuppose the bounded linear map $\\widetilde T:H\\to Y$ and the datum $\\widetilde r\\in Y$ satisfy\n\\[\n \\norm{T-\\widetilde T}\\le\\eta,\\qquad\n \\norm{r-\\widetilde r}\\le\\delta,\\qquad\\eta,\\delta\\ge0.\n\\]\nFor every $M\\ge0$,\n\\[\n \\abs{\\beta_{T,r}(M)-\\beta_{\\widetilde T,\\widetilde r}(M)}\n \\le\\delta+\\eta M.\n\\]\nIn particular, for $\\norm z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\wideti","\\subsection{How measurement and model uncertainty affect the budget}\n\\begin{theorem}[A robust budget profile]\\label{WR05}\nSuppose the bounded linear map $\\widetilde T:H\\to Y$ and the datum $\\widetilde r\\in Y$ satisfy\n\\[\n \\norm{T-\\widetilde T}\\le\\eta,\\qquad\n \\norm{r-\\widetilde r}\\le\\delta,\\qquad\\eta,\\delta\\ge0.\n\\]\nFor every $M\\ge0$,\n\\[\n \\abs{\\beta_{T,r}(M)-\\beta_{\\widetilde T,\\widetilde r}(M)}\n \\le\\delta+\\eta M.\n\\]\nIn particular, for $\\norm z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\","\\subsection{How measurement and model uncertainty affect the budget}\n\\begin{theorem}[A robust budget profile]\\label{WR05}\nSuppose the bounded linear map $\\widetilde T:H\\to Y$ and the datum $\\widetilde r\\in Y$ satisfy\n\\[\n \\norm{T-\\widetilde T}\\le\\eta,\\qquad\n \\norm{r-\\widetilde r}\\le\\delta,\\qquad\\eta,\\delta\\ge0.\n\\]\nFor every $M\\ge0$,\n\\[\n \\abs{\\beta_{T,r}(M)-\\beta_{\\widetilde T,\\widetilde r}(M)}\n \\le\\delta+\\eta M.\n\\]\nIn particular, for $\\norm z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\norm{\\widetilde r-\\widetilde Ta}\\le\\norm{r-Ta}+c.\n\\]\nThe domain of feasible corrections is identical in t","\\subsection{How measurement and model uncertainty affect the budget}\n\\begin{theorem}[A robust budget profile]\\label{WR05}\nSuppose the bounded linear map $\\widetilde T:H\\to Y$ and the datum $\\widetilde r\\in Y$ satisfy\n\\[\n \\norm{T-\\widetilde T}\\le\\eta,\\qquad\n \\norm{r-\\widetilde r}\\le\\delta,\\qquad\\eta,\\delta\\ge0.\n\\]\nFor every $M\\ge0$,\n\\[\n \\abs{\\beta_{T,r}(M)-\\beta_{\\widetilde T,\\widetilde r}(M)}\n \\le\\delta+\\eta M.\n\\]\nIn particular, for $\\norm z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\norm{\\widetilde r-\\widetilde Ta}\\le\\norm{r-Ta}+c.\n\\]\nThe domain of feasible corrections is identical in the two problems. Ta","\\subsection{How measurement and model uncertainty affect the budget}\n\\begin{theorem}[A robust budget profile]\\label{WR05}\nSuppose the bounded linear map $\\widetilde T:H\\to Y$ and the datum $\\widetilde r\\in Y$ satisfy\n\\[\n \\norm{T-\\widetilde T}\\le\\eta,\\qquad\n \\norm{r-\\widetilde r}\\le\\delta,\\qquad\\eta,\\delta\\ge0.\n\\]\nFor every $M\\ge0$,\n\\[\n \\abs{\\beta_{T,r}(M)-\\beta_{\\widetilde T,\\widetilde r}(M)}\n \\le\\delta+\\eta M.\n\\]\nIn particular, for $\\norm z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\norm{\\widetilde r-\\widetilde Ta}\\le\\norm{r-Ta}+c.\n\\]\nThe domain of feasible corrections is identical in the two problems. Taking the infimum of each inequality over that common domain yields\n\\[…","uncertainty affect the budget}\n\\begin{theorem}[A robust budget profile]\\label{WR05}\nSuppose the bounded linear map $\\widetilde T:H\\to Y$ and the datum $\\widetilde r\\in Y$ satisfy\n\\[\n \\norm{T-\\widetilde T}\\le\\eta,\\qquad\n \\norm{r-\\widetilde r}\\le\\delta,\\qquad\\eta,\\delta\\ge0.\n\\]\nFor every $M\\ge0$,\n\\[\n \\abs{\\beta_{T,r}(M)-\\beta_{\\widetilde T,\\widetilde r}(M)}\n \\le\\delta+\\eta M.\n\\]\nIn particular, for $\\norm z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\norm{\\widetilde r-\\widetilde Ta}\\le\\norm{r-Ta}+c.\n\\]\nThe domain of feasible corrections is identical in the two problems. Taking the infimum of each inequality over that common domain yields\n\\[\n \\beta_{T,r}(M)\\le\\beta_{\\widetilde T…","ct the budget}\n\\begin{theorem}[A robust budget profile]\\label{WR05}\nSuppose the bounded linear map $\\widetilde T:H\\to Y$ and the datum $\\widetilde r\\in Y$ satisfy\n\\[\n \\norm{T-\\widetilde T}\\le\\eta,\\qquad\n \\norm{r-\\widetilde r}\\le\\delta,\\qquad\\eta,\\delta\\ge0.\n\\]\nFor every $M\\ge0$,\n\\[\n \\abs{\\beta_{T,r}(M)-\\beta_{\\widetilde T,\\widetilde r}(M)}\n \\le\\delta+\\eta M.\n\\]\nIn particular, for $\\norm z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\norm{\\widetilde r-\\widetilde Ta}\\le\\norm{r-Ta}+c.\n\\]\nThe domain of feasible corrections is identical in the two problems. Taking the infimum of each inequality over that common domain yields\n\\[\n \\beta_{T,r}(M)\\le\\beta_{\\widetilde T,\\widetilde r}(M…","satisfy\n\\[\n \\norm{T-\\widetilde T}\\le\\eta,\\qquad\n \\norm{r-\\widetilde r}\\le\\delta,\\qquad\\eta,\\delta\\ge0.\n\\]\nFor every $M\\ge0$,\n\\[\n \\abs{\\beta_{T,r}(M)-\\beta_{\\widetilde T,\\widetilde r}(M)}\n \\le\\delta+\\eta M.\n\\]\nIn particular, for $\\norm z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\norm{\\widetilde r-\\widetilde Ta}\\le\\norm{r-Ta}+c.\n\\]\nThe domain of feasible corrections is identical in the two problems. Taking the infimum of each inequality over that common domain yields\n\\[\n \\beta_{T,r}(M)\\le\\beta_{\\widetilde T,\\widetilde r}(M)+c,\n \\qquad\n \\beta_{\\widetilde T,\\widetilde r}(M)\\le\\beta_{T,r}(M)+c.\n\\]\nThese are precisely the absolute-value bound. One could equivalently evaluate at …","\\norm{r-\\widetilde r}\\le\\delta,\\qquad\\eta,\\delta\\ge0.\n\\]\nFor every $M\\ge0$,\n\\[\n \\abs{\\beta_{T,r}(M)-\\beta_{\\widetilde T,\\widetilde r}(M)}\n \\le\\delta+\\eta M.\n\\]\nIn particular, for $\\norm z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\norm{\\widetilde r-\\widetilde Ta}\\le\\norm{r-Ta}+c.\n\\]\nThe domain of feasible corrections is identical in the two problems. Taking the infimum of each inequality over that common domain yields\n\\[\n \\beta_{T,r}(M)\\le\\beta_{\\widetilde T,\\widetilde r}(M)+c,\n \\qquad\n \\beta_{\\widetilde T,\\widetilde r}(M)\\le\\beta_{T,r}(M)+c.\n\\]\nThese are precisely the absolute-value bound. One could equivalently evaluate at the two attained minimizers, but no relation betw…","\\beta_{\\widetilde T,\\widetilde r}(M)}\n \\le\\delta+\\eta M.\n\\]\nIn particular, for $\\norm z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\norm{\\widetilde r-\\widetilde Ta}\\le\\norm{r-Ta}+c.\n\\]\nThe domain of feasible corrections is identical in the two problems. Taking the infimum of each inequality over that common domain yields\n\\[\n \\beta_{T,r}(M)\\le\\beta_{\\widetilde T,\\widetilde r}(M)+c,\n \\qquad\n \\beta_{\\widetilde T,\\widetilde r}(M)\\le\\beta_{T,r}(M)+c.\n\\]\nThese are precisely the absolute-value bound. One could equivalently evaluate at the two attained minimizers, but no relation between those minimizers is needed.","z\\le1$,\n\\[\n \\beta_{T,r}(M)\\ge\n \\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}\n       -\\delta-\\eta M.\n\\]\n\\end{theorem}\n\\begin{proof}\nFix a feasible correction $a$. The two residual vectors have difference\n\\[\n (r-Ta)-(\\widetilde r-\\widetilde Ta)\n =(r-\\widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\norm{\\widetilde r-\\widetilde Ta}\\le\\norm{r-Ta}+c.\n\\]\nThe domain of feasible corrections is identical in the two problems. Taking the infimum of each inequality over that common domain yields\n\\[\n \\beta_{T,r}(M)\\le\\beta_{\\widetilde T,\\widetilde r}(M)+c,\n \\qquad\n \\beta_{\\widetilde T,\\widetilde r}(M)\\le\\beta_{T,r}(M)+c.\n\\]\nThese are precisely the absolute-value bound. One could equivalently evaluate at the two attained minimizers, but no relation between those minimizers is needed.","widetilde r)-(T-\\widetilde T)a,\n\\]\nwhose norm is at most $c=\\delta+\\eta M$. The reverse triangle inequality therefore gives both inequalities\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+c,\n \\qquad\n \\norm{\\widetilde r-\\widetilde Ta}\\le\\norm{r-Ta}+c.\n\\]\nThe domain of feasible corrections is identical in the two problems. Taking the infimum of each inequality over that common domain yields\n\\[\n \\beta_{T,r}(M)\\le\\beta_{\\widetilde T,\\widetilde r}(M)+c,\n \\qquad\n \\beta_{\\widetilde T,\\widetilde r}(M)\\le\\beta_{T,r}(M)+c.\n\\]\nThese are precisely the absolute-value bound. One could equivalently evaluate at the two attained minimizers, but no relation between those minimizers is needed.","For any admissible $z$, Theorem~\\ref{WR04} for the nominal problem gives\n\\[\n \\beta_{\\widetilde T,\\widetilde r}(M)\n \\ge\\langle\\widetilde r,z\\rangle-M\\norm{\\widetilde T^*z}.\n\\]\nCombine this with the lower half of the preceding comparison. The result is a certificate against every correction in the original fixed ball, reduced by the explicit uncertainty allowance. If the correction space itself changed, the common-domain step would fail and this proof would no longer apply.\n\\end{proof}\n\\written{WR05}","This estimate identifies a practical boundary of increasingly large corrections. A model error $\\eta$ is amplified by the correction budget even if the measured residual is tiny. The candidate-level inequality is\n\\[\n \\norm{r-Ta}\\le\\norm{\\widetilde r-\\widetilde Ta}+\\delta+\\eta\\norm a.\n\\]\nThe theorem compares maps on one fixed correction space. It does not cover an uncertain preservation map whose kernel itself changes.","\\fixedplate{33_budget_uncertainty}{The nominal budget profile is surrounded by the rigorous uncertainty band $\\delta+\\eta M$. A large budget can improve the nominal fit while enlarging the uncertainty in that conclusion. The example uses $T=1$, $r=1$, $\\delta=1/20$, and $\\eta=1/10$.}{fig:budget-uncertainty}","\\subsection{The normal equation is derived in the protected space}\nFor $\\alpha>0$, define on $H=\\ker p$\n\\[\n J_\\alpha(a)=\\norm{Ta-r}^2+\\alpha\\norm a^2.\n\\]\nBoth norms are the declared Hilbert norms. All minimizers below are corrections in $H$, so preservation is exact for every parameter, not just in the limit. This is ordinary quadratic Tikhonov regularization applied to the actual protected correction operator. The standard regularization background is discussed in \\cite{clason-inverse}. The proof here does not require $T$ to be compact.","Ta-r}^2+\\alpha\\norm a^2.\n\\]\nBoth norms are the declared Hilbert norms. All minimizers below are corrections in $H$, so preservation is exact for every parameter, not just in the limit. This is ordinary quadratic Tikhonov regularization applied to the actual protected correction operator. The standard regularization background is discussed in \\cite{clason-inverse}. The proof here does not require $T$ to be compact.","\\begin{theorem}[A unique regularized correction]\\label{WR06}\nFor every $\\alpha>0$, there is a unique minimizer\n\\[\n a_\\alpha=(T^*T+\\alpha I)^{-1}T^*r.\n\\]\nIt satisfies $T^*(r-Ta_\\alpha)=\\alpha a_\\alpha$. For every $a\\in H$,\n\\begin{equation}\\label{eq:regularization-energy}\n J_\\alpha(a)-J_\\alpha(a_\\alpha)\n =\\norm{T(a-a_\\alpha)}^2+\\alpha\\norm{a-a_\\alpha}^2.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: the normal operator has an actual bounded inverse.} Let $B=T^*T+\\alpha I:H\\to H$. It is bounded and self-adjoint. For every $h\\in H$,\n\\[\n \\langle Bh,h\\rangle=\\norm{Th}^2+\\alpha\\norm h^2\n \\ge\\alpha\\norm h^2.\n\\]\nIf $h\\ne0$, Cauchy--Schwarz and division by $\\norm h$ give $\\norm{Bh}\\ge\\alpha\\norm h$. The same inequality holds at zero. It implies injectivity of $B$.","y $\\alpha>0$, there is a unique minimizer\n\\[\n a_\\alpha=(T^*T+\\alpha I)^{-1}T^*r.\n\\]\nIt satisfies $T^*(r-Ta_\\alpha)=\\alpha a_\\alpha$. For every $a\\in H$,\n\\begin{equation}\\label{eq:regularization-energy}\n J_\\alpha(a)-J_\\alpha(a_\\alpha)\n =\\norm{T(a-a_\\alpha)}^2+\\alpha\\norm{a-a_\\alpha}^2.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: the normal operator has an actual bounded inverse.} Let $B=T^*T+\\alpha I:H\\to H$. It is bounded and self-adjoint. For every $h\\in H$,\n\\[\n \\langle Bh,h\\rangle=\\norm{Th}^2+\\alpha\\norm h^2\n \\ge\\alpha\\norm h^2.\n\\]\nIf $h\\ne0$, Cauchy--Schwarz and division by $\\norm h$ give $\\norm{Bh}\\ge\\alpha\\norm h$. The same inequality holds at zero. It implies injectivity of $B$.","I)^{-1}T^*r.\n\\]\nIt satisfies $T^*(r-Ta_\\alpha)=\\alpha a_\\alpha$. For every $a\\in H$,\n\\begin{equation}\\label{eq:regularization-energy}\n J_\\alpha(a)-J_\\alpha(a_\\alpha)\n =\\norm{T(a-a_\\alpha)}^2+\\alpha\\norm{a-a_\\alpha}^2.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: the normal operator has an actual bounded inverse.} Let $B=T^*T+\\alpha I:H\\to H$. It is bounded and self-adjoint. For every $h\\in H$,\n\\[\n \\langle Bh,h\\rangle=\\norm{Th}^2+\\alpha\\norm h^2\n \\ge\\alpha\\norm h^2.\n\\]\nIf $h\\ne0$, Cauchy--Schwarz and division by $\\norm h$ give $\\norm{Bh}\\ge\\alpha\\norm h$. The same inequality holds at zero. It implies injectivity of $B$.",".\n\\]\nIt satisfies $T^*(r-Ta_\\alpha)=\\alpha a_\\alpha$. For every $a\\in H$,\n\\begin{equation}\\label{eq:regularization-energy}\n J_\\alpha(a)-J_\\alpha(a_\\alpha)\n =\\norm{T(a-a_\\alpha)}^2+\\alpha\\norm{a-a_\\alpha}^2.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: the normal operator has an actual bounded inverse.} Let $B=T^*T+\\alpha I:H\\to H$. It is bounded and self-adjoint. For every $h\\in H$,\n\\[\n \\langle Bh,h\\rangle=\\norm{Th}^2+\\alpha\\norm h^2\n \\ge\\alpha\\norm h^2.\n\\]\nIf $h\\ne0$, Cauchy--Schwarz and division by $\\norm h$ give $\\norm{Bh}\\ge\\alpha\\norm h$. The same inequality holds at zero. It implies injectivity of $B$.","in{equation}\\label{eq:regularization-energy}\n J_\\alpha(a)-J_\\alpha(a_\\alpha)\n =\\norm{T(a-a_\\alpha)}^2+\\alpha\\norm{a-a_\\alpha}^2.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: the normal operator has an actual bounded inverse.} Let $B=T^*T+\\alpha I:H\\to H$. It is bounded and self-adjoint. For every $h\\in H$,\n\\[\n \\langle Bh,h\\rangle=\\norm{Th}^2+\\alpha\\norm h^2\n \\ge\\alpha\\norm h^2.\n\\]\nIf $h\\ne0$, Cauchy--Schwarz and division by $\\norm h$ give $\\norm{Bh}\\ge\\alpha\\norm h$. The same inequality holds at zero. It implies injectivity of $B$.","gy}\n J_\\alpha(a)-J_\\alpha(a_\\alpha)\n =\\norm{T(a-a_\\alpha)}^2+\\alpha\\norm{a-a_\\alpha}^2.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: the normal operator has an actual bounded inverse.} Let $B=T^*T+\\alpha I:H\\to H$. It is bounded and self-adjoint. For every $h\\in H$,\n\\[\n \\langle Bh,h\\rangle=\\norm{Th}^2+\\alpha\\norm h^2\n \\ge\\alpha\\norm h^2.\n\\]\nIf $h\\ne0$, Cauchy--Schwarz and division by $\\norm h$ give $\\norm{Bh}\\ge\\alpha\\norm h$. The same inequality holds at zero. It implies injectivity of $B$.",")-J_\\alpha(a_\\alpha)\n =\\norm{T(a-a_\\alpha)}^2+\\alpha\\norm{a-a_\\alpha}^2.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: the normal operator has an actual bounded inverse.} Let $B=T^*T+\\alpha I:H\\to H$. It is bounded and self-adjoint. For every $h\\in H$,\n\\[\n \\langle Bh,h\\rangle=\\norm{Th}^2+\\alpha\\norm h^2\n \\ge\\alpha\\norm h^2.\n\\]\nIf $h\\ne0$, Cauchy--Schwarz and division by $\\norm h$ give $\\norm{Bh}\\ge\\alpha\\norm h$. The same inequality holds at zero. It implies injectivity of $B$.","nd{theorem}\n\\begin{proof}\n\\emph{Step 1: the normal operator has an actual bounded inverse.} Let $B=T^*T+\\alpha I:H\\to H$. It is bounded and self-adjoint. For every $h\\in H$,\n\\[\n \\langle Bh,h\\rangle=\\norm{Th}^2+\\alpha\\norm h^2\n \\ge\\alpha\\norm h^2.\n\\]\nIf $h\\ne0$, Cauchy--Schwarz and division by $\\norm h$ give $\\norm{Bh}\\ge\\alpha\\norm h$. The same inequality holds at zero. It implies injectivity of $B$.","To prove closedness of its range, suppose $Bh_n\\to y$. Applying the lower bound to $h_n-h_m$ gives\n\\[\n \\norm{h_n-h_m}\\le\\alpha^{-1}\\norm{Bh_n-Bh_m}.\n\\]\nThus $h_n$ is Cauchy. Completeness of $H$ supplies a limit $h$, and continuity gives $Bh=y$. If $y$ is orthogonal to the range of $B$, then for every $h$,\n\\[\n 0=\\langle Bh,y\\rangle=\\langle h,By\\rangle.\n\\]\nTaking $h=By$ gives $By=0$, and injectivity gives $y=0$. A closed subspace with zero orthogonal complement is the whole Hilbert space, so $\\ran B=H$.","closedness of its range, suppose $Bh_n\\to y$. Applying the lower bound to $h_n-h_m$ gives\n\\[\n \\norm{h_n-h_m}\\le\\alpha^{-1}\\norm{Bh_n-Bh_m}.\n\\]\nThus $h_n$ is Cauchy. Completeness of $H$ supplies a limit $h$, and continuity gives $Bh=y$. If $y$ is orthogonal to the range of $B$, then for every $h$,\n\\[\n 0=\\langle Bh,y\\rangle=\\langle h,By\\rangle.\n\\]\nTaking $h=By$ gives $By=0$, and injectivity gives $y=0$. A closed subspace with zero orthogonal complement is the whole Hilbert space, so $\\ran B=H$.","es\n\\[\n \\norm{h_n-h_m}\\le\\alpha^{-1}\\norm{Bh_n-Bh_m}.\n\\]\nThus $h_n$ is Cauchy. Completeness of $H$ supplies a limit $h$, and continuity gives $Bh=y$. If $y$ is orthogonal to the range of $B$, then for every $h$,\n\\[\n 0=\\langle Bh,y\\rangle=\\langle h,By\\rangle.\n\\]\nTaking $h=By$ gives $By=0$, and injectivity gives $y=0$. A closed subspace with zero orthogonal complement is the whole Hilbert space, so $\\ran B=H$.","The inverse therefore exists. Its linearity follows from uniqueness of preimages, and its bound is $\\norm{B^{-1}y}\\le\\alpha^{-1}\\norm y$. This construction does not assume closedness of $\\ran T$. It uses the positive term $\\alpha I$ in the normal operator.","\\emph{Step 2: define the optimizer and prove its normal equation.} Set $a_\\alpha=B^{-1}T^*r$. Then $Ba_\\alpha=T^*r$, or equivalently\n\\[\n T^*(Ta_\\alpha-r)+\\alpha a_\\alpha=0.\n\\]\nBecause the construction takes place in $H=\\ker p$, it already satisfies the exact protection requirement.","\\emph{Step 3: prove global minimality, rather than just stationarity.} Write any proposed correction as $a=a_\\alpha+h$ and set $s=Ta_\\alpha-r$. Expanding both squared norms gives\n\\[\n\\begin{aligned}\n J_\\alpha(a_\\alpha+h)-J_\\alpha(a_\\alpha)\n &=\\norm{Th}^2+\\alpha\\norm h^2\n     +2\\langle s,Th\\rangle+2\\alpha\\langle a_\\alpha,h\\rangle","\\emph{Step 3: prove global minimality, rather than just stationarity.} Write any proposed correction as $a=a_\\alpha+h$ and set $s=Ta_\\alpha-r$. Expanding both squared norms gives\n\\[\n\\begin{aligned}\n J_\\alpha(a_\\alpha+h)-J_\\alpha(a_\\alpha)\n &=\\norm{Th}^2+\\alpha\\norm h^2\n     +2\\langle s,Th\\rangle+2\\alpha\\langle a_\\alpha,h\\rangle\\\\\n &=\\norm{Th}^2+\\alpha\\norm h^2\n     +2\\langle T^*s+\\alpha a_\\alpha,h\\rangle\\\\\n &=\\norm{Th}^2+\\alpha\\norm h^2.\n\\end{aligned}\n\\]\nThis is the asserted energy identity. It is nonnegative and, since $\\alpha>0$, vanishes only for $h=0$. Therefore $a_\\alpha$ is the unique global minimizer. In the zero correction space, the same construction and identities reduce to its single element, zero.\n\\end{proof}\n\\written{WR06}","&=\\norm{Th}^2+\\alpha\\norm h^2.\n\\end{aligned}\n\\]\nThis is the asserted energy identity. It is nonnegative and, since $\\alpha>0$, vanishes only for $h=0$. Therefore $a_\\alpha$ is the unique global minimizer. In the zero correction space, the same construction and identities reduce to its single element, zero.\n\\end{proof}\n\\written{WR06}","A declared regularization parameter is part of the problem. A calculation that changes $\\alpha$ is solving a different optimization problem, even if the protected data and forward operator stay fixed. On the other hand, the preservation condition itself does not change with $\\alpha$.","\\subsection{Coordinate bases must retain the correction metric}\nIn finite dimensions, choose orthonormal coordinates in the ambient candidate and residual spaces. Let $B$ have linearly independent columns spanning $\\ker p$. The columns of $B$ need not be orthonormal. Write $a=Bc$. The coefficient vector $c$ is a representation of the correction, not necessarily an isometric one. Set\n\\[\n S=dB,\\qquad G=B^{\\mathsf T}B.\n\\]\nThe correct coordinate equation is\n\\begin{equation}\\label{eq:gram-regularization}\n (S^{\\mathsf T}S+\\alpha G)c=S^{\\mathsf T}r.\n\\end{equation}\nThe matrix $G$ is positive definite because the columns are independent. Replacing $G$ by the identity changes the meaning of correction size unless the chosen basis is orthonormal.","dimensions, choose orthonormal coordinates in the ambient candidate and residual spaces. Let $B$ have linearly independent columns spanning $\\ker p$. The columns of $B$ need not be orthonormal. Write $a=Bc$. The coefficient vector $c$ is a representation of the correction, not necessarily an isometric one. Set\n\\[\n S=dB,\\qquad G=B^{\\mathsf T}B.\n\\]\nThe correct coordinate equation is\n\\begin{equation}\\label{eq:gram-regularization}\n (S^{\\mathsf T}S+\\alpha G)c=S^{\\mathsf T}r.\n\\end{equation}\nThe matrix $G$ is positive definite because the columns are independent. Replacing $G$ by the identity changes the meaning of correction size unless the chosen basis is orthonormal.","linearly independent columns spanning $\\ker p$. The columns of $B$ need not be orthonormal. Write $a=Bc$. The coefficient vector $c$ is a representation of the correction, not necessarily an isometric one. Set\n\\[\n S=dB,\\qquad G=B^{\\mathsf T}B.\n\\]\nThe correct coordinate equation is\n\\begin{equation}\\label{eq:gram-regularization}\n (S^{\\mathsf T}S+\\alpha G)c=S^{\\mathsf T}r.\n\\end{equation}\nThe matrix $G$ is positive definite because the columns are independent. Replacing $G$ by the identity changes the meaning of correction size unless the chosen basis is orthonormal.","not be orthonormal. Write $a=Bc$. The coefficient vector $c$ is a representation of the correction, not necessarily an isometric one. Set\n\\[\n S=dB,\\qquad G=B^{\\mathsf T}B.\n\\]\nThe correct coordinate equation is\n\\begin{equation}\\label{eq:gram-regularization}\n (S^{\\mathsf T}S+\\alpha G)c=S^{\\mathsf T}r.\n\\end{equation}\nThe matrix $G$ is positive definite because the columns are independent. Replacing $G$ by the identity changes the meaning of correction size unless the chosen basis is orthonormal.","For example, let $d(a_1,a_2)=a_1$, $p(a_1,a_2)=a_1-a_2$, $r=1$, and $\\alpha=1$. A basis of $\\ker p$ is the column $B=(1,1)^{\\mathsf T}$. Here $S=1$ and $G=2$, so the correction is $(1/3,1/3)$. Penalizing the coordinate with $G=1$ instead gives $(1/2,1/2)$. The latter is not the optimizer for the ambient Euclidean norm. The difference is a metric change, not a discrepancy between equivalent calculations.","\\subsection{Parameter monotonicity and the fixed-budget problem}\n\\begin{proposition}[Regularized answers lie on the budget frontier]\\label{WR07}\nIf $0<\\alpha<\\gamma$, then\n\\[\n \\norm{a_\\alpha}\\ge\\norm{a_\\gamma},\\qquad\n \\norm{r-Ta_\\alpha}\\le\\norm{r-Ta_\\gamma}.\n\\]\nAt $M_\\alpha=\\norm{a_\\alpha}$,\n\\[\n \\beta(M_\\alpha)=\\norm{r-Ta_\\alpha}.\n\\]\nWhen $v_\\alpha=r-Ta_\\alpha\\ne0$, its normalization is an optimizing witness in Theorem~\\ref{WR04} at that budget.\n\\end{proposition}\n\\begin{proof}\nLet $A_\\alpha=\\norm{a_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nS","\\subsection{Parameter monotonicity and the fixed-budget problem}\n\\begin{proposition}[Regularized answers lie on the budget frontier]\\label{WR07}\nIf $0<\\alpha<\\gamma$, then\n\\[\n \\norm{a_\\alpha}\\ge\\norm{a_\\gamma},\\qquad\n \\norm{r-Ta_\\alpha}\\le\\norm{r-Ta_\\gamma}.\n\\]\nAt $M_\\alpha=\\norm{a_\\alpha}$,\n\\[\n \\beta(M_\\alpha)=\\norm{r-Ta_\\alpha}.\n\\]\nWhen $v_\\alpha=r-Ta_\\alpha\\ne0$, its normalization is an optimizing witness in Theorem~\\ref{WR04} at that budget.\n\\end{proposition}\n\\begin{proof}\nLet $A_\\alpha=\\norm{a_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to","\\subsection{Parameter monotonicity and the fixed-budget problem}\n\\begin{proposition}[Regularized answers lie on the budget frontier]\\label{WR07}\nIf $0<\\alpha<\\gamma$, then\n\\[\n \\norm{a_\\alpha}\\ge\\norm{a_\\gamma},\\qquad\n \\norm{r-Ta_\\alpha}\\le\\norm{r-Ta_\\gamma}.\n\\]\nAt $M_\\alpha=\\norm{a_\\alpha}$,\n\\[\n \\beta(M_\\alpha)=\\norm{r-Ta_\\alpha}.\n\\]\nWhen $v_\\alpha=r-Ta_\\alpha\\ne0$, its normalization is an optimizing witness in Theorem~\\ref{WR04} at that budget.\n\\end{proposition}\n\\begin{proof}\nLet $A_\\alpha=\\norm{a_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to obtain\n\\[\n R_\\alpha-R_\\gamma\\l","\\subsection{Parameter monotonicity and the fixed-budget problem}\n\\begin{proposition}[Regularized answers lie on the budget frontier]\\label{WR07}\nIf $0<\\alpha<\\gamma$, then\n\\[\n \\norm{a_\\alpha}\\ge\\norm{a_\\gamma},\\qquad\n \\norm{r-Ta_\\alpha}\\le\\norm{r-Ta_\\gamma}.\n\\]\nAt $M_\\alpha=\\norm{a_\\alpha}$,\n\\[\n \\beta(M_\\alpha)=\\norm{r-Ta_\\alpha}.\n\\]\nWhen $v_\\alpha=r-Ta_\\alpha\\ne0$, its normalization is an optimizing witness in Theorem~\\ref{WR04} at that budget.\n\\end{proposition}\n\\begin{proof}\nLet $A_\\alpha=\\norm{a_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to obtain\n\\[\n R_\\alpha-R_\\gamma\\le\\alpha(A_\\gamma-A_\\alpha)\\le0.\n\\]\nTaking s","\\subsection{Parameter monotonicity and the fixed-budget problem}\n\\begin{proposition}[Regularized answers lie on the budget frontier]\\label{WR07}\nIf $0<\\alpha<\\gamma$, then\n\\[\n \\norm{a_\\alpha}\\ge\\norm{a_\\gamma},\\qquad\n \\norm{r-Ta_\\alpha}\\le\\norm{r-Ta_\\gamma}.\n\\]\nAt $M_\\alpha=\\norm{a_\\alpha}$,\n\\[\n \\beta(M_\\alpha)=\\norm{r-Ta_\\alpha}.\n\\]\nWhen $v_\\alpha=r-Ta_\\alpha\\ne0$, its normalization is an optimizing witness in Theorem~\\ref{WR04} at that budget.\n\\end{proposition}\n\\begin{proof}\nLet $A_\\alpha=\\norm{a_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to obtain\n\\[\n R_\\alpha-R_\\gamma\\le\\alpha(A_\\gamma-A_\\alpha)\\le0.\n\\]\nTaking square roots, which preserves orde","gularized answers lie on the budget frontier]\\label{WR07}\nIf $0<\\alpha<\\gamma$, then\n\\[\n \\norm{a_\\alpha}\\ge\\norm{a_\\gamma},\\qquad\n \\norm{r-Ta_\\alpha}\\le\\norm{r-Ta_\\gamma}.\n\\]\nAt $M_\\alpha=\\norm{a_\\alpha}$,\n\\[\n \\beta(M_\\alpha)=\\norm{r-Ta_\\alpha}.\n\\]\nWhen $v_\\alpha=r-Ta_\\alpha\\ne0$, its normalization is an optimizing witness in Theorem~\\ref{WR04} at that budget.\n\\end{proposition}\n\\begin{proof}\nLet $A_\\alpha=\\norm{a_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to obtain\n\\[\n R_\\alpha-R_\\gamma\\le\\alpha(A_\\gamma-A_\\alpha)\\le0.\n\\]\nTaking square roots, which preserves order for nonnegative numbers, proves both monotonicity statements.","et frontier]\\label{WR07}\nIf $0<\\alpha<\\gamma$, then\n\\[\n \\norm{a_\\alpha}\\ge\\norm{a_\\gamma},\\qquad\n \\norm{r-Ta_\\alpha}\\le\\norm{r-Ta_\\gamma}.\n\\]\nAt $M_\\alpha=\\norm{a_\\alpha}$,\n\\[\n \\beta(M_\\alpha)=\\norm{r-Ta_\\alpha}.\n\\]\nWhen $v_\\alpha=r-Ta_\\alpha\\ne0$, its normalization is an optimizing witness in Theorem~\\ref{WR04} at that budget.\n\\end{proposition}\n\\begin{proof}\nLet $A_\\alpha=\\norm{a_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to obtain\n\\[\n R_\\alpha-R_\\gamma\\le\\alpha(A_\\gamma-A_\\alpha)\\le0.\n\\]\nTaking square roots, which preserves order for nonnegative numbers, proves both monotonicity statements.","a_\\alpha}\\le\\norm{r-Ta_\\gamma}.\n\\]\nAt $M_\\alpha=\\norm{a_\\alpha}$,\n\\[\n \\beta(M_\\alpha)=\\norm{r-Ta_\\alpha}.\n\\]\nWhen $v_\\alpha=r-Ta_\\alpha\\ne0$, its normalization is an optimizing witness in Theorem~\\ref{WR04} at that budget.\n\\end{proposition}\n\\begin{proof}\nLet $A_\\alpha=\\norm{a_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to obtain\n\\[\n R_\\alpha-R_\\gamma\\le\\alpha(A_\\gamma-A_\\alpha)\\le0.\n\\]\nTaking square roots, which preserves order for nonnegative numbers, proves both monotonicity statements.","orem~\\ref{WR04} at that budget.\n\\end{proposition}\n\\begin{proof}\nLet $A_\\alpha=\\norm{a_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to obtain\n\\[\n R_\\alpha-R_\\gamma\\le\\alpha(A_\\gamma-A_\\alpha)\\le0.\n\\]\nTaking square roots, which preserves order for nonnegative numbers, proves both monotonicity statements.","egin{proof}\nLet $A_\\alpha=\\norm{a_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to obtain\n\\[\n R_\\alpha-R_\\gamma\\le\\alpha(A_\\gamma-A_\\alpha)\\le0.\n\\]\nTaking square roots, which preserves order for nonnegative numbers, proves both monotonicity statements.","_\\alpha}^2$ and $R_\\alpha=\\norm{r-Ta_\\alpha}^2$. Comparing the two admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to obtain\n\\[\n R_\\alpha-R_\\gamma\\le\\alpha(A_\\gamma-A_\\alpha)\\le0.\n\\]\nTaking square roots, which preserves order for nonnegative numbers, proves both monotonicity statements.","admissible minimizers in each other's objectives gives\n\\[\n R_\\alpha+\\alpha A_\\alpha\\le R_\\gamma+\\alpha A_\\gamma,\n \\qquad R_\\gamma+\\gamma A_\\gamma\\le R_\\alpha+\\gamma A_\\alpha.\n\\]\nAdd these inequalities and cancel the residual terms. The result is\n\\[\n (\\gamma-\\alpha)(A_\\alpha-A_\\gamma)\\ge0.\n\\]\nSince $\\gamma>\\alpha$, it follows that $A_\\alpha\\ge A_\\gamma$. Return to the first comparison to obtain\n\\[\n R_\\alpha-R_\\gamma\\le\\alpha(A_\\gamma-A_\\alpha)\\le0.\n\\]\nTaking square roots, which preserves order for nonnegative numbers, proves both monotonicity statements.","At $M_\\alpha=\\norm{a_\\alpha}$, the vector $a_\\alpha$ is budget-feasible. If another correction $a$ in that ball had strictly smaller residual norm, then\n\\[\n J_\\alpha(a)\n =\\norm{r-Ta}^2+\\alpha\\norm a^2\n <\\norm{r-Ta_\\alpha}^2+\\alpha\\norm{a_\\alpha}^2,\n\\]\ncontradicting its defining minimality. Thus it attains the budget frontier.","For the witness statement, let $v=r-Ta_\\alpha\\ne0$. The normal equation gives $T^*v=\\alpha a_\\alpha$. Also $r=v+Ta_\\alpha$, hence\n\\[\n \\langle r,v\\rangle\n =\\norm v^2+\\langle a_\\alpha,T^*v\\rangle\n =\\norm v^2+\\alpha\\norm{a_\\alpha}^2.\n\\]\nFor $z=v/\\norm v$, the support penalty is\n\\[\n M_\\alpha\\norm{T^*z}\n =\\frac{\\alpha\\norm{a_\\alpha}^2}{\\norm v}.\n\\]\nSubtracting it from $\\langle r,z\\rangle$ yields $\\norm v=\\beta(M_\\alpha)$, so this is an optimizing witness. This calculation also covers a zero correction $a_\\alpha$ with nonzero residual. If $v=0$, normalization is unavailable, but the zero witness attains the zero budget residual.\n\\end{proof}\n\\written{WR07}","lso $r=v+Ta_\\alpha$, hence\n\\[\n \\langle r,v\\rangle\n =\\norm v^2+\\langle a_\\alpha,T^*v\\rangle\n =\\norm v^2+\\alpha\\norm{a_\\alpha}^2.\n\\]\nFor $z=v/\\norm v$, the support penalty is\n\\[\n M_\\alpha\\norm{T^*z}\n =\\frac{\\alpha\\norm{a_\\alpha}^2}{\\norm v}.\n\\]\nSubtracting it from $\\langle r,z\\rangle$ yields $\\norm v=\\beta(M_\\alpha)$, so this is an optimizing witness. This calculation also covers a zero correction $a_\\alpha$ with nonzero residual. If $v=0$, normalization is unavailable, but the zero witness attains the zero budget residual.\n\\end{proof}\n\\written{WR07}","e r,v\\rangle\n =\\norm v^2+\\langle a_\\alpha,T^*v\\rangle\n =\\norm v^2+\\alpha\\norm{a_\\alpha}^2.\n\\]\nFor $z=v/\\norm v$, the support penalty is\n\\[\n M_\\alpha\\norm{T^*z}\n =\\frac{\\alpha\\norm{a_\\alpha}^2}{\\norm v}.\n\\]\nSubtracting it from $\\langle r,z\\rangle$ yields $\\norm v=\\beta(M_\\alpha)$, so this is an optimizing witness. This calculation also covers a zero correction $a_\\alpha$ with nonzero residual. If $v=0$, normalization is unavailable, but the zero witness attains the zero budget residual.\n\\end{proof}\n\\written{WR07}","This links a penalized calculation to a hard-budget certificate. It does not say that every larger budget must produce a different answer. Once the least-squares floor is attained, excess budget can be irrelevant. If $T^*r=0$, every regularized correction is zero and the entire profile is constant.","\\subsection{The limit separates projection from actual realizability}\nKeep $C=\\overline{\\ran T}$, $b=P_Cr$, and $q=\\norm{r-b}$. Since $r-b\\in\\ker T^*$, the regularized corrections depend on $b$, not on the component that is orthogonal to the closure of all attainable residuals.","\\begin{theorem}[The regularization limit alternative]\\label{WR08}\nAs $\\alpha\\downarrow0$,\n\\[\n Ta_\\alpha\\longrightarrow b,\\qquad\n \\norm{r-Ta_\\alpha}\\longrightarrow q,\\qquad\n \\alpha\\norm{a_\\alpha}^2\\longrightarrow0.\n\\]\nIf $b\\in\\ran T$, let $a^\\dagger$ be its unique minimum norm preimage. Then\n\\[\n a_\\alpha\\longrightarrow a^\\dagger\\quad\\text{in norm}.\n\\]\nIf $b\\notin\\ran T$, then $\\norm{a_\\alpha}\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: identify the unavoidable component.} Since $r-b\\perp\\overline{\\ran T}$, the adjoint identity gives $T^*(r-b)=0$. Thus $a_\\alpha(r)=a_\\alpha(b)$. For every correction the orthogonal decomposition is\n\\[\n \\norm{r-Ta}^2=q^2+\\norm{b-Ta}^2.\n\\]\nIn particular both the residual square and the minimum objective are at least $q^2$.","Ta_\\alpha\\longrightarrow b,\\qquad\n \\norm{r-Ta_\\alpha}\\longrightarrow q,\\qquad\n \\alpha\\norm{a_\\alpha}^2\\longrightarrow0.\n\\]\nIf $b\\in\\ran T$, let $a^\\dagger$ be its unique minimum norm preimage. Then\n\\[\n a_\\alpha\\longrightarrow a^\\dagger\\quad\\text{in norm}.\n\\]\nIf $b\\notin\\ran T$, then $\\norm{a_\\alpha}\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: identify the unavoidable component.} Since $r-b\\perp\\overline{\\ran T}$, the adjoint identity gives $T^*(r-b)=0$. Thus $a_\\alpha(r)=a_\\alpha(b)$. For every correction the orthogonal decomposition is\n\\[\n \\norm{r-Ta}^2=q^2+\\norm{b-Ta}^2.\n\\]\nIn particular both the residual square and the minimum objective are at least $q^2$.","longrightarrow q,\\qquad\n \\alpha\\norm{a_\\alpha}^2\\longrightarrow0.\n\\]\nIf $b\\in\\ran T$, let $a^\\dagger$ be its unique minimum norm preimage. Then\n\\[\n a_\\alpha\\longrightarrow a^\\dagger\\quad\\text{in norm}.\n\\]\nIf $b\\notin\\ran T$, then $\\norm{a_\\alpha}\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: identify the unavoidable component.} Since $r-b\\perp\\overline{\\ran T}$, the adjoint identity gives $T^*(r-b)=0$. Thus $a_\\alpha(r)=a_\\alpha(b)$. For every correction the orthogonal decomposition is\n\\[\n \\norm{r-Ta}^2=q^2+\\norm{b-Ta}^2.\n\\]\nIn particular both the residual square and the minimum objective are at least $q^2$.","quad\n \\alpha\\norm{a_\\alpha}^2\\longrightarrow0.\n\\]\nIf $b\\in\\ran T$, let $a^\\dagger$ be its unique minimum norm preimage. Then\n\\[\n a_\\alpha\\longrightarrow a^\\dagger\\quad\\text{in norm}.\n\\]\nIf $b\\notin\\ran T$, then $\\norm{a_\\alpha}\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: identify the unavoidable component.} Since $r-b\\perp\\overline{\\ran T}$, the adjoint identity gives $T^*(r-b)=0$. Thus $a_\\alpha(r)=a_\\alpha(b)$. For every correction the orthogonal decomposition is\n\\[\n \\norm{r-Ta}^2=q^2+\\norm{b-Ta}^2.\n\\]\nIn particular both the residual square and the minimum objective are at least $q^2$.","$ be its unique minimum norm preimage. Then\n\\[\n a_\\alpha\\longrightarrow a^\\dagger\\quad\\text{in norm}.\n\\]\nIf $b\\notin\\ran T$, then $\\norm{a_\\alpha}\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: identify the unavoidable component.} Since $r-b\\perp\\overline{\\ran T}$, the adjoint identity gives $T^*(r-b)=0$. Thus $a_\\alpha(r)=a_\\alpha(b)$. For every correction the orthogonal decomposition is\n\\[\n \\norm{r-Ta}^2=q^2+\\norm{b-Ta}^2.\n\\]\nIn particular both the residual square and the minimum objective are at least $q^2$.","ran T$, then $\\norm{a_\\alpha}\\to\\infty$.\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: identify the unavoidable component.} Since $r-b\\perp\\overline{\\ran T}$, the adjoint identity gives $T^*(r-b)=0$. Thus $a_\\alpha(r)=a_\\alpha(b)$. For every correction the orthogonal decomposition is\n\\[\n \\norm{r-Ta}^2=q^2+\\norm{b-Ta}^2.\n\\]\nIn particular both the residual square and the minimum objective are at least $q^2$.","\\emph{Step 2: take the small-parameter limit in the correct order.} Let $\\epsilon>0$. Since $b$ lies in the closure of the range, choose a single $a\\in H$ with $\\norm{r-Ta}^2<q^2+\\epsilon$. This $a$ is chosen before $\\alpha$. Minimality gives\n\\[\n q^2\\le\\norm{r-Ta_\\alpha}^2\n \\le J_\\alpha(a_\\alpha)\n \\le q^2+\\epsilon+\\alpha\\norm a^2.\n\\]\nFor sufficiently small $\\alpha$, the last term is below $q^2+2\\epsilon$ (and if $a=0$ it is already below $q^2+\\epsilon$). Hence both intervening quantities tend to $q^2$. Their difference is $\\alpha\\norm{a_\\alpha}^2$, which tends to zero. The orthogonal decomposition now implies $\\norm{Ta_\\alpha-b}^2\\to0$, and taking square roots gives the first two limits.","e the small-parameter limit in the correct order.} Let $\\epsilon>0$. Since $b$ lies in the closure of the range, choose a single $a\\in H$ with $\\norm{r-Ta}^2<q^2+\\epsilon$. This $a$ is chosen before $\\alpha$. Minimality gives\n\\[\n q^2\\le\\norm{r-Ta_\\alpha}^2\n \\le J_\\alpha(a_\\alpha)\n \\le q^2+\\epsilon+\\alpha\\norm a^2.\n\\]\nFor sufficiently small $\\alpha$, the last term is below $q^2+2\\epsilon$ (and if $a=0$ it is already below $q^2+\\epsilon$). Hence both intervening quantities tend to $q^2$. Their difference is $\\alpha\\norm{a_\\alpha}^2$, which tends to zero. The orthogonal decomposition now implies $\\norm{Ta_\\alpha-b}^2\\to0$, and taking square roots gives the first two limits.","t in the correct order.} Let $\\epsilon>0$. Since $b$ lies in the closure of the range, choose a single $a\\in H$ with $\\norm{r-Ta}^2<q^2+\\epsilon$. This $a$ is chosen before $\\alpha$. Minimality gives\n\\[\n q^2\\le\\norm{r-Ta_\\alpha}^2\n \\le J_\\alpha(a_\\alpha)\n \\le q^2+\\epsilon+\\alpha\\norm a^2.\n\\]\nFor sufficiently small $\\alpha$, the last term is below $q^2+2\\epsilon$ (and if $a=0$ it is already below $q^2+\\epsilon$). Hence both intervening quantities tend to $q^2$. Their difference is $\\alpha\\norm{a_\\alpha}^2$, which tends to zero. The orthogonal decomposition now implies $\\norm{Ta_\\alpha-b}^2\\to0$, and taking square roots gives the first two limits.","n the closure of the range, choose a single $a\\in H$ with $\\norm{r-Ta}^2<q^2+\\epsilon$. This $a$ is chosen before $\\alpha$. Minimality gives\n\\[\n q^2\\le\\norm{r-Ta_\\alpha}^2\n \\le J_\\alpha(a_\\alpha)\n \\le q^2+\\epsilon+\\alpha\\norm a^2.\n\\]\nFor sufficiently small $\\alpha$, the last term is below $q^2+2\\epsilon$ (and if $a=0$ it is already below $q^2+\\epsilon$). Hence both intervening quantities tend to $q^2$. Their difference is $\\alpha\\norm{a_\\alpha}^2$, which tends to zero. The orthogonal decomposition now implies $\\norm{Ta_\\alpha-b}^2\\to0$, and taking square roots gives the first two limits.","choose a single $a\\in H$ with $\\norm{r-Ta}^2<q^2+\\epsilon$. This $a$ is chosen before $\\alpha$. Minimality gives\n\\[\n q^2\\le\\norm{r-Ta_\\alpha}^2\n \\le J_\\alpha(a_\\alpha)\n \\le q^2+\\epsilon+\\alpha\\norm a^2.\n\\]\nFor sufficiently small $\\alpha$, the last term is below $q^2+2\\epsilon$ (and if $a=0$ it is already below $q^2+\\epsilon$). Hence both intervening quantities tend to $q^2$. Their difference is $\\alpha\\norm{a_\\alpha}^2$, which tends to zero. The orthogonal decomposition now implies $\\norm{Ta_\\alpha-b}^2\\to0$, and taking square roots gives the first two limits.","alpha$. Minimality gives\n\\[\n q^2\\le\\norm{r-Ta_\\alpha}^2\n \\le J_\\alpha(a_\\alpha)\n \\le q^2+\\epsilon+\\alpha\\norm a^2.\n\\]\nFor sufficiently small $\\alpha$, the last term is below $q^2+2\\epsilon$ (and if $a=0$ it is already below $q^2+\\epsilon$). Hence both intervening quantities tend to $q^2$. Their difference is $\\alpha\\norm{a_\\alpha}^2$, which tends to zero. The orthogonal decomposition now implies $\\norm{Ta_\\alpha-b}^2\\to0$, and taking square roots gives the first two limits.","\\emph{Step 3: when the projected datum is attained, obtain a uniform bound.} Suppose $b\\in\\ran T$. The preimage set is a nonempty closed affine subset, so projection of zero supplies its unique minimum-norm element $a^\\dagger$. Comparing with it and subtracting $q^2$ gives\n\\[\n \\norm{Ta_\\alpha-b}^2+\\alpha\\norm{a_\\alpha}^2\n \\le\\alpha\\norm{a^\\dagger}^2.\n\\]\nThus $\\norm{a_\\alpha}\\le\\norm{a^\\dagger}$ for every $\\alpha>0$. This bound does not follow merely from the previously proved vanishing weighted penalty.","p 3: when the projected datum is attained, obtain a uniform bound.} Suppose $b\\in\\ran T$. The preimage set is a nonempty closed affine subset, so projection of zero supplies its unique minimum-norm element $a^\\dagger$. Comparing with it and subtracting $q^2$ gives\n\\[\n \\norm{Ta_\\alpha-b}^2+\\alpha\\norm{a_\\alpha}^2\n \\le\\alpha\\norm{a^\\dagger}^2.\n\\]\nThus $\\norm{a_\\alpha}\\le\\norm{a^\\dagger}$ for every $\\alpha>0$. This bound does not follow merely from the previously proved vanishing weighted penalty.","\\emph{Step 4: strengthen weak convergence to norm convergence.} Take an arbitrary sequence $\\alpha_n\\downarrow0$, or more generally positive $\\alpha_n\\to0$. The bound supplies a weakly convergent subsequence $a_{\\alpha_{n_k}}\\rightharpoonup a$. Since its images converge in norm to $b$, the adjoint pairing argument of Theorem~\\ref{WR01} yields $Ta=b$. Weak lower semicontinuity gives\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{\\alpha_{n_k}}}\n \\le\\norm{a^\\dagger}.\n\\]\nMinimality and uniqueness of $a^\\dagger$ force $a=a^\\dagger$. Moreover the same lower bound, together with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longr","\\emph{Step 4: strengthen weak convergence to norm convergence.} Take an arbitrary sequence $\\alpha_n\\downarrow0$, or more generally positive $\\alpha_n\\to0$. The bound supplies a weakly convergent subsequence $a_{\\alpha_{n_k}}\\rightharpoonup a$. Since its images converge in norm to $b$, the adjoint pairing argument of Theorem~\\ref{WR01} yields $Ta=b$. Weak lower semicontinuity gives\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{\\alpha_{n_k}}}\n \\le\\norm{a^\\dagger}.\n\\]\nMinimality and uniqueness of $a^\\dagger$ force $a=a^\\dagger$. Moreover the same lower bound, together with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence conver","\\emph{Step 4: strengthen weak convergence to norm convergence.} Take an arbitrary sequence $\\alpha_n\\downarrow0$, or more generally positive $\\alpha_n\\to0$. The bound supplies a weakly convergent subsequence $a_{\\alpha_{n_k}}\\rightharpoonup a$. Since its images converge in norm to $b$, the adjoint pairing argument of Theorem~\\ref{WR01} yields $Ta=b$. Weak lower semicontinuity gives\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{\\alpha_{n_k}}}\n \\le\\norm{a^\\dagger}.\n\\]\nMinimality and uniqueness of $a^\\dagger$ force $a=a^\\dagger$. Moreover the same lower bound, together with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there w","\\emph{Step 4: strengthen weak convergence to norm convergence.} Take an arbitrary sequence $\\alpha_n\\downarrow0$, or more generally positive $\\alpha_n\\to0$. The bound supplies a weakly convergent subsequence $a_{\\alpha_{n_k}}\\rightharpoonup a$. Since its images converge in norm to $b$, the adjoint pairing argument of Theorem~\\ref{WR01} yields $Ta=b$. Weak lower semicontinuity gives\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{\\alpha_{n_k}}}\n \\le\\norm{a^\\dagger}.\n\\]\nMinimality and uniqueness of $a^\\dagger$ force $a=a^\\dagger$. Moreover the same lower bound, together with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\a","\\emph{Step 4: strengthen weak convergence to norm convergence.} Take an arbitrary sequence $\\alpha_n\\downarrow0$, or more generally positive $\\alpha_n\\to0$. The bound supplies a weakly convergent subsequence $a_{\\alpha_{n_k}}\\rightharpoonup a$. Since its images converge in norm to $b$, the adjoint pairing argument of Theorem~\\ref{WR01} yields $Ta=b$. Weak lower semicontinuity gives\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{\\alpha_{n_k}}}\n \\le\\norm{a^\\dagger}.\n\\]\nMinimality and uniqueness of $a^\\dagger$ force $a=a^\\dagger$. Moreover the same lower bound, together with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\alpha_n\\to0$ whose corrections all stayed at distance at least $\\ep","\\emph{Step 4: strengthen weak convergence to norm convergence.} Take an arbitrary sequence $\\alpha_n\\downarrow0$, or more generally positive $\\alpha_n\\to0$. The bound supplies a weakly convergent subsequence $a_{\\alpha_{n_k}}\\rightharpoonup a$. Since its images converge in norm to $b$, the adjoint pairing argument of Theorem~\\ref{WR01} yields $Ta=b$. Weak lower semicontinuity gives\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{\\alpha_{n_k}}}\n \\le\\norm{a^\\dagger}.\n\\]\nMinimality and uniqueness of $a^\\dagger$ force $a=a^\\dagger$. Moreover the same lower bound, together with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\alpha_n\\to0$ whose corrections all stayed at distance at least $\\epsilon$ from $a^\\dagger$. The argument just given w…","nce $\\alpha_n\\downarrow0$, or more generally positive $\\alpha_n\\to0$. The bound supplies a weakly convergent subsequence $a_{\\alpha_{n_k}}\\rightharpoonup a$. Since its images converge in norm to $b$, the adjoint pairing argument of Theorem~\\ref{WR01} yields $Ta=b$. Weak lower semicontinuity gives\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{\\alpha_{n_k}}}\n \\le\\norm{a^\\dagger}.\n\\]\nMinimality and uniqueness of $a^\\dagger$ force $a=a^\\dagger$. Moreover the same lower bound, together with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\alpha_n\\to0$ whose corrections all stayed at distance at least $\\epsilon$ from $a^\\dagger$. The argument just given would produce a strongly convergent subsequence of them, a contradiction. This proves th…","arrow0$, or more generally positive $\\alpha_n\\to0$. The bound supplies a weakly convergent subsequence $a_{\\alpha_{n_k}}\\rightharpoonup a$. Since its images converge in norm to $b$, the adjoint pairing argument of Theorem~\\ref{WR01} yields $Ta=b$. Weak lower semicontinuity gives\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{\\alpha_{n_k}}}\n \\le\\norm{a^\\dagger}.\n\\]\nMinimality and uniqueness of $a^\\dagger$ force $a=a^\\dagger$. Moreover the same lower bound, together with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\alpha_n\\to0$ whose corrections all stayed at distance at least $\\epsilon$ from $a^\\dagger$. The argument just given would produce a strongly convergent subsequence of them, a contradiction. This proves the limit for the en…","}\\rightharpoonup a$. Since its images converge in norm to $b$, the adjoint pairing argument of Theorem~\\ref{WR01} yields $Ta=b$. Weak lower semicontinuity gives\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{\\alpha_{n_k}}}\n \\le\\norm{a^\\dagger}.\n\\]\nMinimality and uniqueness of $a^\\dagger$ force $a=a^\\dagger$. Moreover the same lower bound, together with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\alpha_n\\to0$ whose corrections all stayed at distance at least $\\epsilon$ from $a^\\dagger$. The argument just given would produce a strongly convergent subsequence of them, a contradiction. This proves the limit for the entire parameter family.","in norm to $b$, the adjoint pairing argument of Theorem~\\ref{WR01} yields $Ta=b$. Weak lower semicontinuity gives\n\\[\n \\norm a\\le\\liminf_k\\norm{a_{\\alpha_{n_k}}}\n \\le\\norm{a^\\dagger}.\n\\]\nMinimality and uniqueness of $a^\\dagger$ force $a=a^\\dagger$. Moreover the same lower bound, together with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\alpha_n\\to0$ whose corrections all stayed at distance at least $\\epsilon$ from $a^\\dagger$. The argument just given would produce a strongly convergent subsequence of them, a contradiction. This proves the limit for the entire parameter family.","ther with the uniform upper bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\alpha_n\\to0$ whose corrections all stayed at distance at least $\\epsilon$ from $a^\\dagger$. The argument just given would produce a strongly convergent subsequence of them, a contradiction. This proves the limit for the entire parameter family.","bound, gives convergence of the norms to $\\norm{a^\\dagger}$. In a real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\alpha_n\\to0$ whose corrections all stayed at distance at least $\\epsilon$ from $a^\\dagger$. The argument just given would produce a strongly convergent subsequence of them, a contradiction. This proves the limit for the entire parameter family.","real Hilbert space,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\alpha_n\\to0$ whose corrections all stayed at distance at least $\\epsilon$ from $a^\\dagger$. The argument just given would produce a strongly convergent subsequence of them, a contradiction. This proves the limit for the entire parameter family.","ace,\n\\[\n \\norm{a_{\\alpha_{n_k}}-a^\\dagger}^2\n =\\norm{a_{\\alpha_{n_k}}}^2+\\norm{a^\\dagger}^2\n       -2\\langle a_{\\alpha_{n_k}},a^\\dagger\\rangle\\longrightarrow0.\n\\]\nThus this subsequence converges strongly. If the full positive-parameter family failed to converge strongly, there would be an $\\epsilon>0$ and parameters $\\alpha_n\\to0$ whose corrections all stayed at distance at least $\\epsilon$ from $a^\\dagger$. The argument just given would produce a strongly convergent subsequence of them, a contradiction. This proves the limit for the entire parameter family.","\\emph{Step 5: in the unattained case the norms tend to infinity.} If $b\\notin\\ran T$ and the norms failed to tend to infinity, there would exist $M<\\infty$ and positive $\\alpha_n\\to0$ with $\\norm{a_{\\alpha_n}}\\le M$. Their images still tend to $b$ by Step 2. Theorem~\\ref{WR01} would then put $b$ in the actual range, a contradiction. This proves divergence of the norms, not just unboundedness somewhere along the path.\n\\end{proof}\n\\written{WR08}","The vanishing penalty $\\alpha\\norm{a_\\alpha}^2$ is not a boundedness certificate for $a_\\alpha$. Its coefficient is tending to zero. In the escaping case, the penalty and residual can both tend to zero while the actual corrections diverge.","\\begin{center}\n\\begin{tabularx}{\\linewidth}{@{}p{.17\\linewidth}p{.26\\linewidth}X@{}}\n\\toprule Residual floor & Is $b$ attained? & Limit of the protected corrections","midrule\n$q=0$ & Yes & Strong convergence to the nearest exact protected correction.","[3pt]\n$q=0$ & No & Residual vanishes, correction norms diverge, no exact protected answer.\\\\[3pt]\n$q>0$ & Yes & Strong convergence to the least-squares correction of minimum norm, not an exact answer.\\\\[3pt]\n$q>0$ & No & A positive residual floor and diverging correction norms occur together.\\\\\\bottomrule\n\\end{tabularx}\n\\end{center}\n\nThe last case is not hypothetical. Let $Tu=(u_n/n)$ on $\\ell^2$, define $Su=(Tu,0)$ into $\\ell^2\\oplus\\R$, and take $r=((1/n),1)$. Then $b=((1/n),0)$ is in the closure but not the range, while $q=1$. The same escaping corrections as before coexist with an irreducible orthogonal residual. Positive floor and unbounded correction cost are independent features.\n\n\\subsection{Noise is controlled at a fixed parameter}\n\\fixedplate{34_noise_filter}{For one singular value $\\sigma$, the response to residual-data noise is $\\sigma/(\\sigma^2+\\alpha)$. Its maximum occurs at $\\sigma=\\sqrt\\alpha$ and equals $1/(2\\sqrt\\alpha)$. The plot evaluates this formula and the universal horizontal bounds at three explicit positive parameters.}{fig:noise-filter}\n\n\\begin{theorem}[Dat…","\\begin{theorem}[Data sensitivity and a convergent parameter choice]\\label{WR09}\nFor two residual data vectors $r,\\widetilde r$ with $\\norm{r-\\widetilde r}\\le\\delta$ and the same $T,\\alpha$,\n\\[\n \\norm{a_\\alpha(r)-a_\\alpha(\\widetilde r)}\n \\le\\frac{\\delta}{2\\sqrt\\alpha},\\qquad\n \\norm{T(a_\\alpha(r)-a_\\alpha(\\widetilde r))}\\le\\delta.\n\\]\nThe coefficient $1/(2\\sqrt\\alpha)$ is sharp over bounded linear Hilbert operators. If $b=P_Cr\\in\\ran T$, $\\alpha(\\delta)\\to0$, and $\\delta/\\sqrt{\\alpha(\\delta)}\\to0$, then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t","sensitivity and a convergent parameter choice]\\label{WR09}\nFor two residual data vectors $r,\\widetilde r$ with $\\norm{r-\\widetilde r}\\le\\delta$ and the same $T,\\alpha$,\n\\[\n \\norm{a_\\alpha(r)-a_\\alpha(\\widetilde r)}\n \\le\\frac{\\delta}{2\\sqrt\\alpha},\\qquad\n \\norm{T(a_\\alpha(r)-a_\\alpha(\\widetilde r))}\\le\\delta.\n\\]\nThe coefficient $1/(2\\sqrt\\alpha)$ is sharp over bounded linear Hilbert operators. If $b=P_Cr\\in\\ran T$, $\\alpha(\\delta)\\to0$, and $\\delta/\\sqrt{\\alpha(\\delta)}\\to0$, then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true.…","onvergent parameter choice]\\label{WR09}\nFor two residual data vectors $r,\\widetilde r$ with $\\norm{r-\\widetilde r}\\le\\delta$ and the same $T,\\alpha$,\n\\[\n \\norm{a_\\alpha(r)-a_\\alpha(\\widetilde r)}\n \\le\\frac{\\delta}{2\\sqrt\\alpha},\\qquad\n \\norm{T(a_\\alpha(r)-a_\\alpha(\\widetilde r))}\\le\\delta.\n\\]\nThe coefficient $1/(2\\sqrt\\alpha)$ is sharp over bounded linear Hilbert operators. If $b=P_Cr\\in\\ran T$, $\\alpha(\\delta)\\to0$, and $\\delta/\\sqrt{\\alpha(\\delta)}\\to0$, then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the sca…","]\\label{WR09}\nFor two residual data vectors $r,\\widetilde r$ with $\\norm{r-\\widetilde r}\\le\\delta$ and the same $T,\\alpha$,\n\\[\n \\norm{a_\\alpha(r)-a_\\alpha(\\widetilde r)}\n \\le\\frac{\\delta}{2\\sqrt\\alpha},\\qquad\n \\norm{T(a_\\alpha(r)-a_\\alpha(\\widetilde r))}\\le\\delta.\n\\]\nThe coefficient $1/(2\\sqrt\\alpha)$ is sharp over bounded linear Hilbert operators. If $b=P_Cr\\in\\ran T$, $\\alpha(\\delta)\\to0$, and $\\delta/\\sqrt{\\alpha(\\delta)}\\to0$, then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alph…","e r}\\le\\delta$ and the same $T,\\alpha$,\n\\[\n \\norm{a_\\alpha(r)-a_\\alpha(\\widetilde r)}\n \\le\\frac{\\delta}{2\\sqrt\\alpha},\\qquad\n \\norm{T(a_\\alpha(r)-a_\\alpha(\\widetilde r))}\\le\\delta.\n\\]\nThe coefficient $1/(2\\sqrt\\alpha)$ is sharp over bounded linear Hilbert operators. If $b=P_Cr\\in\\ran T$, $\\alpha(\\delta)\\to0$, and $\\delta/\\sqrt{\\alpha(\\delta)}\\to0$, then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}…","and the same $T,\\alpha$,\n\\[\n \\norm{a_\\alpha(r)-a_\\alpha(\\widetilde r)}\n \\le\\frac{\\delta}{2\\sqrt\\alpha},\\qquad\n \\norm{T(a_\\alpha(r)-a_\\alpha(\\widetilde r))}\\le\\delta.\n\\]\nThe coefficient $1/(2\\sqrt\\alpha)$ is sharp over bounded linear Hilbert operators. If $b=P_Cr\\in\\ran T$, $\\alpha(\\delta)\\to0$, and $\\delta/\\sqrt{\\alpha(\\delta)}\\to0$, then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\…","))}\\le\\delta.\n\\]\nThe coefficient $1/(2\\sqrt\\alpha)$ is sharp over bounded linear Hilbert operators. If $b=P_Cr\\in\\ran T$, $\\alpha(\\delta)\\to0$, and $\\delta/\\sqrt{\\alpha(\\delta)}\\to0$, then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T…","\\alpha)$ is sharp over bounded linear Hilbert operators. If $b=P_Cr\\in\\ran T$, $\\alpha(\\delta)\\to0$, and $\\delta/\\sqrt{\\alpha(\\delta)}\\to0$, then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse a…","Cr\\in\\ran T$, $\\alpha(\\delta)\\to0$, and $\\delta/\\sqrt{\\alpha(\\delta)}\\to0$, then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","$\\delta/\\sqrt{\\alpha(\\delta)}\\to0$, then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.",", then the noisy corrections converge strongly to $a^\\dagger$ as $\\delta\\to0$.\n\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","\\end{theorem}\n\\begin{proof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","roof}\nBoth corrections are constructed at the same positive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","ive parameter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","ter and in the same Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","me Hilbert correction space. Let $h=a_\\alpha(\\widetilde r)-a_\\alpha(r)$ and $e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","$e=\\widetilde r-r$. Subtract the two normal equations to obtain\n\\[\n (T^*T+\\alpha I)h=T^*e.\n\\]\nPair with $h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","h$ and use the adjoint identity:\n\\[\n \\norm{Th}^2+\\alpha\\norm h^2\n =\\langle e,Th\\rangle\\le\\delta\\norm{Th}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","}.\n\\]\nPut $t=\\norm{Th}\\ge0$. If $t>0$, dropping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","ping the nonnegative penalty and dividing by $t$ gives $t\\le\\delta$. If $t=0$, that inequality is already true. Completing the scalar square gives\n\\[\n \\alpha\\norm h^2\\le\\delta t-t^2\n =\\frac{\\delta^2}{4}-(t-\\delta/2)^2\n \\le\\frac{\\delta^2}{4}.\n\\]\nBecause $\\alpha>0$, taking square roots proves the correction bound. This also explains why the coefficient is $1/(2\\sqrt\\alpha)$ rather than the coarser $\\norm T/\\alpha$ obtained by bounding the inverse alone.","For sharpness at a fixed parameter, take $H=Y=\\R$ and $T=\\sqrt\\alpha\\,I$. The solution map is multiplication by\n\\[\n \\frac{\\sqrt\\alpha}{\\alpha+\\alpha}=\\frac1{2\\sqrt\\alpha}.\n\\]\nAny nonzero data perturbation attains the first bound. This is worst-case sharpness across operators at that parameter, not a claim of equality for every fixed operator and every parameter.","For the limiting statement, insert the exact-data correction at the selected parameter:\n\\[\n \\norm{a_{\\alpha(\\delta)}(\\widetilde r)-a^\\dagger}\n \\le\\frac{\\delta}{2\\sqrt{\\alpha(\\delta)}}\n       +\\norm{a_{\\alpha(\\delta)}(r)-a^\\dagger}.\n\\]\nThe first term tends to zero by the noise/parameter hypothesis. The second tends to zero by the attained-projection case of Theorem~\\ref{WR08}, because the selected parameter tends to zero. Both statements require the same fixed operator and Hilbert metrics. The conclusion is a least-squares correction when the unavoidable residual component is nonzero.\n\\end{proof}\n\\written{WR09}","The theorem does not manufacture an exact solution when $r$ is outside the range. For the positive-floor attainable case it converges to a least-squares correction. For the unattainable projected datum it cannot produce a bounded exact-data limit. The range hypothesis must therefore precede the parameter-choice conclusion.","\\subsection{The entire regularized family is explicit}\nUse $Tu=(u_n/n)$ and $r=w=(1/n)$ as in Section~\\ref{sec:escape}. Choose $\\alpha=m^{-2}$ for an integer $m\\ge1$. The normal equation in each coordinate gives\n\\[\n a^{(m)}_n=\\frac{m^2}{m^2+n^2},\\qquad\n v^{(m)}_n=r_n-(Ta^{(m)})_n=\\frac{n}{m^2+n^2}.\n\\]\nThe repaired ambient candidate is $(-a^{(m)},1)$. Its protected coordinate is one for every $m$.","\\begin{proposition}[Energy, residual, and certified tails]\\label{WR10}\nThe displayed $a^{(m)}$ belongs to $\\ell^2$ and is the unique regularized correction at $\\alpha=m^{-2}$. Set\n\\[\n A_m=\\norm{a^{(m)}}^2=\\sum_{n\\ge1}\\frac{m^4}{(m^2+n^2)^2},\\qquad\n B_m=\\norm{v^{(m)}}^2=\\sum_{n\\ge1}\\frac{n^2}{(m^2+n^2)^2}.\n\\]\nThen\n\\begin{equation}\\label{eq:soft-bounds}\n \\frac m4\\le A_m\\le\\frac{4m}{3},\\qquad\n \\frac1{4(m+1)}\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with","\\begin{proposition}[Energy, residual, and certified tails]\\label{WR10}\nThe displayed $a^{(m)}$ belongs to $\\ell^2$ and is the unique regularized correction at $\\alpha=m^{-2}$. Set\n\\[\n A_m=\\norm{a^{(m)}}^2=\\sum_{n\\ge1}\\frac{m^4}{(m^2+n^2)^2},\\qquad\n B_m=\\norm{v^{(m)}}^2=\\sum_{n\\ge1}\\frac{n^2}{(m^2+n^2)^2}.\n\\]\nThen\n\\begin{equation}\\label{eq:soft-bounds}\n \\frac m4\\le A_m\\le\\frac{4m}{3},\\qquad\n \\frac1{4(m+1)}\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/","\\begin{proposition}[Energy, residual, and certified tails]\\label{WR10}\nThe displayed $a^{(m)}$ belongs to $\\ell^2$ and is the unique regularized correction at $\\alpha=m^{-2}$. Set\n\\[\n A_m=\\norm{a^{(m)}}^2=\\sum_{n\\ge1}\\frac{m^4}{(m^2+n^2)^2},\\qquad\n B_m=\\norm{v^{(m)}}^2=\\sum_{n\\ge1}\\frac{n^2}{(m^2+n^2)^2}.\n\\]\nThen\n\\begin{equation}\\label{eq:soft-bounds}\n \\frac m4\\le A_m\\le\\frac{4m}{3},\\qquad\n \\frac1{4(m+1)}\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2","\\begin{proposition}[Energy, residual, and certified tails]\\label{WR10}\nThe displayed $a^{(m)}$ belongs to $\\ell^2$ and is the unique regularized correction at $\\alpha=m^{-2}$. Set\n\\[\n A_m=\\norm{a^{(m)}}^2=\\sum_{n\\ge1}\\frac{m^4}{(m^2+n^2)^2},\\qquad\n B_m=\\norm{v^{(m)}}^2=\\sum_{n\\ge1}\\frac{n^2}{(m^2+n^2)^2}.\n\\]\nThen\n\\begin{equation}\\label{eq:soft-bounds}\n \\frac m4\\le A_m\\le\\frac{4m}{3},\\qquad\n \\frac1{4(m+1)}\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^","\\begin{proposition}[Energy, residual, and certified tails]\\label{WR10}\nThe displayed $a^{(m)}$ belongs to $\\ell^2$ and is the unique regularized correction at $\\alpha=m^{-2}$. Set\n\\[\n A_m=\\norm{a^{(m)}}^2=\\sum_{n\\ge1}\\frac{m^4}{(m^2+n^2)^2},\\qquad\n B_m=\\norm{v^{(m)}}^2=\\sum_{n\\ge1}\\frac{n^2}{(m^2+n^2)^2}.\n\\]\nThen\n\\begin{equation}\\label{eq:soft-bounds}\n \\frac m4\\le A_m\\le\\frac{4m}{3},\\qquad\n \\frac1{4(m+1)}\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^{(m)}=\\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\\ref{WR06} therefore identif…","s]\\label{WR10}\nThe displayed $a^{(m)}$ belongs to $\\ell^2$ and is the unique regularized correction at $\\alpha=m^{-2}$. Set\n\\[\n A_m=\\norm{a^{(m)}}^2=\\sum_{n\\ge1}\\frac{m^4}{(m^2+n^2)^2},\\qquad\n B_m=\\norm{v^{(m)}}^2=\\sum_{n\\ge1}\\frac{n^2}{(m^2+n^2)^2}.\n\\]\nThen\n\\begin{equation}\\label{eq:soft-bounds}\n \\frac m4\\le A_m\\le\\frac{4m}{3},\\qquad\n \\frac1{4(m+1)}\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^{(m)}=\\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\\ref{WR06} therefore identifies this actual infinite vector as the unique minimizer,…","belongs to $\\ell^2$ and is the unique regularized correction at $\\alpha=m^{-2}$. Set\n\\[\n A_m=\\norm{a^{(m)}}^2=\\sum_{n\\ge1}\\frac{m^4}{(m^2+n^2)^2},\\qquad\n B_m=\\norm{v^{(m)}}^2=\\sum_{n\\ge1}\\frac{n^2}{(m^2+n^2)^2}.\n\\]\nThen\n\\begin{equation}\\label{eq:soft-bounds}\n \\frac m4\\le A_m\\le\\frac{4m}{3},\\qquad\n \\frac1{4(m+1)}\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^{(m)}=\\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\\ref{WR06} therefore identifies this actual infinite vector as the unique minimizer, rather than just a coordinatewise cand…","is the unique regularized correction at $\\alpha=m^{-2}$. Set\n\\[\n A_m=\\norm{a^{(m)}}^2=\\sum_{n\\ge1}\\frac{m^4}{(m^2+n^2)^2},\\qquad\n B_m=\\norm{v^{(m)}}^2=\\sum_{n\\ge1}\\frac{n^2}{(m^2+n^2)^2}.\n\\]\nThen\n\\begin{equation}\\label{eq:soft-bounds}\n \\frac m4\\le A_m\\le\\frac{4m}{3},\\qquad\n \\frac1{4(m+1)}\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^{(m)}=\\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\\ref{WR06} therefore identifies this actual infinite vector as the unique minimizer, rather than just a coordinatewise candidate.","}^2=\\sum_{n\\ge1}\\frac{n^2}{(m^2+n^2)^2}.\n\\]\nThen\n\\begin{equation}\\label{eq:soft-bounds}\n \\frac m4\\le A_m\\le\\frac{4m}{3},\\qquad\n \\frac1{4(m+1)}\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^{(m)}=\\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\\ref{WR06} therefore identifies this actual infinite vector as the unique minimizer, rather than just a coordinatewise candidate.","}\\frac{n^2}{(m^2+n^2)^2}.\n\\]\nThen\n\\begin{equation}\\label{eq:soft-bounds}\n \\frac m4\\le A_m\\le\\frac{4m}{3},\\qquad\n \\frac1{4(m+1)}\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^{(m)}=\\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\\ref{WR06} therefore identifies this actual infinite vector as the unique minimizer, rather than just a coordinatewise candidate.","\\le B_m\\le\\frac2m.\n\\end{equation}\nFor the sums $A_{m,K},B_{m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^{(m)}=\\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\\ref{WR06} therefore identifies this actual infinite vector as the unique minimizer, rather than just a coordinatewise candidate.","m,K}$ through any integer $K\\ge m$,\n\\begin{equation}\\label{eq:soft-tails}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^{(m)}=\\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\\ref{WR06} therefore identifies this actual infinite vector as the unique minimizer, rather than just a coordinatewise candidate.","ils}\n 0\\le A_m-A_{m,K}\\le\\frac{m^4}{3K^3},\\qquad\n 0\\le B_m-B_{m,K}\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^{(m)}=\\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\\ref{WR06} therefore identifies this actual infinite vector as the unique minimizer, rather than just a coordinatewise candidate.","\\le\\frac1K.\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nFor $n>m$, one has $(a_n^{(m)})^2\\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\\in\\ell^2$, hence the residual sequence is also in $\\ell^2$. The operator is self-adjoint, and the coordinate equality\n\\[\n \\frac1n\\frac{n}{m^2+n^2}\n =m^{-2}\\frac{m^2}{m^2+n^2}\n\\]\nproves $T^*v^{(m)}=\\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\\ref{WR06} therefore identifies this actual infinite vector as the unique minimizer, rather than just a coordinatewise candidate.","For $1\\le n\\le m$, $m^2/(m^2+n^2)\\ge1/2$. There are $m$ such coordinates, so $A_m\\ge m/4$. For the upper bound, the first $m$ squares are at most one, and the decreasing function $x^{-4}$ gives\n\\[\n \\sum_{n>m}\\frac1{n^4}\\le\\int_m^\\infty x^{-4}\\,dx=\\frac1{3m^3}.\n\\]\nThus $A_m\\le m+m^4/(3m^3)=4m/3$.","For $n\\le m$, the residual square is at most $n^2/m^4$. The sum-of-squares estimate\n\\[\n \\sum_{n=1}^m n^2=\\frac{m(m+1)(2m+1)}6\\le m^3\n\\]\nholds for $m\\ge1$, so this part contributes at most $1/m$. For $n>m$, each residual square is at most $1/n^2$, whose tail is at most $1/m$. This proves $B_m\\le2/m$.","For the lower bound, $n>m$ implies $m^2+n^2\\le2n^2$, hence\n\\[\n \\frac{n^2}{(m^2+n^2)^2}\\ge\\frac1{4n^2}.\n\\]\nThe other direction of the decreasing-function integral comparison gives\n\\[\n \\sum_{n>m}\\frac1{n^2}\\ge\\int_{m+1}^\\infty x^{-2}\\,dx=\\frac1{m+1}.\n\\]\nThis proves $B_m\\ge1/(4(m+1))$.","Finally, for $K\\ge m$, both omitted tails consist of nonnegative terms. Bound the first by $m^4\\sum_{n>K}n^{-4}$ and the second by $\\sum_{n>K}n^{-2}$. The same upper integral comparisons give the two certified remainders in \\eqref{eq:soft-tails}. No finite truncation is substituted for an equality of infinite energies.\n\\end{proof}\n\\written{WR10}","Thus the correction norm grows on the order of $\\sqrt m$, the residual norm decays on the order of $m^{-1/2}$, and the penalty $\\alpha A_m=A_m/m^2$ tends to zero. The smooth correction avoids the hard discontinuity at a prefix boundary, but it still approaches the forbidden constant sequence coordinate by coordinate.","\\fixedplate{35_regularized_escape}{At each fixed coordinate, the soft correction approaches one as $m$ tends to infinity. Its transition moves outward as $m$ increases. The chart compares three exact coordinate formulas $m^2/(m^2+n^2)$, not finite approximations to an unspecified optimization procedure. Every displayed sequence has finite energy, but the family's energies are unbounded.}{fig:regularized-escape}","\\begin{theorem}[A sharp budget asymptotic for the diagonal datum]\\label{WR11}\nAs $m\\to\\infty$,\n\\[\n \\frac{A_m}{m}\\longrightarrow\\frac\\pi4,\\qquad\n mB_m\\longrightarrow\\frac\\pi4.\n\\]\nFor the exact budget profile of this infinite problem,\n\\begin{equation}\\label{eq:sharp-budget}\n \\lim_{M\\to\\infty}M\\beta_{T,w}(M)=\\frac\\pi4.\n\\end{equation}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f","\\begin{theorem}[A sharp budget asymptotic for the diagonal datum]\\label{WR11}\nAs $m\\to\\infty$,\n\\[\n \\frac{A_m}{m}\\longrightarrow\\frac\\pi4,\\qquad\n mB_m\\longrightarrow\\frac\\pi4.\n\\]\nFor the exact budget profile of this infinite problem,\n\\begin{equation}\\label{eq:sharp-budget}\n \\lim_{M\\to\\infty}M\\beta_{T,w}(M)=\\frac\\pi4.\n\\end{equation}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1","\\begin{theorem}[A sharp budget asymptotic for the diagonal datum]\\label{WR11}\nAs $m\\to\\infty$,\n\\[\n \\frac{A_m}{m}\\longrightarrow\\frac\\pi4,\\qquad\n mB_m\\longrightarrow\\frac\\pi4.\n\\]\nFor the exact budget profile of this infinite problem,\n\\begin{equation}\\label{eq:sharp-budget}\n \\lim_{M\\to\\infty}M\\beta_{T,w}(M)=\\frac\\pi4.\n\\end{equation}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length be","\\begin{theorem}[A sharp budget asymptotic for the diagonal datum]\\label{WR11}\nAs $m\\to\\infty$,\n\\[\n \\frac{A_m}{m}\\longrightarrow\\frac\\pi4,\\qquad\n mB_m\\longrightarrow\\frac\\pi4.\n\\]\nFor the exact budget profile of this infinite problem,\n\\begin{equation}\\label{eq:sharp-budget}\n \\lim_{M\\to\\infty}M\\beta_{T,w}(M)=\\frac\\pi4.\n\\end{equation}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term te","\\begin{theorem}[A sharp budget asymptotic for the diagonal datum]\\label{WR11}\nAs $m\\to\\infty$,\n\\[\n \\frac{A_m}{m}\\longrightarrow\\frac\\pi4,\\qquad\n mB_m\\longrightarrow\\frac\\pi4.\n\\]\nFor the exact budget profile of this infinite problem,\n\\begin{equation}\\label{eq:sharp-budget}\n \\lim_{M\\to\\infty}M\\beta_{T,w}(M)=\\frac\\pi4.\n\\end{equation}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The s…","mB_m\\longrightarrow\\frac\\pi4.\n\\]\nFor the exact budget profile of this infinite problem,\n\\begin{equation}\\label{eq:sharp-budget}\n \\lim_{M\\to\\infty}M\\beta_{T,w}(M)=\\frac\\pi4.\n\\end{equation}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","infinite problem,\n\\begin{equation}\\label{eq:sharp-budget}\n \\lim_{M\\to\\infty}M\\beta_{T,w}(M)=\\frac\\pi4.\n\\end{equation}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","bel{eq:sharp-budget}\n \\lim_{M\\to\\infty}M\\beta_{T,w}(M)=\\frac\\pi4.\n\\end{equation}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","y}M\\beta_{T,w}(M)=\\frac\\pi4.\n\\end{equation}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.",")=\\frac\\pi4.\n\\end{equation}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","}\nThe hard prefix correction at budget $M=\\sqrt N$ instead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","stead has\n\\[\n \\sqrt N\\,\\norm{w-T\\mathbf1_N}\\longrightarrow1.\n\\]\n\\end{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","nd{theorem}\n\\begin{proof}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","}\n\\emph{Step 1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","1: control the infinite Riemann sums uniformly.} Define\n\\[\n f(x)=\\frac1{(1+x^2)^2},\\qquad\n g(x)=\\frac{x^2}{(1+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","+x^2)^2}.\n\\]\nThen $A_m/m=m^{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","{-1}\\sum_{n\\ge1}f(n/m)$ and $mB_m=m^{-1}\\sum_{n\\ge1}g(n/m)$. Fix $L\\ge2$ and put $N=\\lfloor mL\\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.",". On $[0,L]$, both functions are uniformly continuous. If $\\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives\n\\[\n \\left|\\frac1m\\sum_{n=1}^N f(n/m)-\\int_0^{N/m}f(x)\\,dx\\right|\n \\le L\\omega_f(1/m).\n\\]\nThe missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.","The tails require a bound independent of sufficiently fine discretization. Since $N\\ge mL/2$, the inequalities $f(x)\\le x^{-4}$ and $g(x)\\le x^{-2}$ give\n\\[\n \\frac1m\\sum_{n>N}f(n/m)\n \\le m^3\\sum_{n>N}n^{-4}\\le\\frac{m^3}{3N^3}\n \\le\\frac8{3L^3},\n\\]\n\\[\n \\frac1m\\sum_{n>N}g(n/m)\n \\le m\\sum_{n>N}n^{-2}\\le\\frac mN\\le\\frac2L.\n\\]\nThe integral tails are at most $1/(3L^3)$ and $1/L$. Given an error tolerance, choose $L$ first to control both discrete and integral tails, then choose $m$ to control the compact-interval sums. This justifies passing from finite-interval Riemann convergence to the full half-line sums. No uncontrolled interchange of two infinite limits is involved.","require a bound independent of sufficiently fine discretization. Since $N\\ge mL/2$, the inequalities $f(x)\\le x^{-4}$ and $g(x)\\le x^{-2}$ give\n\\[\n \\frac1m\\sum_{n>N}f(n/m)\n \\le m^3\\sum_{n>N}n^{-4}\\le\\frac{m^3}{3N^3}\n \\le\\frac8{3L^3},\n\\]\n\\[\n \\frac1m\\sum_{n>N}g(n/m)\n \\le m\\sum_{n>N}n^{-2}\\le\\frac mN\\le\\frac2L.\n\\]\nThe integral tails are at most $1/(3L^3)$ and $1/L$. Given an error tolerance, choose $L$ first to control both discrete and integral tails, then choose $m$ to control the compact-interval sums. This justifies passing from finite-interval Riemann convergence to the full half-line sums. No uncontrolled interchange of two infinite limits is involved.","nce $N\\ge mL/2$, the inequalities $f(x)\\le x^{-4}$ and $g(x)\\le x^{-2}$ give\n\\[\n \\frac1m\\sum_{n>N}f(n/m)\n \\le m^3\\sum_{n>N}n^{-4}\\le\\frac{m^3}{3N^3}\n \\le\\frac8{3L^3},\n\\]\n\\[\n \\frac1m\\sum_{n>N}g(n/m)\n \\le m\\sum_{n>N}n^{-2}\\le\\frac mN\\le\\frac2L.\n\\]\nThe integral tails are at most $1/(3L^3)$ and $1/L$. Given an error tolerance, choose $L$ first to control both discrete and integral tails, then choose $m$ to control the compact-interval sums. This justifies passing from finite-interval Riemann convergence to the full half-line sums. No uncontrolled interchange of two infinite limits is involved.","\\emph{Step 2: evaluate the integrals.} The substitution $x=\\tan\\theta$ gives $dx=\\sec^2\\theta\\,d\\theta$ and $1+x^2=\\sec^2\\theta$, so\n\\[\n \\int_0^\\infty f(x)\\,dx=\\int_0^{\\pi/2}\\cos^2\\theta\\,d\\theta=\\frac\\pi4,\n \\qquad\n \\int_0^\\infty g(x)\\,dx=\\int_0^{\\pi/2}\\sin^2\\theta\\,d\\theta=\\frac\\pi4.\n\\]\nStep 1 now proves the two stated asymptotics.","\\emph{Step 3: move from regularization parameters to arbitrary budgets.} Set $M_m=\\sqrt{A_m}$. Every positive coordinate $m^2/(m^2+n^2)$ increases strictly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, pr","\\emph{Step 3: move from regularization parameters to arbitrary budgets.} Set $M_m=\\sqrt{A_m}$. Every positive coordinate $m^2/(m^2+n^2)$ increases strictly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely con","\\emph{Step 3: move from regularization parameters to arbitrary budgets.} Set $M_m=\\sqrt{A_m}$. Every positive coordinate $m^2/(m^2+n^2)$ increases strictly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subse","\\emph{Step 3: move from regularization parameters to arbitrary budgets.} Set $M_m=\\sqrt{A_m}$. Every positive coordinate $m^2/(m^2+n^2)$ increases strictly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of","\\emph{Step 3: move from regularization parameters to arbitrary budgets.} Set $M_m=\\sqrt{A_m}$. Every positive coordinate $m^2/(m^2+n^2)$ increases strictly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","move from regularization parameters to arbitrary budgets.} Set $M_m=\\sqrt{A_m}$. Every positive coordinate $m^2/(m^2+n^2)$ increases strictly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","eters to arbitrary budgets.} Set $M_m=\\sqrt{A_m}$. Every positive coordinate $m^2/(m^2+n^2)$ increases strictly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","^2)$ increases strictly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","es strictly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","ly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","increases strictly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","ctly. Since $A_m/m\\to c=\\pi/4>0$, these budgets tend to infinity and\n\\[\n \\left(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","(\\frac{M_{m+1}}{M_m}\\right)^2\n =\\frac{A_{m+1}/(m+1)}{A_m/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","/m}\\frac{m+1}{m}\\longrightarrow1.\n\\]\nProposition~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","ion~\\ref{WR07} identifies $\\beta(M_m)=\\sqrt{B_m}$. Thus\n\\[\n M_m\\beta(M_m)=\\sqrt{(A_m/m)(mB_m)}\\longrightarrow c.\n\\]\nFor every sufficiently large $M$, choose $m$ with $M_m\\le M\\le M_{m+1}$. Monotonicity of $\\beta$ gives\n\\[\n M_m\\beta(M_{m+1})\\le M\\beta(M)\\le M_{m+1}\\beta(M_m).\n\\]\nThe lower bound is $(M_m/M_{m+1})[M_{m+1}\\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.","\\emph{Step 4: compare the hard-prefix method.} That correction has norm $\\sqrt N$ and residual square $\\sum_{n>N}n^{-2}$. Integral comparison yields\n\\[\n \\frac{N}{N+1}\\le N\\sum_{n>N}\\frac1{n^2}\\le1.\n\\]\nTaking the limit and then a square root gives its leading product one. The two constants are consequently compared at large actual correction budgets, rather than at arbitrarily equated parameter labels.\n\\end{proof}\n\\written{WR11}","At large matched correction budgets, the optimal soft repair therefore has about $\\pi/4$ times the hard-prefix residual. This improves the efficiency of approximation, not the existence result. Both residuals tend to zero only as the budget tends to infinity. The limiting constant is a calculation for this declared model, not a universal regularization efficiency factor.","\\fixedplate{36_budget_product}{The plotted intervals enclose the actual infinite quantities $\\sqrt{A_mB_m}$, using finite sums plus the proved tails. The corresponding hard-prefix intervals enclose $\\sqrt{N\\sum_{n>N}n^{-2}}$ with $N=m$. These have different natural budget sequences. Their limits give the matched-budget asymptotic constants $\\pi/4$ and one.}{fig:budget-product}","\\subsection{The same operator has an attainable datum}\nChange only the residual to $b_n=1/n^2$. Its minimum norm preimage is $a^\\dagger_n=1/n\\in\\ell^2$. At $\\alpha=m^{-2}$, the regularized correction is\n\\[\n c^{(m)}_n=\\frac{m^2}{n(m^2+n^2)},\\qquad\n a^\\dagger_n-c^{(m)}_n=\\frac{n}{m^2+n^2}.\n\\]\nThe bias square is exactly the previously computed $B_m$. This links the two cases without conflating their existence hypotheses: the escaping residual of one problem is the regularization bias of an attainable problem.","\\begin{corollary}[An explicit noisy attainable repair]\\label{WR12}\nFor the attainable datum $b=(1/n^2)$, suppose $\\norm{\\widetilde b-b}\\le\\delta$ with $0<\\delta\\le1$. Choose\n\\[\n m=\\lceil\\delta^{-2/3}\\rceil,\\qquad\\alpha=m^{-2}.\n\\]\nThen the noisy regularized correction satisfies\n\\[\n \\norm{a_\\alpha(\\widetilde b)-a^\\dagger}\n \\le\\sqrt{\\frac2m}+\\frac{\\delta m}{2}\n \\le(\\sqrt2+1)\\delta^{1/3}.\n\\]\n\\end{corollary}\n\\begin{proof}\nThe exact-data bias has coordinates $n/(m^2+n^2)$, so its squared norm is precisely $B_m$. By Proposition~\\ref{WR10},\n\\[\n \\norm{a_{m^{-2}}(b)-a^\\dagger}\\le\\sqrt{2/m}.\n\\]\nTheorem~\\ref{WR09} at the same parameter gives\n\\[\n \\norm{a_{m^{-2}}(\\widetilde b)-a_{m^{-2}}(b)}\n \\le\\frac{\\delta}{2\\sqrt{m^{-2}}}=\\frac{\\delta m}{2}.\n\\]\nThe triangle inequality proves the first bound.","el{WR12}\nFor the attainable datum $b=(1/n^2)$, suppose $\\norm{\\widetilde b-b}\\le\\delta$ with $0<\\delta\\le1$. Choose\n\\[\n m=\\lceil\\delta^{-2/3}\\rceil,\\qquad\\alpha=m^{-2}.\n\\]\nThen the noisy regularized correction satisfies\n\\[\n \\norm{a_\\alpha(\\widetilde b)-a^\\dagger}\n \\le\\sqrt{\\frac2m}+\\frac{\\delta m}{2}\n \\le(\\sqrt2+1)\\delta^{1/3}.\n\\]\n\\end{corollary}\n\\begin{proof}\nThe exact-data bias has coordinates $n/(m^2+n^2)$, so its squared norm is precisely $B_m$. By Proposition~\\ref{WR10},\n\\[\n \\norm{a_{m^{-2}}(b)-a^\\dagger}\\le\\sqrt{2/m}.\n\\]\nTheorem~\\ref{WR09} at the same parameter gives\n\\[\n \\norm{a_{m^{-2}}(\\widetilde b)-a_{m^{-2}}(b)}\n \\le\\frac{\\delta}{2\\sqrt{m^{-2}}}=\\frac{\\delta m}{2}.\n\\]\nThe triangle inequality proves the first bound.","uppose $\\norm{\\widetilde b-b}\\le\\delta$ with $0<\\delta\\le1$. Choose\n\\[\n m=\\lceil\\delta^{-2/3}\\rceil,\\qquad\\alpha=m^{-2}.\n\\]\nThen the noisy regularized correction satisfies\n\\[\n \\norm{a_\\alpha(\\widetilde b)-a^\\dagger}\n \\le\\sqrt{\\frac2m}+\\frac{\\delta m}{2}\n \\le(\\sqrt2+1)\\delta^{1/3}.\n\\]\n\\end{corollary}\n\\begin{proof}\nThe exact-data bias has coordinates $n/(m^2+n^2)$, so its squared norm is precisely $B_m$. By Proposition~\\ref{WR10},\n\\[\n \\norm{a_{m^{-2}}(b)-a^\\dagger}\\le\\sqrt{2/m}.\n\\]\nTheorem~\\ref{WR09} at the same parameter gives\n\\[\n \\norm{a_{m^{-2}}(\\widetilde b)-a_{m^{-2}}(b)}\n \\le\\frac{\\delta}{2\\sqrt{m^{-2}}}=\\frac{\\delta m}{2}.\n\\]\nThe triangle inequality proves the first bound.","lta$ with $0<\\delta\\le1$. Choose\n\\[\n m=\\lceil\\delta^{-2/3}\\rceil,\\qquad\\alpha=m^{-2}.\n\\]\nThen the noisy regularized correction satisfies\n\\[\n \\norm{a_\\alpha(\\widetilde b)-a^\\dagger}\n \\le\\sqrt{\\frac2m}+\\frac{\\delta m}{2}\n \\le(\\sqrt2+1)\\delta^{1/3}.\n\\]\n\\end{corollary}\n\\begin{proof}\nThe exact-data bias has coordinates $n/(m^2+n^2)$, so its squared norm is precisely $B_m$. By Proposition~\\ref{WR10},\n\\[\n \\norm{a_{m^{-2}}(b)-a^\\dagger}\\le\\sqrt{2/m}.\n\\]\nTheorem~\\ref{WR09} at the same parameter gives\n\\[\n \\norm{a_{m^{-2}}(\\widetilde b)-a_{m^{-2}}(b)}\n \\le\\frac{\\delta}{2\\sqrt{m^{-2}}}=\\frac{\\delta m}{2}.\n\\]\nThe triangle inequality proves the first bound.","oisy regularized correction satisfies\n\\[\n \\norm{a_\\alpha(\\widetilde b)-a^\\dagger}\n \\le\\sqrt{\\frac2m}+\\frac{\\delta m}{2}\n \\le(\\sqrt2+1)\\delta^{1/3}.\n\\]\n\\end{corollary}\n\\begin{proof}\nThe exact-data bias has coordinates $n/(m^2+n^2)$, so its squared norm is precisely $B_m$. By Proposition~\\ref{WR10},\n\\[\n \\norm{a_{m^{-2}}(b)-a^\\dagger}\\le\\sqrt{2/m}.\n\\]\nTheorem~\\ref{WR09} at the same parameter gives\n\\[\n \\norm{a_{m^{-2}}(\\widetilde b)-a_{m^{-2}}(b)}\n \\le\\frac{\\delta}{2\\sqrt{m^{-2}}}=\\frac{\\delta m}{2}.\n\\]\nThe triangle inequality proves the first bound.","Because $0<\\delta\\le1$, the number $u=\\delta^{-2/3}$ is at least one. The ceiling satisfies $u\\le m<u+1\\le2u$. Therefore\n\\[\n \\sqrt{2/m}\\le\\sqrt2\\,\\delta^{1/3},\n \\qquad \\frac{\\delta m}{2}\\le\\delta u=\\delta^{1/3}.\n\\]\nAdding these estimates proves the displayed rate with every ceiling factor accounted for. The positive-noise condition is needed to define this particular finite integer choice. The statement does not prescribe an infinite $m$ at $\\delta=0$, nor does it use a probabilistic assumption on the perturbation.\n\\end{proof}\n\\written{WR12}","delta^{-2/3}$ is at least one. The ceiling satisfies $u\\le m<u+1\\le2u$. Therefore\n\\[\n \\sqrt{2/m}\\le\\sqrt2\\,\\delta^{1/3},\n \\qquad \\frac{\\delta m}{2}\\le\\delta u=\\delta^{1/3}.\n\\]\nAdding these estimates proves the displayed rate with every ceiling factor accounted for. The positive-noise condition is needed to define this particular finite integer choice. The statement does not prescribe an infinite $m$ at $\\delta=0$, nor does it use a probabilistic assumption on the perturbation.\n\\end{proof}\n\\written{WR12}","-2/3}$ is at least one. The ceiling satisfies $u\\le m<u+1\\le2u$. Therefore\n\\[\n \\sqrt{2/m}\\le\\sqrt2\\,\\delta^{1/3},\n \\qquad \\frac{\\delta m}{2}\\le\\delta u=\\delta^{1/3}.\n\\]\nAdding these estimates proves the displayed rate with every ceiling factor accounted for. The positive-noise condition is needed to define this particular finite integer choice. The statement does not prescribe an infinite $m$ at $\\delta=0$, nor does it use a probabilistic assumption on the perturbation.\n\\end{proof}\n\\written{WR12}","\\subsection{There are three different residuals}\nKeep $H=\\ker p$, $T=d|_H$, and a residual datum $r$. Fix $\\alpha>0$. Write $a_\\alpha$ for the unique minimizer of Theorem~\\ref{WR06} and let $a\\in H$ be any proposed correction. Three quantities now have different meanings:\n\\[\n v=r-Ta,\\qquad\n g=T^*(Ta-r)+\\alpha a,\\qquad\n e=a-a_\\alpha.\n\\]\nThe vector $v$ is the fit residual. The vector $g$ is the normal-equation residual, equal to one half of the derivative of the quadratic objective after identifying the dual with $H$. The vector $e$ is the unknown correction error. The exact optimizer generally has $v\\ne0$ but always has $g=0$.","Write $a_\\alpha$ for the unique minimizer of Theorem~\\ref{WR06} and let $a\\in H$ be any proposed correction. Three quantities now have different meanings:\n\\[\n v=r-Ta,\\qquad\n g=T^*(Ta-r)+\\alpha a,\\qquad\n e=a-a_\\alpha.\n\\]\nThe vector $v$ is the fit residual. The vector $g$ is the normal-equation residual, equal to one half of the derivative of the quadratic objective after identifying the dual with $H$. The vector $e$ is the unknown correction error. The exact optimizer generally has $v\\ne0$ but always has $g=0$.","for the unique minimizer of Theorem~\\ref{WR06} and let $a\\in H$ be any proposed correction. Three quantities now have different meanings:\n\\[\n v=r-Ta,\\qquad\n g=T^*(Ta-r)+\\alpha a,\\qquad\n e=a-a_\\alpha.\n\\]\nThe vector $v$ is the fit residual. The vector $g$ is the normal-equation residual, equal to one half of the derivative of the quadratic objective after identifying the dual with $H$. The vector $e$ is the unknown correction error. The exact optimizer generally has $v\\ne0$ but always has $g=0$.","proposed correction. Three quantities now have different meanings:\n\\[\n v=r-Ta,\\qquad\n g=T^*(Ta-r)+\\alpha a,\\qquad\n e=a-a_\\alpha.\n\\]\nThe vector $v$ is the fit residual. The vector $g$ is the normal-equation residual, equal to one half of the derivative of the quadratic objective after identifying the dual with $H$. The vector $e$ is the unknown correction error. The exact optimizer generally has $v\\ne0$ but always has $g=0$.","hree quantities now have different meanings:\n\\[\n v=r-Ta,\\qquad\n g=T^*(Ta-r)+\\alpha a,\\qquad\n e=a-a_\\alpha.\n\\]\nThe vector $v$ is the fit residual. The vector $g$ is the normal-equation residual, equal to one half of the derivative of the quadratic objective after identifying the dual with $H$. The vector $e$ is the unknown correction error. The exact optimizer generally has $v\\ne0$ but always has $g=0$.","The adjoint in $g$ is the restricted adjoint on $H$, not the unrestricted ambient adjoint. Equivalently,\n\\[\n T^*v=P_Hd^*v.\n\\]\nThus the correction certificate must measure the part of the normal equation that belongs to the permitted correction space. A numerical program that forgets this projection can incorrectly reject an exact protected optimizer.","\\subsection{A posteriori correction and fit bounds}\n\\begin{theorem}[Normal residual as a solver certificate]\\label{WR13}\nFor every proposed $a\\in H$,\n\\begin{align}\n \\norm{a-a_\\alpha}&\\le\\frac{\\norm g}{\\alpha},\\label{eq:solver-error}","\\subsection{A posteriori correction and fit bounds}\n\\begin{theorem}[Normal residual as a solver certificate]\\label{WR13}\nFor every proposed $a\\in H$,\n\\begin{align}\n \\norm{a-a_\\alpha}&\\le\\frac{\\norm g}{\\alpha},\\label{eq:solver-error}\\\\\n \\norm{T(a-a_\\alpha)}&\\le\\frac{\\norm g}{2\\sqrt\\alpha},\\label{eq:solver-fit}\\\\\n 0\\le J_\\alpha(a)-J_\\alpha(a_\\alpha)&\\le\\frac{\\norm g^2}{\\alpha}.\n \\label{eq:solver-objective}\n\\end{align}\n\\end{theorem}\n\\begin{proof}\nLet $e=a-a_\\alpha$ and $B=T^*T+\\alpha I$. The normal equation of the constructed optimizer gives\n\\[\n Be=(T^*T+\\alpha I)a-T^*r=g.\n\\]\nPairing with $e$ yields the exact identity\n\\begin{equation}\\label{eq:solver-energy}\n \\norm{Te}^2+\\alpha\\norm e^2=\\langle g,e\\rangle\n \\le\\norm g\\norm e.\n\\end{equation}\nIf $e=0$, all three conclusions hold with zero error. Otherwise drop the first nonnegative term and divide by $\\alpha\\norm e>0$ to obtain $\\norm e\\le\\norm g/\\alpha$.","0\\le J_\\alpha(a)-J_\\alpha(a_\\alpha)&\\le\\frac{\\norm g^2}{\\alpha}.\n \\label{eq:solver-objective}\n\\end{align}\n\\end{theorem}\n\\begin{proof}\nLet $e=a-a_\\alpha$ and $B=T^*T+\\alpha I$. The normal equation of the constructed optimizer gives\n\\[\n Be=(T^*T+\\alpha I)a-T^*r=g.\n\\]\nPairing with $e$ yields the exact identity\n\\begin{equation}\\label{eq:solver-energy}\n \\norm{Te}^2+\\alpha\\norm e^2=\\langle g,e\\rangle\n \\le\\norm g\\norm e.\n\\end{equation}\nIf $e=0$, all three conclusions hold with zero error. Otherwise drop the first nonnegative term and divide by $\\alpha\\norm e>0$ to obtain $\\norm e\\le\\norm g/\\alpha$.","le\\frac{\\norm g^2}{\\alpha}.\n \\label{eq:solver-objective}\n\\end{align}\n\\end{theorem}\n\\begin{proof}\nLet $e=a-a_\\alpha$ and $B=T^*T+\\alpha I$. The normal equation of the constructed optimizer gives\n\\[\n Be=(T^*T+\\alpha I)a-T^*r=g.\n\\]\nPairing with $e$ yields the exact identity\n\\begin{equation}\\label{eq:solver-energy}\n \\norm{Te}^2+\\alpha\\norm e^2=\\langle g,e\\rangle\n \\le\\norm g\\norm e.\n\\end{equation}\nIf $e=0$, all three conclusions hold with zero error. Otherwise drop the first nonnegative term and divide by $\\alpha\\norm e>0$ to obtain $\\norm e\\le\\norm g/\\alpha$.","_\\alpha$ and $B=T^*T+\\alpha I$. The normal equation of the constructed optimizer gives\n\\[\n Be=(T^*T+\\alpha I)a-T^*r=g.\n\\]\nPairing with $e$ yields the exact identity\n\\begin{equation}\\label{eq:solver-energy}\n \\norm{Te}^2+\\alpha\\norm e^2=\\langle g,e\\rangle\n \\le\\norm g\\norm e.\n\\end{equation}\nIf $e=0$, all three conclusions hold with zero error. Otherwise drop the first nonnegative term and divide by $\\alpha\\norm e>0$ to obtain $\\norm e\\le\\norm g/\\alpha$.","I$. The normal equation of the constructed optimizer gives\n\\[\n Be=(T^*T+\\alpha I)a-T^*r=g.\n\\]\nPairing with $e$ yields the exact identity\n\\begin{equation}\\label{eq:solver-energy}\n \\norm{Te}^2+\\alpha\\norm e^2=\\langle g,e\\rangle\n \\le\\norm g\\norm e.\n\\end{equation}\nIf $e=0$, all three conclusions hold with zero error. Otherwise drop the first nonnegative term and divide by $\\alpha\\norm e>0$ to obtain $\\norm e\\le\\norm g/\\alpha$.","To obtain the sharper fitted-data bound, retain the quadratic dependence on $\\norm e$ instead of substituting the first estimate. Multiplying \\eqref{eq:solver-energy} by $\\alpha$ and rearranging gives\n\\[\n\\begin{aligned}\n \\alpha\\norm{Te}^2\n &\\le\\alpha\\norm g\\norm e-\\alpha^2\\norm e^2","To obtain the sharper fitted-data bound, retain the quadratic dependence on $\\norm e$ instead of substituting the first estimate. Multiplying \\eqref{eq:solver-energy} by $\\alpha$ and rearranging gives\n\\[\n\\begin{aligned}\n \\alpha\\norm{Te}^2\n &\\le\\alpha\\norm g\\norm e-\\alpha^2\\norm e^2\\\\\n &=\\frac{\\norm g^2}{4}\n     -\\left(\\alpha\\norm e-\\frac{\\norm g}{2}\\right)^2\n \\le\\frac{\\norm g^2}{4}.\n\\end{aligned}\n\\]\nDivision by $\\alpha$ and square roots give the factor $1/(2\\sqrt\\alpha)$.","&=\\frac{\\norm g^2}{4}\n     -\\left(\\alpha\\norm e-\\frac{\\norm g}{2}\\right)^2\n \\le\\frac{\\norm g^2}{4}.\n\\end{aligned}\n\\]\nDivision by $\\alpha$ and square roots give the factor $1/(2\\sqrt\\alpha)$.","By Theorem~\\ref{WR06} the objective gap is exactly $\\norm{Te}^2+\\alpha\\norm e^2$, hence by \\eqref{eq:solver-energy} it is also $\\langle g,e\\rangle$. It is nonnegative and satisfies\n\\[\n J_\\alpha(a)-J_\\alpha(a_\\alpha)\n \\le\\norm g\\norm e\\le\\frac{\\norm g^2}{\\alpha}.\n\\]\nEvery certificate quantity on the right is computed from the proposal $a$ and the declared problem. The unknown optimizer was used to prove the guarantee, but it is not required to evaluate the certificate.\n\\end{proof}\n\\written{WR13}","\\paragraph{Sharp scalar examples.} For sharpness of the correction-error bound, take $T=0$ on $\\R$ and a nonzero error. Then $g=\\alpha e$. For sharpness of the fitted-data bound, take $T=\\sqrt\\alpha\\,I$ on $\\R$, so $g=2\\alpha e$ and $\\norm{Te}=\\norm g/(2\\sqrt\\alpha)$. These examples establish worst-case sharpness, not simultaneous equality for every operator.","For a fixed raw vector $f$, the corresponding repaired candidates are $f-a$ and $f-a_\\alpha$. Their distance is exactly $\\norm e$. Both preserve $pf$ because both corrections belong to $H$. The theorem therefore controls a genuine protected candidate error, not just an intermediate coordinate vector.","\\fixedplate{37_solver_amplification}{A fixed normal-equation tolerance has different consequences at different regularization parameters. The curves are universal certified upper bounds for a declared normal-residual norm, not observed errors of a particular experiment. The correction estimate grows as $\\alpha^{-1}$ while the fitted-data estimate grows as $\\alpha^{-1/2}$.}{fig:solver-amplification}","\\subsection{Why a vanishing normal residual is not enough}\nLet $T=0$ on $\\R$, let $r=0$, and take $\\alpha_n=n^{-2}$ for $n\\ge1$. The exact optimizer is zero for every such $n$. The fixed proposal $a_n=1$ instead has\n\\[\n v_n=0,\\qquad g_n=\\alpha_n\\longrightarrow0,\\qquad\n J_{\\alpha_n}(a_n)=\\alpha_n\\longrightarrow0,\n \\qquad\\abs{a_n-a_{\\alpha_n}}=1.\n\\]\nThus the fit residual, the normal residual, and even the absolute objective value can all tend to zero without convergence to the selected optimizer. The certificate correctly reports $\\abs{g_n}/\\alpha_n=1$.","act optimizer is zero for every such $n$. The fixed proposal $a_n=1$ instead has\n\\[\n v_n=0,\\qquad g_n=\\alpha_n\\longrightarrow0,\\qquad\n J_{\\alpha_n}(a_n)=\\alpha_n\\longrightarrow0,\n \\qquad\\abs{a_n-a_{\\alpha_n}}=1.\n\\]\nThus the fit residual, the normal residual, and even the absolute objective value can all tend to zero without convergence to the selected optimizer. The certificate correctly reports $\\abs{g_n}/\\alpha_n=1$.","\\subsection{A dual certificate does not require solving the normal equation}\nFor an arbitrary $z\\in Y$, define\n\\[\n \\mathcal D_\\alpha(z)=2\\langle r,z\\rangle-\\norm z^2\n                    -\\frac{\\norm{T^*z}^2}{\\alpha}.\n\\]\nUnlike the fixed-budget witness in Theorem~\\ref{WR04}, this dual witness has no unit-norm constraint. Its quadratic penalties supply the control.","\\begin{theorem}[Exact regularized primal and dual gap]\\label{WR14}\nFor every $a\\in H$ and $z\\in Y$,\n\\begin{equation}\\label{eq:regularized-duality-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(z)\n =\\norm{Ta-r+z}^2+\n   \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}.\n\\end{equation}\nConsequently $\\mathcal D_\\alpha(z)\\le J_\\alpha(a)$ for every pair. At\n$a=a_\\alpha$ and $z=z_\\alpha=r-Ta_\\alpha$, equality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm","\\begin{theorem}[Exact regularized primal and dual gap]\\label{WR14}\nFor every $a\\in H$ and $z\\in Y$,\n\\begin{equation}\\label{eq:regularized-duality-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(z)\n =\\norm{Ta-r+z}^2+\n   \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}.\n\\end{equation}\nConsequently $\\mathcal D_\\alpha(z)\\le J_\\alpha(a)$ for every pair. At\n$a=a_\\alpha$ and $z=z_\\alpha=r-Ta_\\alpha$, equality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =","\\begin{theorem}[Exact regularized primal and dual gap]\\label{WR14}\nFor every $a\\in H$ and $z\\in Y$,\n\\begin{equation}\\label{eq:regularized-duality-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(z)\n =\\norm{Ta-r+z}^2+\n   \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}.\n\\end{equation}\nConsequently $\\mathcal D_\\alpha(z)\\le J_\\alpha(a)$ for every pair. At\n$a=a_\\alpha$ and $z=z_\\alpha=r-Ta_\\alpha$, equality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\fra","\\begin{theorem}[Exact regularized primal and dual gap]\\label{WR14}\nFor every $a\\in H$ and $z\\in Y$,\n\\begin{equation}\\label{eq:regularized-duality-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(z)\n =\\norm{Ta-r+z}^2+\n   \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}.\n\\end{equation}\nConsequently $\\mathcal D_\\alpha(z)\\le J_\\alpha(a)$ for every pair. At\n$a=a_\\alpha$ and $z=z_\\alpha=r-Ta_\\alpha$, equality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a","\\begin{theorem}[Exact regularized primal and dual gap]\\label{WR14}\nFor every $a\\in H$ and $z\\in Y$,\n\\begin{equation}\\label{eq:regularized-duality-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(z)\n =\\norm{Ta-r+z}^2+\n   \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}.\n\\end{equation}\nConsequently $\\mathcal D_\\alpha(z)\\le J_\\alpha(a)$ for every pair. At\n$a=a_\\alpha$ and $z=z_\\alpha=r-Ta_\\alpha$, equality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a\\rangle.\n\\]\nThe adjoint identity","\\begin{theorem}[Exact regularized primal and dual gap]\\label{WR14}\nFor every $a\\in H$ and $z\\in Y$,\n\\begin{equation}\\label{eq:regularized-duality-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(z)\n =\\norm{Ta-r+z}^2+\n   \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}.\n\\end{equation}\nConsequently $\\mathcal D_\\alpha(z)\\le J_\\alpha(a)$ for every pair. At\n$a=a_\\alpha$ and $z=z_\\alpha=r-Ta_\\alpha$, equality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a\\rangle.\n\\]\nThe adjoint identity cancels $2\\langle Ta,z\\rangl","}\nFor every $a\\in H$ and $z\\in Y$,\n\\begin{equation}\\label{eq:regularized-duality-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(z)\n =\\norm{Ta-r+z}^2+\n   \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}.\n\\end{equation}\nConsequently $\\mathcal D_\\alpha(z)\\le J_\\alpha(a)$ for every pair. At\n$a=a_\\alpha$ and $z=z_\\alpha=r-Ta_\\alpha$, equality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a\\rangle.\n\\]\nThe adjoint identity cancels $2\\langle Ta,z\\rangle$ against the last term. The remaining expression is\n\\[\n \\norm{Ta-r}^2+\\alpha\\norm a^2\n -\\lef…","thcal D_\\alpha(z)\n =\\norm{Ta-r+z}^2+\n   \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}.\n\\end{equation}\nConsequently $\\mathcal D_\\alpha(z)\\le J_\\alpha(a)$ for every pair. At\n$a=a_\\alpha$ and $z=z_\\alpha=r-Ta_\\alpha$, equality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a\\rangle.\n\\]\nThe adjoint identity cancels $2\\langle Ta,z\\rangle$ against the last term. The remaining expression is\n\\[\n \\norm{Ta-r}^2+\\alpha\\norm a^2\n -\\left(2\\langle r,z\\rangle-\\norm z^2\n                    -\\alpha^{-1}\\norm{T^*z}^2\\right),\n\\]\nwhich is the …","{Ta-r+z}^2+\n   \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}.\n\\end{equation}\nConsequently $\\mathcal D_\\alpha(z)\\le J_\\alpha(a)$ for every pair. At\n$a=a_\\alpha$ and $z=z_\\alpha=r-Ta_\\alpha$, equality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a\\rangle.\n\\]\nThe adjoint identity cancels $2\\langle Ta,z\\rangle$ against the last term. The remaining expression is\n\\[\n \\norm{Ta-r}^2+\\alpha\\norm a^2\n -\\left(2\\langle r,z\\rangle-\\norm z^2\n                    -\\alpha^{-1}\\norm{T^*z}^2\\right),\n\\]\nwhich is the asserted gap. Its nonnega…","$ and $z=z_\\alpha=r-Ta_\\alpha$, equality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a\\rangle.\n\\]\nThe adjoint identity cancels $2\\langle Ta,z\\rangle$ against the last term. The remaining expression is\n\\[\n \\norm{Ta-r}^2+\\alpha\\norm a^2\n -\\left(2\\langle r,z\\rangle-\\norm z^2\n                    -\\alpha^{-1}\\norm{T^*z}^2\\right),\n\\]\nwhich is the asserted gap. Its nonnegativity gives weak duality for every pair, without any constraint on the witness norm.","ality holds, so the dual maximum is attained and equals the primal minimum. Moreover,\n\\[\n \\alpha\\norm{a-a_\\alpha}^2\n \\le J_\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a\\rangle.\n\\]\nThe adjoint identity cancels $2\\langle Ta,z\\rangle$ against the last term. The remaining expression is\n\\[\n \\norm{Ta-r}^2+\\alpha\\norm a^2\n -\\left(2\\langle r,z\\rangle-\\norm z^2\n                    -\\alpha^{-1}\\norm{T^*z}^2\\right),\n\\]\nwhich is the asserted gap. Its nonnegativity gives weak duality for every pair, without any constraint on the witness norm.","\\alpha(a)-\\mathcal D_\\alpha(z).\n\\]\nFor the easily computed choice $z=r-Ta$,\n\\begin{equation}\\label{eq:residual-dual-gap}\n J_\\alpha(a)-\\mathcal D_\\alpha(r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a\\rangle.\n\\]\nThe adjoint identity cancels $2\\langle Ta,z\\rangle$ against the last term. The remaining expression is\n\\[\n \\norm{Ta-r}^2+\\alpha\\norm a^2\n -\\left(2\\langle r,z\\rangle-\\norm z^2\n                    -\\alpha^{-1}\\norm{T^*z}^2\\right),\n\\]\nwhich is the asserted gap. Its nonnegativity gives weak duality for every pair, without any constraint on the witness norm.","r-Ta)=\\frac{\\norm g^2}{\\alpha}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nFor arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a\\rangle.\n\\]\nThe adjoint identity cancels $2\\langle Ta,z\\rangle$ against the last term. The remaining expression is\n\\[\n \\norm{Ta-r}^2+\\alpha\\norm a^2\n -\\left(2\\langle r,z\\rangle-\\norm z^2\n                    -\\alpha^{-1}\\norm{T^*z}^2\\right),\n\\]\nwhich is the asserted gap. Its nonnegativity gives weak duality for every pair, without any constraint on the witness norm.","For arbitrary $a,z$, expand the first square as\n\\[\n \\norm{Ta-r+z}^2\n =\\norm{Ta-r}^2+\\norm z^2+2\\langle Ta,z\\rangle-2\\langle r,z\\rangle,\n\\]\nand the scaled second square as\n\\[\n \\frac{\\norm{T^*z-\\alpha a}^2}{\\alpha}\n =\\frac{\\norm{T^*z}^2}{\\alpha}+\\alpha\\norm a^2\n       -2\\langle T^*z,a\\rangle.\n\\]\nThe adjoint identity cancels $2\\langle Ta,z\\rangle$ against the last term. The remaining expression is\n\\[\n \\norm{Ta-r}^2+\\alpha\\norm a^2\n -\\left(2\\langle r,z\\rangle-\\norm z^2\n                    -\\alpha^{-1}\\norm{T^*z}^2\\right),\n\\]\nwhich is the asserted gap. Its nonnegativity gives weak duality for every pair, without any constraint on the witness norm.","Set $a=a_\\alpha$ and $z_\\alpha=r-Ta_\\alpha$. The first square is zero by this definition. The normal equation gives $T^*z_\\alpha=\\alpha a_\\alpha$, so the second is zero as well. Therefore\n\\[\n \\mathcal D_\\alpha(z_\\alpha)=J_\\alpha(a_\\alpha).\n\\]\nFor every other $z$, weak duality with $a_\\alpha$ gives $\\mathcal D_\\alpha(z)\\le J_\\alpha(a_\\alpha)$. This proves attainment of the dual maximum and equality with the primal minimum directly.","For a general proposed pair $(a,z)$, decompose its gap as\n\\[\n J_\\alpha(a)-\\mathcal D_\\alpha(z)\n =[J_\\alpha(a)-J_\\alpha(a_\\alpha)]\n  +[J_\\alpha(a_\\alpha)-\\mathcal D_\\alpha(z)].\n\\]\nThe second bracket is nonnegative by weak duality, while Theorem~\\ref{WR06} bounds the first below by $\\alpha\\norm{a-a_\\alpha}^2$. This proves the correction-error certificate.","Finally take $z=r-Ta$. The first square in the gap is zero, while $T^*z-\\alpha a=-g$. The second becomes $\\norm g^2/\\alpha$, proving the residual-witness formula. A small duality gap controls optimizer error only after division by the specified positive parameter.\n\\end{proof}\n\\written{WR14}","\\subsection{A protected example with every certificate calculated}\nTake\n\\[\n d(x,y)=x,\\qquad p(x,y)=x-y,\\qquad f=(1,0),\\qquad r=df=1,\\qquad\\alpha=1.\n\\]\nEvery allowed correction has the form $(t,t)$. The objective is\n\\[\n J(t,t)=(t-1)^2+2t^2,\n\\]\nso $a_\\alpha=(1/3,1/3)$ and $J(a_\\alpha)=2/3$. Suppose a procedure returns only $a=(1/4,1/4)$. Then\n\\[\n v=\\frac34,\\qquad T^*v=(3/8,3/8),\\qquad\n g=(-1/8,-1/8),\\qquad\\norm g^2=\\frac1{32}.\n\\]\nThe actual correction-error square is $1/72$, while the certified upper bound is $1/32$. The actual fitted-error square is $1/144$, while the sharper bound in Theorem~\\ref{WR13} is $1/128$.","\\[\n d(x,y)=x,\\qquad p(x,y)=x-y,\\qquad f=(1,0),\\qquad r=df=1,\\qquad\\alpha=1.\n\\]\nEvery allowed correction has the form $(t,t)$. The objective is\n\\[\n J(t,t)=(t-1)^2+2t^2,\n\\]\nso $a_\\alpha=(1/3,1/3)$ and $J(a_\\alpha)=2/3$. Suppose a procedure returns only $a=(1/4,1/4)$. Then\n\\[\n v=\\frac34,\\qquad T^*v=(3/8,3/8),\\qquad\n g=(-1/8,-1/8),\\qquad\\norm g^2=\\frac1{32}.\n\\]\nThe actual correction-error square is $1/72$, while the certified upper bound is $1/32$. The actual fitted-error square is $1/144$, while the sharper bound in Theorem~\\ref{WR13} is $1/128$.",",0),\\qquad r=df=1,\\qquad\\alpha=1.\n\\]\nEvery allowed correction has the form $(t,t)$. The objective is\n\\[\n J(t,t)=(t-1)^2+2t^2,\n\\]\nso $a_\\alpha=(1/3,1/3)$ and $J(a_\\alpha)=2/3$. Suppose a procedure returns only $a=(1/4,1/4)$. Then\n\\[\n v=\\frac34,\\qquad T^*v=(3/8,3/8),\\qquad\n g=(-1/8,-1/8),\\qquad\\norm g^2=\\frac1{32}.\n\\]\nThe actual correction-error square is $1/72$, while the certified upper bound is $1/32$. The actual fitted-error square is $1/144$, while the sharper bound in Theorem~\\ref{WR13} is $1/128$.","allowed correction has the form $(t,t)$. The objective is\n\\[\n J(t,t)=(t-1)^2+2t^2,\n\\]\nso $a_\\alpha=(1/3,1/3)$ and $J(a_\\alpha)=2/3$. Suppose a procedure returns only $a=(1/4,1/4)$. Then\n\\[\n v=\\frac34,\\qquad T^*v=(3/8,3/8),\\qquad\n g=(-1/8,-1/8),\\qquad\\norm g^2=\\frac1{32}.\n\\]\nThe actual correction-error square is $1/72$, while the certified upper bound is $1/32$. The actual fitted-error square is $1/144$, while the sharper bound in Theorem~\\ref{WR13} is $1/128$.","\\[\n J(t,t)=(t-1)^2+2t^2,\n\\]\nso $a_\\alpha=(1/3,1/3)$ and $J(a_\\alpha)=2/3$. Suppose a procedure returns only $a=(1/4,1/4)$. Then\n\\[\n v=\\frac34,\\qquad T^*v=(3/8,3/8),\\qquad\n g=(-1/8,-1/8),\\qquad\\norm g^2=\\frac1{32}.\n\\]\nThe actual correction-error square is $1/72$, while the certified upper bound is $1/32$. The actual fitted-error square is $1/144$, while the sharper bound in Theorem~\\ref{WR13} is $1/128$.","The proposed objective is $11/16$. Its actual excess over the optimum is $1/48$. The residual witness gives\n\\[\n \\mathcal D_1(3/4)=\\frac{21}{32},\\qquad\n J(a)-\\mathcal D_1(3/4)=\\frac1{32}.\n\\]\nThe duality gap is an upper bound for suboptimality, not necessarily the exact suboptimality. Both repaired candidates preserve $p(f)=1$.","\\fixedplate{38_objective_certificate}{The objective on the actual protected line is $3t^2-2t+1$. The proposed correction at $t=1/4$, exact minimizer at $t=1/3$, and residual-witness lower bound are distinct. The vertical gap to the dual level is the checkable certificate. The smaller gap to the minimum is the true suboptimality.}{fig:objective-certificate}","\\subsection{The residual of the coordinate equation is a covector}\nIn finite dimensions, use orthonormal coordinates in the ambient candidate and residual spaces. Let the columns of $B$ form an arbitrary basis of $H=\\ker p$, not necessarily an orthonormal one. Write $a=Bc$, $S=dB$, and $G=B^{\\mathsf T}B$. As before, $G$ is positive definite. For an inexact coefficient vector $c$, the residual of the coordinate normal equation is\n\\[\n q=(S^{\\mathsf T}S+\\alpha G)c-S^{\\mathsf T}r.\n\\]\nThe vector $q$ stores pairings against basis vectors. It is not the coordinate vector of the Hilbert normal residual unless the basis is orthonormal.","rdinate equation is a covector}\nIn finite dimensions, use orthonormal coordinates in the ambient candidate and residual spaces. Let the columns of $B$ form an arbitrary basis of $H=\\ker p$, not necessarily an orthonormal one. Write $a=Bc$, $S=dB$, and $G=B^{\\mathsf T}B$. As before, $G$ is positive definite. For an inexact coefficient vector $c$, the residual of the coordinate normal equation is\n\\[\n q=(S^{\\mathsf T}S+\\alpha G)c-S^{\\mathsf T}r.\n\\]\nThe vector $q$ stores pairings against basis vectors. It is not the coordinate vector of the Hilbert normal residual unless the basis is orthonormal.","onormal coordinates in the ambient candidate and residual spaces. Let the columns of $B$ form an arbitrary basis of $H=\\ker p$, not necessarily an orthonormal one. Write $a=Bc$, $S=dB$, and $G=B^{\\mathsf T}B$. As before, $G$ is positive definite. For an inexact coefficient vector $c$, the residual of the coordinate normal equation is\n\\[\n q=(S^{\\mathsf T}S+\\alpha G)c-S^{\\mathsf T}r.\n\\]\nThe vector $q$ stores pairings against basis vectors. It is not the coordinate vector of the Hilbert normal residual unless the basis is orthonormal.","\\begin{proposition}[The invariant norm of a coordinate certificate]\\label{WR15}\nThe actual normal residual is\n\\begin{equation}\\label{eq:covector-norm}\n g=BG^{-1}q,\\qquad \\norm g^2=q^{\\mathsf T}G^{-1}q.\n\\end{equation}\nFor any invertible change of basis $B'=BC$, the same proposed correction has\n\\[\n c'=C^{-1}c,\\quad G'=C^{\\mathsf T}GC,\\quad q'=C^{\\mathsf T}q,\n\\]\nand\n\\[\n (q')^{\\mathsf T}(G')^{-1}q'=q^{\\mathsf T}G^{-1}q.\n\\]\n\\end{proposition}\n\\begin{proof}\nBecause $B$ has independent columns, for every nonzero coefficient vector $c$,\n\\[\n c^{\\mathsf T}Gc=c^{\\mathsf T}B^{\\mathsf T}Bc=\\norm{Bc}^2>0.\n\\]\nThus $G$ is positive definite and invertible. The matrix $P=BG^{-1}B^{\\mathsf T}$ is symmetric and obeys\n\\[\n P^2=BG^{-1}(B^{\\mathsf T}B)G^{-1}B^{\\mathsf T}=P,\n \\qquad PB=B.\n\\]\nIts range is precisely the column space $H$, so it is the orthogonal projection onto the entire permitted correction space.","te]\\label{WR15}\nThe actual normal residual is\n\\begin{equation}\\label{eq:covector-norm}\n g=BG^{-1}q,\\qquad \\norm g^2=q^{\\mathsf T}G^{-1}q.\n\\end{equation}\nFor any invertible change of basis $B'=BC$, the same proposed correction has\n\\[\n c'=C^{-1}c,\\quad G'=C^{\\mathsf T}GC,\\quad q'=C^{\\mathsf T}q,\n\\]\nand\n\\[\n (q')^{\\mathsf T}(G')^{-1}q'=q^{\\mathsf T}G^{-1}q.\n\\]\n\\end{proposition}\n\\begin{proof}\nBecause $B$ has independent columns, for every nonzero coefficient vector $c$,\n\\[\n c^{\\mathsf T}Gc=c^{\\mathsf T}B^{\\mathsf T}Bc=\\norm{Bc}^2>0.\n\\]\nThus $G$ is positive definite and invertible. The matrix $P=BG^{-1}B^{\\mathsf T}$ is symmetric and obeys\n\\[\n P^2=BG^{-1}(B^{\\mathsf T}B)G^{-1}B^{\\mathsf T}=P,\n \\qquad PB=B.\n\\]\nIts range is precisely the column space $H$, so it is the orthogonal projection onto the entire permitted correction space.","el{eq:covector-norm}\n g=BG^{-1}q,\\qquad \\norm g^2=q^{\\mathsf T}G^{-1}q.\n\\end{equation}\nFor any invertible change of basis $B'=BC$, the same proposed correction has\n\\[\n c'=C^{-1}c,\\quad G'=C^{\\mathsf T}GC,\\quad q'=C^{\\mathsf T}q,\n\\]\nand\n\\[\n (q')^{\\mathsf T}(G')^{-1}q'=q^{\\mathsf T}G^{-1}q.\n\\]\n\\end{proposition}\n\\begin{proof}\nBecause $B$ has independent columns, for every nonzero coefficient vector $c$,\n\\[\n c^{\\mathsf T}Gc=c^{\\mathsf T}B^{\\mathsf T}Bc=\\norm{Bc}^2>0.\n\\]\nThus $G$ is positive definite and invertible. The matrix $P=BG^{-1}B^{\\mathsf T}$ is symmetric and obeys\n\\[\n P^2=BG^{-1}(B^{\\mathsf T}B)G^{-1}B^{\\mathsf T}=P,\n \\qquad PB=B.\n\\]\nIts range is precisely the column space $H$, so it is the orthogonal projection onto the entire permitted correction space.","covector-norm}\n g=BG^{-1}q,\\qquad \\norm g^2=q^{\\mathsf T}G^{-1}q.\n\\end{equation}\nFor any invertible change of basis $B'=BC$, the same proposed correction has\n\\[\n c'=C^{-1}c,\\quad G'=C^{\\mathsf T}GC,\\quad q'=C^{\\mathsf T}q,\n\\]\nand\n\\[\n (q')^{\\mathsf T}(G')^{-1}q'=q^{\\mathsf T}G^{-1}q.\n\\]\n\\end{proposition}\n\\begin{proof}\nBecause $B$ has independent columns, for every nonzero coefficient vector $c$,\n\\[\n c^{\\mathsf T}Gc=c^{\\mathsf T}B^{\\mathsf T}Bc=\\norm{Bc}^2>0.\n\\]\nThus $G$ is positive definite and invertible. The matrix $P=BG^{-1}B^{\\mathsf T}$ is symmetric and obeys\n\\[\n P^2=BG^{-1}(B^{\\mathsf T}B)G^{-1}B^{\\mathsf T}=P,\n \\qquad PB=B.\n\\]\nIts range is precisely the column space $H$, so it is the orthogonal projection onto the entire permitted correction space.","equation}\nFor any invertible change of basis $B'=BC$, the same proposed correction has\n\\[\n c'=C^{-1}c,\\quad G'=C^{\\mathsf T}GC,\\quad q'=C^{\\mathsf T}q,\n\\]\nand\n\\[\n (q')^{\\mathsf T}(G')^{-1}q'=q^{\\mathsf T}G^{-1}q.\n\\]\n\\end{proposition}\n\\begin{proof}\nBecause $B$ has independent columns, for every nonzero coefficient vector $c$,\n\\[\n c^{\\mathsf T}Gc=c^{\\mathsf T}B^{\\mathsf T}Bc=\\norm{Bc}^2>0.\n\\]\nThus $G$ is positive definite and invertible. The matrix $P=BG^{-1}B^{\\mathsf T}$ is symmetric and obeys\n\\[\n P^2=BG^{-1}(B^{\\mathsf T}B)G^{-1}B^{\\mathsf T}=P,\n \\qquad PB=B.\n\\]\nIts range is precisely the column space $H$, so it is the orthogonal projection onto the entire permitted correction space.",", the same proposed correction has\n\\[\n c'=C^{-1}c,\\quad G'=C^{\\mathsf T}GC,\\quad q'=C^{\\mathsf T}q,\n\\]\nand\n\\[\n (q')^{\\mathsf T}(G')^{-1}q'=q^{\\mathsf T}G^{-1}q.\n\\]\n\\end{proposition}\n\\begin{proof}\nBecause $B$ has independent columns, for every nonzero coefficient vector $c$,\n\\[\n c^{\\mathsf T}Gc=c^{\\mathsf T}B^{\\mathsf T}Bc=\\norm{Bc}^2>0.\n\\]\nThus $G$ is positive definite and invertible. The matrix $P=BG^{-1}B^{\\mathsf T}$ is symmetric and obeys\n\\[\n P^2=BG^{-1}(B^{\\mathsf T}B)G^{-1}B^{\\mathsf T}=P,\n \\qquad PB=B.\n\\]\nIts range is precisely the column space $H$, so it is the orthogonal projection onto the entire permitted correction space.",",\\quad G'=C^{\\mathsf T}GC,\\quad q'=C^{\\mathsf T}q,\n\\]\nand\n\\[\n (q')^{\\mathsf T}(G')^{-1}q'=q^{\\mathsf T}G^{-1}q.\n\\]\n\\end{proposition}\n\\begin{proof}\nBecause $B$ has independent columns, for every nonzero coefficient vector $c$,\n\\[\n c^{\\mathsf T}Gc=c^{\\mathsf T}B^{\\mathsf T}Bc=\\norm{Bc}^2>0.\n\\]\nThus $G$ is positive definite and invertible. The matrix $P=BG^{-1}B^{\\mathsf T}$ is symmetric and obeys\n\\[\n P^2=BG^{-1}(B^{\\mathsf T}B)G^{-1}B^{\\mathsf T}=P,\n \\qquad PB=B.\n\\]\nIts range is precisely the column space $H$, so it is the orthogonal projection onto the entire permitted correction space.",".\n\\]\n\\end{proposition}\n\\begin{proof}\nBecause $B$ has independent columns, for every nonzero coefficient vector $c$,\n\\[\n c^{\\mathsf T}Gc=c^{\\mathsf T}B^{\\mathsf T}Bc=\\norm{Bc}^2>0.\n\\]\nThus $G$ is positive definite and invertible. The matrix $P=BG^{-1}B^{\\mathsf T}$ is symmetric and obeys\n\\[\n P^2=BG^{-1}(B^{\\mathsf T}B)G^{-1}B^{\\mathsf T}=P,\n \\qquad PB=B.\n\\]\nIts range is precisely the column space $H$, so it is the orthogonal projection onto the entire permitted correction space.","The restricted Hilbert normal residual is\n\\[\n g=P d^{\\mathsf T}(dBc-r)+\\alpha Bc.\n\\]\nSubstitute $S=dB$, $G=B^{\\mathsf T}B$ and the expression for $P$:\n\\[\n g=BG^{-1}\\bigl(S^{\\mathsf T}(Sc-r)+\\alpha Gc\\bigr)=BG^{-1}q.\n\\]\nEquivalently $q=B^{\\mathsf T}g$ records the pairings of $g$ with the basis columns. Its squared Hilbert norm is\n\\[\n g^{\\mathsf T}g\n =q^{\\mathsf T}G^{-1}B^{\\mathsf T}BG^{-1}q\n =q^{\\mathsf T}G^{-1}q,\n\\]\nusing symmetry of $G^{-1}$. This proves both coordinate formulas.","is\n\\[\n g=P d^{\\mathsf T}(dBc-r)+\\alpha Bc.\n\\]\nSubstitute $S=dB$, $G=B^{\\mathsf T}B$ and the expression for $P$:\n\\[\n g=BG^{-1}\\bigl(S^{\\mathsf T}(Sc-r)+\\alpha Gc\\bigr)=BG^{-1}q.\n\\]\nEquivalently $q=B^{\\mathsf T}g$ records the pairings of $g$ with the basis columns. Its squared Hilbert norm is\n\\[\n g^{\\mathsf T}g\n =q^{\\mathsf T}G^{-1}B^{\\mathsf T}BG^{-1}q\n =q^{\\mathsf T}G^{-1}q,\n\\]\nusing symmetry of $G^{-1}$. This proves both coordinate formulas.","For the basis change $B'=BC$, the same ambient vector is represented by $c'=C^{-1}c$. Also $S'=SC$ and $G'=C^{\\mathsf T}GC$. Substitute these formulas into the new coordinate normal equation:\n\\[\n\\begin{aligned}\n q'&=((S')^{\\mathsf T}S'+\\alpha G')c'-(S')^{\\mathsf T}r","For the basis change $B'=BC$, the same ambient vector is represented by $c'=C^{-1}c$. Also $S'=SC$ and $G'=C^{\\mathsf T}GC$. Substitute these formulas into the new coordinate normal equation:\n\\[\n\\begin{aligned}\n q'&=((S')^{\\mathsf T}S'+\\alpha G')c'-(S')^{\\mathsf T}r\\\\\n &=C^{\\mathsf T}(S^{\\mathsf T}S+\\alpha G)CC^{-1}c\n      -C^{\\mathsf T}S^{\\mathsf T}r=C^{\\mathsf T}q.\n\\end{aligned}\n\\]\nThe inverse metric is $(G')^{-1}=C^{-1}G^{-1}(C^{\\mathsf T})^{-1}$. Hence\n\\[\n (q')^{\\mathsf T}(G')^{-1}q'\n =q^{\\mathsf T}C C^{-1}G^{-1}(C^{\\mathsf T})^{-1}C^{\\mathsf T}q\n =q^{\\mathsf T}G^{-1}q.\n\\]\nFor a zero-dimensional kernel these are empty matrices and zero vectors, and the resulting scalar norm is zero. The spanning assumption is indispensable: a basis of a proper subspace would compute a projection to a different correction problem.\n\\end{proof}\n\\written{WR15}","&=C^{\\mathsf T}(S^{\\mathsf T}S+\\alpha G)CC^{-1}c\n      -C^{\\mathsf T}S^{\\mathsf T}r=C^{\\mathsf T}q.\n\\end{aligned}\n\\]\nThe inverse metric is $(G')^{-1}=C^{-1}G^{-1}(C^{\\mathsf T})^{-1}$. Hence\n\\[\n (q')^{\\mathsf T}(G')^{-1}q'\n =q^{\\mathsf T}C C^{-1}G^{-1}(C^{\\mathsf T})^{-1}C^{\\mathsf T}q\n =q^{\\mathsf T}G^{-1}q.\n\\]\nFor a zero-dimensional kernel these are empty matrices and zero vectors, and the resulting scalar norm is zero. The spanning assumption is indispensable: a basis of a proper subspace would compute a projection to a different correction problem.\n\\end{proof}\n\\written{WR15}","In the previous example, $B=(1,1)^{\\mathsf T}$, $G=2$, and the proposal $c=1/4$ gives $q=-1/4$. Thus $q^{\\mathsf T}G^{-1}q=1/32$, as required. Replacing $B$ by $sB$ changes $q$ to $sq$ and $G$ to $s^2G$. The Euclidean norm of the coordinate covector can be made arbitrarily small or large by changing the nonzero scalar $s$. The actual correction, normal residual, and certified error do not change.","\\fixedplate{39_covector_metric}{Rescaling the basis of the same protected line rescales the raw coordinate covector. Its Euclidean norm is therefore not a representation-invariant stopping criterion. The Gram-corrected norm stays equal to $1/\\sqrt{32}$ for every displayed scale. All systems represent the same proposed correction.}{fig:covector-metric}","\\subsection{Separate data error, numerical error, and regularization bias}\nLet $r$ be the intended datum and $\\widetilde r$ the measured datum. Suppose\n\\[\n \\norm{\\widetilde r-r}\\le\\delta,\n \\qquad \\norm{T^*(Ta-\\widetilde r)+\\alpha a}\\le\\varepsilon.\n\\]\nThe maps $T,p$ and their Hilbert metrics remain fixed. By Theorem~\\ref{WR13} and the fixed-parameter part of Theorem~\\ref{WR09},\n\\begin{equation}\\label{eq:combined-solver-data}\n \\norm{a-a_\\alpha(r)}\n \\le \\frac{\\varepsilon}{\\alpha}+\\frac{\\delta}{2\\sqrt\\alpha}.\n\\end{equation}\nThe first term is the solver's contribution. The second is the worst-case propagation of data error. They should not be collapsed into one unqualified residual tolerance.","\\begin{theorem}[A consistent inexact stopping rule]\\label{WR16}\nAssume the projected intended datum belongs to the actual range of $T$, with minimum-norm preimage $a^\\dagger$. Along a sequence of positive parameters and measured problems, suppose\n\\[\n \\alpha_n\\to0,\\qquad\n \\frac{\\delta_n}{\\sqrt{\\alpha_n}}\\to0,\\qquad\n \\frac{\\varepsilon_n}{\\alpha_n}\\to0.\n\\]\nIf the proposed corrections $a_n\\in H$ satisfy the displayed data and normal-residual bounds, then $a_n\\to a^\\dagger$ in norm. At every index,\n\\[\n \\norm{a_n-a^\\dagger}\n \\le\\frac{\\varepsilon_n}{\\alpha_n}\n      +\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}\n      +\\norm{a_{\\alpha_n}(r)-a^\\dagger}.\n\\]\n\\end{theorem}\n\\begin{proof}\nAt index $n$, distinguish two exact minimizers: let $\\widetilde a_n=a_{\\alpha_n}(\\widetilde r_n)$ solve the measured problem and let $a_n^0=a_{\\alpha_n}(r)$ solv","\\begin{theorem}[A consistent inexact stopping rule]\\label{WR16}\nAssume the projected intended datum belongs to the actual range of $T$, with minimum-norm preimage $a^\\dagger$. Along a sequence of positive parameters and measured problems, suppose\n\\[\n \\alpha_n\\to0,\\qquad\n \\frac{\\delta_n}{\\sqrt{\\alpha_n}}\\to0,\\qquad\n \\frac{\\varepsilon_n}{\\alpha_n}\\to0.\n\\]\nIf the proposed corrections $a_n\\in H$ satisfy the displayed data and normal-residual bounds, then $a_n\\to a^\\dagger$ in norm. At every index,\n\\[\n \\norm{a_n-a^\\dagger}\n \\le\\frac{\\varepsilon_n}{\\alpha_n}\n      +\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}\n      +\\norm{a_{\\alpha_n}(r)-a^\\dagger}.\n\\]\n\\end{theorem}\n\\begin{proof}\nAt index $n$, distinguish two exact minimizers: let $\\widetilde a_n=a_{\\alpha_n}(\\widetilde r_n)$ solve the measured problem and let $a_n^0=a_{\\alpha_n}(r)$ solve the intended-data problem. All three c","\\begin{theorem}[A consistent inexact stopping rule]\\label{WR16}\nAssume the projected intended datum belongs to the actual range of $T$, with minimum-norm preimage $a^\\dagger$. Along a sequence of positive parameters and measured problems, suppose\n\\[\n \\alpha_n\\to0,\\qquad\n \\frac{\\delta_n}{\\sqrt{\\alpha_n}}\\to0,\\qquad\n \\frac{\\varepsilon_n}{\\alpha_n}\\to0.\n\\]\nIf the proposed corrections $a_n\\in H$ satisfy the displayed data and normal-residual bounds, then $a_n\\to a^\\dagger$ in norm. At every index,\n\\[\n \\norm{a_n-a^\\dagger}\n \\le\\frac{\\varepsilon_n}{\\alpha_n}\n      +\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}\n      +\\norm{a_{\\alpha_n}(r)-a^\\dagger}.\n\\]\n\\end{theorem}\n\\begin{proof}\nAt index $n$, distinguish two exact minimizers: let $\\widetilde a_n=a_{\\alpha_n}(\\widetilde r_n)$ solve the measured problem and let $a_n^0=a_{\\alpha_n}(r)$ solve the intended-data problem. All three corrections, including the proposal $a_n$, belong to the same space $H$.","l{WR16}\nAssume the projected intended datum belongs to the actual range of $T$, with minimum-norm preimage $a^\\dagger$. Along a sequence of positive parameters and measured problems, suppose\n\\[\n \\alpha_n\\to0,\\qquad\n \\frac{\\delta_n}{\\sqrt{\\alpha_n}}\\to0,\\qquad\n \\frac{\\varepsilon_n}{\\alpha_n}\\to0.\n\\]\nIf the proposed corrections $a_n\\in H$ satisfy the displayed data and normal-residual bounds, then $a_n\\to a^\\dagger$ in norm. At every index,\n\\[\n \\norm{a_n-a^\\dagger}\n \\le\\frac{\\varepsilon_n}{\\alpha_n}\n      +\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}\n      +\\norm{a_{\\alpha_n}(r)-a^\\dagger}.\n\\]\n\\end{theorem}\n\\begin{proof}\nAt index $n$, distinguish two exact minimizers: let $\\widetilde a_n=a_{\\alpha_n}(\\widetilde r_n)$ solve the measured problem and let $a_n^0=a_{\\alpha_n}(r)$ solve the intended-data problem. All three corrections, including the proposal $a_n$, belong to the same space $H$.","elongs to the actual range of $T$, with minimum-norm preimage $a^\\dagger$. Along a sequence of positive parameters and measured problems, suppose\n\\[\n \\alpha_n\\to0,\\qquad\n \\frac{\\delta_n}{\\sqrt{\\alpha_n}}\\to0,\\qquad\n \\frac{\\varepsilon_n}{\\alpha_n}\\to0.\n\\]\nIf the proposed corrections $a_n\\in H$ satisfy the displayed data and normal-residual bounds, then $a_n\\to a^\\dagger$ in norm. At every index,\n\\[\n \\norm{a_n-a^\\dagger}\n \\le\\frac{\\varepsilon_n}{\\alpha_n}\n      +\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}\n      +\\norm{a_{\\alpha_n}(r)-a^\\dagger}.\n\\]\n\\end{theorem}\n\\begin{proof}\nAt index $n$, distinguish two exact minimizers: let $\\widetilde a_n=a_{\\alpha_n}(\\widetilde r_n)$ solve the measured problem and let $a_n^0=a_{\\alpha_n}(r)$ solve the intended-data problem. All three corrections, including the proposal $a_n$, belong to the same space $H$.","ta_n}{\\sqrt{\\alpha_n}}\\to0,\\qquad\n \\frac{\\varepsilon_n}{\\alpha_n}\\to0.\n\\]\nIf the proposed corrections $a_n\\in H$ satisfy the displayed data and normal-residual bounds, then $a_n\\to a^\\dagger$ in norm. At every index,\n\\[\n \\norm{a_n-a^\\dagger}\n \\le\\frac{\\varepsilon_n}{\\alpha_n}\n      +\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}\n      +\\norm{a_{\\alpha_n}(r)-a^\\dagger}.\n\\]\n\\end{theorem}\n\\begin{proof}\nAt index $n$, distinguish two exact minimizers: let $\\widetilde a_n=a_{\\alpha_n}(\\widetilde r_n)$ solve the measured problem and let $a_n^0=a_{\\alpha_n}(r)$ solve the intended-data problem. All three corrections, including the proposal $a_n$, belong to the same space $H$.","arepsilon_n}{\\alpha_n}\\to0.\n\\]\nIf the proposed corrections $a_n\\in H$ satisfy the displayed data and normal-residual bounds, then $a_n\\to a^\\dagger$ in norm. At every index,\n\\[\n \\norm{a_n-a^\\dagger}\n \\le\\frac{\\varepsilon_n}{\\alpha_n}\n      +\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}\n      +\\norm{a_{\\alpha_n}(r)-a^\\dagger}.\n\\]\n\\end{theorem}\n\\begin{proof}\nAt index $n$, distinguish two exact minimizers: let $\\widetilde a_n=a_{\\alpha_n}(\\widetilde r_n)$ solve the measured problem and let $a_n^0=a_{\\alpha_n}(r)$ solve the intended-data problem. All three corrections, including the proposal $a_n$, belong to the same space $H$.","displayed data and normal-residual bounds, then $a_n\\to a^\\dagger$ in norm. At every index,\n\\[\n \\norm{a_n-a^\\dagger}\n \\le\\frac{\\varepsilon_n}{\\alpha_n}\n      +\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}\n      +\\norm{a_{\\alpha_n}(r)-a^\\dagger}.\n\\]\n\\end{theorem}\n\\begin{proof}\nAt index $n$, distinguish two exact minimizers: let $\\widetilde a_n=a_{\\alpha_n}(\\widetilde r_n)$ solve the measured problem and let $a_n^0=a_{\\alpha_n}(r)$ solve the intended-data problem. All three corrections, including the proposal $a_n$, belong to the same space $H$.","a_n-a^\\dagger}\n \\le\\frac{\\varepsilon_n}{\\alpha_n}\n      +\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}\n      +\\norm{a_{\\alpha_n}(r)-a^\\dagger}.\n\\]\n\\end{theorem}\n\\begin{proof}\nAt index $n$, distinguish two exact minimizers: let $\\widetilde a_n=a_{\\alpha_n}(\\widetilde r_n)$ solve the measured problem and let $a_n^0=a_{\\alpha_n}(r)$ solve the intended-data problem. All three corrections, including the proposal $a_n$, belong to the same space $H$.","ilon_n}{\\alpha_n}\n      +\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}\n      +\\norm{a_{\\alpha_n}(r)-a^\\dagger}.\n\\]\n\\end{theorem}\n\\begin{proof}\nAt index $n$, distinguish two exact minimizers: let $\\widetilde a_n=a_{\\alpha_n}(\\widetilde r_n)$ solve the measured problem and let $a_n^0=a_{\\alpha_n}(r)$ solve the intended-data problem. All three corrections, including the proposal $a_n$, belong to the same space $H$.","The supplied normal-residual bound is for the measured datum. Therefore Theorem~\\ref{WR13} applies first to that problem and gives\n\\[\n \\norm{a_n-\\widetilde a_n}\\le\\frac{\\varepsilon_n}{\\alpha_n}.\n\\]\nThe pointwise data estimate in Theorem~\\ref{WR09}, at the same parameter, then gives\n\\[\n \\norm{\\widetilde a_n-a_n^0}\\le\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}.\n\\]\nInsert these two intermediate vectors between $a_n$ and $a^\\dagger$ in the triangle inequality to obtain the displayed three-term estimate.","e supplied normal-residual bound is for the measured datum. Therefore Theorem~\\ref{WR13} applies first to that problem and gives\n\\[\n \\norm{a_n-\\widetilde a_n}\\le\\frac{\\varepsilon_n}{\\alpha_n}.\n\\]\nThe pointwise data estimate in Theorem~\\ref{WR09}, at the same parameter, then gives\n\\[\n \\norm{\\widetilde a_n-a_n^0}\\le\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}.\n\\]\nInsert these two intermediate vectors between $a_n$ and $a^\\dagger$ in the triangle inequality to obtain the displayed three-term estimate.","normal-residual bound is for the measured datum. Therefore Theorem~\\ref{WR13} applies first to that problem and gives\n\\[\n \\norm{a_n-\\widetilde a_n}\\le\\frac{\\varepsilon_n}{\\alpha_n}.\n\\]\nThe pointwise data estimate in Theorem~\\ref{WR09}, at the same parameter, then gives\n\\[\n \\norm{\\widetilde a_n-a_n^0}\\le\\frac{\\delta_n}{2\\sqrt{\\alpha_n}}.\n\\]\nInsert these two intermediate vectors between $a_n$ and $a^\\dagger$ in the triangle inequality to obtain the displayed three-term estimate.","The first term tends to zero by the solver-tolerance hypothesis and the second by the data/parameter hypothesis. Since the projected intended datum is attained and $\\alpha_n\\to0$, the last term tends to zero by Theorem~\\ref{WR08}. For a direct epsilon argument, make each term smaller than one third of a prescribed tolerance beyond its own threshold and take the maximum of the three thresholds. This gives norm convergence of the proposals.","Protection is not merely asymptotic: $a_n\\in\\ker p$ implies $p(f-a_n)=pf$ at every index. The proof compares no changing operators or changing kernels. Both the pointwise certificate and the convergence conclusion have formal counterparts under these hypotheses.\n\\end{proof}\n\\written{WR16}","\\subsection{Retaining the explicit diagonal rate with an inexact solver}\nFor the attainable datum of Corollary~\\ref{WR12} and $0<\\delta\\le1$, take $m=\\lceil\\delta^{-2/3}\\rceil$ and $\\alpha=m^{-2}$ as before. If an inexact solver satisfies\n\\[\n \\varepsilon\\le\\kappa\\alpha\\delta^{1/3}\n \\quad(\\kappa\\ge0),\n\\]\nthen the same argument gives\n\\[\n \\norm{a-a^\\dagger}\\le(\\sqrt2+1+\\kappa)\\delta^{1/3}.\n\\]\nThus the inner solve can retain the proven outer accuracy if its normal residual decreases on the scale $\\alpha\\delta^{1/3}$. Since $\\alpha$ is on the scale $\\delta^{4/3}$, the corresponding normal-residual scale is $\\delta^{5/3}$, with the stated ceiling factors retained in the exact rule. This is a sufficient rule for this problem, not a universal optimal stopping law.","of Corollary~\\ref{WR12} and $0<\\delta\\le1$, take $m=\\lceil\\delta^{-2/3}\\rceil$ and $\\alpha=m^{-2}$ as before. If an inexact solver satisfies\n\\[\n \\varepsilon\\le\\kappa\\alpha\\delta^{1/3}\n \\quad(\\kappa\\ge0),\n\\]\nthen the same argument gives\n\\[\n \\norm{a-a^\\dagger}\\le(\\sqrt2+1+\\kappa)\\delta^{1/3}.\n\\]\nThus the inner solve can retain the proven outer accuracy if its normal residual decreases on the scale $\\alpha\\delta^{1/3}$. Since $\\alpha$ is on the scale $\\delta^{4/3}$, the corresponding normal-residual scale is $\\delta^{5/3}$, with the stated ceiling factors retained in the exact rule. This is a sufficient rule for this problem, not a universal optimal stopping law.","$0<\\delta\\le1$, take $m=\\lceil\\delta^{-2/3}\\rceil$ and $\\alpha=m^{-2}$ as before. If an inexact solver satisfies\n\\[\n \\varepsilon\\le\\kappa\\alpha\\delta^{1/3}\n \\quad(\\kappa\\ge0),\n\\]\nthen the same argument gives\n\\[\n \\norm{a-a^\\dagger}\\le(\\sqrt2+1+\\kappa)\\delta^{1/3}.\n\\]\nThus the inner solve can retain the proven outer accuracy if its normal residual decreases on the scale $\\alpha\\delta^{1/3}$. Since $\\alpha$ is on the scale $\\delta^{4/3}$, the corresponding normal-residual scale is $\\delta^{5/3}$, with the stated ceiling factors retained in the exact rule. This is a sufficient rule for this problem, not a universal optimal stopping law.","lceil\\delta^{-2/3}\\rceil$ and $\\alpha=m^{-2}$ as before. If an inexact solver satisfies\n\\[\n \\varepsilon\\le\\kappa\\alpha\\delta^{1/3}\n \\quad(\\kappa\\ge0),\n\\]\nthen the same argument gives\n\\[\n \\norm{a-a^\\dagger}\\le(\\sqrt2+1+\\kappa)\\delta^{1/3}.\n\\]\nThus the inner solve can retain the proven outer accuracy if its normal residual decreases on the scale $\\alpha\\delta^{1/3}$. Since $\\alpha$ is on the scale $\\delta^{4/3}$, the corresponding normal-residual scale is $\\delta^{5/3}$, with the stated ceiling factors retained in the exact rule. This is a sufficient rule for this problem, not a universal optimal stopping law.","an inexact solver satisfies\n\\[\n \\varepsilon\\le\\kappa\\alpha\\delta^{1/3}\n \\quad(\\kappa\\ge0),\n\\]\nthen the same argument gives\n\\[\n \\norm{a-a^\\dagger}\\le(\\sqrt2+1+\\kappa)\\delta^{1/3}.\n\\]\nThus the inner solve can retain the proven outer accuracy if its normal residual decreases on the scale $\\alpha\\delta^{1/3}$. Since $\\alpha$ is on the scale $\\delta^{4/3}$, the corresponding normal-residual scale is $\\delta^{5/3}$, with the stated ceiling factors retained in the exact rule. This is a sufficient rule for this problem, not a universal optimal stopping law.","The declaration \\path{DiagonalRegularization.diagonal_inexact_rounded_rate} formalizes this specialization for the actual infinite diagonal operator. Its hypotheses bound both the data error and the normal residual evaluated against the measured datum. It retains the literal ceiling in $m$ and derives the exact-data bias estimate within the model.","\\fixedplate{40_stopping_schedule}{For the attainable diagonal example, the three displayed terms are the proved bias bound, propagated data-noise bound, and a solver-error budget chosen to preserve the same $\\delta^{1/3}$ rate. The ceiling in $m=\\lceil\\delta^{-2/3}\\rceil$ is evaluated explicitly. The plot shows bounds, not measurements from a trained system.}{fig:stopping-schedule}","\\subsection{The executable verification contract}\nThe program takes $d,p,f,\\alpha$ and a proposed ambient correction, with optional measured residuals, a full kernel basis, and an independent dual witness. It rejects a correction that changes the protected data and checks the basis's dimensions, rank, and kernel membership. It does not silently project a rejected proposal into another problem.","The output records $f-a$, protection before and after, measured fit, actual model residual, the Gram-corrected normal certificate, objective values, and squared error bounds. Verification recomputes the record without calling an optimizer. Independent determinant-based tests compare the bounds with actual errors in small ambient constrained systems. A certificate establishes consistency with its included problem, not authentication that this is the externally intended task.","\\chap{Mathematical meaning and the remaining boundaries}\\label{sec:meaning}\n\\subsection{A relative theory can be complete without being absolute}\nThe strict space \\(\\Gamma(F)\\) is canonical relative to its diagram: its universal property determines it up to the appropriate unique isomorphism. The additive obstruction is canonical relative to its admissible corrections. A marked realization problem is invariant under the specified comparison equivalences. A repair can be canonical relative to a protected observable, or orthogonal relative to a fixed inner product.","\\subsection{Internal realization and intended objects}\nSuppose a separate category \\(\\mathcal S\\) contains the intended semantic objects of an application, and a realization construction gives a comparison functor\n\\[\n \\mathcal R:\\mathcal S\\longrightarrow\\mathcal K,\n\\]\nwhere \\(\\mathcal K\\) is the relevant category or groupoid of coherent data. Defining \\(\\mathcal K\\) does not establish that \\(\\mathcal R\\) is faithful, full, or essentially surjective. These are new mathematical questions about the actual semantic comparison.","Faithfulness asks whether distinct semantic arrows with the same source and target can become indistinguishable. Fullness asks whether every arrow between the images of two semantic objects comes from an arrow between those objects. Essential surjectivity asks whether every object of \\(\\mathcal K\\) is isomorphic to the image of an actual semantic object. Depending on the application, one may seek only a restricted image or a classified obstruction to belonging to it.","\\subsection{Every finite test can pass while the infinite problem fails}\n\\begin{example}[Finite prefixes without a global point]\\label{KS54}\nLet \\(X_n=\\{m\\in\\N:m\\ge n\\}\\), with bonding maps given by inclusion. Every level is nonempty, and every finite prefix has a coherent point. The inverse limit is nevertheless empty.\n\\end{example}\n\\begin{proof}\nFor a fixed terminal level \\(N\\), the integer \\(N\\) belongs to every fiber \\(X_n\\) with \\(n\\le N\\). Assign that same integer at each such level. Every restriction is an inclusion, so it preserves the assigned integer and all finite coherence equations hold. This proves finite solvability with the quantifiers in the order ``for every \\(N\\), there exists a prefix.''","st can pass while the infinite problem fails}\n\\begin{example}[Finite prefixes without a global point]\\label{KS54}\nLet \\(X_n=\\{m\\in\\N:m\\ge n\\}\\), with bonding maps given by inclusion. Every level is nonempty, and every finite prefix has a coherent point. The inverse limit is nevertheless empty.\n\\end{example}\n\\begin{proof}\nFor a fixed terminal level \\(N\\), the integer \\(N\\) belongs to every fiber \\(X_n\\) with \\(n\\le N\\). Assign that same integer at each such level. Every restriction is an inclusion, so it preserves the assigned integer and all finite coherence equations hold. This proves finite solvability with the quantifiers in the order ``for every \\(N\\), there exists a prefix.''","hile the infinite problem fails}\n\\begin{example}[Finite prefixes without a global point]\\label{KS54}\nLet \\(X_n=\\{m\\in\\N:m\\ge n\\}\\), with bonding maps given by inclusion. Every level is nonempty, and every finite prefix has a coherent point. The inverse limit is nevertheless empty.\n\\end{example}\n\\begin{proof}\nFor a fixed terminal level \\(N\\), the integer \\(N\\) belongs to every fiber \\(X_n\\) with \\(n\\le N\\). Assign that same integer at each such level. Every restriction is an inclusion, so it preserves the assigned integer and all finite coherence equations hold. This proves finite solvability with the quantifiers in the order ``for every \\(N\\), there exists a prefix.''","verse limit is nevertheless empty.\n\\end{example}\n\\begin{proof}\nFor a fixed terminal level \\(N\\), the integer \\(N\\) belongs to every fiber \\(X_n\\) with \\(n\\le N\\). Assign that same integer at each such level. Every restriction is an inclusion, so it preserves the assigned integer and all finite coherence equations hold. This proves finite solvability with the quantifiers in the order ``for every \\(N\\), there exists a prefix.''","Suppose instead that \\(x=(x_n)\\) is a global point. Equality along each inclusion says that the underlying natural numbers of \\(x_{n+1}\\) and \\(x_n\\) are equal. Induction therefore makes every underlying coordinate equal to the number \\(m\\) underlying \\(x_0\\). Membership of \\(x_{m+1}\\) in \\(X_{m+1}\\) now gives \\(m\\ge m+1\\), a contradiction. Thus no global point exists.","There is no conflict with a gluing theorem for prefixes. The prefix chosen above for length \\(N+1\\) restricts to the value \\(N+1\\), not to our chosen value \\(N\\) on the shorter prefix. Nor could another choice of one solution for every length give a matching family: its value at level zero would force the same forbidden global integer. Separate existence at every length is strictly weaker than existence of a compatible family of prefix solutions.\n\\end{proof}\n\\source{KS54}\nThis counterexample is formalized in \\code{FinitePrefixBoundary.lean} using the original resolution and partial-section types. It explains why inverse-limit existence needs more than separate finite solvability. Finiteness, compactness, surjectivity, or another coherence principle must be proved in the relevant applica","There is no conflict with a gluing theorem for prefixes. The prefix chosen above for length \\(N+1\\) restricts to the value \\(N+1\\), not to our chosen value \\(N\\) on the shorter prefix. Nor could another choice of one solution for every length give a matching family: its value at level zero would force the same forbidden global integer. Separate existence at every length is strictly weaker than existence of a compatible family of prefix solutions.\n\\end{proof}\n\\source{KS54}\nThis counterexample is formalized in \\code{FinitePrefixBoundary.lean} using the original resolution and partial-section types. It explains why inverse-limit existence needs more than separate finite solvability. Finiteness, compactness, surjectivity, or another coherence principle must be proved in the relevant application. The example does not cont","There is no conflict with a gluing theorem for prefixes. The prefix chosen above for length \\(N+1\\) restricts to the value \\(N+1\\), not to our chosen value \\(N\\) on the shorter prefix. Nor could another choice of one solution for every length give a matching family: its value at level zero would force the same forbidden global integer. Separate existence at every length is strictly weaker than existence of a compatible family of prefix solutions.\n\\end{proof}\n\\source{KS54}\nThis counterexample is formalized in \\code{FinitePrefixBoundary.lean} using the original resolution and partial-section types. It explains why inverse-limit existence needs more than separate finite solvability. Finiteness, compactness, surjectivity, or another coherence principle must be proved in the relevant application. The example does not contradict the strict gluing theore","\\begin{proposition}[Finite fibers and persistent images]\\label{KS58}\nLet \\((X_n,\\rho_n)\\) be a sequential inverse system of nonempty finite sets. Its inverse limit is nonempty. More precisely, if \\(\\rho_{n,m}:X_m\\to X_n\\) is composite restriction, then\n\\[\n \\im(\\operatorname{ev}_n)=Y_n,\n \\qquad Y_n=\\bigcap_{m\\ge n}\\im\\rho_{n,m}.\n\\]\nThus a prescribed value at level \\(n\\) extends precisely when it survives every later image.\n\\end{proposition}\n\\begin{proof}\nWrite \\(I_{n,m}=\\rho_{n,m}(X_m)\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible","\\begin{proposition}[Finite fibers and persistent images]\\label{KS58}\nLet \\((X_n,\\rho_n)\\) be a sequential inverse system of nonempty finite sets. Its inverse limit is nonempty. More precisely, if \\(\\rho_{n,m}:X_m\\to X_n\\) is composite restriction, then\n\\[\n \\im(\\operatorname{ev}_n)=Y_n,\n \\qquad Y_n=\\bigcap_{m\\ge n}\\im\\rho_{n,m}.\n\\]\nThus a prescribed value at level \\(n\\) extends precisely when it survives every later image.\n\\end{proposition}\n\\begin{proof}\nWrite \\(I_{n,m}=\\rho_{n,m}(X_m)\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible. There is an index \\(s_n\\ge n\\) such that \\(I_{n,m}=I_{n,s_n}\\) for all \\(m\\ge s_n\\). Consequently \\(Y_n=I_{n,s_n}\\) is nonempty.","8}\nLet \\((X_n,\\rho_n)\\) be a sequential inverse system of nonempty finite sets. Its inverse limit is nonempty. More precisely, if \\(\\rho_{n,m}:X_m\\to X_n\\) is composite restriction, then\n\\[\n \\im(\\operatorname{ev}_n)=Y_n,\n \\qquad Y_n=\\bigcap_{m\\ge n}\\im\\rho_{n,m}.\n\\]\nThus a prescribed value at level \\(n\\) extends precisely when it survives every later image.\n\\end{proposition}\n\\begin{proof}\nWrite \\(I_{n,m}=\\rho_{n,m}(X_m)\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible. There is an index \\(s_n\\ge n\\) such that \\(I_{n,m}=I_{n,s_n}\\) for all \\(m\\ge s_n\\). Consequently \\(Y_n=I_{n,s_n}\\) is nonempty.","uential inverse system of nonempty finite sets. Its inverse limit is nonempty. More precisely, if \\(\\rho_{n,m}:X_m\\to X_n\\) is composite restriction, then\n\\[\n \\im(\\operatorname{ev}_n)=Y_n,\n \\qquad Y_n=\\bigcap_{m\\ge n}\\im\\rho_{n,m}.\n\\]\nThus a prescribed value at level \\(n\\) extends precisely when it survives every later image.\n\\end{proposition}\n\\begin{proof}\nWrite \\(I_{n,m}=\\rho_{n,m}(X_m)\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible. There is an index \\(s_n\\ge n\\) such that \\(I_{n,m}=I_{n,s_n}\\) for all \\(m\\ge s_n\\). Consequently \\(Y_n=I_{n,s_n}\\) is nonempty.","ystem of nonempty finite sets. Its inverse limit is nonempty. More precisely, if \\(\\rho_{n,m}:X_m\\to X_n\\) is composite restriction, then\n\\[\n \\im(\\operatorname{ev}_n)=Y_n,\n \\qquad Y_n=\\bigcap_{m\\ge n}\\im\\rho_{n,m}.\n\\]\nThus a prescribed value at level \\(n\\) extends precisely when it survives every later image.\n\\end{proposition}\n\\begin{proof}\nWrite \\(I_{n,m}=\\rho_{n,m}(X_m)\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible. There is an index \\(s_n\\ge n\\) such that \\(I_{n,m}=I_{n,s_n}\\) for all \\(m\\ge s_n\\). Consequently \\(Y_n=I_{n,s_n}\\) is nonempty.","nonempty. More precisely, if \\(\\rho_{n,m}:X_m\\to X_n\\) is composite restriction, then\n\\[\n \\im(\\operatorname{ev}_n)=Y_n,\n \\qquad Y_n=\\bigcap_{m\\ge n}\\im\\rho_{n,m}.\n\\]\nThus a prescribed value at level \\(n\\) extends precisely when it survives every later image.\n\\end{proposition}\n\\begin{proof}\nWrite \\(I_{n,m}=\\rho_{n,m}(X_m)\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible. There is an index \\(s_n\\ge n\\) such that \\(I_{n,m}=I_{n,s_n}\\) for all \\(m\\ge s_n\\). Consequently \\(Y_n=I_{n,s_n}\\) is nonempty.","ev}_n)=Y_n,\n \\qquad Y_n=\\bigcap_{m\\ge n}\\im\\rho_{n,m}.\n\\]\nThus a prescribed value at level \\(n\\) extends precisely when it survives every later image.\n\\end{proposition}\n\\begin{proof}\nWrite \\(I_{n,m}=\\rho_{n,m}(X_m)\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible. There is an index \\(s_n\\ge n\\) such that \\(I_{n,m}=I_{n,s_n}\\) for all \\(m\\ge s_n\\). Consequently \\(Y_n=I_{n,s_n}\\) is nonempty.","ery later image.\n\\end{proposition}\n\\begin{proof}\nWrite \\(I_{n,m}=\\rho_{n,m}(X_m)\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible. There is an index \\(s_n\\ge n\\) such that \\(I_{n,m}=I_{n,s_n}\\) for all \\(m\\ge s_n\\). Consequently \\(Y_n=I_{n,s_n}\\) is nonempty.","roposition}\n\\begin{proof}\nWrite \\(I_{n,m}=\\rho_{n,m}(X_m)\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible. There is an index \\(s_n\\ge n\\) such that \\(I_{n,m}=I_{n,s_n}\\) for all \\(m\\ge s_n\\). Consequently \\(Y_n=I_{n,s_n}\\) is nonempty.","e \\(I_{n,m}=\\rho_{n,m}(X_m)\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible. There is an index \\(s_n\\ge n\\) such that \\(I_{n,m}=I_{n,s_n}\\) for all \\(m\\ge s_n\\). Consequently \\(Y_n=I_{n,s_n}\\) is nonempty.","\\) for \\(m\\ge n\\). Each \\(I_{n,m}\\) is nonempty. The composition law gives\n\\[\n I_{n,m+1}=\\rho_{n,m}\\bigl(\\rho_m(X_{m+1})\\bigr)\\subseteq I_{n,m}.\n\\]\nThese are subsets of the finite set \\(X_n\\). Every strict decrease removes an element permanently, so only finitely many strict decreases are possible. There is an index \\(s_n\\ge n\\) such that \\(I_{n,m}=I_{n,s_n}\\) for all \\(m\\ge s_n\\). Consequently \\(Y_n=I_{n,s_n}\\) is nonempty.","We next prove that restriction induces a \\emph{surjection} \\(Y_{n+1}\\to Y_n\\), even though \\(X_{n+1}\\to X_n\\) need not be surjective. For membership, let \\(z\\in Y_{n+1}\\). Given \\(m\\ge n+1\\), write \\(z=\\rho_{n+1,m}(w)\\) for some \\(w\\in X_m\\). Then \\(\\rho_n z=\\rho_{n,m}(w)\\). Together with the automatic condition at \\(m=n\\), this shows \\(\\rho_nz\\in Y_n\\).","For surjectivity, fix \\(y\\in Y_n\\) and choose\n\\[\n m\\ge\\max\\{s_n,s_{n+1},n+1\\}.\n\\]\nSince \\(y\\in I_{n,m}\\), there is \\(w\\in X_m\\) with \\(\\rho_{n,m}(w)=y\\). Put \\(z=\\rho_{n+1,m}(w)\\). Stabilization at level \\(n+1\\) gives \\(z\\in Y_{n+1}\\), and composition gives \\(\\rho_nz=y\\).","Choose \\(x_0\\in Y_0\\) and recursively choose \\(x_{n+1}\\in Y_{n+1}\\) above \\(x_n\\). The preceding surjectivity justifies each step. Adjacent coherence implies every composite coherence equation, so this gives a point. This is an existence construction using countably many choices, not an asserted effective procedure for locating the stabilization indices.","Finally, fix a level \\(n\\). Every global point satisfies \\(x_n=\\rho_{n,m}(x_m)\\) for each \\(m\\ge n\\), hence \\(x_n\\in Y_n\\). Conversely, start with any \\(y\\in Y_n\\), lift recursively above level \\(n\\) through the surjections just proved, and define the lower coordinates by \\(x_i=\\rho_{i,n}(y)\\). The composite restriction laws verify coherence across and below level \\(n\\). The resulting point evaluates to \\(y\\). These two inclusions prove the asserted equality of evaluation images, not merely nonemptiness of the inverse limit.\n\\end{proof}\n\\source{KS58}\nThis is the standard finite-set inverse-limit principle, here with its sequential persistent-image proof~\\cite{stacks-limits}. Its formalization in \\code{FiniteRealiza","Finally, fix a level \\(n\\). Every global point satisfies \\(x_n=\\rho_{n,m}(x_m)\\) for each \\(m\\ge n\\), hence \\(x_n\\in Y_n\\). Conversely, start with any \\(y\\in Y_n\\), lift recursively above level \\(n\\) through the surjections just proved, and define the lower coordinates by \\(x_i=\\rho_{i,n}(y)\\). The composite restriction laws verify coherence across and below level \\(n\\). The resulting point evaluates to \\(y\\). These two inclusions prove the asserted equality of evaluation images, not merely nonemptiness of the inverse limit.\n\\end{proof}\n\\source{KS58}\nThis is the standard finite-set inverse-limit principle, here with its sequential persistent-image proof~\\cite{stacks-limits}. Its formalization in \\code{FiniteRealizationExistence.lean} constructs an explicit equivalence","Finally, fix a level \\(n\\). Every global point satisfies \\(x_n=\\rho_{n,m}(x_m)\\) for each \\(m\\ge n\\), hence \\(x_n\\in Y_n\\). Conversely, start with any \\(y\\in Y_n\\), lift recursively above level \\(n\\) through the surjections just proved, and define the lower coordinates by \\(x_i=\\rho_{i,n}(y)\\). The composite restriction laws verify coherence across and below level \\(n\\). The resulting point evaluates to \\(y\\). These two inclusions prove the asserted equality of evaluation images, not merely nonemptiness of the inverse limit.\n\\end{proof}\n\\source{KS58}\nThis is the standard finite-set inverse-limit principle, here with its sequential persistent-image proof~\\cite{stacks-limits}. Its formalization in \\code{FiniteRealizationExistence.lean} constructs an explicit equivalence with the pinned lib","Finally, fix a level \\(n\\). Every global point satisfies \\(x_n=\\rho_{n,m}(x_m)\\) for each \\(m\\ge n\\), hence \\(x_n\\in Y_n\\). Conversely, start with any \\(y\\in Y_n\\), lift recursively above level \\(n\\) through the surjections just proved, and define the lower coordinates by \\(x_i=\\rho_{i,n}(y)\\). The composite restriction laws verify coherence across and below level \\(n\\). The resulting point evaluates to \\(y\\). These two inclusions prove the asserted equality of evaluation images, not merely nonemptiness of the inverse limit.\n\\end{proof}\n\\source{KS58}\nThis is the standard finite-set inverse-limit principle, here with its sequential persistent-image proof~\\cite{stacks-limits}. Its formalization in \\code{FiniteRealizationExistence.lean} constructs an explicit equivalence with the pinned library's section type, t","Finally, fix a level \\(n\\). Every global point satisfies \\(x_n=\\rho_{n,m}(x_m)\\) for each \\(m\\ge n\\), hence \\(x_n\\in Y_n\\). Conversely, start with any \\(y\\in Y_n\\), lift recursively above level \\(n\\) through the surjections just proved, and define the lower coordinates by \\(x_i=\\rho_{i,n}(y)\\). The composite restriction laws verify coherence across and below level \\(n\\). The resulting point evaluates to \\(y\\). These two inclusions prove the asserted equality of evaluation images, not merely nonemptiness of the inverse limit.\n\\end{proof}\n\\source{KS58}\nThis is the standard finite-set inverse-limit principle, here with its sequential persistent-image proof~\\cite{stacks-limits}. Its formalization in \\code{FiniteRealizationExistence.lean} constructs an explicit equivalence with the pinned library's section type, then applies the finite cofiltered-system","Finally, fix a level \\(n\\). Every global point satisfies \\(x_n=\\rho_{n,m}(x_m)\\) for each \\(m\\ge n\\), hence \\(x_n\\in Y_n\\). Conversely, start with any \\(y\\in Y_n\\), lift recursively above level \\(n\\) through the surjections just proved, and define the lower coordinates by \\(x_i=\\rho_{i,n}(y)\\). The composite restriction laws verify coherence across and below level \\(n\\). The resulting point evaluates to \\(y\\). These two inclusions prove the asserted equality of evaluation images, not merely nonemptiness of the inverse limit.\n\\end{proof}\n\\source{KS58}\nThis is the standard finite-set inverse-limit principle, here with its sequential persistent-image proof~\\cite{stacks-limits}. Its formalization in \\code{FiniteRealizationExistence.lean} constructs an explicit equivalence with the pinned library's section type, then applies the finite cofiltered-system and eventual-range theorems. The pri","Finally, fix a level \\(n\\). Every global point satisfies \\(x_n=\\rho_{n,m}(x_m)\\) for each \\(m\\ge n\\), hence \\(x_n\\in Y_n\\). Conversely, start with any \\(y\\in Y_n\\), lift recursively above level \\(n\\) through the surjections just proved, and define the lower coordinates by \\(x_i=\\rho_{i,n}(y)\\). The composite restriction laws verify coherence across and below level \\(n\\). The resulting point evaluates to \\(y\\). These two inclusions prove the asserted equality of evaluation images, not merely nonemptiness of the inverse limit.\n\\end{proof}\n\\source{KS58}\nThis is the standard finite-set inverse-limit principle, here with its sequential persistent-image proof~\\cite{stacks-limits}. Its formalization in \\code{FiniteRealizationExistence.lean} constructs an explicit equivalence with the pinned library's section type, then applies the finite cofiltered-system and eventual-range theorems. The printed sequential proof and this formal route establish the same stated criterion. Both use ordin","Finally, fix a level \\(n\\). Every global point satisfies \\(x_n=\\rho_{n,m}(x_m)\\) for each \\(m\\ge n\\), hence \\(x_n\\in Y_n\\). Conversely, start with any \\(y\\in Y_n\\), lift recursively above level \\(n\\) through the surjections just proved, and define the lower coordinates by \\(x_i=\\rho_{i,n}(y)\\). The composite restriction laws verify coherence across and below level \\(n\\). The resulting point evaluates to \\(y\\). These two inclusions prove the asserted equality of evaluation images, not merely nonemptiness of the inverse limit.\n\\end{proof}\n\\source{KS58}\nThis is the standard finite-set inverse-limit principle, here with its sequential persistent-image proof~\\cite{stacks-limits}. Its formalization in \\code{FiniteRealizationExistence.lean} constructs an explicit equivalence with the pinned library's section type, then applies the finite cofiltered-system and eventual-range theorems. The printed sequential proof and this formal route establish the same stated criterion. Both use ordinary classical choice. Surjective bonding maps on arbitrary nonempty sets give","fix a level \\(n\\). Every global point satisfies \\(x_n=\\rho_{n,m}(x_m)\\) for each \\(m\\ge n\\), hence \\(x_n\\in Y_n\\). Conversely, start with any \\(y\\in Y_n\\), lift recursively above level \\(n\\) through the surjections just proved, and define the lower coordinates by \\(x_i=\\rho_{i,n}(y)\\). The composite restriction laws verify coherence across and below level \\(n\\). The resulting point evaluates to \\(y\\). These two inclusions prove the asserted equality of evaluation images, not merely nonemptiness of the inverse limit.\n\\end{proof}\n\\source{KS58}\nThis is the standard finite-set inverse-limit principle, here with its sequential persistent-image proof~\\cite{stacks-limits}. Its formalization in \\code{FiniteRealizationExistence.lean} constructs an explicit equivalence with the pinned library's section type, then applies the finite cofiltered-system and eventual-range theorems. The printed sequential proof and this formal route establish the same stated criterion. Both use ordinary classical choice. Surjective bonding maps on arbitrary nonempty sets give a second sufficient condition by direct","\\subsection{A finite computation that exposes the preservation contract}\nConsider a raw vector \\(x=(2,0,5)\\in\\R^3\\) with defect \\(d(x)=x_1-x_2\\). Unconstrained orthogonal repair gives \\((1,1,5)\\), with squared error two. If the first and third coordinates must be preserved, the unique exact repair is \\((2,2,5)\\), with squared error four.","For a finite linear system over a field, the same question admits independently checkable positive and negative certificates. Let \\(A\\in\\K^{m\\times n}\\) encode compatibility, \\(P\\in\\K^{p\\times n}\\) the protected values, and \\(x\\in\\K^n\\) the raw candidate. Let the columns of \\(N\\) be a basis of \\(\\ker P\\). The correction equation is\n\\[\n ANz=Ax,\\qquad y=x-Nz.\n\\]\nIt is equivalent to the stacked system \\(Ay=0\\), \\(Py=Px\\).","a finite linear system over a field, the same question admits independently checkable positive and negative certificates. Let \\(A\\in\\K^{m\\times n}\\) encode compatibility, \\(P\\in\\K^{p\\times n}\\) the protected values, and \\(x\\in\\K^n\\) the raw candidate. Let the columns of \\(N\\) be a basis of \\(\\ker P\\). The correction equation is\n\\[\n ANz=Ax,\\qquad y=x-Nz.\n\\]\nIt is equivalent to the stacked system \\(Ay=0\\), \\(Py=Px\\).","inear system over a field, the same question admits independently checkable positive and negative certificates. Let \\(A\\in\\K^{m\\times n}\\) encode compatibility, \\(P\\in\\K^{p\\times n}\\) the protected values, and \\(x\\in\\K^n\\) the raw candidate. Let the columns of \\(N\\) be a basis of \\(\\ker P\\). The correction equation is\n\\[\n ANz=Ax,\\qquad y=x-Nz.\n\\]\nIt is equivalent to the stacked system \\(Ay=0\\), \\(Py=Px\\).","\\begin{proposition}[A finite certificate alternative]\\label{KS57}\nExactly one of the following occurs:\n\\begin{enumerate}\n\\item There is \\(y\\) with \\(Ay=0\\) and \\(Py=Px\\). The full answer set is \\(y+H\\), where \\(H=\\ker A\\cap\\ker P\\).\n\\item There are row vectors \\(\\lambda\\in\\K^{1\\times m}\\) and \\(\\mu\\in\\K^{1\\times p}\\) with\n\\[\n \\lambda A=\\mu P,\\qquad\\lambda Ax=1.\n\\]\n\\end{enumerate}\nThe negative witness annihilates every admissible correction defect, since \\(\\lambda AN=\\mu PN=0\\). The ambiguity dimension in the positive case is\n\\[\n \\dim H=n-\\operatorname{rank}\\begin{pmatrix}A","KS57}\nExactly one of the following occurs:\n\\begin{enumerate}\n\\item There is \\(y\\) with \\(Ay=0\\) and \\(Py=Px\\). The full answer set is \\(y+H\\), where \\(H=\\ker A\\cap\\ker P\\).\n\\item There are row vectors \\(\\lambda\\in\\K^{1\\times m}\\) and \\(\\mu\\in\\K^{1\\times p}\\) with\n\\[\n \\lambda A=\\mu P,\\qquad\\lambda Ax=1.\n\\]\n\\end{enumerate}\nThe negative witness annihilates every admissible correction defect, since \\(\\lambda AN=\\mu PN=0\\). The ambiguity dimension in the positive case is\n\\[\n \\dim H=n-\\operatorname{rank}\\begin{pmatrix}A","is \\(y\\) with \\(Ay=0\\) and \\(Py=Px\\). The full answer set is \\(y+H\\), where \\(H=\\ker A\\cap\\ker P\\).\n\\item There are row vectors \\(\\lambda\\in\\K^{1\\times m}\\) and \\(\\mu\\in\\K^{1\\times p}\\) with\n\\[\n \\lambda A=\\mu P,\\qquad\\lambda Ax=1.\n\\]\n\\end{enumerate}\nThe negative witness annihilates every admissible correction defect, since \\(\\lambda AN=\\mu PN=0\\). The ambiguity dimension in the positive case is\n\\[\n \\dim H=n-\\operatorname{rank}\\begin{pmatrix}A\\\\P\\end{pmatrix}.\n\\]\n\\end{proposition}\n\\begin{proof}\nSet \\(T=\\begin{pmatrix}A","P\\end{pmatrix}.\n\\]\n\\end{proposition}\n\\begin{proof}\nSet \\(T=\\begin{pmatrix}A\\\\P\\end{pmatrix}\\) and \\(b=\\begin{pmatrix}0","P\\end{pmatrix}\\) and \\(b=\\begin{pmatrix}0\\\\Px\\end{pmatrix}\\). A vector \\(y\\) is an exact protected answer if and only if \\(Ty=b\\). Perform Gaussian row operations on the augmented matrix, keeping their product \\(U\\). Each elementary row operation is invertible over the given field. Thus \\(Ty=b\\) and \\(UTy=Ub\\) have exactly the same solutions.","Px\\end{pmatrix}\\). A vector \\(y\\) is an exact protected answer if and only if \\(Ty=b\\). Perform Gaussian row operations on the augmented matrix, keeping their product \\(U\\). Each elementary row operation is invertible over the given field. Thus \\(Ty=b\\) and \\(UTy=Ub\\) have exactly the same solutions.","If no row of the reduced augmented system has zero left side and nonzero right side, assign arbitrary values to its free variables and solve for its pivot variables. This supplies a solution \\(y_0\\). For any vector \\(y\\),\n\\[\n Ty=b\\quad\\Longleftrightarrow\\quad T(y-y_0)=0.\n\\]\nTranslation by \\(y_0\\) is therefore a bijection from \\(\\ker T\\) to the answer set, with inverse \\(y\\mapsto y-y_0\\). Rank--nullity applied to the actual linear map \\(T:\\K^n\\to\\K^{m+p}\\) gives the ambiguity dimension. Dependent prescription rows do not alter this argument.","m has zero left side and nonzero right side, assign arbitrary values to its free variables and solve for its pivot variables. This supplies a solution \\(y_0\\). For any vector \\(y\\),\n\\[\n Ty=b\\quad\\Longleftrightarrow\\quad T(y-y_0)=0.\n\\]\nTranslation by \\(y_0\\) is therefore a bijection from \\(\\ker T\\) to the answer set, with inverse \\(y\\mapsto y-y_0\\). Rank--nullity applied to the actual linear map \\(T:\\K^n\\to\\K^{m+p}\\) gives the ambiguity dimension. Dependent prescription rows do not alter this argument.","If the reduced system is inconsistent, choose a row with zero left side and right side \\(\\alpha\\ne0\\). Let \\(w\\) be the corresponding row of \\(U\\), and split it according to the two blocks as \\(w=(u,v)\\). The retained row operations give the literal identities\n\\[\n uA+vP=0,\\qquad vPx=\\alpha.\n\\]\nDivision is permitted because \\(\\alpha\\) is a nonzero field element. Define \\(\\lambda=-\\alpha^{-1}u\\) and \\(\\mu=\\alpha^{-1}v\\). Multiplying the displayed equations by \\(\\alpha^{-1}\\) gives\n\\[\n \\lambda A=\\mu P,\\qquad \\lambda Ax=\\mu Px=1.\n\\]\nFor every admissible correction \\(a\\in\\ker P\\), these same row equations imply \\(\\lambda Aa=\\mu Pa=0\\). The witness therefore annihilates the entire correction image while taking value one on the required defect.","the reduced system is inconsistent, choose a row with zero left side and right side \\(\\alpha\\ne0\\). Let \\(w\\) be the corresponding row of \\(U\\), and split it according to the two blocks as \\(w=(u,v)\\). The retained row operations give the literal identities\n\\[\n uA+vP=0,\\qquad vPx=\\alpha.\n\\]\nDivision is permitted because \\(\\alpha\\) is a nonzero field element. Define \\(\\lambda=-\\alpha^{-1}u\\) and \\(\\mu=\\alpha^{-1}v\\). Multiplying the displayed equations by \\(\\alpha^{-1}\\) gives\n\\[\n \\lambda A=\\mu P,\\qquad \\lambda Ax=\\mu Px=1.\n\\]\nFor every admissible correction \\(a\\in\\ker P\\), these same row equations imply \\(\\lambda Aa=\\mu Pa=0\\). The witness therefore annihilates the entire correction image while taking value one on the required defect.","ft side and right side \\(\\alpha\\ne0\\). Let \\(w\\) be the corresponding row of \\(U\\), and split it according to the two blocks as \\(w=(u,v)\\). The retained row operations give the literal identities\n\\[\n uA+vP=0,\\qquad vPx=\\alpha.\n\\]\nDivision is permitted because \\(\\alpha\\) is a nonzero field element. Define \\(\\lambda=-\\alpha^{-1}u\\) and \\(\\mu=\\alpha^{-1}v\\). Multiplying the displayed equations by \\(\\alpha^{-1}\\) gives\n\\[\n \\lambda A=\\mu P,\\qquad \\lambda Ax=\\mu Px=1.\n\\]\nFor every admissible correction \\(a\\in\\ker P\\), these same row equations imply \\(\\lambda Aa=\\mu Pa=0\\). The witness therefore annihilates the entire correction image while taking value one on the required defect.","\\(\\alpha\\ne0\\). Let \\(w\\) be the corresponding row of \\(U\\), and split it according to the two blocks as \\(w=(u,v)\\). The retained row operations give the literal identities\n\\[\n uA+vP=0,\\qquad vPx=\\alpha.\n\\]\nDivision is permitted because \\(\\alpha\\) is a nonzero field element. Define \\(\\lambda=-\\alpha^{-1}u\\) and \\(\\mu=\\alpha^{-1}v\\). Multiplying the displayed equations by \\(\\alpha^{-1}\\) gives\n\\[\n \\lambda A=\\mu P,\\qquad \\lambda Ax=\\mu Px=1.\n\\]\nFor every admissible correction \\(a\\in\\ker P\\), these same row equations imply \\(\\lambda Aa=\\mu Pa=0\\). The witness therefore annihilates the entire correction image while taking value one on the required defect.","g to the two blocks as \\(w=(u,v)\\). The retained row operations give the literal identities\n\\[\n uA+vP=0,\\qquad vPx=\\alpha.\n\\]\nDivision is permitted because \\(\\alpha\\) is a nonzero field element. Define \\(\\lambda=-\\alpha^{-1}u\\) and \\(\\mu=\\alpha^{-1}v\\). Multiplying the displayed equations by \\(\\alpha^{-1}\\) gives\n\\[\n \\lambda A=\\mu P,\\qquad \\lambda Ax=\\mu Px=1.\n\\]\nFor every admissible correction \\(a\\in\\ker P\\), these same row equations imply \\(\\lambda Aa=\\mu Pa=0\\). The witness therefore annihilates the entire correction image while taking value one on the required defect.","d row operations give the literal identities\n\\[\n uA+vP=0,\\qquad vPx=\\alpha.\n\\]\nDivision is permitted because \\(\\alpha\\) is a nonzero field element. Define \\(\\lambda=-\\alpha^{-1}u\\) and \\(\\mu=\\alpha^{-1}v\\). Multiplying the displayed equations by \\(\\alpha^{-1}\\) gives\n\\[\n \\lambda A=\\mu P,\\qquad \\lambda Ax=\\mu Px=1.\n\\]\nFor every admissible correction \\(a\\in\\ker P\\), these same row equations imply \\(\\lambda Aa=\\mu Pa=0\\). The witness therefore annihilates the entire correction image while taking value one on the required defect.","The alternatives cannot occur together. If \\(y\\) were an exact protected answer in the presence of such a witness, then\n\\[\n 1=\\lambda Ax=\\mu Px=\\mu Py=\\lambda Ay=0.\n\\]\nA field has \\(1\\ne0\\), so this is impossible. Consistency or inconsistency of the reduced system proves exhaustiveness, and the contradiction proves exclusivity. Empty row or column blocks cause no extra division: normalization is used only in the inconsistent case, where the nonzero scalar has already been obtained. The proof does not transfer to integral coefficients by silently allowing the same division.\n\\end{proof}\n\\source{KS57}\nThe formal proof gives these literal row-vector equations over an arbitrary field. It uses algebraic dual separation of the stacked image ra","The alternatives cannot occur together. If \\(y\\) were an exact protected answer in the presence of such a witness, then\n\\[\n 1=\\lambda Ax=\\mu Px=\\mu Py=\\lambda Ay=0.\n\\]\nA field has \\(1\\ne0\\), so this is impossible. Consistency or inconsistency of the reduced system proves exhaustiveness, and the contradiction proves exclusivity. Empty row or column blocks cause no extra division: normalization is used only in the inconsistent case, where the nonzero scalar has already been obtained. The proof does not transfer to integral coefficients by silently allowing the same division.\n\\end{proof}\n\\source{KS57}\nThe formal proof gives these literal row-vector equations over an arbitrary field. It uses algebraic dual separation of the stacked image rather than verifying the row-reduction program. The affine answer equivalence and rank-nullity formula are proved separa","The alternatives cannot occur together. If \\(y\\) were an exact protected answer in the presence of such a witness, then\n\\[\n 1=\\lambda Ax=\\mu Px=\\mu Py=\\lambda Ay=0.\n\\]\nA field has \\(1\\ne0\\), so this is impossible. Consistency or inconsistency of the reduced system proves exhaustiveness, and the contradiction proves exclusivity. Empty row or column blocks cause no extra division: normalization is used only in the inconsistent case, where the nonzero scalar has already been obtained. The proof does not transfer to integral coefficients by silently allowing the same division.\n\\end{proof}\n\\source{KS57}\nThe formal proof gives these literal row-vector equations over an arbitrary field. It uses algebraic dual separation of the stacked image rather than verifying the row-reduction program. The affine answer equivalence and rank-nullity formula are proved separately. Thus a normalize","\\subsection{One equation, three preservation problems}\nFor the following three cases work over \\(\\Q\\), with \\(A=(1,-1,0)\\) and \\(x=(2,0,5)\\). With \\(P(x)=(x_1,x_3)\\), the unique answer is \\((2,2,5)\\). With \\(P(x)=x_3\\), the entire answer set is \\(\\{(t,t,5):t\\in\\Q\\}\\). With \\(P(x)=(x_1,x_2)\\), no answer exists. In this last case\n\\[\n \\lambda=(1/2),\\qquad\\mu=(1/2,-1/2)\n\\]\nis a normalized negative certificate over \\(\\Q\\) or \\(\\R\\): \\(\\lambda A=\\mu P\\) and \\(\\lambda Ax=1\\).","equation, three preservation problems}\nFor the following three cases work over \\(\\Q\\), with \\(A=(1,-1,0)\\) and \\(x=(2,0,5)\\). With \\(P(x)=(x_1,x_3)\\), the unique answer is \\((2,2,5)\\). With \\(P(x)=x_3\\), the entire answer set is \\(\\{(t,t,5):t\\in\\Q\\}\\). With \\(P(x)=(x_1,x_2)\\), no answer exists. In this last case\n\\[\n \\lambda=(1/2),\\qquad\\mu=(1/2,-1/2)\n\\]\nis a normalized negative certificate over \\(\\Q\\) or \\(\\R\\): \\(\\lambda A=\\mu P\\) and \\(\\lambda Ax=1\\).","three preservation problems}\nFor the following three cases work over \\(\\Q\\), with \\(A=(1,-1,0)\\) and \\(x=(2,0,5)\\). With \\(P(x)=(x_1,x_3)\\), the unique answer is \\((2,2,5)\\). With \\(P(x)=x_3\\), the entire answer set is \\(\\{(t,t,5):t\\in\\Q\\}\\). With \\(P(x)=(x_1,x_2)\\), no answer exists. In this last case\n\\[\n \\lambda=(1/2),\\qquad\\mu=(1/2,-1/2)\n\\]\nis a normalized negative certificate over \\(\\Q\\) or \\(\\R\\): \\(\\lambda A=\\mu P\\) and \\(\\lambda Ax=1\\).","eservation problems}\nFor the following three cases work over \\(\\Q\\), with \\(A=(1,-1,0)\\) and \\(x=(2,0,5)\\). With \\(P(x)=(x_1,x_3)\\), the unique answer is \\((2,2,5)\\). With \\(P(x)=x_3\\), the entire answer set is \\(\\{(t,t,5):t\\in\\Q\\}\\). With \\(P(x)=(x_1,x_2)\\), no answer exists. In this last case\n\\[\n \\lambda=(1/2),\\qquad\\mu=(1/2,-1/2)\n\\]\nis a normalized negative certificate over \\(\\Q\\) or \\(\\R\\): \\(\\lambda A=\\mu P\\) and \\(\\lambda Ax=1\\).","the following three cases work over \\(\\Q\\), with \\(A=(1,-1,0)\\) and \\(x=(2,0,5)\\). With \\(P(x)=(x_1,x_3)\\), the unique answer is \\((2,2,5)\\). With \\(P(x)=x_3\\), the entire answer set is \\(\\{(t,t,5):t\\in\\Q\\}\\). With \\(P(x)=(x_1,x_2)\\), no answer exists. In this last case\n\\[\n \\lambda=(1/2),\\qquad\\mu=(1/2,-1/2)\n\\]\nis a normalized negative certificate over \\(\\Q\\) or \\(\\R\\): \\(\\lambda A=\\mu P\\) and \\(\\lambda Ax=1\\).","The package implements this finite interface over \\(\\Q\\) in \\code{examples/protected_repair.py}. It returns a particular exact answer and an ambiguity basis, or a normalized dual witness. Its verifier checks the certificate equations directly. A Gaussian particular answer is \\emph{not} advertised as the nearest answer of Theorem~\\ref{KS56}. That is a separate geometric selection.","Exact field elimination does not solve the integer parity problem by changing its coefficient domain. Over \\(\\Q\\), \\(2m=1\\) has the solution \\(1/2\\). Over \\(\\Z\\), it has none. The rational interface rejects floating-point inputs rather than silently treating an approximate decimal as exact arithmetic. Its finite tests establish the tested implementation behavior, not an arbitrary-input Lean refinement theorem.","its view equations to its obstruction, its ambiguity, and its numerical repair. The following calculations instantiate the preceding principles. They do not introduce a second, silently enlarged formal theory.}\n\\section{A closed signal loop}\\label{sec:signal-loop}\n\\subsection{Prescribed differences as affine view transitions}\nConsider four scalar views arranged around a cycle. Their values are \\(x_0,x_1,x_2,x_3\\), and the prescribed differences are \\(r_0,r_1,r_2,r_3\\). Indices are taken modulo four. The compatibility equations are\n\\begin{equation}\\label{eq:cycle-difference}\n x_{i+1}-x_i=r_i.\n\\end{equation}\nWork over \\(\\Q\\) or \\(\\R\\). These equations can be presented as a strict set-valued view system: an oriented edge has pullback \\(z\\mapsto z-r_i\\), and a path has the composite affine pullback. Use the category of paths, retaining the full loop as a path rather than imposing that it is an identity arrow in advance.","nd its numerical repair. The following calculations instantiate the preceding principles. They do not introduce a second, silently enlarged formal theory.}\n\\section{A closed signal loop}\\label{sec:signal-loop}\n\\subsection{Prescribed differences as affine view transitions}\nConsider four scalar views arranged around a cycle. Their values are \\(x_0,x_1,x_2,x_3\\), and the prescribed differences are \\(r_0,r_1,r_2,r_3\\). Indices are taken modulo four. The compatibility equations are\n\\begin{equation}\\label{eq:cycle-difference}\n x_{i+1}-x_i=r_i.\n\\end{equation}\nWork over \\(\\Q\\) or \\(\\R\\). These equations can be presented as a strict set-valued view system: an oriented edge has pullback \\(z\\mapsto z-r_i\\), and a path has the composite affine pullback. Use the category of paths, retaining the full loop as a path rather than imposing that it is an identity arrow in advance.","not introduce a second, silently enlarged formal theory.}\n\\section{A closed signal loop}\\label{sec:signal-loop}\n\\subsection{Prescribed differences as affine view transitions}\nConsider four scalar views arranged around a cycle. Their values are \\(x_0,x_1,x_2,x_3\\), and the prescribed differences are \\(r_0,r_1,r_2,r_3\\). Indices are taken modulo four. The compatibility equations are\n\\begin{equation}\\label{eq:cycle-difference}\n x_{i+1}-x_i=r_i.\n\\end{equation}\nWork over \\(\\Q\\) or \\(\\R\\). These equations can be presented as a strict set-valued view system: an oriented edge has pullback \\(z\\mapsto z-r_i\\), and a path has the composite affine pullback. Use the category of paths, retaining the full loop as a path rather than imposing that it is an identity arrow in advance.","ed signal loop}\\label{sec:signal-loop}\n\\subsection{Prescribed differences as affine view transitions}\nConsider four scalar views arranged around a cycle. Their values are \\(x_0,x_1,x_2,x_3\\), and the prescribed differences are \\(r_0,r_1,r_2,r_3\\). Indices are taken modulo four. The compatibility equations are\n\\begin{equation}\\label{eq:cycle-difference}\n x_{i+1}-x_i=r_i.\n\\end{equation}\nWork over \\(\\Q\\) or \\(\\R\\). These equations can be presented as a strict set-valued view system: an oriented edge has pullback \\(z\\mapsto z-r_i\\), and a path has the composite affine pullback. Use the category of paths, retaining the full loop as a path rather than imposing that it is an identity arrow in advance.","loop}\\label{sec:signal-loop}\n\\subsection{Prescribed differences as affine view transitions}\nConsider four scalar views arranged around a cycle. Their values are \\(x_0,x_1,x_2,x_3\\), and the prescribed differences are \\(r_0,r_1,r_2,r_3\\). Indices are taken modulo four. The compatibility equations are\n\\begin{equation}\\label{eq:cycle-difference}\n x_{i+1}-x_i=r_i.\n\\end{equation}\nWork over \\(\\Q\\) or \\(\\R\\). These equations can be presented as a strict set-valued view system: an oriented edge has pullback \\(z\\mapsto z-r_i\\), and a path has the composite affine pullback. Use the category of paths, retaining the full loop as a path rather than imposing that it is an identity arrow in advance.","calar views arranged around a cycle. Their values are \\(x_0,x_1,x_2,x_3\\), and the prescribed differences are \\(r_0,r_1,r_2,r_3\\). Indices are taken modulo four. The compatibility equations are\n\\begin{equation}\\label{eq:cycle-difference}\n x_{i+1}-x_i=r_i.\n\\end{equation}\nWork over \\(\\Q\\) or \\(\\R\\). These equations can be presented as a strict set-valued view system: an oriented edge has pullback \\(z\\mapsto z-r_i\\), and a path has the composite affine pullback. Use the category of paths, retaining the full loop as a path rather than imposing that it is an identity arrow in advance.","Every edge map is bijective. Around the cycle their composite is translation by \\(-\\sum_i r_i\\). A global realization must be fixed by that translation. Thus the example makes local invertibility and global compatibility visibly different without needing a noncommutative group.","\\subsection{The complete solvability calculation}\nSumming \\eqref{eq:cycle-difference} gives the necessary condition\n\\begin{equation}\\label{eq:cycle-obstruction}\n r_0+r_1+r_2+r_3=0.\n\\end{equation}\nIt is sufficient as well. Given an anchor \\(a\\), set\n\\[\n x_0=a,\\quad x_1=a+r_0,\\quad\n x_2=a+r_0+r_1,\\quad x_3=a+r_0+r_1+r_2.\n\\]\nThe first three equations hold by construction, and the last is exactly \\eqref{eq:cycle-obstruction}. Every solution has this form because the first three equations recursively determine it from \\(x_0\\). Therefore a compatible residual has a one-dimensional family of primitives, with constant shifts as its ambiguity. Prescribing the anchor removes that ambiguity.","n{equation}\\label{eq:cycle-obstruction}\n r_0+r_1+r_2+r_3=0.\n\\end{equation}\nIt is sufficient as well. Given an anchor \\(a\\), set\n\\[\n x_0=a,\\quad x_1=a+r_0,\\quad\n x_2=a+r_0+r_1,\\quad x_3=a+r_0+r_1+r_2.\n\\]\nThe first three equations hold by construction, and the last is exactly \\eqref{eq:cycle-obstruction}. Every solution has this form because the first three equations recursively determine it from \\(x_0\\). Therefore a compatible residual has a one-dimensional family of primitives, with constant shifts as its ambiguity. Prescribing the anchor removes that ambiguity.","In matrix form, the difference operator is\n\\[\n D=\\begin{pmatrix}\n -1&1&0&0\\\\\n 0&-1&1&0\\\\\n 0&0&-1&1\\\\\n 1&0&0&-1\n \\end{pmatrix},\\qquad\n \\ker D=\\operatorname{span}\\{(1,1,1,1)\\}.\n\\]\nThe explicit construction proves\n\\[\n \\ran D=\\ker\\ell,\\qquad \\ell(r)=\\sum_i r_i.\n\\]\nThus \\(\\ell\\) is a complete feasibility test for this prescribed-difference problem. It annihilates every actual difference family, but takes a nonzero value on an incompatible one. This is the finite obstruction-certificate pattern of Proposition~\\ref{KS15} and Proposition~\\ref{KS57} in a fully visible model.","1&0&0&-1\n \\end{pmatrix},\\qquad\n \\ker D=\\operatorname{span}\\{(1,1,1,1)\\}.\n\\]\nThe explicit construction proves\n\\[\n \\ran D=\\ker\\ell,\\qquad \\ell(r)=\\sum_i r_i.\n\\]\nThus \\(\\ell\\) is a complete feasibility test for this prescribed-difference problem. It annihilates every actual difference family, but takes a nonzero value on an incompatible one. This is the finite obstruction-certificate pattern of Proposition~\\ref{KS15} and Proposition~\\ref{KS57} in a fully visible model.","\\subsection{Repairing an incompatible residual is a new problem}\nTake\n\\[\n r=(1,0,0,0).\n\\]\nEach of the first three edge equations can be satisfied, but the full sum is one. There is no coherent closed-loop signal with exactly these four differences. Computing a long list of partial values does not remove the missing wraparound equation.","To repair the residual, now regard the vector of edge data as a candidate in its own right. Its defect is \\(\\ell(r)\\). The exact subspace is the zero-sum hyperplane. With the ordinary Euclidean norm and no protected edge coordinates, the nearest compatible residual is\n\\[\n \\widehat r=r-\\frac{\\ell(r)}4\\mathbf1\n =\\left(\\frac34,-\\frac14,-\\frac14,-\\frac14\\right).\n\\]\nThe correction is perpendicular to the hyperplane, so the projection theorem proves it is the unique nearest vector there. Its squared size is \\(1/4\\). This is a repair of the edge data, not a solution of the original incompatible equations with those data unchanged.","ell(r)\\). The exact subspace is the zero-sum hyperplane. With the ordinary Euclidean norm and no protected edge coordinates, the nearest compatible residual is\n\\[\n \\widehat r=r-\\frac{\\ell(r)}4\\mathbf1\n =\\left(\\frac34,-\\frac14,-\\frac14,-\\frac14\\right).\n\\]\nThe correction is perpendicular to the hyperplane, so the projection theorem proves it is the unique nearest vector there. Its squared size is \\(1/4\\). This is a repair of the edge data, not a solution of the original incompatible equations with those data unchanged.","Anchor the repaired signal at zero. Its primitive is\n\\begin{equation}\\label{eq:cycle-anchor-primitive}\n \\widehat x=\\left(0,\\frac34,\\frac12,\\frac14\\right).\n\\end{equation}\nThe next partial sum is zero, so it closes. If instead the desired representative has mean zero, subtract the mean \\(3/8\\):\n\\[\n \\widehat x_{\\rm mean}\n =\\left(-\\frac38,\\frac38,\\frac18,-\\frac18\\right).\n\\]\nBoth primitives realize exactly \\(\\widehat r\\). They differ by an element of \\(\\ker D\\). The anchor condition and the mean condition are two different selections from the same affine solution fiber.","the repaired signal at zero. Its primitive is\n\\begin{equation}\\label{eq:cycle-anchor-primitive}\n \\widehat x=\\left(0,\\frac34,\\frac12,\\frac14\\right).\n\\end{equation}\nThe next partial sum is zero, so it closes. If instead the desired representative has mean zero, subtract the mean \\(3/8\\):\n\\[\n \\widehat x_{\\rm mean}\n =\\left(-\\frac38,\\frac38,\\frac18,-\\frac18\\right).\n\\]\nBoth primitives realize exactly \\(\\widehat r\\). They differ by an element of \\(\\ker D\\). The anchor condition and the mean condition are two different selections from the same affine solution fiber.","\\begin{equation}\\label{eq:cycle-anchor-primitive}\n \\widehat x=\\left(0,\\frac34,\\frac12,\\frac14\\right).\n\\end{equation}\nThe next partial sum is zero, so it closes. If instead the desired representative has mean zero, subtract the mean \\(3/8\\):\n\\[\n \\widehat x_{\\rm mean}\n =\\left(-\\frac38,\\frac38,\\frac18,-\\frac18\\right).\n\\]\nBoth primitives realize exactly \\(\\widehat r\\). They differ by an element of \\(\\ker D\\). The anchor condition and the mean condition are two different selections from the same affine solution fiber.","\\subsection{Protecting three edges forces a different repair}\nSuppose the first three values \\((r_0,r_1,r_2)=(1,0,0)\\) are required to remain fixed. The admissible corrections then affect only the final edge. Compatibility forces\n\\[\n \\widetilde r=(1,0,0,-1),\\qquad\n \\widetilde x=(0,1,1,1)\n\\]\nfor anchor zero. This is the unique repair preserving those three edges. Its squared correction size is one rather than \\(1/4\\). The larger error is not a failure of nearest repair: the feasible set has changed.","otecting three edges forces a different repair}\nSuppose the first three values \\((r_0,r_1,r_2)=(1,0,0)\\) are required to remain fixed. The admissible corrections then affect only the final edge. Compatibility forces\n\\[\n \\widetilde r=(1,0,0,-1),\\qquad\n \\widetilde x=(0,1,1,1)\n\\]\nfor anchor zero. This is the unique repair preserving those three edges. Its squared correction size is one rather than \\(1/4\\). The larger error is not a failure of nearest repair: the feasible set has changed.","Here the general protected construction is completely explicit. With candidate space \\(E=\\R^4\\), defect \\(d=\\ell\\), and protected observable \\(p(r)=(r_0,r_1,r_2)\\), the allowed direction is the last coordinate axis. The restricted operator \\(d|_{\\ker p}\\) has gain one. Without protection, the least gain for distance to \\(\\ker\\ell\\) is \\(1/2\\), since \\(\\norm{\\ell}=2\\). The different constants measure different correction spaces.","\\subsection{Where the packet and transfer viewpoints enter}\nThe raw edge family contains information about four transitions. Its zero-sum constraint links all four. Summing a block of differences transfers information to a coarser transition, while retaining the differences within the block preserves the detail that the sum omits. For example,\n\\[\n r_0+r_1=x_2-x_0\n\\]\ncontains the two-step transition but does not determine \\(r_0\\) and \\(r_1\\) separately. Replacing \\((r_0,r_1)\\) by \\((r_0+t,r_1-t)\\) leaves that coarse transition unchanged.","t and transfer viewpoints enter}\nThe raw edge family contains information about four transitions. Its zero-sum constraint links all four. Summing a block of differences transfers information to a coarser transition, while retaining the differences within the block preserves the detail that the sum omits. For example,\n\\[\n r_0+r_1=x_2-x_0\n\\]\ncontains the two-step transition but does not determine \\(r_0\\) and \\(r_1\\) separately. Replacing \\((r_0,r_1)\\) by \\((r_0+t,r_1-t)\\) leaves that coarse transition unchanged.","er viewpoints enter}\nThe raw edge family contains information about four transitions. Its zero-sum constraint links all four. Summing a block of differences transfers information to a coarser transition, while retaining the differences within the block preserves the detail that the sum omits. For example,\n\\[\n r_0+r_1=x_2-x_0\n\\]\ncontains the two-step transition but does not determine \\(r_0\\) and \\(r_1\\) separately. Replacing \\((r_0,r_1)\\) by \\((r_0+t,r_1-t)\\) leaves that coarse transition unchanged.","dge family contains information about four transitions. Its zero-sum constraint links all four. Summing a block of differences transfers information to a coarser transition, while retaining the differences within the block preserves the detail that the sum omits. For example,\n\\[\n r_0+r_1=x_2-x_0\n\\]\ncontains the two-step transition but does not determine \\(r_0\\) and \\(r_1\\) separately. Replacing \\((r_0,r_1)\\) by \\((r_0+t,r_1-t)\\) leaves that coarse transition unchanged.","ns information about four transitions. Its zero-sum constraint links all four. Summing a block of differences transfers information to a coarser transition, while retaining the differences within the block preserves the detail that the sum omits. For example,\n\\[\n r_0+r_1=x_2-x_0\n\\]\ncontains the two-step transition but does not determine \\(r_0\\) and \\(r_1\\) separately. Replacing \\((r_0,r_1)\\) by \\((r_0+t,r_1-t)\\) leaves that coarse transition unchanged.","\\subsection{The connection and its exact families}\nLet the fiber at each of three vertices be \\(\\R^2\\). Use identity transport along the route \\(0\\to1\\to2\\), and let the closing edge carry the reflection\n\\[\n J=\\begin{pmatrix}1&0","\\subsection{The connection and its exact families}\nLet the fiber at each of three vertices be \\(\\R^2\\). Use identity transport along the route \\(0\\to1\\to2\\), and let the closing edge carry the reflection\n\\[\n J=\\begin{pmatrix}1&0\\\\0&-1\\end{pmatrix}.\n\\]\nFor one chosen orientation of each edge, the residuals are\n\\[\n r_{01}=x_1-x_0,\\qquad r_{12}=x_2-x_1,\\qquad\n r_{20}=Jx_0-x_2.\n\\]\nThe reverse transports are inverses, so this defines an inverse-consistent isometric connection. Exactness first forces \\(x_1=x_0\\) and \\(x_2=x_0\\), and the remaining edge then forces \\(Jx_0=x_0\\). Hence every exact family is\n\\begin{equation}\\label{eq:reflection-exact}\n ((a,0),(a,0),(a,0)),\\qquad a\\in\\R.\n\\end{equation}\nThe loop is not the identity on the whole fiber. Its fixed space is the first coordinate axis. Every anchor on that axis extends uniquely, and every anchor with nonzero second coordinate is obstructed if its full value is to be preserved.","0&-1\\end{pmatrix}.\n\\]\nFor one chosen orientation of each edge, the residuals are\n\\[\n r_{01}=x_1-x_0,\\qquad r_{12}=x_2-x_1,\\qquad\n r_{20}=Jx_0-x_2.\n\\]\nThe reverse transports are inverses, so this defines an inverse-consistent isometric connection. Exactness first forces \\(x_1=x_0\\) and \\(x_2=x_0\\), and the remaining edge then forces \\(Jx_0=x_0\\). Hence every exact family is\n\\begin{equation}\\label{eq:reflection-exact}\n ((a,0),(a,0),(a,0)),\\qquad a\\in\\R.\n\\end{equation}\nThe loop is not the identity on the whole fiber. Its fixed space is the first coordinate axis. Every anchor on that axis extends uniquely, and every anchor with nonzero second coordinate is obstructed if its full value is to be preserved.","The three-edge residual also gives a direct telescope:\n\\[\n r_{01}+r_{12}+r_{20}=(J-I)x_0.\n\\]\nIts second component is \\(-2(x_0)_2\\). In this chosen route normalization, the loop residual determines the transverse anchor component. It does not constrain the first component, which is the exact freedom seen in \\eqref{eq:reflection-exact}.","\\subsection{Three different questions for the same raw data}\nTake\n\\[\n x_0=(2,1),\\qquad x_1=(4,-1),\\qquad x_2=(3,2).\n\\]\nThe measured oriented residuals are\n\\[\n r_{01}=(2,-2),\\qquad r_{12}=(-1,3),\\qquad r_{20}=(-1,-3).\n\\]\nPreserving the entire original anchor is impossible, because \\((2,1)\\) is not fixed by \\(J\\). The earlier obstruction criterion therefore gives a negative answer to that particular preservation problem.","Preserving only the feasible projection of the anchor gives a different answer. Orthogonally project \\((2,1)\\) to \\((2,0)\\), then reconstruct along the tree. The result is\n\\[\n P_{\\rm anchor}x=((2,0),(2,0),(2,0)).\n\\]\nIt is the unique exact family preserving the first component of the original anchor. It does not preserve its second component, and makes no claim to do so.","For global nearest-point repair, declare the Euclidean product norm on \\((\\R^2)^3\\). The closest member of \\eqref{eq:reflection-exact} uses the mean of the three first coordinates, namely three:\n\\[\n P_{\\rm nearest}x=((3,0),(3,0),(3,0)).\n\\]\nTo verify this without appealing only to a general theorem, write the squared error at parameter \\(a\\):\n\\[\n (2-a)^2+(4-a)^2+(3-a)^2+1^2+(-1)^2+2^2\n =3(a-3)^2+8.\n\\]\nIt is minimized uniquely at \\(a=3\\). The anchor-based answer has squared error eleven. The globally nearest answer has squared error eight. Their difference has squared norm three, giving the expected Pythagorean identity \\(11=8+3\\).","eclare the Euclidean product norm on \\((\\R^2)^3\\). The closest member of \\eqref{eq:reflection-exact} uses the mean of the three first coordinates, namely three:\n\\[\n P_{\\rm nearest}x=((3,0),(3,0),(3,0)).\n\\]\nTo verify this without appealing only to a general theorem, write the squared error at parameter \\(a\\):\n\\[\n (2-a)^2+(4-a)^2+(3-a)^2+1^2+(-1)^2+2^2\n =3(a-3)^2+8.\n\\]\nIt is minimized uniquely at \\(a=3\\). The anchor-based answer has squared error eleven. The globally nearest answer has squared error eight. Their difference has squared norm three, giving the expected Pythagorean identity \\(11=8+3\\).",",(3,0)).\n\\]\nTo verify this without appealing only to a general theorem, write the squared error at parameter \\(a\\):\n\\[\n (2-a)^2+(4-a)^2+(3-a)^2+1^2+(-1)^2+2^2\n =3(a-3)^2+8.\n\\]\nIt is minimized uniquely at \\(a=3\\). The anchor-based answer has squared error eleven. The globally nearest answer has squared error eight. Their difference has squared norm three, giving the expected Pythagorean identity \\(11=8+3\\).","\\subsection{An exact conditioning calculation}\nFor this paragraph use one orientation per edge and the Euclidean norm on the six residual coordinates. Group candidate coordinates by fiber direction. The two scalar blocks of the defect operator are\n\\[\n A=\\begin{pmatrix}-1&1&0\\\\0&-1&1\\\\1&0&-1\\end{pmatrix},\\qquad\n B=\\begin{pmatrix}-1&1&0\\\\0&-1&1\\\\-1&0&-1\\end{pmatrix}.\n\\]\nThe ordinary cycle block satisfies\n\\[\n A^{\\mathsf T}A=3I-\\mathbf1\\mathbf1^{\\mathsf T}.\n\\]\nIt is zero on the constant line and equals \\(3I\\) on its orthogonal complement. The reflected block satisfies\n\\[\n B^{\\mathsf T}B=I+vv^{\\mathsf T},\\qquad v=(1,-1,1)^{\\mathsf T}.\n\\]\nSince \\(\\norm v^2=3\\), this matrix has eigenvalue four in the direction of \\(v\\) and eigenvalue one on its orthogonal complement. In particular it is positive definite.","-1&0&-1\\end{pmatrix}.\n\\]\nThe ordinary cycle block satisfies\n\\[\n A^{\\mathsf T}A=3I-\\mathbf1\\mathbf1^{\\mathsf T}.\n\\]\nIt is zero on the constant line and equals \\(3I\\) on its orthogonal complement. The reflected block satisfies\n\\[\n B^{\\mathsf T}B=I+vv^{\\mathsf T},\\qquad v=(1,-1,1)^{\\mathsf T}.\n\\]\nSince \\(\\norm v^2=3\\), this matrix has eigenvalue four in the direction of \\(v\\) and eigenvalue one on its orthogonal complement. In particular it is positive definite.","These two identities show that the least nonnegative gain for the full defect-to-distance estimate in this Euclidean model is one. The lower bound is attained, for example, on a candidate with zero first components and second-component vector \\((1,1,0)\\), which lies in \\(v^\\perp\\). The squared input distance and squared residual are both two.","For the displayed raw candidate, the squared residual is\n\\[\n (2^2+(-2)^2)+((-1)^2+3^2)+((-1)^2+(-3)^2)=28.\n\\]\nIts squared nearest-point error is eight, consistent with the estimate. The certificate is universal for this fixed matrix because the Gram identities hold for all vectors, not because that particular example passed.","If both orientations are retained with a Euclidean sum-of-squares residual norm, each isometric edge contributes its squared residual norm twice. The residual norm is multiplied by \\(\\sqrt2\\), and the least gain becomes \\(1/\\sqrt2\\). With the maximum residual norm used in the general graph theorem, that duplication does not change the maximum. This is a concrete illustration of why the norm convention must travel with a certificate.","midrule\\endhead\n\\(\\V,F_v,f^*\\) & A declared view category, its fibers, and contravariant restrictions.","\\(\\Gamma(F)\\) & The strict set of coherent global families.","\\(\\Gamma_U(F),\\Ext_U(s)\\) & Partial sections on a full patch and their global extension fiber.","\\(\\delta,\\partial,d,D\\) & Respectively the view differential, tower differential, a model residual, and a Hilbert defect operator. Each target is declared where used.","\\(p,p_U,p_L\\) & A protected observable, restriction to prescribed views, and an innovation observable depending on the splitting.","\\(\\mathfrak o_U,H_U\\) & The existence obstruction and the group of invisible coherent changes. Additive and nonabelian conventions are stated separately.","\\(C_j,J_j,Q_N,S_N\\) & Coarsener, chosen lift, restored coarse remainder, and retained finite reconstruction.","\\(c_f,a_v\\) & Coherent nonabelian transition labels and gauge values.","\\(\\mathfrak K_G,J_{\\rm comp}\\) & The cocycle patch groupoid and its composition-support topology.","\\(\\mathcal E_b(R)\\) & Realizations equipped with an actual identification of their boundary with \\(b\\).","\\(T_i,K_{ij},R_p\\) & Transport to the root, the corresponding root loop operator, and reconstruction from the root.","\\(\\mathcal H,q,P\\) & Root residual, an orthogonal kernel projector, and the explicitly defined model repair. A graph repair need not be orthogonal in the global maximum norm.","\\(\\pi_{W,U},H_U\\) & The map forgetting part of the prescription and the corresponding invisible coherent ambiguity.","\\(\\phi,\\operatorname{EventualHom}\\) & A monotone source schedule and transport classes modulo eventual equality.","\\(\\chi_A,M=SA-BC\\) & Split exact/residual coordinates and the mismatch between chosen repair lifts.","\\(\\omega,Q_n,C_n,b_n\\) & A residual, its prefix-product rate, movement coefficient, and summable step budget. Here $Q_n$ is not the dyadic remainder operator $Q_N$.","\\(\\nu_B,d_W,d_W^{\\mathrm{sym}}\\) & Residual floor, extended directed transport distance, and its symmetric maximum.","\\(c(x,y),\\kappa(x,y)\\) & The two-bit pair phase and its antisymmetric scalar-gauge invariant.","\\(H,T,r\\) & In the budget chapters, $H=\\ker p$ with its inherited Hilbert metric, $T=d|_H$, and $r=df$ for the true raw residual. Measured data are declared separately.","\\(\\beta_{T,r}(M),a_\\alpha\\) & Minimum residual at correction budget $M$, and the quadratic regularized correction.","\\(C,b,q\\) & Closure of the correction image, its projection of the residual, and the irreducible residual floor.","\\(v,g,e,\\mathcal D_\\alpha\\) & Fit residual, Hilbert normal residual, optimizer error, and regularized dual objective in the solver chapters.","\\(B,G,q\\) & Kernel basis, its Gram matrix, and the coordinate normal covector. This $q$ is not the scalar residual floor used earlier.","\\(\\simeq,\\simeq_L\\) & A specified equivalence, and a continuous linear equivalence. Neither notation alone asserts an isometry.","\\subsection*{1. A stable limit does not make every seed persistent}\nAt every level take \\(X_n=\\{0,1\\}\\), with each bonding map sending both elements to zero. Find the inverse limit and the image of evaluation at every level. Does nonemptiness of every finite fiber make the seed one extendable?","\\textbf{Solution.} For a coherent family, \\(x_n=\\rho_n(x_{n+1})=0\\) for every \\(n\\). The all-zero family is coherent and is therefore the only point. Every evaluation image is \\(\\{0\\}\\). Equivalently, the image of restriction from any strictly later level is already \\(\\{0\\}\\), so the persistent image is \\(\\{0\\}\\). The seed one never extends. This is compatible with Proposition~\\ref{KS58}, which proves nonemptiness and characterizes the persistent image, not surjectivity of every evaluation.","\\subsection*{2. A residual chart can change without changing the defect}\nLet \\(d:\\R^2\\to\\R\\) be \\(d(u,r)=r\\). For \\(a\\in\\R\\), let \\(A_a(r)=(ar,r)\\). Compute the exact repair \\(P_a\\), compare it with \\(P_b\\), and decide when they commute.","\\textbf{Solution.} Since \\(dA_a=I\\),\n\\[\n P_a(u,r)=(u-ar,0),\\qquad\n P_b(u,r)-P_a(u,r)=((a-b)r,0).\n\\]\nBoth fix the horizontal exact axis. Thus \\(P_aP_b=P_b\\) and \\(P_bP_a=P_a\\). They commute exactly when they are equal, which occurs exactly when \\(a=b\\), as can be checked at input \\((0,1)\\). Every repair has zero residual. The different answers reflect the different right inverses, as in Theorem~\\ref{KS65}.","\\subsection*{3. Pairwise invertibility does not supply a triple law}\nLet \\(t\\) be a nonidentity transposition in \\(S_3\\). Suppose three proposed overlap comparisons require frames \\(b_0,b_1,b_2\\) satisfying\n\\[\n b_1=tb_0,\\qquad b_2=tb_1,\\qquad b_2=tb_0.\n\\]\nEvery comparison is individually invertible. Can the three frames exist?","\\textbf{Solution.} The first two equations give \\(b_2=t^2b_0=b_0\\). The third then gives \\(b_0=tb_0\\), and right cancellation gives \\(t=1\\), contrary to the assumption. Invertibility of every proposed overlap map does not prove the composition law needed for descent. Theorem~\\ref{KS34} applies to data satisfying that law, not to arbitrary lists of pairwise isomorphisms.","\\subsection*{4. Zero noisy output is not a certificate of reference exactness}\nTake \\(D(a,b)=a\\), \\(\\widetilde D(a,b)=a+\\tfrac1{10}b\\), and input \\(x=(-10,100)\\). The measured operator returns zero exactly. Find the distance to the reference kernel and evaluate the reference uncertainty bound.","\\textbf{Solution.} The reference kernel is the vertical axis, so the distance is ten. Its residual gain is one, and the operator error norm is \\(1/10\\). With \\(y=0\\) and \\(\\eta=0\\), Proposition~\\ref{KS69} gives\n\\[\n \\dist(x,\\ker D)\\le\\frac1{10}\\sqrt{10^2+100^2}=\\sqrt{101}.\n\\]\nThe true distance ten is below this bound. A formula omitting the operator-error term would instead claim a zero distance, which is false. The input is exact for the measured model and not for the reference one.","\\subsection*{5. A prefix computation can hide a wraparound defect}\nOn a four-cycle, take proposed differences \\((2,-1,3,-3)\\) and anchor zero. Integrate the first three edges. Which equation fails, and what changes when only the final edge is allowed to move?","\\textbf{Solution.} The first three edges give values \\((0,2,1,4)\\). Their required last difference is \\(x_0-x_3=-4\\), not \\(-3\\). The total prescribed sum is one, so the original loop has no realization. Keeping the first three edges forces the unique correction \\((2,-1,3,-4)\\), whose sum is zero. This is a protected repair of the edge data. It is not a realization of the original data without modification.","\\subsection*{6. Residual decay can hide unprotected drift}\nIn maximum-norm $\\R^3$, use $p(s,u,r)=s$ and $d(s,u,r)=r$. Compare the paths $(1,0,(n+1)^{-1})$ and $(1,H_n,(n+1)^{-1})$, where $H_n=\\sum_{j=1}^{n}1/j$. Which hypotheses distinguish them?","\\textbf{Solution.} Both preserve $p=1$ and have the same residual ratios $(n+1)/(n+2)$. The first moves by $1/((n+1)(n+2))$ and converges to $(1,0,0)$. The second moves by $1/(n+1)$ and has an unbounded second coordinate. In the variable-rate theorem, the first has $C_n=1/(n+2)$ and a summable weighted budget. The second has $C_n=1$, making that budget harmonic and divergent. The difference is movement, not residual decay or semantic preservation.","\\subsection*{7. How a finite observation changes the apparent rate}\nLet $\\eta_n=1/(n+1)$. Find its profile after sampling at $\\phi(k)=2^k-1$. Does the new profile prove exponential convergence in the original index?","\\textbf{Solution.} The sampled value is $\\eta_{\\phi(k)}=2^{-k}$. Recovering the original index gives $k=\\log_2(n+1)$ on the sampled stages, so the value is still $1/(n+1)$. The schedule preserves vanishing but changes the meaning of the horizontal axis. No faster original process has been proved. Proposition~\\ref{KS75} is a convergence equivalence under a monotone cofinal schedule, not a rate-invariance theorem.","\\subsection*{8. What does a zero floor fail to certify?}\nFor the diagonal model of Section~\\ref{sec:escape}, can an exact protected answer exist in $\\ell^2\\oplus\\R$? Does a coherent family of exact finite prefixes exist? Explain why the answers differ without contradicting finite-fiber existence.","\\textbf{Solution.} Exactness with protected value one forces every sequence coordinate to equal minus one. This is not square summable, so there is no protected Hilbert answer. Each finite prefix nevertheless has the unique solution $(-1,\\ldots,-1;1)$, and these solutions are coherent under truncation. Their inverse limit contains the all-minus-one sequence in the full product. The finite-fiber theorem supplies that coherent point. It does not prove that the point is in the image of the square-summable semantic space. The correction image is dense but does not contain the raw residual, so its floor is zero and its algebraic class is nonzero.","\\subsection*{9. A phase witness without enumerating every gauge}\nFor $c(x,y)=x_2y_1$ on $\\mathbf F_2^2$, show that no scalar coboundary equals $c$. Then compute the square of the lifted label $(0,e_1+e_2)$ in the eight-element extension.","\\textbf{Solution.} Scalar coboundaries $a(x)+a(y)-a(x+y)$ are symmetric, but $c(e_1,e_2)=0$ and $c(e_2,e_1)=1$. Thus no scalar coboundary equals $c$. For $x=e_1+e_2$, $c(x,x)=1$, so $(0,x)^2=(1,0)$ and its fourth power is the identity. The corresponding matrix is $XZ$, whose square is $-I$. The additional phase bit records precisely the sign lost by projecting back to the four-label group.","\\subsection*{10. The finite energy identity}\nIn $\\R^2$, first prescribe $a_1+a_2=1$, then also prescribe $a_1=3/4$. Find the two minimum norm corrections and check the exact nested-affine identity.","\\textbf{Solution.} The first correction is $(1/2,1/2)$, with energy $1/2$. The second is $(3/4,1/4)$, with energy $5/8$. Their difference has squared norm $1/8=5/8-1/2$. The identity uses the affine nature of both constraint sets. For the nested closed convex sets $[1,\\infty)$ and $[2,\\infty)$ in $\\R$, the minimum points instead have squared difference one and energy difference three.","\\subsection*{11. A basis is not automatically an isometry}\nIn Euclidean $\\R^2$, let $p(a)=a_1-2a_2$, $d(a)=a_1$, $r=1$, and $\\alpha=1$. Compute the protected regularized correction using the basis $(2,1)$ of the kernel. What changes if the Gram matrix is omitted?","\\textbf{Solution.} The coordinate $c$ gives $a=(2c,c)$, with $S=2$ and $G=5$. The normal equation is $9c=2$, so $a=(4/9,2/9)$. Its objective is $5/9$. Replacing $G$ by one gives $c=2/5$ and $a=(4/5,2/5)$, whose actual ambient objective is $21/25>5/9$. That replacement solves a different metric problem.","\\subsection*{12. A robust negative certificate}\nA nominal scalar system has $T=1$, $r=1$. The operator error is at most $1/10$ and the data error at most $1/20$. Can a correction of norm at most $1/2$ achieve true residual at most $3/10$? What does the same test say at budget two?","\\textbf{Solution.} At budget $1/2$, the nominal optimum is $1/2$. Subtracting $1/20+(1/10)(1/2)$ leaves $2/5>3/10$, so the proposed accuracy is excluded for every admissible true model. At budget two, the lower estimate is negative and clips to zero. It excludes nothing. The disappearance of this certificate is not proof of exact feasibility.","\\subsection*{13. A smaller coordinate residual without a better correction}\nFor the protected example in Section~\\ref{sec:inexact-repair}, keep the same correction $(1/4,1/4)$ but replace the kernel basis $B=(1,1)^{\\mathsf T}$ by $B'=B/100$. Compute the coordinate normal residual and its actual Hilbert norm.","\\textbf{Solution.} In the original basis, $c=1/4$, $G=2$ and $q=3c-1=-1/4$. In the new basis, $c'=25$, $G'=1/5000$ and $q'=-1/400$. The coordinate residual is one hundred times smaller, but\n\\[\n (q')^{\\mathsf T}(G')^{-1}q'=\\frac{5000}{160000}=\\frac1{32}\n =q^{\\mathsf T}G^{-1}q.\n\\]\nThe physical correction and the certificate have not improved. The changed number records a changed basis.","\\subsection*{14. How small must the solver residual be?}\nFor $n\\ge1$, suppose $\\alpha_n=n^{-2}$, the measured-data error is bounded by $n^{-2}$, and the exact-data regularized corrections converge to the desired minimum-norm answer. Does a solver normal-residual tolerance $n^{-2}$ guarantee convergence of the proposals? Give a sufficient replacement.","\\textbf{Solution.} The noise term is at most $1/(2n)$, but the solver term is at most one. Thus the displayed tolerance does not provide the required guarantee. The scalar counterexample $T=0$, $r=0$, $a_n=1$ meets it and does not approach the exact optimizer. Replacing it by $n^{-3}$ makes the solver term at most $1/n$, so the triangle inequality and the assumed vanishing bias prove convergence. This is a sufficient worst-case rule, not a necessary condition for every particular operator.","\\subsection*{15. Exact optimization of measured data is not exact realization}\nLet $d$ be the identity on $\\R$, protect no nonzero observable, take raw input $f=2$, measured residual $r=1$, and $\\alpha=1$. Find the exact measured-data optimizer and compare the fit residual with the actual repaired residual.","\\textbf{Solution.} The objective $(a-1)^2+a^2$ is minimized at $a=1/2$. Its normal residual is zero, so the solver certificate is exact. The measured fit residual is $r-da=1/2$. The actual repaired candidate is $f-a=3/2$, whose model residual is $3/2$. The difference is $df-r=1$. A certificate about solving the measured optimization problem does not remove measurement error or the regularization bias.","[4pt]\nA stable residual-to-distance bound & The required closed range or bounded inverse in the declared normed model. In a protected problem, use the restriction to $\\ker p$. & \\ref{KS43}, \\ref{KS56}","[4pt]\n\\ref{WR02} & Nested affine minimum corrections & Verified\\\\[4pt]\n\\ref{WR03} & Attainment and shape of the budget profile & Verified\\\\[4pt]\n\\ref{WR04} & Sharp budget certificate & Verified\\\\[4pt]\n\\ref{WR05} & Robust budget comparison & Verified\\\\[4pt]\n\\ref{WR06} & Unique regularized correction & Verified\\\\[4pt]\n\\ref{WR07} & Parameter and budget comparison & Verified\\\\[4pt]\n\\ref{WR08} & Regularization limit alternative & Verified\\\\[4pt]\n\\ref{WR09} & Data sensitivity and convergent parameter choice & Verified\\\\[4pt]\n\\ref{WR10} & Infinite diagonal family and certified tails & Verified\\\\[4pt]\n\\ref{WR11} & Sharp diagonal budget asymptotic & Verified\\\\[4pt]\n\\ref{WR12} & Rounded noisy diagonal rate & Verified\\\\[4pt]\n\\ref{WR13} & Normal-residual solver certificate & Verified\\\\[4pt]\n\\ref{WR14} & Regularized primal and dual gap & Verified\\\\[4pt]\n\\ref{WR15} & Coordinate certificate and arbitrary basis change & Verified\\\\[4pt]\n\\ref{WR16} & Consistent inexact stopping rule & Verified\\\\\n\\bottomrule\\end{longtable}\n\n\\subsection{Bounded realization and budget geometry}\\label{app:wr-source-map}\nT…","-1}q.\n\\]\nEvery invertible basis change is allowed. The changed coordinates describe the same correction, the residual transforms as a covector, and the displayed squared norm is invariant. The proof includes the zero-dimensional correction space. It does not replace the Hilbert norm by a coordinate supremum norm or assume that the basis is orthonormal. These metric identities hold for every real $\\alpha$. The optimizer estimates still require positive $\\alpha$.","anged coordinates describe the same correction, the residual transforms as a covector, and the displayed squared norm is invariant. The proof includes the zero-dimensional correction space. It does not replace the Hilbert norm by a coordinate supremum norm or assume that the basis is orthonormal. These metric identities hold for every real $\\alpha$. The optimizer estimates still require positive $\\alpha$.","\\subsection{The infinite diagonal example}\nProposition~\\ref{WR10}, Theorem~\\ref{WR11}, and Corollary~\\ref{WR12} have formal counterparts in the namespace \\path{KomorebiSpace.DiagonalRegularization}. The formal space is the real Hilbert space \\code{lp} at exponent two. Its natural index starts at zero, so formal coordinate $j$ denotes printed coordinate $n=j+1$. The operator, data, corrections, residuals, and hard prefixes are elements or maps of that actual infinite space.","The first pair identifies the universal regularized correction, proves it is the unique global minimizer, equates the infinite squared norms with $A_m$ and $B_m$, and proves all four global bounds and both nonnegative omitted-tail bounds. The asymptotic theorem proves both scalar limits, the budget-profile limit $\\pi/4$ over all real $M$ tending to infinity, and the hard-prefix comparison with constant one. Its budget conclusion is not restricted to the regularization mesh.","The noisy theorem uses the actual attainable datum and its minimum-norm preimage. It retains the literal ceiling in $m=\\lceil\\delta^{-2/3}\\rceil$ and proves both displayed inequalities for every Hilbert perturbation within the stated noise bound. The bias estimate is derived within the model. The inexact rate adds the stated solver tolerance to this same model and parameter choice.","The range obstruction uses actual unit coordinate vectors with residual norm $1/(j+1)$ and applies the general unit-witness theorem. The hard-prefix images converge in Hilbert norm to $w$, placing it in the closure of the range. An exact preimage would have every coordinate equal to one, contradicting square summability. These declarations formalize the concrete example following Theorem~\\ref{KS43}. Exact finite enclosures remain separate illustrations of the infinite results.","\\subsection{Projection onto a closed convex set}\n\\begin{lemma}[Existence, uniqueness, and the variational inequality]\\label{lem:projection-tool}\nLet \\(H\\) be a real Hilbert space, let \\(C\\subseteq H\\) be nonempty, closed, and convex, and fix \\(x\\in H\\). There is a unique \\(c\\in C\\) minimizing \\(\\norm{x-c}\\). It is characterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-","\\subsection{Projection onto a closed convex set}\n\\begin{lemma}[Existence, uniqueness, and the variational inequality]\\label{lem:projection-tool}\nLet \\(H\\) be a real Hilbert space, let \\(C\\subseteq H\\) be nonempty, closed, and convex, and fix \\(x\\in H\\). There is a unique \\(c\\in C\\) minimizing \\(\\norm{x-c}\\). It is characterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4","\\subsection{Projection onto a closed convex set}\n\\begin{lemma}[Existence, uniqueness, and the variational inequality]\\label{lem:projection-tool}\nLet \\(H\\) be a real Hilbert space, let \\(C\\subseteq H\\) be nonempty, closed, and convex, and fix \\(x\\in H\\). There is a unique \\(c\\in C\\) minimizing \\(\\norm{x-c}\\). It is characterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_","\\subsection{Projection onto a closed convex set}\n\\begin{lemma}[Existence, uniqueness, and the variational inequality]\\label{lem:projection-tool}\nLet \\(H\\) be a real Hilbert space, let \\(C\\subseteq H\\) be nonempty, closed, and convex, and fix \\(x\\in H\\). There is a unique \\(c\\in C\\) minimizing \\(\\norm{x-c}\\). It is characterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]","\\subsection{Projection onto a closed convex set}\n\\begin{lemma}[Existence, uniqueness, and the variational inequality]\\label{lem:projection-tool}\nLet \\(H\\) be a real Hilbert space, let \\(C\\subseteq H\\) be nonempty, closed, and convex, and fix \\(x\\in H\\). There is a unique \\(c\\in C\\) minimizing \\(\\norm{x-c}\\). It is characterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to ze","\\subsection{Projection onto a closed convex set}\n\\begin{lemma}[Existence, uniqueness, and the variational inequality]\\label{lem:projection-tool}\nLet \\(H\\) be a real Hilbert space, let \\(C\\subseteq H\\) be nonempty, closed, and convex, and fix \\(x\\in H\\). There is a unique \\(c\\in C\\) minimizing \\(\\norm{x-c}\\). It is characterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\…","t}\n\\begin{lemma}[Existence, uniqueness, and the variational inequality]\\label{lem:projection-tool}\nLet \\(H\\) be a real Hilbert space, let \\(C\\subseteq H\\) be nonempty, closed, and convex, and fix \\(x\\in H\\). There is a unique \\(c\\in C\\) minimizing \\(\\norm{x-c}\\). It is characterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and co…","nequality]\\label{lem:projection-tool}\nLet \\(H\\) be a real Hilbert space, let \\(C\\subseteq H\\) be nonempty, closed, and convex, and fix \\(x\\in H\\). There is a unique \\(c\\in C\\) minimizing \\(\\norm{x-c}\\). It is characterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing seq…","jection-tool}\nLet \\(H\\) be a real Hilbert space, let \\(C\\subseteq H\\) be nonempty, closed, and convex, and fix \\(x\\in H\\). There is a unique \\(c\\in C\\) minimizing \\(\\norm{x-c}\\). It is characterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary cl…","d convex, and fix \\(x\\in H\\). There is a unique \\(c\\in C\\) minimizing \\(\\norm{x-c}\\). It is characterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary closed set need not have been Cauchy. The midpoint was the essential use of convexity.","acterized by\n\\begin{equation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary closed set need not have been Cauchy. The midpoint was the essential use of convexity.","quation}\\label{eq:projection-variational}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary closed set need not have been Cauchy. The midpoint was the essential use of convexity.","}\n \\langle x-c,z-c\\rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary closed set need not have been Cauchy. The midpoint was the essential use of convexity.","rangle\\le0\\qquad(z\\in C).\n\\end{equation}\nIf \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary closed set need not have been Cauchy. The midpoint was the essential use of convexity.","If \\(C=c_0+L\\) is a closed affine subspace with direction space \\(L\\), the condition is \\(x-c\\perp L\\).\n\\end{lemma}\n\\begin{proof}\n\\textbf{Produce a Cauchy minimizing sequence.} Put \\(d=\\inf_{z\\in C}\\norm{x-z}\\). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary closed set need not have been Cauchy. The midpoint was the essential use of convexity.","). Nonemptiness makes this a finite nonnegative real number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary closed set need not have been Cauchy. The midpoint was the essential use of convexity.","l number. Choose \\(c_n\\in C\\) so that \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary closed set need not have been Cauchy. The midpoint was the essential use of convexity.","hat \\(\\norm{x-c_n}\\to d\\). Convexity places \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary closed set need not have been Cauchy. The midpoint was the essential use of convexity.","ces \\((c_n+c_m)/2\\) in \\(C\\). The parallelogram identity gives\n\\[\n \\norm{c_n-c_m}^2\n =2\\norm{x-c_n}^2+2\\norm{x-c_m}^2\n   -4\\left\\|x-\\frac{c_n+c_m}{2}\\right\\|^2\n \\le2\\norm{x-c_n}^2+2\\norm{x-c_m}^2-4d^2.\n\\]\nThe right side tends to zero as both indices tend to infinity. Thus \\((c_n)\\) is Cauchy. Completeness supplies a limit \\(c\\in H\\), closedness puts it in \\(C\\), and continuity of the norm gives \\(\\norm{x-c}=d\\). A minimizing sequence in an arbitrary closed set need not have been Cauchy. The midpoint was the essential use of convexity.","\\textbf{Prove uniqueness.} If \\(c,c'\\in C\\) both attain \\(d\\), the same midpoint inequality gives\n\\[\n \\norm{c-c'}^2\\le2d^2+2d^2-4d^2=0.\n\\]\nTherefore \\(c=c'\\).","\\textbf{Derive the variational inequality.} For \\(z\\in C\\) and \\(0<t\\le1\\), convexity makes \\(c+t(z-c)\\) feasible. Minimality and the inner-product expansion yield\n\\[\n 0\\le\\norm{x-c-t(z-c)}^2-\\norm{x-c}^2\n   =-2t\\langle x-c,z-c\\rangle+t^2\\norm{z-c}^2.\n\\]\nDivide by the positive number \\(t\\) and let \\(t\\downarrow0\\). This gives \\eqref{eq:projection-variational}. Conversely, if that inequality holds, then for every \\(z\\in C\\),\n\\[\n \\norm{x-z}^2\n =\\norm{x-c}^2-2\\langle x-c,z-c\\rangle+\\norm{z-c}^2\n \\ge\\norm{x-c}^2.\n\\]\nIt therefore characterizes the unique minimizer.","variational inequality.} For \\(z\\in C\\) and \\(0<t\\le1\\), convexity makes \\(c+t(z-c)\\) feasible. Minimality and the inner-product expansion yield\n\\[\n 0\\le\\norm{x-c-t(z-c)}^2-\\norm{x-c}^2\n   =-2t\\langle x-c,z-c\\rangle+t^2\\norm{z-c}^2.\n\\]\nDivide by the positive number \\(t\\) and let \\(t\\downarrow0\\). This gives \\eqref{eq:projection-variational}. Conversely, if that inequality holds, then for every \\(z\\in C\\),\n\\[\n \\norm{x-z}^2\n =\\norm{x-c}^2-2\\langle x-c,z-c\\rangle+\\norm{z-c}^2\n \\ge\\norm{x-c}^2.\n\\]\nIt therefore characterizes the unique minimizer.","inequality.} For \\(z\\in C\\) and \\(0<t\\le1\\), convexity makes \\(c+t(z-c)\\) feasible. Minimality and the inner-product expansion yield\n\\[\n 0\\le\\norm{x-c-t(z-c)}^2-\\norm{x-c}^2\n   =-2t\\langle x-c,z-c\\rangle+t^2\\norm{z-c}^2.\n\\]\nDivide by the positive number \\(t\\) and let \\(t\\downarrow0\\). This gives \\eqref{eq:projection-variational}. Conversely, if that inequality holds, then for every \\(z\\in C\\),\n\\[\n \\norm{x-z}^2\n =\\norm{x-c}^2-2\\langle x-c,z-c\\rangle+\\norm{z-c}^2\n \\ge\\norm{x-c}^2.\n\\]\nIt therefore characterizes the unique minimizer.","For an affine subspace, \\(c+h\\) and \\(c-h\\) are feasible for every \\(h\\in L\\). Applying the inequality with both signs gives \\(\\langle x-c,h\\rangle=0\\). Orthogonality conversely implies the same variational inequality. In this case the preceding expansion becomes the Pythagorean equality, not merely an inequality.\n\\end{proof}","For a closed linear subspace \\(L\\), denote its projection by \\(P_L\\). The lemma gives a unique decomposition\n\\[\n x=P_Lx+(x-P_Lx),\\qquad P_Lx\\in L,\\quad x-P_Lx\\in L^\\perp.\n\\]\nAdding two such decompositions, or multiplying one by a scalar, proves linearity by uniqueness. The Pythagorean identity gives \\(\\norm{P_Lx}\\le\\norm{x}\\), so \\(P_L\\) is bounded. It fixes \\(L\\), annihilates \\(L^\\perp\\), and satisfies \\(P_L^2=P_L\\). For \\(x,y\\in H\\), orthogonality gives\n\\[\n \\langle P_Lx,y\\rangle=\\langle P_Lx,P_Ly\\rangle\n =\\langle x,P_Ly\\rangle.\n\\]\nThus \\(P_L\\) is self-adjoint. These properties are conclusions of the construction, not extra data assumed for a proposed projection.","sed linear subspace \\(L\\), denote its projection by \\(P_L\\). The lemma gives a unique decomposition\n\\[\n x=P_Lx+(x-P_Lx),\\qquad P_Lx\\in L,\\quad x-P_Lx\\in L^\\perp.\n\\]\nAdding two such decompositions, or multiplying one by a scalar, proves linearity by uniqueness. The Pythagorean identity gives \\(\\norm{P_Lx}\\le\\norm{x}\\), so \\(P_L\\) is bounded. It fixes \\(L\\), annihilates \\(L^\\perp\\), and satisfies \\(P_L^2=P_L\\). For \\(x,y\\in H\\), orthogonality gives\n\\[\n \\langle P_Lx,y\\rangle=\\langle P_Lx,P_Ly\\rangle\n =\\langle x,P_Ly\\rangle.\n\\]\nThus \\(P_L\\) is self-adjoint. These properties are conclusions of the construction, not extra data assumed for a proposed projection.","\\(L\\), denote its projection by \\(P_L\\). The lemma gives a unique decomposition\n\\[\n x=P_Lx+(x-P_Lx),\\qquad P_Lx\\in L,\\quad x-P_Lx\\in L^\\perp.\n\\]\nAdding two such decompositions, or multiplying one by a scalar, proves linearity by uniqueness. The Pythagorean identity gives \\(\\norm{P_Lx}\\le\\norm{x}\\), so \\(P_L\\) is bounded. It fixes \\(L\\), annihilates \\(L^\\perp\\), and satisfies \\(P_L^2=P_L\\). For \\(x,y\\in H\\), orthogonality gives\n\\[\n \\langle P_Lx,y\\rangle=\\langle P_Lx,P_Ly\\rangle\n =\\langle x,P_Ly\\rangle.\n\\]\nThus \\(P_L\\) is self-adjoint. These properties are conclusions of the construction, not extra data assumed for a proposed projection.","(P_L\\). The lemma gives a unique decomposition\n\\[\n x=P_Lx+(x-P_Lx),\\qquad P_Lx\\in L,\\quad x-P_Lx\\in L^\\perp.\n\\]\nAdding two such decompositions, or multiplying one by a scalar, proves linearity by uniqueness. The Pythagorean identity gives \\(\\norm{P_Lx}\\le\\norm{x}\\), so \\(P_L\\) is bounded. It fixes \\(L\\), annihilates \\(L^\\perp\\), and satisfies \\(P_L^2=P_L\\). For \\(x,y\\in H\\), orthogonality gives\n\\[\n \\langle P_Lx,y\\rangle=\\langle P_Lx,P_Ly\\rangle\n =\\langle x,P_Ly\\rangle.\n\\]\nThus \\(P_L\\) is self-adjoint. These properties are conclusions of the construction, not extra data assumed for a proposed projection.","\\quad x-P_Lx\\in L^\\perp.\n\\]\nAdding two such decompositions, or multiplying one by a scalar, proves linearity by uniqueness. The Pythagorean identity gives \\(\\norm{P_Lx}\\le\\norm{x}\\), so \\(P_L\\) is bounded. It fixes \\(L\\), annihilates \\(L^\\perp\\), and satisfies \\(P_L^2=P_L\\). For \\(x,y\\in H\\), orthogonality gives\n\\[\n \\langle P_Lx,y\\rangle=\\langle P_Lx,P_Ly\\rangle\n =\\langle x,P_Ly\\rangle.\n\\]\nThus \\(P_L\\) is self-adjoint. These properties are conclusions of the construction, not extra data assumed for a proposed projection.","\\subsection{Riesz representation and the adjoint}\n\\begin{lemma}[Representing a bounded functional]\\label{lem:riesz-tool}\nFor every bounded real linear functional \\(\\ell:H\\to\\R\\) on a real Hilbert space, there is a unique \\(u\\in H\\) with\n\\[\n \\ell(h)=\\langle h,u\\rangle\\quad(h\\in H),\\qquad \\norm{\\ell}=\\norm{u}.\n\\]\n\\end{lemma}\n\\begin{proof}\nIf \\(\\ell=0\\), take \\(u=0\\). Otherwise choose \\(x\\) with \\(\\ell(x)\\ne0\\). Boundedness makes \\(L=\\ker\\ell\\) closed. Use Lemma~\\ref{lem:projection-tool} to form \\(v=x-P_Lx\\). Then \\(v\\perp L\\) and \\(\\ell(v)=\\ell(x)\\ne0\\), in particular \\(v\\ne0\\).","and the adjoint}\n\\begin{lemma}[Representing a bounded functional]\\label{lem:riesz-tool}\nFor every bounded real linear functional \\(\\ell:H\\to\\R\\) on a real Hilbert space, there is a unique \\(u\\in H\\) with\n\\[\n \\ell(h)=\\langle h,u\\rangle\\quad(h\\in H),\\qquad \\norm{\\ell}=\\norm{u}.\n\\]\n\\end{lemma}\n\\begin{proof}\nIf \\(\\ell=0\\), take \\(u=0\\). Otherwise choose \\(x\\) with \\(\\ell(x)\\ne0\\). Boundedness makes \\(L=\\ker\\ell\\) closed. Use Lemma~\\ref{lem:projection-tool} to form \\(v=x-P_Lx\\). Then \\(v\\perp L\\) and \\(\\ell(v)=\\ell(x)\\ne0\\), in particular \\(v\\ne0\\).","abel{lem:riesz-tool}\nFor every bounded real linear functional \\(\\ell:H\\to\\R\\) on a real Hilbert space, there is a unique \\(u\\in H\\) with\n\\[\n \\ell(h)=\\langle h,u\\rangle\\quad(h\\in H),\\qquad \\norm{\\ell}=\\norm{u}.\n\\]\n\\end{lemma}\n\\begin{proof}\nIf \\(\\ell=0\\), take \\(u=0\\). Otherwise choose \\(x\\) with \\(\\ell(x)\\ne0\\). Boundedness makes \\(L=\\ker\\ell\\) closed. Use Lemma~\\ref{lem:projection-tool} to form \\(v=x-P_Lx\\). Then \\(v\\perp L\\) and \\(\\ell(v)=\\ell(x)\\ne0\\), in particular \\(v\\ne0\\).","}\nFor every bounded real linear functional \\(\\ell:H\\to\\R\\) on a real Hilbert space, there is a unique \\(u\\in H\\) with\n\\[\n \\ell(h)=\\langle h,u\\rangle\\quad(h\\in H),\\qquad \\norm{\\ell}=\\norm{u}.\n\\]\n\\end{lemma}\n\\begin{proof}\nIf \\(\\ell=0\\), take \\(u=0\\). Otherwise choose \\(x\\) with \\(\\ell(x)\\ne0\\). Boundedness makes \\(L=\\ker\\ell\\) closed. Use Lemma~\\ref{lem:projection-tool} to form \\(v=x-P_Lx\\). Then \\(v\\perp L\\) and \\(\\ell(v)=\\ell(x)\\ne0\\), in particular \\(v\\ne0\\).","ed real linear functional \\(\\ell:H\\to\\R\\) on a real Hilbert space, there is a unique \\(u\\in H\\) with\n\\[\n \\ell(h)=\\langle h,u\\rangle\\quad(h\\in H),\\qquad \\norm{\\ell}=\\norm{u}.\n\\]\n\\end{lemma}\n\\begin{proof}\nIf \\(\\ell=0\\), take \\(u=0\\). Otherwise choose \\(x\\) with \\(\\ell(x)\\ne0\\). Boundedness makes \\(L=\\ker\\ell\\) closed. Use Lemma~\\ref{lem:projection-tool} to form \\(v=x-P_Lx\\). Then \\(v\\perp L\\) and \\(\\ell(v)=\\ell(x)\\ne0\\), in particular \\(v\\ne0\\).","\\) on a real Hilbert space, there is a unique \\(u\\in H\\) with\n\\[\n \\ell(h)=\\langle h,u\\rangle\\quad(h\\in H),\\qquad \\norm{\\ell}=\\norm{u}.\n\\]\n\\end{lemma}\n\\begin{proof}\nIf \\(\\ell=0\\), take \\(u=0\\). Otherwise choose \\(x\\) with \\(\\ell(x)\\ne0\\). Boundedness makes \\(L=\\ker\\ell\\) closed. Use Lemma~\\ref{lem:projection-tool} to form \\(v=x-P_Lx\\). Then \\(v\\perp L\\) and \\(\\ell(v)=\\ell(x)\\ne0\\), in particular \\(v\\ne0\\).","For any \\(h\\in H\\), the vector \\(h-\\ell(h)v/\\ell(v)\\) belongs to \\(L\\). Pairing it with \\(v\\) therefore gives\n\\[\n \\langle h,v\\rangle=\\frac{\\ell(h)}{\\ell(v)}\\norm{v}^2.\n\\]\nConsequently \\(u=\\ell(v)v/\\norm{v}^2\\) represents \\(\\ell\\). If \\(u'\\) also represents it, pairing the difference of the two formulas with \\(h=u-u'\\) gives \\(\\norm{u-u'}^2=0\\), hence uniqueness.","Cauchy--Schwarz gives \\(|\\ell(h)|\\le\\norm{u}\\norm{h}\\), so \\(\\norm{\\ell}\\le\\norm{u}\\). If \\(u\\ne0\\), the unit vector \\(h=u/\\norm{u}\\) gives the reverse inequality. The zero case was already included.\n\\end{proof}","Let \\(T:H\\to R\\) be a bounded linear map between real Hilbert spaces. For fixed \\(z\\in R\\), the functional \\(h\\mapsto\\langle Th,z\\rangle\\) is bounded by \\(\\norm{T}\\norm{z}\\norm{h}\\). The lemma therefore defines a unique vector \\(T^*z\\in H\\) with\n\\[\n \\langle Th,z\\rangle=\\langle h,T^*z\\rangle.\n\\]\nUniqueness of representatives proves that \\(T^*\\) is linear. The same bound proves \\(\\norm{T^*z}\\le\\norm{T}\\norm{z}\\). Conversely,\n\\[\n \\norm{Th}=\\sup_{\\norm{z}\\le1}|\\langle Th,z\\rangle|\n \\le\\norm{T^*}\\norm{h},\n\\]\nwhere the first equality follows by testing \\(z=Th/\\norm{Th}\\) when \\(Th\\ne0\\). Hence \\(\\norm{T^*}=\\norm{T}\\).","linear map between real Hilbert spaces. For fixed \\(z\\in R\\), the functional \\(h\\mapsto\\langle Th,z\\rangle\\) is bounded by \\(\\norm{T}\\norm{z}\\norm{h}\\). The lemma therefore defines a unique vector \\(T^*z\\in H\\) with\n\\[\n \\langle Th,z\\rangle=\\langle h,T^*z\\rangle.\n\\]\nUniqueness of representatives proves that \\(T^*\\) is linear. The same bound proves \\(\\norm{T^*z}\\le\\norm{T}\\norm{z}\\). Conversely,\n\\[\n \\norm{Th}=\\sup_{\\norm{z}\\le1}|\\langle Th,z\\rangle|\n \\le\\norm{T^*}\\norm{h},\n\\]\nwhere the first equality follows by testing \\(z=Th/\\norm{Th}\\) when \\(Th\\ne0\\). Hence \\(\\norm{T^*}=\\norm{T}\\).","\\norm{T}\\norm{z}\\norm{h}\\). The lemma therefore defines a unique vector \\(T^*z\\in H\\) with\n\\[\n \\langle Th,z\\rangle=\\langle h,T^*z\\rangle.\n\\]\nUniqueness of representatives proves that \\(T^*\\) is linear. The same bound proves \\(\\norm{T^*z}\\le\\norm{T}\\norm{z}\\). Conversely,\n\\[\n \\norm{Th}=\\sup_{\\norm{z}\\le1}|\\langle Th,z\\rangle|\n \\le\\norm{T^*}\\norm{h},\n\\]\nwhere the first equality follows by testing \\(z=Th/\\norm{Th}\\) when \\(Th\\ne0\\). Hence \\(\\norm{T^*}=\\norm{T}\\).","\\). The lemma therefore defines a unique vector \\(T^*z\\in H\\) with\n\\[\n \\langle Th,z\\rangle=\\langle h,T^*z\\rangle.\n\\]\nUniqueness of representatives proves that \\(T^*\\) is linear. The same bound proves \\(\\norm{T^*z}\\le\\norm{T}\\norm{z}\\). Conversely,\n\\[\n \\norm{Th}=\\sup_{\\norm{z}\\le1}|\\langle Th,z\\rangle|\n \\le\\norm{T^*}\\norm{h},\n\\]\nwhere the first equality follows by testing \\(z=Th/\\norm{Th}\\) when \\(Th\\ne0\\). Hence \\(\\norm{T^*}=\\norm{T}\\).","efore defines a unique vector \\(T^*z\\in H\\) with\n\\[\n \\langle Th,z\\rangle=\\langle h,T^*z\\rangle.\n\\]\nUniqueness of representatives proves that \\(T^*\\) is linear. The same bound proves \\(\\norm{T^*z}\\le\\norm{T}\\norm{z}\\). Conversely,\n\\[\n \\norm{Th}=\\sup_{\\norm{z}\\le1}|\\langle Th,z\\rangle|\n \\le\\norm{T^*}\\norm{h},\n\\]\nwhere the first equality follows by testing \\(z=Th/\\norm{Th}\\) when \\(Th\\ne0\\). Hence \\(\\norm{T^*}=\\norm{T}\\).","The defining pairing also gives the identity\n\\begin{equation}\\label{eq:adjoint-range-orthogonal}\n (\\ran T)^\\perp=\\ker T^*.\n\\end{equation}\nIndeed, \\(z\\) is orthogonal to every \\(Th\\) exactly when \\(\\langle h,T^*z\\rangle=0\\) for every \\(h\\), which is equivalent to \\(T^*z=0\\). Orthogonal complements do not change on taking closure. The projection decomposition then gives\n\\[\n \\overline{\\ran T}=(\\ker T^*)^\\perp.\n\\]\nThis is why a self-adjoint injective operator has dense range. It does not yet say its range is closed. In the regularized normal-operator proof, the coercive lower bound supplies that separate fact.","\\begin{lemma}[Baire's complete-space principle]\\label{lem:baire-tool}\nA nonempty complete metric space cannot be the union of countably many closed sets with empty interior.\n\\end{lemma}\n\\begin{proof}\nSuppose \\(F_1,F_2,\\ldots\\) are closed with empty interior. Choose a nonempty open ball whose closed ball \\(B_1\\), of positive radius at most \\(1/2\\), avoids \\(F_1\\). Such a choice is possible because the complement of \\(F_1\\) is open and meets every nonempty open set. Recursively choose a closed ball \\(B_{n+1}\\) of positive radius at most \\(2^{-(n+1)}\\), contained in the open ball with the same center and radius as \\(B_n\\), and disjoint from \\(F_{n+1}\\). The intersection of that open ball with the complement of \\(F_{n+1}\\) is nonempty and open, so it contains a sufficiently small closed ball. This also covers the case of isolated points.","s complete-space principle]\\label{lem:baire-tool}\nA nonempty complete metric space cannot be the union of countably many closed sets with empty interior.\n\\end{lemma}\n\\begin{proof}\nSuppose \\(F_1,F_2,\\ldots\\) are closed with empty interior. Choose a nonempty open ball whose closed ball \\(B_1\\), of positive radius at most \\(1/2\\), avoids \\(F_1\\). Such a choice is possible because the complement of \\(F_1\\) is open and meets every nonempty open set. Recursively choose a closed ball \\(B_{n+1}\\) of positive radius at most \\(2^{-(n+1)}\\), contained in the open ball with the same center and radius as \\(B_n\\), and disjoint from \\(F_{n+1}\\). The intersection of that open ball with the complement of \\(F_{n+1}\\) is nonempty and open, so it contains a sufficiently small closed ball. This also covers the case of isolated points.","annot be the union of countably many closed sets with empty interior.\n\\end{lemma}\n\\begin{proof}\nSuppose \\(F_1,F_2,\\ldots\\) are closed with empty interior. Choose a nonempty open ball whose closed ball \\(B_1\\), of positive radius at most \\(1/2\\), avoids \\(F_1\\). Such a choice is possible because the complement of \\(F_1\\) is open and meets every nonempty open set. Recursively choose a closed ball \\(B_{n+1}\\) of positive radius at most \\(2^{-(n+1)}\\), contained in the open ball with the same center and radius as \\(B_n\\), and disjoint from \\(F_{n+1}\\). The intersection of that open ball with the complement of \\(F_{n+1}\\) is nonempty and open, so it contains a sufficiently small closed ball. This also covers the case of isolated points.","osed sets with empty interior.\n\\end{lemma}\n\\begin{proof}\nSuppose \\(F_1,F_2,\\ldots\\) are closed with empty interior. Choose a nonempty open ball whose closed ball \\(B_1\\), of positive radius at most \\(1/2\\), avoids \\(F_1\\). Such a choice is possible because the complement of \\(F_1\\) is open and meets every nonempty open set. Recursively choose a closed ball \\(B_{n+1}\\) of positive radius at most \\(2^{-(n+1)}\\), contained in the open ball with the same center and radius as \\(B_n\\), and disjoint from \\(F_{n+1}\\). The intersection of that open ball with the complement of \\(F_{n+1}\\) is nonempty and open, so it contains a sufficiently small closed ball. This also covers the case of isolated points.","ots\\) are closed with empty interior. Choose a nonempty open ball whose closed ball \\(B_1\\), of positive radius at most \\(1/2\\), avoids \\(F_1\\). Such a choice is possible because the complement of \\(F_1\\) is open and meets every nonempty open set. Recursively choose a closed ball \\(B_{n+1}\\) of positive radius at most \\(2^{-(n+1)}\\), contained in the open ball with the same center and radius as \\(B_n\\), and disjoint from \\(F_{n+1}\\). The intersection of that open ball with the complement of \\(F_{n+1}\\) is nonempty and open, so it contains a sufficiently small closed ball. This also covers the case of isolated points.","interior. Choose a nonempty open ball whose closed ball \\(B_1\\), of positive radius at most \\(1/2\\), avoids \\(F_1\\). Such a choice is possible because the complement of \\(F_1\\) is open and meets every nonempty open set. Recursively choose a closed ball \\(B_{n+1}\\) of positive radius at most \\(2^{-(n+1)}\\), contained in the open ball with the same center and radius as \\(B_n\\), and disjoint from \\(F_{n+1}\\). The intersection of that open ball with the complement of \\(F_{n+1}\\) is nonempty and open, so it contains a sufficiently small closed ball. This also covers the case of isolated points.","sitive radius at most \\(1/2\\), avoids \\(F_1\\). Such a choice is possible because the complement of \\(F_1\\) is open and meets every nonempty open set. Recursively choose a closed ball \\(B_{n+1}\\) of positive radius at most \\(2^{-(n+1)}\\), contained in the open ball with the same center and radius as \\(B_n\\), and disjoint from \\(F_{n+1}\\). The intersection of that open ball with the complement of \\(F_{n+1}\\) is nonempty and open, so it contains a sufficiently small closed ball. This also covers the case of isolated points.","The centers form a Cauchy sequence: all centers after stage \\(n\\) lie in \\(B_n\\), whose diameter is at most \\(2^{1-n}\\). Completeness gives a limit. For each fixed \\(n\\), a tail of centers lies in the closed set \\(B_n\\), so the limit lies there. It therefore belongs to no \\(F_n\\), contradicting that these sets cover the space.\n\\end{proof}","\\begin{lemma}[Continuous inverse for Banach spaces]\\label{lem:inverse-tool}\nLet \\(S:X\\to Y\\) be a bounded linear surjection between real Banach spaces. There is a constant \\(C<\\infty\\) such that every \\(y\\in Y\\) has a preimage \\(x\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) a","\\begin{lemma}[Continuous inverse for Banach spaces]\\label{lem:inverse-tool}\nLet \\(S:X\\to Y\\) be a bounded linear surjection between real Banach spaces. There is a constant \\(C<\\infty\\) such that every \\(y\\in Y\\) has a preimage \\(x\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtractin","\\begin{lemma}[Continuous inverse for Banach spaces]\\label{lem:inverse-tool}\nLet \\(S:X\\to Y\\) be a bounded linear surjection between real Banach spaces. There is a constant \\(C<\\infty\\) such that every \\(y\\in Y\\) has a preimage \\(x\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h","\\begin{lemma}[Continuous inverse for Banach spaces]\\label{lem:inverse-tool}\nLet \\(S:X\\to Y\\) be a bounded linear surjection between real Banach spaces. There is a constant \\(C<\\infty\\) such that every \\(y\\in Y\\) has a preimage \\(x\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)","\\begin{lemma}[Continuous inverse for Banach spaces]\\label{lem:inverse-tool}\nLet \\(S:X\\to Y\\) be a bounded linear surjection between real Banach spaces. There is a constant \\(C<\\infty\\) such that every \\(y\\in Y\\) has a preimage \\(x\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a","\\begin{lemma}[Continuous inverse for Banach spaces]\\label{lem:inverse-tool}\nLet \\(S:X\\to Y\\) be a bounded linear surjection between real Banach spaces. There is a constant \\(C<\\infty\\) such that every \\(y\\in Y\\) has a preimage \\(x\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\","\\begin{lemma}[Continuous inverse for Banach spaces]\\label{lem:inverse-tool}\nLet \\(S:X\\to Y\\) be a bounded linear surjection between real Banach spaces. There is a constant \\(C<\\infty\\) such that every \\(y\\in Y\\) has a preimage \\(x\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\over","emma}[Continuous inverse for Banach spaces]\\label{lem:inverse-tool}\nLet \\(S:X\\to Y\\) be a bounded linear surjection between real Banach spaces. There is a constant \\(C<\\infty\\) such that every \\(y\\in Y\\) has a preimage \\(x\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives a…",":inverse-tool}\nLet \\(S:X\\to Y\\) be a bounded linear surjection between real Banach spaces. There is a constant \\(C<\\infty\\) such that every \\(y\\in Y\\) has a preimage \\(x\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actua…","reimage \\(x\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","\\) with \\(\\norm{x}\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","\\le C\\norm{y}\\). If \\(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","(S\\) is also injective, its inverse is bounded linear.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","r.\n\\end{lemma}\n\\begin{proof}\nIf \\(Y=\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","\\{0\\}\\), take the zero preimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","reimage. Otherwise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","wise write \\(B_X\\) for the closed unit ball. Surjectivity gives\n\\[\n Y=\\bigcup_{n\\ge1}\\overline{S(nB_X)}.\n\\]\nBy Lemma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","mma~\\ref{lem:baire-tool}, at least one of the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","f the closed sets on the right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","right has nonempty interior. Thus there are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","ere are \\(n\\ge1\\), \\(y_0\\in Y\\), and \\(\\rho>0\\) such that\n\\[\n B(y_0,\\rho)\\subseteq\\overline{S(nB_X)}.\n\\]\nHere \\(B(y_0,\\rho)\\) is an open ball. If \\(\\norm{h}<\\rho\\), both \\(y_0+h\\) and \\(y_0\\) belong to this closure. Choose approximating image sequences for both. Subtracting their preimages shows \\(h\\in\\overline{S(2nB_X)}\\). Scaling by \\(2n\\) gives a constant \\(c=\\rho/(2n)>0\\) with\n\\begin{equation}\\label{eq:inverse-closure-ball}\n B(0,c)\\subseteq\\overline{S(B_X)}.\n\\end{equation}\nThis only gives approximate preimages. We next turn them into an actual preimage with a summable budget.","For a nonzero residual \\(r\\in Y\\), set \\(\\lambda=2\\norm{r}/c\\). The vector \\(r/\\lambda\\) has norm \\(c/2\\), so \\eqref{eq:inverse-closure-ball} supplies \\(z\\in B_X\\) with\n\\[\n \\norm{Sz-r/\\lambda}\\le c/4.\n\\]\nThen \\(u=\\lambda z\\) satisfies\n\\[\n \\norm{u}\\le 2\\norm{r}/c,\\qquad\n \\norm{r-Su}\\le\\norm{r}/2.\n\\]\nIf the residual is zero, choose \\(u=0\\). Start from \\(r_0=y\\), repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}","For a nonzero residual \\(r\\in Y\\), set \\(\\lambda=2\\norm{r}/c\\). The vector \\(r/\\lambda\\) has norm \\(c/2\\), so \\eqref{eq:inverse-closure-ball} supplies \\(z\\in B_X\\) with\n\\[\n \\norm{Sz-r/\\lambda}\\le c/4.\n\\]\nThen \\(u=\\lambda z\\) satisfies\n\\[\n \\norm{u}\\le 2\\norm{r}/c,\\qquad\n \\norm{r-Su}\\le\\norm{r}/2.\n\\]\nIf the residual is zero, choose \\(u=0\\). Start from \\(r_0=y\\), repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\","For a nonzero residual \\(r\\in Y\\), set \\(\\lambda=2\\norm{r}/c\\). The vector \\(r/\\lambda\\) has norm \\(c/2\\), so \\eqref{eq:inverse-closure-ball} supplies \\(z\\in B_X\\) with\n\\[\n \\norm{Sz-r/\\lambda}\\le c/4.\n\\]\nThen \\(u=\\lambda z\\) satisfies\n\\[\n \\norm{u}\\le 2\\norm{r}/c,\\qquad\n \\norm{r-Su}\\le\\norm{r}/2.\n\\]\nIf the residual is zero, choose \\(u=0\\). Start from \\(r_0=y\\), repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\(Sx=y\\). Thus we may take","For a nonzero residual \\(r\\in Y\\), set \\(\\lambda=2\\norm{r}/c\\). The vector \\(r/\\lambda\\) has norm \\(c/2\\), so \\eqref{eq:inverse-closure-ball} supplies \\(z\\in B_X\\) with\n\\[\n \\norm{Sz-r/\\lambda}\\le c/4.\n\\]\nThen \\(u=\\lambda z\\) satisfies\n\\[\n \\norm{u}\\le 2\\norm{r}/c,\\qquad\n \\norm{r-Su}\\le\\norm{r}/2.\n\\]\nIf the residual is zero, choose \\(u=0\\). Start from \\(r_0=y\\), repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\(Sx=y\\). Thus we may take \\(C=4/c\\).","r a nonzero residual \\(r\\in Y\\), set \\(\\lambda=2\\norm{r}/c\\). The vector \\(r/\\lambda\\) has norm \\(c/2\\), so \\eqref{eq:inverse-closure-ball} supplies \\(z\\in B_X\\) with\n\\[\n \\norm{Sz-r/\\lambda}\\le c/4.\n\\]\nThen \\(u=\\lambda z\\) satisfies\n\\[\n \\norm{u}\\le 2\\norm{r}/c,\\qquad\n \\norm{r-Su}\\le\\norm{r}/2.\n\\]\nIf the residual is zero, choose \\(u=0\\). Start from \\(r_0=y\\), repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\(Sx=y\\). Thus we may take \\(C=4/c\\).","\\lambda=2\\norm{r}/c\\). The vector \\(r/\\lambda\\) has norm \\(c/2\\), so \\eqref{eq:inverse-closure-ball} supplies \\(z\\in B_X\\) with\n\\[\n \\norm{Sz-r/\\lambda}\\le c/4.\n\\]\nThen \\(u=\\lambda z\\) satisfies\n\\[\n \\norm{u}\\le 2\\norm{r}/c,\\qquad\n \\norm{r-Su}\\le\\norm{r}/2.\n\\]\nIf the residual is zero, choose \\(u=0\\). Start from \\(r_0=y\\), repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\(Sx=y\\). Thus we may take \\(C=4/c\\).","h\n\\[\n \\norm{Sz-r/\\lambda}\\le c/4.\n\\]\nThen \\(u=\\lambda z\\) satisfies\n\\[\n \\norm{u}\\le 2\\norm{r}/c,\\qquad\n \\norm{r-Su}\\le\\norm{r}/2.\n\\]\nIf the residual is zero, choose \\(u=0\\). Start from \\(r_0=y\\), repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\(Sx=y\\). Thus we may take \\(C=4/c\\).","]\nThen \\(u=\\lambda z\\) satisfies\n\\[\n \\norm{u}\\le 2\\norm{r}/c,\\qquad\n \\norm{r-Su}\\le\\norm{r}/2.\n\\]\nIf the residual is zero, choose \\(u=0\\). Start from \\(r_0=y\\), repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\(Sx=y\\). Thus we may take \\(C=4/c\\).","norm{r}/c,\\qquad\n \\norm{r-Su}\\le\\norm{r}/2.\n\\]\nIf the residual is zero, choose \\(u=0\\). Start from \\(r_0=y\\), repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\(Sx=y\\). Thus we may take \\(C=4/c\\).","residual is zero, choose \\(u=0\\). Start from \\(r_0=y\\), repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\(Sx=y\\). Thus we may take \\(C=4/c\\).",", repeat this construction, and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\(Sx=y\\). Thus we may take \\(C=4/c\\).","and define \\(r_{j+1}=r_j-Su_j\\). Induction yields\n\\[\n \\norm{r_j}\\le2^{-j}\\norm{y},\\qquad\n \\sum_{j\\ge0}\\norm{u_j}\\le\\frac{4}{c}\\norm{y}.\n\\]\nThe series of preimages converges in \\(X\\) by completeness, to a vector \\(x\\) with the displayed norm bound. Continuity of \\(S\\), together with the finite telescoping equation\n\\[\n S\\left(\\sum_{j=0}^{N-1}u_j\\right)=y-r_N,\n\\]\nproves \\(Sx=y\\). Thus we may take \\(C=4/c\\).","If \\(S\\) is bijective, the preimage is unique. Uniqueness makes the inverse additive and homogeneous, and the bound just obtained makes it continuous. In the merely surjective case the construction does not assert a linear choice of preimage. In particular it does not assert that every Banach-space surjection has a bounded linear right inverse.\n\\end{proof}","For example, in Theorem~\\ref{KS43} the map to be inverted has source \\((\\ker D)^\\perp\\) and target \\(\\ran D\\). Both are closed subspaces of the stated Banach spaces. The inverse lemma applies only after the proof has established those closedness and bijectivity facts. It would not apply with an unclosed range equipped with its inherited norm.","\\subsection{Neumann inversion with both inverse identities}\n\\begin{lemma}[A strict perturbation radius]\\label{lem:neumann-tool}\nLet \\(W\\) be a real Banach space and \\(T:W\\to W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\no","\\subsection{Neumann inversion with both inverse identities}\n\\begin{lemma}[A strict perturbation radius]\\label{lem:neumann-tool}\nLet \\(W\\) be a real Banach space and \\(T:W\\to W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Ta","\\subsection{Neumann inversion with both inverse identities}\n\\begin{lemma}[A strict perturbation radius]\\label{lem:neumann-tool}\nLet \\(W\\) be a real Banach space and \\(T:W\\to W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves o","\\subsection{Neumann inversion with both inverse identities}\n\\begin{lemma}[A strict perturbation radius]\\label{lem:neumann-tool}\nLet \\(W\\) be a real Banach space and \\(T:W\\to W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm conver","\\subsection{Neumann inversion with both inverse identities}\n\\begin{lemma}[A strict perturbation radius]\\label{lem:neumann-tool}\nLet \\(W\\) be a real Banach space and \\(T:W\\to W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the st","\\subsection{Neumann inversion with both inverse identities}\n\\begin{lemma}[A strict perturbation radius]\\label{lem:neumann-tool}\nLet \\(W\\) be a real Banach space and \\(T:W\\to W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","es}\n\\begin{lemma}[A strict perturbation radius]\\label{lem:neumann-tool}\nLet \\(W\\) be a real Banach space and \\(T:W\\to W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","l{lem:neumann-tool}\nLet \\(W\\) be a real Banach space and \\(T:W\\to W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","tool}\nLet \\(W\\) be a real Banach space and \\(T:W\\to W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","Banach space and \\(T:W\\to W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","o W\\) bounded linear, with \\(\\norm{T}\\le\\theta<1\\) and \\(\\theta\\ge0\\). Then\n\\[\n Z=\\sum_{j=0}^{\\infty}T^j\n\\]\nexists in operator norm and satisfies\n\\[\n (I-T)Z=Z(I-T)=I,\\qquad \\norm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","rm{Z}\\le\\frac1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","c1{1-\\theta}.\n\\]\n\\end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","end{lemma}\n\\begin{proof}\nFirst recall why the space of bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","bounded operators on \\(W\\) is complete. An operator-norm Cauchy sequence \\(A_n\\) makes \\(A_nx\\) Cauchy for each \\(x\\), and hence defines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","fines \\(Ax=\\lim_nA_nx\\). Pointwise limits preserve linearity. The sequence of operator norms is bounded, so the pointwise limit is a bounded operator. For a fixed sufficiently large \\(n\\), pass \\(m\\to\\infty\\) in \\(\\norm{(A_n-A_m)x}\\le\\varepsilon\\norm{x}\\) to obtain \\(\\norm{(A_n-A)x}\\le\\varepsilon\\norm{x}\\). Taking the supremum over the unit ball proves operator-norm convergence. Completeness of \\(W\\) is the step that produced each pointwise limit.","Submultiplicativity gives \\(\\norm{T^j}\\le\\theta^j\\), with the convention \\(T^0=I\\). The finite partial sums are therefore operator-norm Cauchy, since for \\(M>N\\),\n\\[\n \\left\\|\\sum_{j=N+1}^{M}T^j\\right\\|\n \\le\\sum_{j=N+1}^{M}\\theta^j\\le\\frac{\\theta^{N+1}}{1-\\theta}.\n\\]\nTheir limit \\(Z\\) has norm at most \\(\\sum_{j\\ge0}\\theta^j=1/(1-\\theta)\\). The operator identities for finite sums are\n\\[\n (I-T)\\sum_{j=0}^{N}T^j=I-T^{N+1},\\qquad\n \\left(\\sum_{j=0}^{N}T^j\\right)(I-T)=I-T^{N+1}.\n\\]\nMultiplication by a fixed bounded operator is continuous in operator norm, and \\(\\norm{T^{N+1}}\\le\\theta^{N+1}\\to0\\). Passing to the limit proves both inverse identities. When \\(W\\) is the zero space the same statements hold, with its unique identity map. When \\(\\theta","Submultiplicativity gives \\(\\norm{T^j}\\le\\theta^j\\), with the convention \\(T^0=I\\). The finite partial sums are therefore operator-norm Cauchy, since for \\(M>N\\),\n\\[\n \\left\\|\\sum_{j=N+1}^{M}T^j\\right\\|\n \\le\\sum_{j=N+1}^{M}\\theta^j\\le\\frac{\\theta^{N+1}}{1-\\theta}.\n\\]\nTheir limit \\(Z\\) has norm at most \\(\\sum_{j\\ge0}\\theta^j=1/(1-\\theta)\\). The operator identities for finite sums are\n\\[\n (I-T)\\sum_{j=0}^{N}T^j=I-T^{N+1},\\qquad\n \\left(\\sum_{j=0}^{N}T^j\\right)(I-T)=I-T^{N+1}.\n\\]\nMultiplication by a fixed bounded operator is continuous in operator norm, and \\(\\norm{T^{N+1}}\\le\\theta^{N+1}\\to0\\). Passing to the limit proves both inverse identities. When \\(W\\) is the zero space the same statements hold, with its unique identity map. When \\(\\theta=1\\), the summable tail and dec","Submultiplicativity gives \\(\\norm{T^j}\\le\\theta^j\\), with the convention \\(T^0=I\\). The finite partial sums are therefore operator-norm Cauchy, since for \\(M>N\\),\n\\[\n \\left\\|\\sum_{j=N+1}^{M}T^j\\right\\|\n \\le\\sum_{j=N+1}^{M}\\theta^j\\le\\frac{\\theta^{N+1}}{1-\\theta}.\n\\]\nTheir limit \\(Z\\) has norm at most \\(\\sum_{j\\ge0}\\theta^j=1/(1-\\theta)\\). The operator identities for finite sums are\n\\[\n (I-T)\\sum_{j=0}^{N}T^j=I-T^{N+1},\\qquad\n \\left(\\sum_{j=0}^{N}T^j\\right)(I-T)=I-T^{N+1}.\n\\]\nMultiplication by a fixed bounded operator is continuous in operator norm, and \\(\\norm{T^{N+1}}\\le\\theta^{N+1}\\to0\\). Passing to the limit proves both inverse identities. When \\(W\\) is the zero space the same statements hold, with its unique identity map. When \\(\\theta=1\\), the summable tail and decay used here disappear, so the proof supplies no endpoint conclusion.\n\\end{proo","Submultiplicativity gives \\(\\norm{T^j}\\le\\theta^j\\), with the convention \\(T^0=I\\). The finite partial sums are therefore operator-norm Cauchy, since for \\(M>N\\),\n\\[\n \\left\\|\\sum_{j=N+1}^{M}T^j\\right\\|\n \\le\\sum_{j=N+1}^{M}\\theta^j\\le\\frac{\\theta^{N+1}}{1-\\theta}.\n\\]\nTheir limit \\(Z\\) has norm at most \\(\\sum_{j\\ge0}\\theta^j=1/(1-\\theta)\\). The operator identities for finite sums are\n\\[\n (I-T)\\sum_{j=0}^{N}T^j=I-T^{N+1},\\qquad\n \\left(\\sum_{j=0}^{N}T^j\\right)(I-T)=I-T^{N+1}.\n\\]\nMultiplication by a fixed bounded operator is continuous in operator norm, and \\(\\norm{T^{N+1}}\\le\\theta^{N+1}\\to0\\). Passing to the limit proves both inverse identities. When \\(W\\) is the zero space the same statements hold, with its unique identity map. When \\(\\theta=1\\), the summable tail and decay used here disappear, so the proof supplies no endpoint conclusion.\n\\end{proof}",",\n\\[\n \\left\\|\\sum_{j=N+1}^{M}T^j\\right\\|\n \\le\\sum_{j=N+1}^{M}\\theta^j\\le\\frac{\\theta^{N+1}}{1-\\theta}.\n\\]\nTheir limit \\(Z\\) has norm at most \\(\\sum_{j\\ge0}\\theta^j=1/(1-\\theta)\\). The operator identities for finite sums are\n\\[\n (I-T)\\sum_{j=0}^{N}T^j=I-T^{N+1},\\qquad\n \\left(\\sum_{j=0}^{N}T^j\\right)(I-T)=I-T^{N+1}.\n\\]\nMultiplication by a fixed bounded operator is continuous in operator norm, and \\(\\norm{T^{N+1}}\\le\\theta^{N+1}\\to0\\). Passing to the limit proves both inverse identities. When \\(W\\) is the zero space the same statements hold, with its unique identity map. When \\(\\theta=1\\), the summable tail and decay used here disappear, so the proof supplies no endpoint conclusion.\n\\end{proof}","eta}.\n\\]\nTheir limit \\(Z\\) has norm at most \\(\\sum_{j\\ge0}\\theta^j=1/(1-\\theta)\\). The operator identities for finite sums are\n\\[\n (I-T)\\sum_{j=0}^{N}T^j=I-T^{N+1},\\qquad\n \\left(\\sum_{j=0}^{N}T^j\\right)(I-T)=I-T^{N+1}.\n\\]\nMultiplication by a fixed bounded operator is continuous in operator norm, and \\(\\norm{T^{N+1}}\\le\\theta^{N+1}\\to0\\). Passing to the limit proves both inverse identities. When \\(W\\) is the zero space the same statements hold, with its unique identity map. When \\(\\theta=1\\), the summable tail and decay used here disappear, so the proof supplies no endpoint conclusion.\n\\end{proof}","e operator identities for finite sums are\n\\[\n (I-T)\\sum_{j=0}^{N}T^j=I-T^{N+1},\\qquad\n \\left(\\sum_{j=0}^{N}T^j\\right)(I-T)=I-T^{N+1}.\n\\]\nMultiplication by a fixed bounded operator is continuous in operator norm, and \\(\\norm{T^{N+1}}\\le\\theta^{N+1}\\to0\\). Passing to the limit proves both inverse identities. When \\(W\\) is the zero space the same statements hold, with its unique identity map. When \\(\\theta=1\\), the summable tail and decay used here disappear, so the proof supplies no endpoint conclusion.\n\\end{proof}","The central construction is more stringent. For every set $A\\subseteq\\N$, a countable group $G_A$ is placed between that direct sum $D$ and its product completion $P$. All groups have the same torsion subgroup, the same rational quotient, and the same full unmarked finite-quotient system. Nevertheless, their abstract isomorphism types are classified by $A$ modulo finite changes, and an oracle computes a copy precisely when it computes $A$.","nt. For every set $A\\subseteq\\N$, a countable group $G_A$ is placed between that direct sum $D$ and its product completion $P$. All groups have the same torsion subgroup, the same rational quotient, and the same full unmarked finite-quotient system. Nevertheless, their abstract isomorphism types are classified by $A$ modulo finite changes, and an oracle computes a copy precisely when it computes $A$.","\\section*{How to read the book}\nChapters 1--3 separate the realization contracts and develop the first obstruction examples. Chapters 4--8 give the main construction and its proofs. Chapter 4 also describes the general ambient mechanism: rational subspaces of $P/D$ correspond to pure intermediate groups. Chapters 9--10 extract the reconstruction and grammar consequences. Chapter 11 provides calculations that can be followed by hand and reproduced in Python. Chapter 12 returns to the interpretation in Realization Calculus. The appendices contain implementation guidance, a theorem map, a formal proof guide, and a glossary.","\\chapter*{Notation and principal results}\n\\addcontentsline{toc}{chapter}{Notation and principal results}\n\\markboth{Notation and principal results}{}\nAll groups are written additively unless a finite target group is explicitly allowed to be nonabelian. The natural numbers start at zero. Primes are indexed by\n\\[\np_0=2,\\quad p_1=3,\\quad p_2=5,\\quad p_3=7,\\quad\\ldots.\n\\]\n\\begin{longtable}{@{}p{0.23\\textwidth}p{0.71\\textwidth}@{}}\n\\toprule\nSymbol & Meaning","midrule\n$C_p=\\F_p$ & The additive cyclic group of prime order $p$.","$D=\\bigoplus_p C_p$ & Finite-support prime-coordinate vectors.","$P=\\prod_p C_p$ & All prime-coordinate vectors, with product topology.","$q:P\\to V$ & The quotient map to $V=P/D$.","$D_B$ & The subgroup supported on prime indices in $B\\subseteq\\N$.","$H_W=q^{-1}(W)$ & The intermediate group associated to a rational subspace $W\\le V$.","$S_A$ & The prime set $\\{p_{2n+\\one_A(n)}:n\\in\\N\\}$.","$u_A$ & The coordinate indicator of $S_A$ in $P$.","$G_A$ & The group $q^{-1}(\\Q q(u_A))$.","$s_A(r)$ & The specified rational lift of $r q(u_A)$ to $P$.","$c_A(r,t)$ & The finite cocycle $s_A(r)+s_A(t)-s_A(r+t)$.","$e_p$ & The vector supported at $p$ with value one.","$p\\mid_G g$ & There exists $y\\in G$ with $py=g$.","$A\\eqfin B$ & The symmetric difference $A\\triangle B$ is finite.","$\\rad(m)$ & The product of the distinct prime divisors of $m$, with $\\rad(1)=1$.","$\\widehat G$ & The profinite completion, not the Pontryagin dual.","$A\\le_T X$ & Membership in $A$ is computable using oracle $X$.","$K_{\\mathfrak G}(G)$ & Least admissible finite description length, or $+\\infty$.","\\noindent The main statements have the following form:\n\\begin{align*}\n\\Tor(G_A)&=D,&G_A/D&\\cong\\Q,\\\\\nG_A/mG_A&\\cong C_{\\rad(m)},&\\widehat{G_A}&\\cong P,\\\\\nG_A\\cong G_B&\\iff A\\eqfin B,&&\\\\\nG_A\\text{ has an }X\\text{-computable copy}&\\iff A\\le_T X.&&\n\\end{align*}\nThe extension by $D$ does not split. Distinct eventual selector classes do not embed in one another. All of these conclusions are proved within the book, with the exact hypotheses retained.","G_A\\text{ has an }X\\text{-computable copy}&\\iff A\\le_T X.&&\n\\end{align*}\nThe extension by $D$ does not split. Distinct eventual selector classes do not embed in one another. All of these conclusions are proved within the book, with the exact hypotheses retained.","The integer group $\\Z$ illustrates only the first distinction. One generator and its integer multiples produce infinitely many elements. The groups developed below separate the other distinctions as well. Their purpose is not to make infinite objects mysterious. It is to account for the information that a construction actually consumes.","\\begin{definition}[Operator-seed realization]\nLet $M$ be an abelian group and let $\\mathcal O\\subseteq\\End(M)$ be a declared collection of endomorphisms. For $S\\subseteq M$, its operator hull is the subgroup generated by all $T(s)$, where $s\\in S$ and $T$ is a finite composition of members of $\\mathcal O$, including the identity. A finite operator-seed realization requires a finite $S$ whose hull is $M$.\n\\end{definition}\nThe allowed addition and inverse operations are part of the group structure. An arbitrary target-dependent map is not automatically an admissible endomorphism. In particular, one cannot evade a generation obstruction by introducing a map that simply returns the desired answer.","in{definition}[Operator-seed realization]\nLet $M$ be an abelian group and let $\\mathcal O\\subseteq\\End(M)$ be a declared collection of endomorphisms. For $S\\subseteq M$, its operator hull is the subgroup generated by all $T(s)$, where $s\\in S$ and $T$ is a finite composition of members of $\\mathcal O$, including the identity. A finite operator-seed realization requires a finite $S$ whose hull is $M$.\n\\end{definition}\nThe allowed addition and inverse operations are part of the group structure. An arbitrary target-dependent map is not automatically an admissible endomorphism. In particular, one cannot evade a generation obstruction by introducing a map that simply returns the desired answer.","If all endomorphisms are allowed, this is module generation over the unital ring $\\End(M)$. We write $\\srank_{\\End(M)}(M)$ for its least generating cardinal. An infinite value can therefore express a strong obstruction even when the operator collection itself is unrestricted.","ription grammar consists of a countable alphabet, a class of valid finite strings, and a semantics assigning each valid string one group isomorphism type. The semantics need not be computable. Any permitted parameters must have a declared coding convention.\n\\end{definition}\nThis definition allows a finite sentence to specify a noncomputable group. It does not count an arbitrary real or subset of $\\N$ as a free symbol. If such a parameter is admitted, its information must be recorded separately.","\\begin{definition}[Computable copy]\nA computable copy of a countable group is an isomorphic group on a computably named domain, with decidable equality and computable group operations. An $X$-computable copy permits an oracle $X$ in those computations. Infinite groups may equivalently be presented on $\\N$.\n\\end{definition}\nA presentation by generators and relators is a different object. Enumerating the relators need not decide whether two words represent the same element. The effective presentations in this book always include decidable equality in the named copy.","\\begin{definition}[Topological generation]\nFor a topological group $M$, a subset $S$ topologically generates $M$ when the closure of the subgroup generated by $S$ is all of $M$. Its least possible size is denoted $\\trank(M)$. An operator version takes the closure of the operator hull instead.\n\\end{definition}\nThe closure is taken in the stated topology. A change of topology is a change of contract, not an innocuous notational adjustment.","\\begin{proposition}[Countable-code bound]\n\\label{thm:codecount}\nLet $\\mathfrak G$ be a finite-description grammar as above. Then its class of realized group isomorphism types has cardinal at most $\\aleph_0$.\n\\end{proposition}\n\\begin{proof}\nFor each length $n$, the set of strings of length $n$ is countable. The union over $n\\in\\N$ is countable. The valid strings form a subset, and a single-valued image of a countable set is countable.\n\\end{proof}\nThis is a boundary of description, not of existence. It also says nothing about whether a particular named group lies outside a chosen grammar. The later constructions supply structural examples and an explicit relative diagonal argument.","\\begin{definition}[Grammar-relative description complexity]\nFor a fixed binary coding of $\\mathfrak G$, let $\\lvert d\\rvert$ be its bit length and set\n\\[\nK_{\\mathfrak G}(G)=\\inf\\{\\lvert d\\rvert:\\ d\\text{ is valid and }\\operatorname{Realize}_{\\mathfrak G}(d)\\cong G\\},\n\\]\nwith value $+\\infty$ when there is no such description.\n\\end{definition}\nThe value depends on the grammar and the coding. A fixed finite description can be long without being impossible. Conversely, a group with $K_{\\mathfrak G}(G)=+\\infty$ may have a short description in an expanded language.","g of $\\mathfrak G$, let $\\lvert d\\rvert$ be its bit length and set\n\\[\nK_{\\mathfrak G}(G)=\\inf\\{\\lvert d\\rvert:\\ d\\text{ is valid and }\\operatorname{Realize}_{\\mathfrak G}(d)\\cong G\\},\n\\]\nwith value $+\\infty$ when there is no such description.\n\\end{definition}\nThe value depends on the grammar and the coding. A fixed finite description can be long without being impossible. Conversely, a group with $K_{\\mathfrak G}(G)=+\\infty$ may have a short description in an expanded language.","\\begin{boundary}[No uncharged infinite parameters]\nThe expression $G=q^{-1}(\\Q q(u))$ is short. If $u$ is an arbitrary infinite coordinate vector, that expression is not a finite specification of $G$ without a specification of $u$. Counting its surrounding symbols does not count the information in the parameter.\n\\end{boundary}\nThis boundary will become exact: the computational information in a selected $u_A$ is precisely what is needed to compute the corresponding group.","boundary}[No uncharged infinite parameters]\nThe expression $G=q^{-1}(\\Q q(u))$ is short. If $u$ is an arbitrary infinite coordinate vector, that expression is not a finite specification of $G$ without a specification of $u$. Counting its surrounding symbols does not count the information in the parameter.\n\\end{boundary}\nThis boundary will become exact: the computational information in a selected $u_A$ is precisely what is needed to compute the corresponding group.","\\begin{theorem}[Reachability obstruction]\n\\label{thm:reach}\nLet $M=\\bigoplus_{i\\in I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachab","\\begin{theorem}[Reachability obstruction]\n\\label{thm:reach}\nLet $M=\\bigoplus_{i\\in I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coo","\\begin{theorem}[Reachability obstruction]\n\\label{thm:reach}\nLet $M=\\bigoplus_{i\\in I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final as","\\begin{theorem}[Reachability obstruction]\n\\label{thm:reach}\nLet $M=\\bigoplus_{i\\in I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final assertion follows by taking the projection onto an omi","\\begin{theorem}[Reachability obstruction]\n\\label{thm:reach}\nLet $M=\\bigoplus_{i\\in I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final assertion follows by taking the projection onto an omitted nonzero packet.\n\\end","\\begin{theorem}[Reachability obstruction]\n\\label{thm:reach}\nLet $M=\\bigoplus_{i\\in I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final assertion follows by taking the projection onto an omitted nonzero packet.\n\\end{proof}\nReachability is necessary, not suff","\\begin{theorem}[Reachability obstruction]\n\\label{thm:reach}\nLet $M=\\bigoplus_{i\\in I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final assertion follows by taking the projection onto an omitted nonzero packet.\n\\end{proof}\nReachability is necessary, not sufficient.","\\begin{theorem}[Reachability obstruction]\n\\label{thm:reach}\nLet $M=\\bigoplus_{i\\in I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final assertion follows by taking the projection onto an omitted nonzero packet.\n\\end{proof}\nReachability is necessary, not sufficient. A permitted path may carry zero action on the part","\\begin{theorem}[Reachability obstruction]\n\\label{thm:reach}\nLet $M=\\bigoplus_{i\\in I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final assertion follows by taking the projection onto an omitted nonzero packet.\n\\end{proof}\nReachability is necessary, not sufficient. A permitted path may carry zero action on the particular seed, or its image ma…","\\bigoplus_{i\\in I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final assertion follows by taking the projection onto an omitted nonzero packet.\n\\end{proof}\nReachability is necessary, not sufficient. A permitted path may carry zero action on the particular seed, or its image may be a proper subgroup of the target packet. The theorem therefore …","n I}M_i$ be a direct sum of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final assertion follows by taking the projection onto an omitted nonzero packet.\n\\end{proof}\nReachability is necessary, not sufficient. A permitted path may carry zero action on the particular seed, or its image may be a proper subgroup of the target packet. The theorem therefore gives a non-ge…","um of abelian groups with coordinate projections $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final assertion follows by taking the projection onto an omitted nonzero packet.\n\\end{proof}\nReachability is necessary, not sufficient. A permitted path may carry zero action on the particular seed, or its image may be a proper subgroup of the target packet. The theorem therefore gives a non-generation certificate b…","ctions $P_i$. Suppose every primitive operator $T$ is an endomorphism and its transfer block $P_jTP_i$ vanishes unless $i\\to j$ is an allowed edge of a directed graph $E$. If $F$ is the union of the supports of a seed family $S$, then\n\\[\n\\Hull_{\\mathcal O}(S)\\subseteq\n\\bigoplus_{j\\in\\operatorname{Reach}_E(F)}M_j.\n\\]\nIf $M_j\\ne0$ and $j$ is not reachable, $P_j$ annihilates the entire hull but not $M$.\n\\end{theorem}\n\\begin{proof}\nAn element of the direct sum has finite support. Applying one primitive operator can move its support only along allowed edges. Induction on word length confines every operator image of a seed to packets reachable by a finite path. Finite sums and inverses cannot introduce a coordinate outside that set. The final assertion follows by taking the projection onto an omitted nonzero packet.\n\\end{proof}\nReachability is necessary, not sufficient. A permitted path may carry zero action on the particular seed, or its image may be a proper subgroup of the target packet. The theorem therefore gives a non-generation certificate but does not assert generation whenever the","The distinction is easiest to express in terms of an embedding\n\\[\nG\\longrightarrow \\varprojlim G/N.\n\\]\nResidual finiteness makes this map injective. The inverse limit includes all compatible tuples. The image of the original $G$ can nevertheless be a proper dense subgroup. Forgetting that image can forget the object even while retaining enough coordinates to distinguish every element once the object is supplied.","\\section{Finite stages and two connecting systems}\nFor $N\\ge1$, define\n\\[\nA_N=\\prod_{i<N}C_{p_i},\\qquad m_N=\\prod_{i<N}p_i.\n\\]\nThe Chinese remainder map sends $k\\bmod m_N$ to $(k\\bmod p_i)_{i<N}$. It is an isomorphism, so $A_N$ is cyclic and its all-one vector is a generator.","\\begin{center}\n\\begin{tabular}{@{}cccc@{}}\n\\toprule\n$N$&Prime factors&$m_N$&Finite group","midrule\n1&$2$&2&$C_2$","2&$2,3$&6&$C_6$","3&$2,3,5$&30&$C_{30}$","4&$2,3,5,7$&210&$C_{210}$","5&$2,3,5,7,11$&2310&$C_{2310}$","Zero extension gives injections $A_N\\to A_{N+1}$. Forgetting the last coordinate gives surjections $A_{N+1}\\to A_N$. These are different diagrams. Their limits are\n\\[\nD=\\varinjlim_N A_N=\\bigoplus_p C_p,\n\\qquad\nP=\\varprojlim_N A_N=\\prod_p C_p.\n\\]\nThe first group contains only finite-support vectors. The second contains arbitrary compatible coordinate assignments.","Give each $C_p$ its discrete topology and give $P$ the product topology. A basic neighborhood fixes only finitely many coordinates. We give every subgroup of $P$ the induced topology when discussing topological generation, unless explicitly stated otherwise.","\\section{Torsion and endomorphisms}\n\\begin{theorem}[The torsion subgroup of the product]\n\\label{thm:torP}\nThe subgroup $D$ is dense and proper in $P$, and\n\\[\n\\Tor(P)=D.\n\\]\n\\end{theorem}\n\\begin{proof}\nA prescribed finite coordinate pattern is realized by a finite-support vector. Thus $D$ meets every basic open set and is dense. A finite-support vector is annihilated by the product of the finitely many primes in its support, so it is torsion.","Conversely, suppose $mx=0$ for a positive integer $m$. Whenever $p\\nmid m$, multiplication by $m$ is invertible in $C_p$, and hence $x_p=0$. The support of $x$ is contained in the finite set of prime divisors of $m$. Thus $x\\in D$. The all-one vector has infinite support, so it belongs to $P\\setminus D$.\n\\end{proof}","\\begin{proposition}[Diagonal endomorphisms of the direct sum]\n\\label{thm:endD}\nEvery endomorphism of $D$ acts by a scalar on each prime coordinate, and\n\\[\n\\End(D)\\cong\\prod_p\\F_p\n\\]\nas rings.\n\\end{proposition}\n\\begin{proof}\nThe subgroup annihilated by a prime $p$ is exactly the $C_p$ coordinate. Every endomorphism preserves it, because $px=0$ implies $pT(x)=T(px)=0$. Its restriction to that cyclic group is multiplication by a unique scalar in $\\F_p$. An arbitrary element of $D$ is a finite sum of such primary elements, so these restrictions determine the map. Conversely, any family of coordinate scalars preserves finite support and defines an endomorphism. Addition and composition correspond to coordinate addition and multiplication.\n\\end{proof}","e direct sum]\n\\label{thm:endD}\nEvery endomorphism of $D$ acts by a scalar on each prime coordinate, and\n\\[\n\\End(D)\\cong\\prod_p\\F_p\n\\]\nas rings.\n\\end{proposition}\n\\begin{proof}\nThe subgroup annihilated by a prime $p$ is exactly the $C_p$ coordinate. Every endomorphism preserves it, because $px=0$ implies $pT(x)=T(px)=0$. Its restriction to that cyclic group is multiplication by a unique scalar in $\\F_p$. An arbitrary element of $D$ is a finite sum of such primary elements, so these restrictions determine the map. Conversely, any family of coordinate scalars preserves finite support and defines an endomorphism. Addition and composition correspond to coordinate addition and multiplication.\n\\end{proof}","very endomorphism of $D$ acts by a scalar on each prime coordinate, and\n\\[\n\\End(D)\\cong\\prod_p\\F_p\n\\]\nas rings.\n\\end{proposition}\n\\begin{proof}\nThe subgroup annihilated by a prime $p$ is exactly the $C_p$ coordinate. Every endomorphism preserves it, because $px=0$ implies $pT(x)=T(px)=0$. Its restriction to that cyclic group is multiplication by a unique scalar in $\\F_p$. An arbitrary element of $D$ is a finite sum of such primary elements, so these restrictions determine the map. Conversely, any family of coordinate scalars preserves finite support and defines an endomorphism. Addition and composition correspond to coordinate addition and multiplication.\n\\end{proof}","\\begin{corollary}[An infinite seed obstruction]\n\\label{thm:rankD}\nThe endomorphism-seed rank of $D$ is $\\aleph_0$. No finite set topologically generates $D$, even after all endomorphisms are admitted.\n\\end{corollary}\n\\begin{proof}\nThe union of the supports of finitely many seeds is finite. Diagonal endomorphisms preserve its complement as zero coordinates. Any omitted coordinate projection is a certificate from Theorem~\\ref{thm:reach}. Its kernel is closed in the induced topology, so taking closure does not help. The countably many vectors $e_p$ generate $D$ algebraically, proving the exact rank.\n\\end{proof}\nNotice that this conclusion does not depend on restricting the number of endomorphisms. Even the full endomorphism ring cannot create a missing prime component from the available see","\\begin{corollary}[An infinite seed obstruction]\n\\label{thm:rankD}\nThe endomorphism-seed rank of $D$ is $\\aleph_0$. No finite set topologically generates $D$, even after all endomorphisms are admitted.\n\\end{corollary}\n\\begin{proof}\nThe union of the supports of finitely many seeds is finite. Diagonal endomorphisms preserve its complement as zero coordinates. Any omitted coordinate projection is a certificate from Theorem~\\ref{thm:reach}. Its kernel is closed in the induced topology, so taking closure does not help. The countably many vectors $e_p$ generate $D$ algebraically, proving the exact rank.\n\\end{proof}\nNotice that this conclusion does not depend on restricting the number of endomorphisms. Even the full endomorphism ring cannot create a missing prime component from the available seeds.","rates $D$, even after all endomorphisms are admitted.\n\\end{corollary}\n\\begin{proof}\nThe union of the supports of finitely many seeds is finite. Diagonal endomorphisms preserve its complement as zero coordinates. Any omitted coordinate projection is a certificate from Theorem~\\ref{thm:reach}. Its kernel is closed in the induced topology, so taking closure does not help. The countably many vectors $e_p$ generate $D$ algebraically, proving the exact rank.\n\\end{proof}\nNotice that this conclusion does not depend on restricting the number of endomorphisms. Even the full endomorphism ring cannot create a missing prime component from the available seeds.","after all endomorphisms are admitted.\n\\end{corollary}\n\\begin{proof}\nThe union of the supports of finitely many seeds is finite. Diagonal endomorphisms preserve its complement as zero coordinates. Any omitted coordinate projection is a certificate from Theorem~\\ref{thm:reach}. Its kernel is closed in the induced topology, so taking closure does not help. The countably many vectors $e_p$ generate $D$ algebraically, proving the exact rank.\n\\end{proof}\nNotice that this conclusion does not depend on restricting the number of endomorphisms. Even the full endomorphism ring cannot create a missing prime component from the available seeds.","\\section{A single topological generator after completion}\n\\begin{theorem}[Topological cyclicity]\n\\label{thm:procyclic}\nLet $u=(1,1,1,\\ldots)\\in P$. Then\n\\[\n\\overline{\\langle u\\rangle}=P.\n\\]\nIn particular, $\\trank(P)=1$.\n\\end{theorem}\n\\begin{proof}\nA basic open set prescribes a tuple on finitely many distinct prime coordinates. The Chinese remainder theorem supplies an integer $k$ with the specified residues. The element $ku$ belongs to the given open set. Therefore the cyclic subgroup is dense. The product is nontrivial, so zero generators do not suffice.\n\\end{proof}\nThe generator $u$ is absent from $D$. The topological statement does not exhibit a finite generating family inside the original dense subgroup. This is the precise mechanism by which completion changes the generation contract.","ical generator after completion}\n\\begin{theorem}[Topological cyclicity]\n\\label{thm:procyclic}\nLet $u=(1,1,1,\\ldots)\\in P$. Then\n\\[\n\\overline{\\langle u\\rangle}=P.\n\\]\nIn particular, $\\trank(P)=1$.\n\\end{theorem}\n\\begin{proof}\nA basic open set prescribes a tuple on finitely many distinct prime coordinates. The Chinese remainder theorem supplies an integer $k$ with the specified residues. The element $ku$ belongs to the given open set. Therefore the cyclic subgroup is dense. The product is nontrivial, so zero generators do not suffice.\n\\end{proof}\nThe generator $u$ is absent from $D$. The topological statement does not exhibit a finite generating family inside the original dense subgroup. This is the precise mechanism by which completion changes the generation contract.","In particular, $\\trank(P)=1$.\n\\end{theorem}\n\\begin{proof}\nA basic open set prescribes a tuple on finitely many distinct prime coordinates. The Chinese remainder theorem supplies an integer $k$ with the specified residues. The element $ku$ belongs to the given open set. Therefore the cyclic subgroup is dense. The product is nontrivial, so zero generators do not suffice.\n\\end{proof}\nThe generator $u$ is absent from $D$. The topological statement does not exhibit a finite generating family inside the original dense subgroup. This is the precise mechanism by which completion changes the generation contract.","rank(P)=1$.\n\\end{theorem}\n\\begin{proof}\nA basic open set prescribes a tuple on finitely many distinct prime coordinates. The Chinese remainder theorem supplies an integer $k$ with the specified residues. The element $ku$ belongs to the given open set. Therefore the cyclic subgroup is dense. The product is nontrivial, so zero generators do not suffice.\n\\end{proof}\nThe generator $u$ is absent from $D$. The topological statement does not exhibit a finite generating family inside the original dense subgroup. This is the precise mechanism by which completion changes the generation contract.","\\begin{figure}[tbp]\n\\centering\\includegraphics[width=\\textwidth]{\\figfile{01_two_limits.pdf}}\n\\caption{The finite stages agree, but the arrows do not. Zero extension produces the finite-support group $D$. Coordinate restriction produces the complete product $P$. The labels state algebraic endomorphism-seed rank for $D$ and ordinary topological rank for $P$.}\n\\label{fig:twolimits}\n\\end{figure}","\\section{The complete finite-quotient system}\n\\begin{proposition}[Profinite completion of the direct sum]\n\\label{thm:Dcompletion}\nThe canonical embedding of $D$ into $P$ identifies $P$ with $\\widehat D$.\n\\end{proposition}\n\\begin{proof}\nLet $f:D\\to F$ be a homomorphism to a finite group. The image of a prime-order factor $C_p$ is trivial unless $p$ divides $\\lvert F\\rvert$. Thus $f$ vanishes on all but finitely many prime components and factors through a finite coordinate product. Kernels of finite coordinate projections are therefore cofinal among the finite-index normal subgroups of $D$. Taking their inverse limit gives exactly $P$, with the canonical map supplied by the coordinates.\n\\end{proof}\nThis proof does not assume that $F$ is abelian. The image is abelian because the source is, and the order argument applies to each factor. The definition","\\section{The complete finite-quotient system}\n\\begin{proposition}[Profinite completion of the direct sum]\n\\label{thm:Dcompletion}\nThe canonical embedding of $D$ into $P$ identifies $P$ with $\\widehat D$.\n\\end{proposition}\n\\begin{proof}\nLet $f:D\\to F$ be a homomorphism to a finite group. The image of a prime-order factor $C_p$ is trivial unless $p$ divides $\\lvert F\\rvert$. Thus $f$ vanishes on all but finitely many prime components and factors through a finite coordinate product. Kernels of finite coordinate projections are therefore cofinal among the finite-index normal subgroups of $D$. Taking their inverse limit gives exactly $P$, with the canonical map supplied by the coordinates.\n\\end{proof}\nThis proof does not assume that $F$ is abelian. The image is abelian because the source is, and the order argument applies to each factor. The definition and gene","\\section{The complete finite-quotient system}\n\\begin{proposition}[Profinite completion of the direct sum]\n\\label{thm:Dcompletion}\nThe canonical embedding of $D$ into $P$ identifies $P$ with $\\widehat D$.\n\\end{proposition}\n\\begin{proof}\nLet $f:D\\to F$ be a homomorphism to a finite group. The image of a prime-order factor $C_p$ is trivial unless $p$ divides $\\lvert F\\rvert$. Thus $f$ vanishes on all but finitely many prime components and factors through a finite coordinate product. Kernels of finite coordinate projections are therefore cofinal among the finite-index normal subgroups of $D$. Taking their inverse limit gives exactly $P$, with the canonical map supplied by the coordinates.\n\\end{proof}\nThis proof does not assume that $F$ is abelian. The image is abelian because the source is, and the order argument applies to each factor. The definition and general properties","\\section{The complete finite-quotient system}\n\\begin{proposition}[Profinite completion of the direct sum]\n\\label{thm:Dcompletion}\nThe canonical embedding of $D$ into $P$ identifies $P$ with $\\widehat D$.\n\\end{proposition}\n\\begin{proof}\nLet $f:D\\to F$ be a homomorphism to a finite group. The image of a prime-order factor $C_p$ is trivial unless $p$ divides $\\lvert F\\rvert$. Thus $f$ vanishes on all but finitely many prime components and factors through a finite coordinate product. Kernels of finite coordinate projections are therefore cofinal among the finite-index normal subgroups of $D$. Taking their inverse limit gives exactly $P$, with the canonical map supplied by the coordinates.\n\\end{proof}\nThis proof does not assume that $F$ is abelian. The image is abelian because the source is, and the order argument applies to each factor. The definition and general properties of profinite compl","\\section{The complete finite-quotient system}\n\\begin{proposition}[Profinite completion of the direct sum]\n\\label{thm:Dcompletion}\nThe canonical embedding of $D$ into $P$ identifies $P$ with $\\widehat D$.\n\\end{proposition}\n\\begin{proof}\nLet $f:D\\to F$ be a homomorphism to a finite group. The image of a prime-order factor $C_p$ is trivial unless $p$ divides $\\lvert F\\rvert$. Thus $f$ vanishes on all but finitely many prime components and factors through a finite coordinate product. Kernels of finite coordinate projections are therefore cofinal among the finite-index normal subgroups of $D$. Taking their inverse limit gives exactly $P$, with the canonical map supplied by the coordinates.\n\\end{proof}\nThis proof does not assume that $F$ is abelian. The image is abelian because the source is, and the order argument applies to each factor. The definition and general properties of profinite completion may be found in \\cite{Wilkes}. All needed","\\section{The complete finite-quotient system}\n\\begin{proposition}[Profinite completion of the direct sum]\n\\label{thm:Dcompletion}\nThe canonical embedding of $D$ into $P$ identifies $P$ with $\\widehat D$.\n\\end{proposition}\n\\begin{proof}\nLet $f:D\\to F$ be a homomorphism to a finite group. The image of a prime-order factor $C_p$ is trivial unless $p$ divides $\\lvert F\\rvert$. Thus $f$ vanishes on all but finitely many prime components and factors through a finite coordinate product. Kernels of finite coordinate projections are therefore cofinal among the finite-index normal subgroups of $D$. Taking their inverse limit gives exactly $P$, with the canonical map supplied by the coordinates.\n\\end{proof}\nThis proof does not assume that $F$ is abelian. The image is abelian because the source is, and the order argument applies to each factor. The definition and general properties of profinite completion may be found in \\cite{Wilkes}. All needed cofinality statements in this book are proved for the specific groups under c","\\section{The complete finite-quotient system}\n\\begin{proposition}[Profinite completion of the direct sum]\n\\label{thm:Dcompletion}\nThe canonical embedding of $D$ into $P$ identifies $P$ with $\\widehat D$.\n\\end{proposition}\n\\begin{proof}\nLet $f:D\\to F$ be a homomorphism to a finite group. The image of a prime-order factor $C_p$ is trivial unless $p$ divides $\\lvert F\\rvert$. Thus $f$ vanishes on all but finitely many prime components and factors through a finite coordinate product. Kernels of finite coordinate projections are therefore cofinal among the finite-index normal subgroups of $D$. Taking their inverse limit gives exactly $P$, with the canonical map supplied by the coordinates.\n\\end{proof}\nThis proof does not assume that $F$ is abelian. The image is abelian because the source is, and the order argument applies to each factor. The definition and general properties of profinite completion may be found in \\cite{Wilkes}. All needed cofinality statements in this book are proved for the specific groups under consideration.","widehat D$.\n\\end{proposition}\n\\begin{proof}\nLet $f:D\\to F$ be a homomorphism to a finite group. The image of a prime-order factor $C_p$ is trivial unless $p$ divides $\\lvert F\\rvert$. Thus $f$ vanishes on all but finitely many prime components and factors through a finite coordinate product. Kernels of finite coordinate projections are therefore cofinal among the finite-index normal subgroups of $D$. Taking their inverse limit gives exactly $P$, with the canonical map supplied by the coordinates.\n\\end{proof}\nThis proof does not assume that $F$ is abelian. The image is abelian because the source is, and the order argument applies to each factor. The definition and general properties of profinite completion may be found in \\cite{Wilkes}. All needed cofinality statements in this book are proved for the specific groups under consideration.","t $f:D\\to F$ be a homomorphism to a finite group. The image of a prime-order factor $C_p$ is trivial unless $p$ divides $\\lvert F\\rvert$. Thus $f$ vanishes on all but finitely many prime components and factors through a finite coordinate product. Kernels of finite coordinate projections are therefore cofinal among the finite-index normal subgroups of $D$. Taking their inverse limit gives exactly $P$, with the canonical map supplied by the coordinates.\n\\end{proof}\nThis proof does not assume that $F$ is abelian. The image is abelian because the source is, and the order argument applies to each factor. The definition and general properties of profinite completion may be found in \\cite{Wilkes}. All needed cofinality statements in this book are proved for the specific groups under consideration.","unless $p$ divides $\\lvert F\\rvert$. Thus $f$ vanishes on all but finitely many prime components and factors through a finite coordinate product. Kernels of finite coordinate projections are therefore cofinal among the finite-index normal subgroups of $D$. Taking their inverse limit gives exactly $P$, with the canonical map supplied by the coordinates.\n\\end{proof}\nThis proof does not assume that $F$ is abelian. The image is abelian because the source is, and the order argument applies to each factor. The definition and general properties of profinite completion may be found in \\cite{Wilkes}. All needed cofinality statements in this book are proved for the specific groups under consideration.","\\section{A rational space behind the torsion}\nPut $V=P/D$ and let $q:P\\to V$ be the quotient map.","\\begin{theorem}[Unique division in the ambient quotient]\n\\label{thm:Vrational}\nMultiplication by every positive integer is bijective on $V$. Thus $V$ has a unique rational vector-space structure extending its additive group structure. The product $P$ has no nonzero divisible subgroup.\n\\end{theorem}\n\\begin{proof}\nFix $m\\ge1$ and $x\\in P$. Define $y_p=m^{-1}x_p$ when $p\\nmid m$, and $y_p=0$ when $p\\mid m$. Then $x-my$ has finite support. Thus $q(x)=m q(y)$, proving surjectivity on $V$.","}[Unique division in the ambient quotient]\n\\label{thm:Vrational}\nMultiplication by every positive integer is bijective on $V$. Thus $V$ has a unique rational vector-space structure extending its additive group structure. The product $P$ has no nonzero divisible subgroup.\n\\end{theorem}\n\\begin{proof}\nFix $m\\ge1$ and $x\\in P$. Define $y_p=m^{-1}x_p$ when $p\\nmid m$, and $y_p=0$ when $p\\mid m$. Then $x-my$ has finite support. Thus $q(x)=m q(y)$, proving surjectivity on $V$.","quotient]\n\\label{thm:Vrational}\nMultiplication by every positive integer is bijective on $V$. Thus $V$ has a unique rational vector-space structure extending its additive group structure. The product $P$ has no nonzero divisible subgroup.\n\\end{theorem}\n\\begin{proof}\nFix $m\\ge1$ and $x\\in P$. Define $y_p=m^{-1}x_p$ when $p\\nmid m$, and $y_p=0$ when $p\\mid m$. Then $x-my$ has finite support. Thus $q(x)=m q(y)$, proving surjectivity on $V$.","lication by every positive integer is bijective on $V$. Thus $V$ has a unique rational vector-space structure extending its additive group structure. The product $P$ has no nonzero divisible subgroup.\n\\end{theorem}\n\\begin{proof}\nFix $m\\ge1$ and $x\\in P$. Define $y_p=m^{-1}x_p$ when $p\\nmid m$, and $y_p=0$ when $p\\mid m$. Then $x-my$ has finite support. Thus $q(x)=m q(y)$, proving surjectivity on $V$.","For injectivity, suppose $mx\\in D$. Outside the union of the finite support of $mx$ and the prime divisors of $m$, multiplication by $m$ is invertible and $x_p=0$. Hence $x\\in D$ and $q(x)=0$. Unique division defines $(a/b)v$ as the unique solution $w$ of $bw=av$. The vector-space laws follow by clearing positive denominators and using injectivity of integer multiplication.","Finally, a homomorphism from a divisible group to $C_p$ is zero. For each source element $x$, write $x=py$ and observe that its image is $p$ times an element of $C_p$. Every coordinate of a divisible subgroup of $P$ therefore vanishes, so that subgroup is zero.\n\\end{proof}\nA group with no nonzero divisible subgroup is called \\emph{reduced}. Thus $P$, and every subgroup of $P$, is reduced. Nevertheless $P/D$ is divisible. Quotients can possess division that no subgroup of the original group supports. The nonsplitting theorem later turns this distinc","Finally, a homomorphism from a divisible group to $C_p$ is zero. For each source element $x$, write $x=py$ and observe that its image is $p$ times an element of $C_p$. Every coordinate of a divisible subgroup of $P$ therefore vanishes, so that subgroup is zero.\n\\end{proof}\nA group with no nonzero divisible subgroup is called \\emph{reduced}. Thus $P$, and every subgroup of $P$, is reduced. Nevertheless $P/D$ is divisible. Quotients can possess division that no subgroup of the original group supports. The nonsplitting theorem later turns this distinction into an exact obstruction.","y, a homomorphism from a divisible group to $C_p$ is zero. For each source element $x$, write $x=py$ and observe that its image is $p$ times an element of $C_p$. Every coordinate of a divisible subgroup of $P$ therefore vanishes, so that subgroup is zero.\n\\end{proof}\nA group with no nonzero divisible subgroup is called \\emph{reduced}. Thus $P$, and every subgroup of $P$, is reduced. Nevertheless $P/D$ is divisible. Quotients can possess division that no subgroup of the original group supports. The nonsplitting theorem later turns this distinction into an exact obstruction.","\\section{The entire subgroup lattice}\nFor $B\\subseteq\\N$, set\n\\[\nD_B=\\bigoplus_{i\\in B}C_{p_i}\\le D.\n\\]\nA prime spectrum records which prime orders occur. Define $\\sigma_p(M)=1$ when $M$ contains a nonzero element annihilated by $p$, and set it to zero otherwise. Since $p$ is prime, a nonzero such element has order exactly $p$.","\\begin{theorem}[Subgroup and isomorphism classification]\n\\label{thm:subD}\nThe map $B\\mapsto D_B$ is an order isomorphism from $\\mathcal P(\\N)$ onto $\\Sub(D)$. Every subgroup is fully invariant under all endomorphisms of $D$, and\n\\[\nD_B\\cong D_C\\quad\\Longleftrightarrow\\quad B=C.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $H\\le D$ and let $x\\in H$. Its support is a finite set $F$ of primes. For $p\\in F$, choose by the Chinese remainder theorem an integer $k$ that is one modulo $p$ and zero modulo every other prime in $F$. Then $kx$ is supported only at $p$, where it is nonzero. It generates the entire $C_p$ coordinate. Thus $H$ contains every coordinate factor appearing in one of its elements.","ation]\n\\label{thm:subD}\nThe map $B\\mapsto D_B$ is an order isomorphism from $\\mathcal P(\\N)$ onto $\\Sub(D)$. Every subgroup is fully invariant under all endomorphisms of $D$, and\n\\[\nD_B\\cong D_C\\quad\\Longleftrightarrow\\quad B=C.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $H\\le D$ and let $x\\in H$. Its support is a finite set $F$ of primes. For $p\\in F$, choose by the Chinese remainder theorem an integer $k$ that is one modulo $p$ and zero modulo every other prime in $F$. Then $kx$ is supported only at $p$, where it is nonzero. It generates the entire $C_p$ coordinate. Thus $H$ contains every coordinate factor appearing in one of its elements.","The map $B\\mapsto D_B$ is an order isomorphism from $\\mathcal P(\\N)$ onto $\\Sub(D)$. Every subgroup is fully invariant under all endomorphisms of $D$, and\n\\[\nD_B\\cong D_C\\quad\\Longleftrightarrow\\quad B=C.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $H\\le D$ and let $x\\in H$. Its support is a finite set $F$ of primes. For $p\\in F$, choose by the Chinese remainder theorem an integer $k$ that is one modulo $p$ and zero modulo every other prime in $F$. Then $kx$ is supported only at $p$, where it is nonzero. It generates the entire $C_p$ coordinate. Thus $H$ contains every coordinate factor appearing in one of its elements.","rphism from $\\mathcal P(\\N)$ onto $\\Sub(D)$. Every subgroup is fully invariant under all endomorphisms of $D$, and\n\\[\nD_B\\cong D_C\\quad\\Longleftrightarrow\\quad B=C.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $H\\le D$ and let $x\\in H$. Its support is a finite set $F$ of primes. For $p\\in F$, choose by the Chinese remainder theorem an integer $k$ that is one modulo $p$ and zero modulo every other prime in $F$. Then $kx$ is supported only at $p$, where it is nonzero. It generates the entire $C_p$ coordinate. Thus $H$ contains every coordinate factor appearing in one of its elements.","m $\\mathcal P(\\N)$ onto $\\Sub(D)$. Every subgroup is fully invariant under all endomorphisms of $D$, and\n\\[\nD_B\\cong D_C\\quad\\Longleftrightarrow\\quad B=C.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $H\\le D$ and let $x\\in H$. Its support is a finite set $F$ of primes. For $p\\in F$, choose by the Chinese remainder theorem an integer $k$ that is one modulo $p$ and zero modulo every other prime in $F$. Then $kx$ is supported only at $p$, where it is nonzero. It generates the entire $C_p$ coordinate. Thus $H$ contains every coordinate factor appearing in one of its elements.","Sub(D)$. Every subgroup is fully invariant under all endomorphisms of $D$, and\n\\[\nD_B\\cong D_C\\quad\\Longleftrightarrow\\quad B=C.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $H\\le D$ and let $x\\in H$. Its support is a finite set $F$ of primes. For $p\\in F$, choose by the Chinese remainder theorem an integer $k$ that is one modulo $p$ and zero modulo every other prime in $F$. Then $kx$ is supported only at $p$, where it is nonzero. It generates the entire $C_p$ coordinate. Thus $H$ contains every coordinate factor appearing in one of its elements.","ll endomorphisms of $D$, and\n\\[\nD_B\\cong D_C\\quad\\Longleftrightarrow\\quad B=C.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $H\\le D$ and let $x\\in H$. Its support is a finite set $F$ of primes. For $p\\in F$, choose by the Chinese remainder theorem an integer $k$ that is one modulo $p$ and zero modulo every other prime in $F$. Then $kx$ is supported only at $p$, where it is nonzero. It generates the entire $C_p$ coordinate. Thus $H$ contains every coordinate factor appearing in one of its elements.",", and\n\\[\nD_B\\cong D_C\\quad\\Longleftrightarrow\\quad B=C.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $H\\le D$ and let $x\\in H$. Its support is a finite set $F$ of primes. For $p\\in F$, choose by the Chinese remainder theorem an integer $k$ that is one modulo $p$ and zero modulo every other prime in $F$. Then $kx$ is supported only at $p$, where it is nonzero. It generates the entire $C_p$ coordinate. Thus $H$ contains every coordinate factor appearing in one of its elements.","Let $B$ be the set of indices of these factors. Every element of $H$ lies in $D_B$, and every finite sum of the factors in $B$ lies in $H$. Hence $H=D_B$. Uniqueness follows by testing membership of $e_{p_i}$. Proposition~\\ref{thm:endD} proves full invariance. Finally $\\sigma_{p_i}(D_B)=1$ exactly when $i\\in B$, and these values are isomorphism invariants.\n\\end{proof}\nThe lattice operations have literal set-theoretic descriptions:\n\\[\nD_B+D_C=D_{B\\cup C},\\qquad D_B\\cap D_C=D_{B\\cap C},\n\\qquad D=D_B\\oplus D_{\\N\\setminus B}.\n\\]\nThe infinite support choice is therefore already visible in an elementary algebraic lattice. No exotic group operation is required.","e factors. Every element of $H$ lies in $D_B$, and every finite sum of the factors in $B$ lies in $H$. Hence $H=D_B$. Uniqueness follows by testing membership of $e_{p_i}$. Proposition~\\ref{thm:endD} proves full invariance. Finally $\\sigma_{p_i}(D_B)=1$ exactly when $i\\in B$, and these values are isomorphism invariants.\n\\end{proof}\nThe lattice operations have literal set-theoretic descriptions:\n\\[\nD_B+D_C=D_{B\\cup C},\\qquad D_B\\cap D_C=D_{B\\cap C},\n\\qquad D=D_B\\oplus D_{\\N\\setminus B}.\n\\]\nThe infinite support choice is therefore already visible in an elementary algebraic lattice. No exotic group operation is required.","Give $\\Sub(D)$ the topology inherited from $\\{0,1\\}^D$ by its membership functions. Membership of a vector with finite support depends on finitely many selector bits. Conversely, membership of $e_{p_i}$ recovers the $i$th bit. Consequently the order isomorphism above is also a homeomorphism from Cantor space $\\bits$ to $\\Sub(D)$.","\\begin{corollary}[Finite observations leave continuum many possibilities]\n\\label{thm:finiteviews}\nFix finitely many prime indices and their membership values. There are continuum many pairwise nonisomorphic $D_B$ consistent with those values. A full compatible sequence of finite membership patterns determines one $B$, but no fixed finite prefix does.\n\\end{corollary}\n\\begin{proof}\nThe unspecified bits form a countably infinite set of independent binary choices. Distinct completions give nonisomorphic groups by Theorem~\\ref{thm:subD}. A full compatible family fixes every coordinate and hence fixes the subset.\n\\end{proof}","\\section{Enumeration is enough for a computable copy}\nA subset of $\\N$ is computably enumerable if a program can list exactly its elements. The program need not decide that an element will never appear. All these notions relativize to an oracle $X$.","\\begin{theorem}[The prime-spectrum criterion]\n\\label{thm:spectrumce}\nFor every oracle $X$,\n\\[\nD_B\\text{ has an }X\\text{-computable copy}\n\\quad\\Longleftrightarrow\\quad\nB\\text{ is computably enumerable relative to }X.\n\\]\n\\end{theorem}\n\\begin{proof}\nSuppose first that a copy is computable in $X$. For each $i$, search its domain for a nonzero element $x$ with $p_i x=0$. Each individual test is decidable relative to $X$. Such a witness exists exactly when $i\\in B$. Dovetailing the searches enumerates $B$.","The prime-spectrum criterion]\n\\label{thm:spectrumce}\nFor every oracle $X$,\n\\[\nD_B\\text{ has an }X\\text{-computable copy}\n\\quad\\Longleftrightarrow\\quad\nB\\text{ is computably enumerable relative to }X.\n\\]\n\\end{theorem}\n\\begin{proof}\nSuppose first that a copy is computable in $X$. For each $i$, search its domain for a nonzero element $x$ with $p_i x=0$. Each individual test is decidable relative to $X$. Such a witness exists exactly when $i\\in B$. Dovetailing the searches enumerates $B$.",":spectrumce}\nFor every oracle $X$,\n\\[\nD_B\\text{ has an }X\\text{-computable copy}\n\\quad\\Longleftrightarrow\\quad\nB\\text{ is computably enumerable relative to }X.\n\\]\n\\end{theorem}\n\\begin{proof}\nSuppose first that a copy is computable in $X$. For each $i$, search its domain for a nonzero element $x$ with $p_i x=0$. Each individual test is decidable relative to $X$. Such a witness exists exactly when $i\\in B$. Dovetailing the searches enumerates $B$.","{ has an }X\\text{-computable copy}\n\\quad\\Longleftrightarrow\\quad\nB\\text{ is computably enumerable relative to }X.\n\\]\n\\end{theorem}\n\\begin{proof}\nSuppose first that a copy is computable in $X$. For each $i$, search its domain for a nonzero element $x$ with $p_i x=0$. Each individual test is decidable relative to $X$. Such a witness exists exactly when $i\\in B$. Dovetailing the searches enumerates $B$.","Conversely, suppose $B$ is infinite and enumerable in $X$. List its distinct elements as $b_0,b_1,\\ldots$. To compute $b_j$, wait for the $(j+1)$st distinct enumerated value. This procedure terminates for every $j$. Use canonical finite tuples with the $j$th coordinate in $\\F_{p_{b_j}}$. Coordinatewise addition, normalization, and equality are computable in $X$. If $B$ is finite, $D_B$ is a finite group and has a computable copy. This last existence statement does not require a uniform way to decide when a finite enumeration has finished.\n\\end{proof}\nThis construction is classical. Montalb\\'an gives the prime direct-sum example and its enumeration-degree interpretation explicitly \\cite[Example III.39]{Montalban}.","Conversely, suppose $B$ is infinite and enumerable in $X$. List its distinct elements as $b_0,b_1,\\ldots$. To compute $b_j$, wait for the $(j+1)$st distinct enumerated value. This procedure terminates for every $j$. Use canonical finite tuples with the $j$th coordinate in $\\F_{p_{b_j}}$. Coordinatewise addition, normalization, and equality are computable in $X$. If $B$ is finite, $D_B$ is a finite group and has a computable copy. This last existence statement does not require a uniform way to decide when a finite enumeration has finished.\n\\end{proof}\nThis construction is classical. Montalb\\'an gives the prime direct-sum example and its enumeration-degree interpretation explicitly \\cite[Example III.39]{Montalban}. It is included here because its c","Conversely, suppose $B$ is infinite and enumerable in $X$. List its distinct elements as $b_0,b_1,\\ldots$. To compute $b_j$, wait for the $(j+1)$st distinct enumerated value. This procedure terminates for every $j$. Use canonical finite tuples with the $j$th coordinate in $\\F_{p_{b_j}}$. Coordinatewise addition, normalization, and equality are computable in $X$. If $B$ is finite, $D_B$ is a finite group and has a computable copy. This last existence statement does not require a uniform way to decide when a finite enumeration has finished.\n\\end{proof}\nThis construction is classical. Montalb\\'an gives the prime direct-sum example and its enumeration-degree interpretation explicitly \\cite[Example III.39]{Montalban}. It is included here because its contrast with the later common-completion family","Conversely, suppose $B$ is infinite and enumerable in $X$. List its distinct elements as $b_0,b_1,\\ldots$. To compute $b_j$, wait for the $(j+1)$st distinct enumerated value. This procedure terminates for every $j$. Use canonical finite tuples with the $j$th coordinate in $\\F_{p_{b_j}}$. Coordinatewise addition, normalization, and equality are computable in $X$. If $B$ is finite, $D_B$ is a finite group and has a computable copy. This last existence statement does not require a uniform way to decide when a finite enumeration has finished.\n\\end{proof}\nThis construction is classical. Montalb\\'an gives the prime direct-sum example and its enumeration-degree interpretation explicitly \\cite[Example III.39]{Montalban}. It is included here because its contrast with the later common-completion family is mathematically","Conversely, suppose $B$ is infinite and enumerable in $X$. List its distinct elements as $b_0,b_1,\\ldots$. To compute $b_j$, wait for the $(j+1)$st distinct enumerated value. This procedure terminates for every $j$. Use canonical finite tuples with the $j$th coordinate in $\\F_{p_{b_j}}$. Coordinatewise addition, normalization, and equality are computable in $X$. If $B$ is finite, $D_B$ is a finite group and has a computable copy. This last existence statement does not require a uniform way to decide when a finite enumeration has finished.\n\\end{proof}\nThis construction is classical. Montalb\\'an gives the prime direct-sum example and its enumeration-degree interpretation explicitly \\cite[Example III.39]{Montalban}. It is included here because its contrast with the later common-completion family is mathematically informative.","\\section{A finite definition without computable equality}\nLet $K=\\{n:\\text{program }n\\text{ halts on input }n\\}$ be the diagonal halting set. It is enumerable. It is not decidable: a decision procedure would permit a program that halts on its own index precisely when the decision procedure says it does not. Its complement is not enumerable either, since simultaneous enumeration of a set and its complement decides membership.","Define\n\\[\nB_K=\\{2n:n\\in\\N\\}\\cup\\{2n+1:n\\notin K\\},\n\\qquad H=D_{B_K}.\n\\]\nThis is an explicit mathematical definition of a countably infinite, locally cyclic abelian group.","\\begin{corollary}\n\\label{thm:noncompD}\nThe group $H$ has no computable copy. It nevertheless has an effectively enumerable infinite list of generators and relators.\n\\end{corollary}\n\\begin{proof}\nIf $B_K$ were enumerable, filtering its odd entries would enumerate the complement of $K$. Theorem~\\ref{thm:spectrumce} therefore excludes a computable copy.","For the relator description, introduce a commuting generator $x_i$ for each prime index, impose $p_i x_i=0$, and impose $x_{2n+1}=0$ whenever program $n$ is observed to halt on input $n$. The resulting quotient retains exactly the prime factors indexed by $B_K$. All relators are effectively enumerable. Deciding whether an odd generator has become trivial is not supplied by this enumeration.\n\\end{proof}\nThe conclusion is not absolute finite indescribability. The displayed definition is finite. What fails is the stronger contract of a countable copy with computable operations and decidable equality.","\\section{A purely finite-state boundary}\nConsider a restricted grammar in which a finite automaton reads the unary word $1^n$ and selects the factor $C_{p_n}$ exactly when it accepts.","\\begin{theorem}[Unary selector classification]\n\\label{thm:unary}\nA group $D_B$ is realizable by this unary-selector grammar if and only if $B$ is ultimately periodic.\n\\end{theorem}\n\\begin{proof}\nOn a unary input, the state after $n$ letters is obtained by iterating a single map of a finite state set. The orbit eventually repeats a state, and from that point it is periodic. Its acceptance sequence is therefore ultimately periodic.","Conversely, any finite prefix followed by a periodic binary pattern is recognized by a finite directed path entering a cycle. The prime-spectrum isomorphism criterion ensures that a different accepted set cannot accidentally realize an isomorphic $D_B$.\n\\end{proof}","\\begin{example}[A computable group outside the unary grammar]\nLet $B=\\{2^k:k\\ge0\\}$. It is decidable and enumerable, so $D_B$ has a computable copy. It is not ultimately periodic. Indeed, if $d>0$ were an eventual period, choose $2^k$ beyond the threshold with $2^k>d$. Periodicity would force $2^k+d$ to be a power of two, although it lies strictly between $2^k$ and $2^{k+1}$. Thus $D_B$ is outside the unary grammar. A binary-input automaton recognizes the strings $10^k$, so changing the input representation removes this particular obstruction.\n\\end{example}","\\{2^k:k\\ge0\\}$. It is decidable and enumerable, so $D_B$ has a computable copy. It is not ultimately periodic. Indeed, if $d>0$ were an eventual period, choose $2^k$ beyond the threshold with $2^k>d$. Periodicity would force $2^k+d$ to be a power of two, although it lies strictly between $2^k$ and $2^{k+1}$. Thus $D_B$ is outside the unary grammar. A binary-input automaton recognizes the strings $10^k$, so changing the input representation removes this particular obstruction.\n\\end{example}","\\section{What the first family does not accomplish}\nThe groups $D_B$ have distinct prime spectra, and those differences are already detectable by finite quotients. They do not yet show a failure of reconstruction from the full unmarked finite-quotient system. They isolate a simpler issue: which prime factors have been admitted in the first place.","The main family keeps every prime torsion factor present. It then places the selector in the divisibility of infinite-order elements instead. As a result, all finite quotients remain fixed while the effective cost of the source group varies. The rational space $P/D$ is the mechanism that permits this stronger separation.","\\section{The preimage construction}\nKeep $D\\subseteq P$ and $V=P/D$ from Chapter~\\ref{ch:ambient}. For a rational vector subspace $W\\le V$, define\n\\[\nH_W=q^{-1}(W).\n\\]\nThis is an actual subgroup of $P$ containing $D$. Its quotient by $D$ is canonically $W$. It is dense because it contains the dense subgroup $D$.","The construction separates the ambient coordinate space from the admissible realization. The quotient $V$ records infinite coordinate tails modulo finite changes. A rational subspace $W$ selects which such tails are allowed. The preimage restores all finite-support corrections.","\\begin{definition}[Purity]\nA subgroup $H$ of an abelian group $M$ is pure when\n\\[\nH\\cap mM=mH\\qquad(m\\ge1).\n\\]\nEquivalently, if an equation $my=h$ with $h\\in H$ has a solution in $M$, then it has a solution in $H$.\n\\end{definition}\nPurity is a standard algebraic notion \\cite{Fuchs}. The proof below uses only the displayed criterion. In particular, no structure theorem for mixed abelian groups is required.","\\begin{theorem}[Pure-preimage correspondence]\n\\label{thm:purecorrespondence}\nThe map\n\\[\nW\\longmapsto H_W=q^{-1}(W)\n\\]\nis an inclusion-preserving bijection between rational subspaces of $V$ and pure subgroups $H$ of $P$ containing $D$.\n\\end{theorem}\n\\begin{proof}\nFirst let $W$ be rational and suppose $h\\in H_W\\cap mP$. Because $W$ is closed under division by $m$, choose $x\\in H_W$ with $q(x)=q(h)/m$. Then $d=h-mx$ belongs to $D$. At a prime dividing $m$, the coordinate of $h$ is zero because $h\\in mP$, and that of $mx$ is also zero. Thus $d_p=0$ whenever $p\\mid m$.","eorem}[Pure-preimage correspondence]\n\\label{thm:purecorrespondence}\nThe map\n\\[\nW\\longmapsto H_W=q^{-1}(W)\n\\]\nis an inclusion-preserving bijection between rational subspaces of $V$ and pure subgroups $H$ of $P$ containing $D$.\n\\end{theorem}\n\\begin{proof}\nFirst let $W$ be rational and suppose $h\\in H_W\\cap mP$. Because $W$ is closed under division by $m$, choose $x\\in H_W$ with $q(x)=q(h)/m$. Then $d=h-mx$ belongs to $D$. At a prime dividing $m$, the coordinate of $h$ is zero because $h\\in mP$, and that of $mx$ is also zero. Thus $d_p=0$ whenever $p\\mid m$.","correspondence]\n\\label{thm:purecorrespondence}\nThe map\n\\[\nW\\longmapsto H_W=q^{-1}(W)\n\\]\nis an inclusion-preserving bijection between rational subspaces of $V$ and pure subgroups $H$ of $P$ containing $D$.\n\\end{theorem}\n\\begin{proof}\nFirst let $W$ be rational and suppose $h\\in H_W\\cap mP$. Because $W$ is closed under division by $m$, choose $x\\in H_W$ with $q(x)=q(h)/m$. Then $d=h-mx$ belongs to $D$. At a prime dividing $m$, the coordinate of $h$ is zero because $h\\in mP$, and that of $mx$ is also zero. Thus $d_p=0$ whenever $p\\mid m$.","hm:purecorrespondence}\nThe map\n\\[\nW\\longmapsto H_W=q^{-1}(W)\n\\]\nis an inclusion-preserving bijection between rational subspaces of $V$ and pure subgroups $H$ of $P$ containing $D$.\n\\end{theorem}\n\\begin{proof}\nFirst let $W$ be rational and suppose $h\\in H_W\\cap mP$. Because $W$ is closed under division by $m$, choose $x\\in H_W$ with $q(x)=q(h)/m$. Then $d=h-mx$ belongs to $D$. At a prime dividing $m$, the coordinate of $h$ is zero because $h\\in mP$, and that of $mx$ is also zero. Thus $d_p=0$ whenever $p\\mid m$.","The map\n\\[\nW\\longmapsto H_W=q^{-1}(W)\n\\]\nis an inclusion-preserving bijection between rational subspaces of $V$ and pure subgroups $H$ of $P$ containing $D$.\n\\end{theorem}\n\\begin{proof}\nFirst let $W$ be rational and suppose $h\\in H_W\\cap mP$. Because $W$ is closed under division by $m$, choose $x\\in H_W$ with $q(x)=q(h)/m$. Then $d=h-mx$ belongs to $D$. At a prime dividing $m$, the coordinate of $h$ is zero because $h\\in mP$, and that of $mx$ is also zero. Thus $d_p=0$ whenever $p\\mid m$.","sto H_W=q^{-1}(W)\n\\]\nis an inclusion-preserving bijection between rational subspaces of $V$ and pure subgroups $H$ of $P$ containing $D$.\n\\end{theorem}\n\\begin{proof}\nFirst let $W$ be rational and suppose $h\\in H_W\\cap mP$. Because $W$ is closed under division by $m$, choose $x\\in H_W$ with $q(x)=q(h)/m$. Then $d=h-mx$ belongs to $D$. At a prime dividing $m$, the coordinate of $h$ is zero because $h\\in mP$, and that of $mx$ is also zero. Thus $d_p=0$ whenever $p\\mid m$.","an inclusion-preserving bijection between rational subspaces of $V$ and pure subgroups $H$ of $P$ containing $D$.\n\\end{theorem}\n\\begin{proof}\nFirst let $W$ be rational and suppose $h\\in H_W\\cap mP$. Because $W$ is closed under division by $m$, choose $x\\in H_W$ with $q(x)=q(h)/m$. Then $d=h-mx$ belongs to $D$. At a prime dividing $m$, the coordinate of $h$ is zero because $h\\in mP$, and that of $mx$ is also zero. Thus $d_p=0$ whenever $p\\mid m$.","bijection between rational subspaces of $V$ and pure subgroups $H$ of $P$ containing $D$.\n\\end{theorem}\n\\begin{proof}\nFirst let $W$ be rational and suppose $h\\in H_W\\cap mP$. Because $W$ is closed under division by $m$, choose $x\\in H_W$ with $q(x)=q(h)/m$. Then $d=h-mx$ belongs to $D$. At a prime dividing $m$, the coordinate of $h$ is zero because $h\\in mP$, and that of $mx$ is also zero. Thus $d_p=0$ whenever $p\\mid m$.","n rational subspaces of $V$ and pure subgroups $H$ of $P$ containing $D$.\n\\end{theorem}\n\\begin{proof}\nFirst let $W$ be rational and suppose $h\\in H_W\\cap mP$. Because $W$ is closed under division by $m$, choose $x\\in H_W$ with $q(x)=q(h)/m$. Then $d=h-mx$ belongs to $D$. At a prime dividing $m$, the coordinate of $h$ is zero because $h\\in mP$, and that of $mx$ is also zero. Thus $d_p=0$ whenever $p\\mid m$.","Define $e_p=m^{-1}d_p$ at every other prime and $e_p=0$ at primes dividing $m$. The vector $e$ has finite support and satisfies $me=d$. Hence $h=m(x+e)$ with $x+e\\in H_W$. This proves purity.","Conversely, let $D\\subseteq H\\subseteq P$ be pure. Its image $q(H)$ is an additive subgroup of $V$. Given $h\\in H$ and $m\\ge1$, let $d\\in D$ agree with $h$ on the primes dividing $m$ and vanish elsewhere. Then $h-d\\in mP\\cap H$. Purity gives $h-d=mx$ for some $x\\in H$, so $q(x)=q(h)/m$. Thus $q(H)$ is closed under rational scalars. Since $D\\subseteq H$, the equality $H=q^{-1}(q(H))$ holds. This proves that the constructions are inverse. Both preserve inclusion.\n\\end{proof}\nThe bottom object is $H_{\\{0\\}}=D$, and the top object is $H_V=P$. Intersections of rational subspaces correspond to intersections of their preimages. Sums correspond to subgroup sums, with $D$ serving as the bottom of the interval of intermediate groups.","Given $h\\in H$ and $m\\ge1$, let $d\\in D$ agree with $h$ on the primes dividing $m$ and vanish elsewhere. Then $h-d\\in mP\\cap H$. Purity gives $h-d=mx$ for some $x\\in H$, so $q(x)=q(h)/m$. Thus $q(H)$ is closed under rational scalars. Since $D\\subseteq H$, the equality $H=q^{-1}(q(H))$ holds. This proves that the constructions are inverse. Both preserve inclusion.\n\\end{proof}\nThe bottom object is $H_{\\{0\\}}=D$, and the top object is $H_V=P$. Intersections of rational subspaces correspond to intersections of their preimages. Sums correspond to subgroup sums, with $D$ serving as the bottom of the interval of intermediate groups.","D$ agree with $h$ on the primes dividing $m$ and vanish elsewhere. Then $h-d\\in mP\\cap H$. Purity gives $h-d=mx$ for some $x\\in H$, so $q(x)=q(h)/m$. Thus $q(H)$ is closed under rational scalars. Since $D\\subseteq H$, the equality $H=q^{-1}(q(H))$ holds. This proves that the constructions are inverse. Both preserve inclusion.\n\\end{proof}\nThe bottom object is $H_{\\{0\\}}=D$, and the top object is $H_V=P$. Intersections of rational subspaces correspond to intersections of their preimages. Sums correspond to subgroup sums, with $D$ serving as the bottom of the interval of intermediate groups.","so $q(x)=q(h)/m$. Thus $q(H)$ is closed under rational scalars. Since $D\\subseteq H$, the equality $H=q^{-1}(q(H))$ holds. This proves that the constructions are inverse. Both preserve inclusion.\n\\end{proof}\nThe bottom object is $H_{\\{0\\}}=D$, and the top object is $H_V=P$. Intersections of rational subspaces correspond to intersections of their preimages. Sums correspond to subgroup sums, with $D$ serving as the bottom of the interval of intermediate groups.","\\section{Root conditions do not depend on the subspace}\n\\begin{theorem}[Ambient root criterion]\n\\label{thm:ambientroot}\nFor every rational subspace $W$, every $h\\in H_W$, and every $m\\ge1$,\n\\[\nh\\in mH_W\\quad\\Longleftrightarrow\\quad\nh_p=0\\text{ for all primes }p\\mid m.\n\\]\n\\end{theorem}\n\\begin{proof}\nMultiplication by $m$ on $P$ is zero at primes dividing $m$ and invertible on every other coordinate. Thus the right-hand condition is equivalent to $h\\in mP$. Apply purity from Theorem~\\ref{thm:purecorrespondence}.\n\\end{proof}\nThe condition is independent of $W$, but the set of admissible $h$ is not. This distinction is the reason identical quotient laws can coexist with different original groups.","pace}\n\\begin{theorem}[Ambient root criterion]\n\\label{thm:ambientroot}\nFor every rational subspace $W$, every $h\\in H_W$, and every $m\\ge1$,\n\\[\nh\\in mH_W\\quad\\Longleftrightarrow\\quad\nh_p=0\\text{ for all primes }p\\mid m.\n\\]\n\\end{theorem}\n\\begin{proof}\nMultiplication by $m$ on $P$ is zero at primes dividing $m$ and invertible on every other coordinate. Thus the right-hand condition is equivalent to $h\\in mP$. Apply purity from Theorem~\\ref{thm:purecorrespondence}.\n\\end{proof}\nThe condition is independent of $W$, but the set of admissible $h$ is not. This distinction is the reason identical quotient laws can coexist with different original groups.","\\in H_W$, and every $m\\ge1$,\n\\[\nh\\in mH_W\\quad\\Longleftrightarrow\\quad\nh_p=0\\text{ for all primes }p\\mid m.\n\\]\n\\end{theorem}\n\\begin{proof}\nMultiplication by $m$ on $P$ is zero at primes dividing $m$ and invertible on every other coordinate. Thus the right-hand condition is equivalent to $h\\in mP$. Apply purity from Theorem~\\ref{thm:purecorrespondence}.\n\\end{proof}\nThe condition is independent of $W$, but the set of admissible $h$ is not. This distinction is the reason identical quotient laws can coexist with different original groups.","h\\in mH_W\\quad\\Longleftrightarrow\\quad\nh_p=0\\text{ for all primes }p\\mid m.\n\\]\n\\end{theorem}\n\\begin{proof}\nMultiplication by $m$ on $P$ is zero at primes dividing $m$ and invertible on every other coordinate. Thus the right-hand condition is equivalent to $h\\in mP$. Apply purity from Theorem~\\ref{thm:purecorrespondence}.\n\\end{proof}\nThe condition is independent of $W$, but the set of admissible $h$ is not. This distinction is the reason identical quotient laws can coexist with different original groups.","\\begin{theorem}[One finite-quotient system for every preimage]\n\\label{thm:generalquotients}\nFor every rational subspace $W$ and every $m\\ge1$, the coordinate map induces an isomorphism\n\\[\nH_W/mH_W\\ \\cong\\ \\prod_{p\\mid m}C_p.\n\\]\nFor $m\\mid n$, the quotient transition map is the coordinate restriction from primes dividing $n$ to primes dividing $m$. These quotients are cofinal among all finite quotients, and\n\\[\n\\widehat{H_W}\\cong P.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe coordinate map onto the displayed finite product is surjective already on $D$. Its kernel is $mH_W$ by Theorem~\\ref{thm:ambientroot}. The transition assertion follows by evaluating the same coordinates.","em}[One finite-quotient system for every preimage]\n\\label{thm:generalquotients}\nFor every rational subspace $W$ and every $m\\ge1$, the coordinate map induces an isomorphism\n\\[\nH_W/mH_W\\ \\cong\\ \\prod_{p\\mid m}C_p.\n\\]\nFor $m\\mid n$, the quotient transition map is the coordinate restriction from primes dividing $n$ to primes dividing $m$. These quotients are cofinal among all finite quotients, and\n\\[\n\\widehat{H_W}\\cong P.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe coordinate map onto the displayed finite product is surjective already on $D$. Its kernel is $mH_W$ by Theorem~\\ref{thm:ambientroot}. The transition assertion follows by evaluating the same coordinates.","e coordinate map induces an isomorphism\n\\[\nH_W/mH_W\\ \\cong\\ \\prod_{p\\mid m}C_p.\n\\]\nFor $m\\mid n$, the quotient transition map is the coordinate restriction from primes dividing $n$ to primes dividing $m$. These quotients are cofinal among all finite quotients, and\n\\[\n\\widehat{H_W}\\cong P.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe coordinate map onto the displayed finite product is surjective already on $D$. Its kernel is $mH_W$ by Theorem~\\ref{thm:ambientroot}. The transition assertion follows by evaluating the same coordinates.","duces an isomorphism\n\\[\nH_W/mH_W\\ \\cong\\ \\prod_{p\\mid m}C_p.\n\\]\nFor $m\\mid n$, the quotient transition map is the coordinate restriction from primes dividing $n$ to primes dividing $m$. These quotients are cofinal among all finite quotients, and\n\\[\n\\widehat{H_W}\\cong P.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe coordinate map onto the displayed finite product is surjective already on $D$. Its kernel is $mH_W$ by Theorem~\\ref{thm:ambientroot}. The transition assertion follows by evaluating the same coordinates.","If $H_W\\to F$ is a homomorphism to a finite group, then the image is abelian and is killed by $\\lvert F\\rvert$. Its kernel therefore contains $\\lvert F\\rvert H_W$. Hence the displayed kernels are cofinal. Their inverse limit is the product of all prime coordinates, which is $P$. Under this identification the completion map is the original inclusion $H_W\\hookrightarrow P$.\n\\end{proof}\nThere is a stronger natural statement. It compares maps to every finite target rather than merely comparing the quotient groups.","\\begin{proposition}[Finite-target equivalence]\n\\label{thm:finitehom}\nFor every finite group $F$, restriction along $D\\hookrightarrow H_W$ gives a natural bijection\n\\[\n\\Hom(H_W,F)\\longrightarrow\\Hom(D,F).\n\\]\n\\end{proposition}\n\\begin{proof}\nPut $m=\\lvert F\\rvert$. Every $h\\in H_W$ agrees modulo $mH_W$ with an element of $D$, namely its finite coordinate vector on primes dividing $m$. Every homomorphism to $F$ kills $mH_W$. Thus two maps agreeing on $D$ agree everywhere.","roposition}[Finite-target equivalence]\n\\label{thm:finitehom}\nFor every finite group $F$, restriction along $D\\hookrightarrow H_W$ gives a natural bijection\n\\[\n\\Hom(H_W,F)\\longrightarrow\\Hom(D,F).\n\\]\n\\end{proposition}\n\\begin{proof}\nPut $m=\\lvert F\\rvert$. Every $h\\in H_W$ agrees modulo $mH_W$ with an element of $D$, namely its finite coordinate vector on primes dividing $m$. Every homomorphism to $F$ kills $mH_W$. Thus two maps agreeing on $D$ agree everywhere.","}[Finite-target equivalence]\n\\label{thm:finitehom}\nFor every finite group $F$, restriction along $D\\hookrightarrow H_W$ gives a natural bijection\n\\[\n\\Hom(H_W,F)\\longrightarrow\\Hom(D,F).\n\\]\n\\end{proposition}\n\\begin{proof}\nPut $m=\\lvert F\\rvert$. Every $h\\in H_W$ agrees modulo $mH_W$ with an element of $D$, namely its finite coordinate vector on primes dividing $m$. Every homomorphism to $F$ kills $mH_W$. Thus two maps agreeing on $D$ agree everywhere.","l{thm:finitehom}\nFor every finite group $F$, restriction along $D\\hookrightarrow H_W$ gives a natural bijection\n\\[\n\\Hom(H_W,F)\\longrightarrow\\Hom(D,F).\n\\]\n\\end{proposition}\n\\begin{proof}\nPut $m=\\lvert F\\rvert$. Every $h\\in H_W$ agrees modulo $mH_W$ with an element of $D$, namely its finite coordinate vector on primes dividing $m$. Every homomorphism to $F$ kills $mH_W$. Thus two maps agreeing on $D$ agree everywhere.","Conversely, a map $D\\to F$ kills every prime factor whose order does not divide $m$. It factors through the finite coordinate product on primes dividing $m$. Composing that finite-coordinate rule with the coordinate projection of $H_W$ extends it. The extension is unique by the first part. Postcomposition with a map of finite groups commutes with restriction, giving naturality.\n\\end{proof}\nThe argument never subtracts maps to a nonabelian group. The assertion is a bijection of sets of homomorphisms, and naturality is with respect to postcomposition.","\\section{Countability, rank, and nonsplitting}\n\\begin{proposition}[Structure of an ambient preimage]\n\\label{thm:HWstructure}\nFor every rational subspace $W$,\n\\[\n\\Tor(H_W)=D,\\qquad H_W/D\\cong W.\n\\]\nThe group $H_W$ is reduced and residually finite. It is countable exactly when $W$ has finite or countably infinite dimension over $\\Q$. If $W\\ne0$, the sequence\n\\[\n0\\longrightarrow D\\longrightarrow H_W\\longrightarrow W\\longrightarrow0\n\\]\ndoes not split.\n\\end{proposition}\n\\begin{proof}\nThe torsion equality follows from $D\\subseteq H_W\\subseteq P$ and Theorem~\\ref{thm:torP}. The quotient assertion is the first isomorphism theorem. Reducedness follows from reducedness of $P$. A nonzero element has a nonzero finite coordinate, which detects it and proves residual finiteness.","ture}\nFor every rational subspace $W$,\n\\[\n\\Tor(H_W)=D,\\qquad H_W/D\\cong W.\n\\]\nThe group $H_W$ is reduced and residually finite. It is countable exactly when $W$ has finite or countably infinite dimension over $\\Q$. If $W\\ne0$, the sequence\n\\[\n0\\longrightarrow D\\longrightarrow H_W\\longrightarrow W\\longrightarrow0\n\\]\ndoes not split.\n\\end{proposition}\n\\begin{proof}\nThe torsion equality follows from $D\\subseteq H_W\\subseteq P$ and Theorem~\\ref{thm:torP}. The quotient assertion is the first isomorphism theorem. Reducedness follows from reducedness of $P$. A nonzero element has a nonzero finite coordinate, which detects it and proves residual finiteness.","hen $W$ has finite or countably infinite dimension over $\\Q$. If $W\\ne0$, the sequence\n\\[\n0\\longrightarrow D\\longrightarrow H_W\\longrightarrow W\\longrightarrow0\n\\]\ndoes not split.\n\\end{proposition}\n\\begin{proof}\nThe torsion equality follows from $D\\subseteq H_W\\subseteq P$ and Theorem~\\ref{thm:torP}. The quotient assertion is the first isomorphism theorem. Reducedness follows from reducedness of $P$. A nonzero element has a nonzero finite coordinate, which detects it and proves residual finiteness.","A rational vector space on a finite or countable basis is countable, since each vector has a finite rational linear expression. Each fiber of $H_W\\to W$ is a coset of the countable group $D$. Thus $H_W$ is countable in this case. Conversely, a quotient of a countable group is countable and has at most a countable basis.","A splitting would embed the nonzero divisible group $W$ into $P$. Theorem~\\ref{thm:Vrational} excludes every such subgroup.\n\\end{proof}\nThe rank of the torsion-free quotient is exactly $\\dim_{\\Q}W$. This gives countable examples of every finite rank and of countably infinite rank with the same finite-quotient system. The rank-one family studied next is chosen because it admits an especially sharp classification and computability theorem.","\\section{What has been fixed and what still varies}\nThe entire product $P$, all finite quotient objects, their transition maps, and the maps to arbitrary finite targets are fixed throughout the construction. The subspace $W$ remains free. This is an ambient realization theorem, not a method for discovering $W$ from those observations.","The preimage formula also identifies the correct role of completion. Every $H_W$ contains the same dense torsion subgroup. Its completion admits every compatible coordinate tuple. Its original image admits only those tuples whose tail class lies in $W$. The missing information is an admissibility condition on the quotient, not a defect in the finite coordinates themselves.","\\section{Choosing a rational line}\nLet $A\\subseteq\\N$, write $a_n=\\one_A(n)$, and define\n\\[\nS_A=\\{p_{2n+a_n}:n\\in\\N\\}.\n\\]\nExactly one prime from each pair is selected. Both $S_A$ and its complement are infinite. Let $u_A\\in P$ be its indicator:\n\\[\n(u_A)_p=\\begin{cases}1,&p\\in S_A,","\\section{Choosing a rational line}\nLet $A\\subseteq\\N$, write $a_n=\\one_A(n)$, and define\n\\[\nS_A=\\{p_{2n+a_n}:n\\in\\N\\}.\n\\]\nExactly one prime from each pair is selected. Both $S_A$ and its complement are infinite. Let $u_A\\in P$ be its indicator:\n\\[\n(u_A)_p=\\begin{cases}1,&p\\in S_A,\\\\0,&p\\notin S_A.\\end{cases}\n\\]\nBecause $u_A$ has infinite support, $q(u_A)\\ne0$. We can therefore define a rational line and its preimage by\n\\[\nW_A=\\Q q(u_A),\\qquad G_A=q^{-1}(W_A).\n\\]\nThis definition produces an actual group for every set $A$. It does not claim an effective method for obtaining the selector bits of an arbitrary $A$.","0,&p\\notin S_A.\\end{cases}\n\\]\nBecause $u_A$ has infinite support, $q(u_A)\\ne0$. We can therefore define a rational line and its preimage by\n\\[\nW_A=\\Q q(u_A),\\qquad G_A=q^{-1}(W_A).\n\\]\nThis definition produces an actual group for every set $A$. It does not claim an effective method for obtaining the selector bits of an arbitrary $A$.","\\begin{figure}[tbp]\n\\centering\\includegraphics[width=\\textwidth]{\\figfile{02_prime_pairs.pdf}}\n\\caption{Each bit selects exactly one prime from its pair. A filled point is selected. The torsion factor $C_p$ is present in $G_A$ whether or not $p$ is selected. Selection controls the infinite-order lift and its divisibility, not the presence of the torsion factor.}\n\\label{fig:primepairs}\n\\end{figure}","\\section{A section with finite arithmetic defects}\nFor a reduced rational $r=a/b$, with $b>0$, define\n\\begin{equation}\n\\label{eq:section}\n(s_A(a/b))_p=\n\\begin{cases}\na b^{-1}\\pmod p,&p\\in S_A\\text{ and }p\\nmid b,","\\section{A section with finite arithmetic defects}\nFor a reduced rational $r=a/b$, with $b>0$, define\n\\begin{equation}\n\\label{eq:section}\n(s_A(a/b))_p=\n\\begin{cases}\na b^{-1}\\pmod p,&p\\in S_A\\text{ and }p\\nmid b,\\\\\n0,&p\\notin S_A\\text{ or }p\\mid b.\n\\end{cases}\n\\end{equation}\nThe inverse is used only when the denominator is a unit in $\\F_p$. The convention at denominator primes is essential. It is a choice of representatives, not a claim that division by $p$ is defined in $C_p$.","0,&p\\notin S_A\\text{ or }p\\mid b.\n\\end{cases}\n\\end{equation}\nThe inverse is used only when the denominator is a unit in $\\F_p$. The convention at denominator primes is essential. It is a choice of representatives, not a claim that division by $p$ is defined in $C_p$.","The difference $b s_A(a/b)-a u_A$ is supported on the prime divisors of $b$. It follows that\n\\[\nq(s_A(r))=r q(u_A).\n\\]\nThus every element of $G_A$ is $s_A(r)+d$ for some $r\\in\\Q$ and $d\\in D$.","\\begin{theorem}[Unique normal form]\n\\label{thm:normalform}\nEvery element of $G_A$ has a unique expression\n\\[\ng=s_A(r)+d,\\qquad (r,d)\\in\\Q\\times D.\n\\]\nConsequently $G_A$ is countably infinite, with a countable normal-form domain that does not depend on the selector.\n\\end{theorem}\n\\begin{proof}\nExistence follows from the preimage definition and the displayed quotient identity. If two forms are equal, applying $q$ gives $(r-r')q(u_A)=0$. The vector $q(u_A)$ is nonzero in a rational vector space, so $r=r'$. The finite-support terms are then equal. Both $\\Q$ and $D$ are countable, while the quotient is the infinite group $\\Q$.\n\\end{proof}\nThe domain can be named without knowing $A$. The group operation on those names will contain the selector information.","em}[Unique normal form]\n\\label{thm:normalform}\nEvery element of $G_A$ has a unique expression\n\\[\ng=s_A(r)+d,\\qquad (r,d)\\in\\Q\\times D.\n\\]\nConsequently $G_A$ is countably infinite, with a countable normal-form domain that does not depend on the selector.\n\\end{theorem}\n\\begin{proof}\nExistence follows from the preimage definition and the displayed quotient identity. If two forms are equal, applying $q$ gives $(r-r')q(u_A)=0$. The vector $q(u_A)$ is nonzero in a rational vector space, so $r=r'$. The finite-support terms are then equal. Both $\\Q$ and $D$ are countable, while the quotient is the infinite group $\\Q$.\n\\end{proof}\nThe domain can be named without knowing $A$. The group operation on those names will contain the selector information.","normal form]\n\\label{thm:normalform}\nEvery element of $G_A$ has a unique expression\n\\[\ng=s_A(r)+d,\\qquad (r,d)\\in\\Q\\times D.\n\\]\nConsequently $G_A$ is countably infinite, with a countable normal-form domain that does not depend on the selector.\n\\end{theorem}\n\\begin{proof}\nExistence follows from the preimage definition and the displayed quotient identity. If two forms are equal, applying $q$ gives $(r-r')q(u_A)=0$. The vector $q(u_A)$ is nonzero in a rational vector space, so $r=r'$. The finite-support terms are then equal. Both $\\Q$ and $D$ are countable, while the quotient is the infinite group $\\Q$.\n\\end{proof}\nThe domain can be named without knowing $A$. The group operation on those names will contain the selector information.","alform}\nEvery element of $G_A$ has a unique expression\n\\[\ng=s_A(r)+d,\\qquad (r,d)\\in\\Q\\times D.\n\\]\nConsequently $G_A$ is countably infinite, with a countable normal-form domain that does not depend on the selector.\n\\end{theorem}\n\\begin{proof}\nExistence follows from the preimage definition and the displayed quotient identity. If two forms are equal, applying $q$ gives $(r-r')q(u_A)=0$. The vector $q(u_A)$ is nonzero in a rational vector space, so $r=r'$. The finite-support terms are then equal. Both $\\Q$ and $D$ are countable, while the quotient is the infinite group $\\Q$.\n\\end{proof}\nThe domain can be named without knowing $A$. The group operation on those names will contain the selector information.","ion\n\\[\ng=s_A(r)+d,\\qquad (r,d)\\in\\Q\\times D.\n\\]\nConsequently $G_A$ is countably infinite, with a countable normal-form domain that does not depend on the selector.\n\\end{theorem}\n\\begin{proof}\nExistence follows from the preimage definition and the displayed quotient identity. If two forms are equal, applying $q$ gives $(r-r')q(u_A)=0$. The vector $q(u_A)$ is nonzero in a rational vector space, so $r=r'$. The finite-support terms are then equal. Both $\\Q$ and $D$ are countable, while the quotient is the infinite group $\\Q$.\n\\end{proof}\nThe domain can be named without knowing $A$. The group operation on those names will contain the selector information.","uently $G_A$ is countably infinite, with a countable normal-form domain that does not depend on the selector.\n\\end{theorem}\n\\begin{proof}\nExistence follows from the preimage definition and the displayed quotient identity. If two forms are equal, applying $q$ gives $(r-r')q(u_A)=0$. The vector $q(u_A)$ is nonzero in a rational vector space, so $r=r'$. The finite-support terms are then equal. Both $\\Q$ and $D$ are countable, while the quotient is the infinite group $\\Q$.\n\\end{proof}\nThe domain can be named without knowing $A$. The group operation on those names will contain the selector information.","_A$ is countably infinite, with a countable normal-form domain that does not depend on the selector.\n\\end{theorem}\n\\begin{proof}\nExistence follows from the preimage definition and the displayed quotient identity. If two forms are equal, applying $q$ gives $(r-r')q(u_A)=0$. The vector $q(u_A)$ is nonzero in a rational vector space, so $r=r'$. The finite-support terms are then equal. Both $\\Q$ and $D$ are countable, while the quotient is the infinite group $\\Q$.\n\\end{proof}\nThe domain can be named without knowing $A$. The group operation on those names will contain the selector information.","in that does not depend on the selector.\n\\end{theorem}\n\\begin{proof}\nExistence follows from the preimage definition and the displayed quotient identity. If two forms are equal, applying $q$ gives $(r-r')q(u_A)=0$. The vector $q(u_A)$ is nonzero in a rational vector space, so $r=r'$. The finite-support terms are then equal. Both $\\Q$ and $D$ are countable, while the quotient is the infinite group $\\Q$.\n\\end{proof}\nThe domain can be named without knowing $A$. The group operation on those names will contain the selector information.","gin{proof}\nExistence follows from the preimage definition and the displayed quotient identity. If two forms are equal, applying $q$ gives $(r-r')q(u_A)=0$. The vector $q(u_A)$ is nonzero in a rational vector space, so $r=r'$. The finite-support terms are then equal. Both $\\Q$ and $D$ are countable, while the quotient is the infinite group $\\Q$.\n\\end{proof}\nThe domain can be named without knowing $A$. The group operation on those names will contain the selector information.","\\section{The cocycle law}\nDefine\n\\[\nc_A(r,t)=s_A(r)+s_A(t)-s_A(r+t).\n\\]\nThe sum is taken in $P$, but its value belongs to $D$.","\\begin{lemma}[A finite support bound]\n\\label{thm:cocyclesupport}\nFor all $r,t\\in\\Q$,\n\\[\n\\supp c_A(r,t)\\subseteq\n\\{p:p\\mid\\den(r)\\den(t)\\}.\n\\]\n\\end{lemma}\n\\begin{proof}\nAt a prime not dividing either denominator, all three rationals have well-defined reductions. On a selected coordinate, reduction respects addition, so the defect is zero. On an unselected coordinate, all section values are zero. The reduced denominator of $r+t$ has no prime divisor absent from the two original denominators.\n\\end{proof}","port bound]\n\\label{thm:cocyclesupport}\nFor all $r,t\\in\\Q$,\n\\[\n\\supp c_A(r,t)\\subseteq\n\\{p:p\\mid\\den(r)\\den(t)\\}.\n\\]\n\\end{lemma}\n\\begin{proof}\nAt a prime not dividing either denominator, all three rationals have well-defined reductions. On a selected coordinate, reduction respects addition, so the defect is zero. On an unselected coordinate, all section values are zero. The reduced denominator of $r+t$ has no prime divisor absent from the two original denominators.\n\\end{proof}","\\begin{theorem}[Explicit group law]\n\\label{thm:cocyclelaw}\nIn normal-form coordinates, addition is\n\\begin{equation}\n\\label{eq:grouplaw}\n(r,d)+(t,e)=(r+t,d+e+c_A(r,t)).\n\\end{equation}\nThe identity is $(0,0)$ and the inverse of $(r,d)$ is $(-r,-d)$. The law is associative and commutative, and the map $(r,d)\\mapsto s_A(r)+d$ is an isomorphism onto the subgroup $G_A\\le P$.\n\\end{theorem}\n\\begin{proof}\nSubstitution into addition in $P$ gives the formula. The section satisfies $s_A(0)=0$ and $s_A(-r)=-s_A(r)$, so $c_A(r,0)=0$ and $c_A(r,-r)=0$. These identities give the stated identity and inverse.","law]\n\\label{thm:cocyclelaw}\nIn normal-form coordinates, addition is\n\\begin{equation}\n\\label{eq:grouplaw}\n(r,d)+(t,e)=(r+t,d+e+c_A(r,t)).\n\\end{equation}\nThe identity is $(0,0)$ and the inverse of $(r,d)$ is $(-r,-d)$. The law is associative and commutative, and the map $(r,d)\\mapsto s_A(r)+d$ is an isomorphism onto the subgroup $G_A\\le P$.\n\\end{theorem}\n\\begin{proof}\nSubstitution into addition in $P$ gives the formula. The section satisfies $s_A(0)=0$ and $s_A(-r)=-s_A(r)$, so $c_A(r,0)=0$ and $c_A(r,-r)=0$. These identities give the stated identity and inverse.","ordinates, addition is\n\\begin{equation}\n\\label{eq:grouplaw}\n(r,d)+(t,e)=(r+t,d+e+c_A(r,t)).\n\\end{equation}\nThe identity is $(0,0)$ and the inverse of $(r,d)$ is $(-r,-d)$. The law is associative and commutative, and the map $(r,d)\\mapsto s_A(r)+d$ is an isomorphism onto the subgroup $G_A\\le P$.\n\\end{theorem}\n\\begin{proof}\nSubstitution into addition in $P$ gives the formula. The section satisfies $s_A(0)=0$ and $s_A(-r)=-s_A(r)$, so $c_A(r,0)=0$ and $c_A(r,-r)=0$. These identities give the stated identity and inverse.","tion is\n\\begin{equation}\n\\label{eq:grouplaw}\n(r,d)+(t,e)=(r+t,d+e+c_A(r,t)).\n\\end{equation}\nThe identity is $(0,0)$ and the inverse of $(r,d)$ is $(-r,-d)$. The law is associative and commutative, and the map $(r,d)\\mapsto s_A(r)+d$ is an isomorphism onto the subgroup $G_A\\le P$.\n\\end{theorem}\n\\begin{proof}\nSubstitution into addition in $P$ gives the formula. The section satisfies $s_A(0)=0$ and $s_A(-r)=-s_A(r)$, so $c_A(r,0)=0$ and $c_A(r,-r)=0$. These identities give the stated identity and inverse.","n}\n\\label{eq:grouplaw}\n(r,d)+(t,e)=(r+t,d+e+c_A(r,t)).\n\\end{equation}\nThe identity is $(0,0)$ and the inverse of $(r,d)$ is $(-r,-d)$. The law is associative and commutative, and the map $(r,d)\\mapsto s_A(r)+d$ is an isomorphism onto the subgroup $G_A\\le P$.\n\\end{theorem}\n\\begin{proof}\nSubstitution into addition in $P$ gives the formula. The section satisfies $s_A(0)=0$ and $s_A(-r)=-s_A(r)$, so $c_A(r,0)=0$ and $c_A(r,-r)=0$. These identities give the stated identity and inverse.","plaw}\n(r,d)+(t,e)=(r+t,d+e+c_A(r,t)).\n\\end{equation}\nThe identity is $(0,0)$ and the inverse of $(r,d)$ is $(-r,-d)$. The law is associative and commutative, and the map $(r,d)\\mapsto s_A(r)+d$ is an isomorphism onto the subgroup $G_A\\le P$.\n\\end{theorem}\n\\begin{proof}\nSubstitution into addition in $P$ gives the formula. The section satisfies $s_A(0)=0$ and $s_A(-r)=-s_A(r)$, so $c_A(r,0)=0$ and $c_A(r,-r)=0$. These identities give the stated identity and inverse.","Symmetry is immediate. The cocycle equation\n\\[\nc_A(r,t)+c_A(r+t,v)=c_A(t,v)+c_A(r,t+v)\n\\]\nfollows by expanding and cancelling the section terms. Expanding the two associations in \\eqref{eq:grouplaw} now proves associativity. Uniqueness of normal forms makes the ambient map bijective, and the addition formula makes it a homomorphism.\n\\end{proof}\nThis is the standard symmetric cocycle construction of an abelian extension \\cite[Section 3.2]{Lupini}. Here the particular section makes each individual correction finitely supported and exactly computable from finitely many selector queries.","\\begin{proposition}[One cocycle value reads one coordinate]\n\\label{thm:cocyclebit}\nFor every prime $p$,\n\\[\nc_A(1/p,(p-1)/p)=-\\one_{S_A}(p)e_p.\n\\]\n\\end{proposition}\n\\begin{proof}\nThe support bound leaves only the coordinate $p$. At that coordinate both fractional lifts are zero, and the lift of their sum, which is one, is $\\one_{S_A}(p)$.\n\\end{proof}\nDropping the cocycle is therefore not a harmless simplification. The direct product operation on $\\Q\\times D$ removes exactly these defects and gives a different, split extension.","ordinate]\n\\label{thm:cocyclebit}\nFor every prime $p$,\n\\[\nc_A(1/p,(p-1)/p)=-\\one_{S_A}(p)e_p.\n\\]\n\\end{proposition}\n\\begin{proof}\nThe support bound leaves only the coordinate $p$. At that coordinate both fractional lifts are zero, and the lift of their sum, which is one, is $\\one_{S_A}(p)$.\n\\end{proof}\nDropping the cocycle is therefore not a harmless simplification. The direct product operation on $\\Q\\times D$ removes exactly these defects and gives a different, split extension.","\\section{Structure of the extension}\nBy Chapter~\\ref{ch:purity}, the group is pure in $P$, reduced, residually finite, and dense. Its torsion is $D$, and the rational coordinate gives the exact sequence\n\\begin{equation}\n\\label{eq:extension}\n0\\longrightarrow D\\longrightarrow G_A\\xrightarrow{\\rho_A}\\Q\\longrightarrow0,\n\\qquad \\rho_A(s_A(r)+d)=r.\n\\end{equation}\nIt does not split.","There is an equivalent cocycle obstruction. Suppose a function $b:\\Q\\to D$ modified the section to an additive map $r\\mapsto s_A(r)+b(r)$. Then\n\\[\nc_A(r,t)+b(r)+b(t)-b(r+t)=0\n\\]\nfor every $r,t$. The modified section would split \\eqref{eq:extension}. Hence no such function exists. The cocycle is the defect of a section into $P$, but it is not removable by a correction valued in $D$.","\\begin{proposition}[Finitely generated subgroups]\n\\label{thm:fgsubgroups}\nEvery finitely generated subgroup of $G_A$ is either finite cyclic or isomorphic to $\\Z\\oplus C_m$ for a squarefree positive integer $m$, with $C_1$ trivial.\n\\end{proposition}\n\\begin{proof}\nLet $L\\le G_A$ be generated by finitely many elements. Its image in $\\Q$ is a finitely generated subgroup of $\\Q$, hence either zero or infinite cyclic. If it is zero, then $L\\le D$ and its finite support places it inside a finite cyclic group.","subgroups]\n\\label{thm:fgsubgroups}\nEvery finitely generated subgroup of $G_A$ is either finite cyclic or isomorphic to $\\Z\\oplus C_m$ for a squarefree positive integer $m$, with $C_1$ trivial.\n\\end{proposition}\n\\begin{proof}\nLet $L\\le G_A$ be generated by finitely many elements. Its image in $\\Q$ is a finitely generated subgroup of $\\Q$, hence either zero or infinite cyclic. If it is zero, then $L\\le D$ and its finite support places it inside a finite cyclic group.","Otherwise choose a lift $x\\in L$ of a generator of its rational image. Subtracting suitable integer multiples of $x$ from the given generators produces finitely many elements of $D$. They generate the kernel of $L\\to\\rho_A(L)$, which is finite cyclic of squarefree order. The subgroup $\\langle x\\rangle$ intersects that torsion kernel trivially, giving the direct sum.\n\\end{proof}\nThe extensions over finitely generated rational images split, but the extension over all of $\\Q$ does not. This is another exact distinction between f","Otherwise choose a lift $x\\in L$ of a generator of its rational image. Subtracting suitable integer multiples of $x$ from the given generators produces finitely many elements of $D$. They generate the kernel of $L\\to\\rho_A(L)$, which is finite cyclic of squarefree order. The subgroup $\\langle x\\rangle$ intersects that torsion kernel trivially, giving the direct sum.\n\\end{proof}\nThe extensions over finitely generated rational images split, but the extension over all of $\\Q$ does not. This is another exact distinction between finite algebraic stages and the global realization.","tracting suitable integer multiples of $x$ from the given generators produces finitely many elements of $D$. They generate the kernel of $L\\to\\rho_A(L)$, which is finite cyclic of squarefree order. The subgroup $\\langle x\\rangle$ intersects that torsion kernel trivially, giving the direct sum.\n\\end{proof}\nThe extensions over finitely generated rational images split, but the extension over all of $\\Q$ does not. This is another exact distinction between finite algebraic stages and the global realization.","\\begin{figure}[H]\n\\centering\\includegraphics[width=0.94\\textwidth]{\\figfile{03_root_certificates.pdf}}\n\\caption{For the marked element $u_A$, one prime in each pair gives an obstruction and the other gives a root. A nonzero coordinate at a prime dividing the multiplier certifies impossibility. The displayed pattern is an exact worked selector, not a statistical sample.}\n\\label{fig:rootcertificates}\n\\end{figure}","\\section{An exact decision criterion}\nThe general purity argument already establishes the root condition. For $G_A$, normal forms make the successful branch constructive.","\\begin{theorem}[Root or obstruction]\n\\label{thm:rootcertificate}\nLet $g=s_A(r)+d\\in G_A$ and let $m\\ge1$. Then\n\\[\ng\\in mG_A\\quad\\Longleftrightarrow\\quad\n g_p=0\\quad\\text{for every }p\\mid m.\n\\]\nIf the condition fails, a prime $p\\mid m$ with $g_p\\ne0$ is a nonexistence certificate. If it holds, a root is obtained by the finite procedure in the proof.\n\\end{theorem}\n\\begin{proof}\nNecessity follows because multiplication by $m$ is zero in every coordinate $C_p$ with $p\\mid m$.","obstruction]\n\\label{thm:rootcertificate}\nLet $g=s_A(r)+d\\in G_A$ and let $m\\ge1$. Then\n\\[\ng\\in mG_A\\quad\\Longleftrightarrow\\quad\n g_p=0\\quad\\text{for every }p\\mid m.\n\\]\nIf the condition fails, a prime $p\\mid m$ with $g_p\\ne0$ is a nonexistence certificate. If it holds, a root is obtained by the finite procedure in the proof.\n\\end{theorem}\n\\begin{proof}\nNecessity follows because multiplication by $m$ is zero in every coordinate $C_p$ with $p\\mid m$.","ificate}\nLet $g=s_A(r)+d\\in G_A$ and let $m\\ge1$. Then\n\\[\ng\\in mG_A\\quad\\Longleftrightarrow\\quad\n g_p=0\\quad\\text{for every }p\\mid m.\n\\]\nIf the condition fails, a prime $p\\mid m$ with $g_p\\ne0$ is a nonexistence certificate. If it holds, a root is obtained by the finite procedure in the proof.\n\\end{theorem}\n\\begin{proof}\nNecessity follows because multiplication by $m$ is zero in every coordinate $C_p$ with $p\\mid m$.","t $g=s_A(r)+d\\in G_A$ and let $m\\ge1$. Then\n\\[\ng\\in mG_A\\quad\\Longleftrightarrow\\quad\n g_p=0\\quad\\text{for every }p\\mid m.\n\\]\nIf the condition fails, a prime $p\\mid m$ with $g_p\\ne0$ is a nonexistence certificate. If it holds, a root is obtained by the finite procedure in the proof.\n\\end{theorem}\n\\begin{proof}\nNecessity follows because multiplication by $m$ is zero in every coordinate $C_p$ with $p\\mid m$.","Assume the condition holds and put $t=r/m$. The vector\n\\[\nh=g-ms_A(t)\n\\]\nhas zero image in $V$ and therefore lies in $D$. Its support is contained in the union of the support of $d$ and the prime divisors of $\\den(r)\\den(t)$. Indeed, at any other selected coordinate, reduction gives $r-mt=0$, and unselected section values are zero.","For $p\\mid m$, the coordinate $h_p$ is zero. At every other prime put $e_p=m^{-1}h_p$, and at primes dividing $m$ put $e_p=0$. Then $e\\in D$ and $me=h$. Consequently\n\\[\ny=s_A(t)+e\\quad\\text{satisfies}\\quad my=g.\n\\]\nEvery nonzero correction coordinate is found in the stated finite support set.\n\\end{proof}\nA failed root is witnessed by one coordinate. A successful root requires a rational division and a finite-support adjustment. This is a simple instance of a realization problem with both positive and negative certificates.","\\begin{proposition}[All roots and their number]\n\\label{thm:rootfiber}\nIf $my=g$ is solvable, its solution set is the coset\n\\[\ny_0+G_A[m],\\qquad\nG_A[m]=\\bigoplus_{p\\mid m}C_p.\n\\]\nIt has exactly $\\rad(m)$ elements.\n\\end{proposition}\n\\begin{proof}\nTwo solutions differ by an element annihilated by $m$, and adding such an element to a solution gives another. The ambient product has a nonzero $m$-torsion coordinate only when $p\\mid m$. All those finite coordinate factors lie in $D\\subseteq G_A$. The kernel is therefore the displayed finite direct sum, whose cardinal is $\\rad(m)$.\n\\end{proof}\nRepeated prime powers do not enlarge the root kernel. For example, a solvable fourth-root equation has two solutions, not four. The squarefree coordinate structure is responsible for th","\\begin{proposition}[All roots and their number]\n\\label{thm:rootfiber}\nIf $my=g$ is solvable, its solution set is the coset\n\\[\ny_0+G_A[m],\\qquad\nG_A[m]=\\bigoplus_{p\\mid m}C_p.\n\\]\nIt has exactly $\\rad(m)$ elements.\n\\end{proposition}\n\\begin{proof}\nTwo solutions differ by an element annihilated by $m$, and adding such an element to a solution gives another. The ambient product has a nonzero $m$-torsion coordinate only when $p\\mid m$. All those finite coordinate factors lie in $D\\subseteq G_A$. The kernel is therefore the displayed finite direct sum, whose cardinal is $\\rad(m)$.\n\\end{proof}\nRepeated prime powers do not enlarge the root kernel. For example, a solvable fourth-root equation has two solutions, not four. The squarefree coordinate structure is responsible for this behavior.","roots and their number]\n\\label{thm:rootfiber}\nIf $my=g$ is solvable, its solution set is the coset\n\\[\ny_0+G_A[m],\\qquad\nG_A[m]=\\bigoplus_{p\\mid m}C_p.\n\\]\nIt has exactly $\\rad(m)$ elements.\n\\end{proposition}\n\\begin{proof}\nTwo solutions differ by an element annihilated by $m$, and adding such an element to a solution gives another. The ambient product has a nonzero $m$-torsion coordinate only when $p\\mid m$. All those finite coordinate factors lie in $D\\subseteq G_A$. The kernel is therefore the displayed finite direct sum, whose cardinal is $\\rad(m)$.\n\\end{proof}\nRepeated prime powers do not enlarge the root kernel. For example, a solvable fourth-root equation has two solutions, not four. The squarefree coordinate structure is responsible for this behavior.","$ is solvable, its solution set is the coset\n\\[\ny_0+G_A[m],\\qquad\nG_A[m]=\\bigoplus_{p\\mid m}C_p.\n\\]\nIt has exactly $\\rad(m)$ elements.\n\\end{proposition}\n\\begin{proof}\nTwo solutions differ by an element annihilated by $m$, and adding such an element to a solution gives another. The ambient product has a nonzero $m$-torsion coordinate only when $p\\mid m$. All those finite coordinate factors lie in $D\\subseteq G_A$. The kernel is therefore the displayed finite direct sum, whose cardinal is $\\rad(m)$.\n\\end{proof}\nRepeated prime powers do not enlarge the root kernel. For example, a solvable fourth-root equation has two solutions, not four. The squarefree coordinate structure is responsible for this behavior.","_p.\n\\]\nIt has exactly $\\rad(m)$ elements.\n\\end{proposition}\n\\begin{proof}\nTwo solutions differ by an element annihilated by $m$, and adding such an element to a solution gives another. The ambient product has a nonzero $m$-torsion coordinate only when $p\\mid m$. All those finite coordinate factors lie in $D\\subseteq G_A$. The kernel is therefore the displayed finite direct sum, whose cardinal is $\\rad(m)$.\n\\end{proof}\nRepeated prime powers do not enlarge the root kernel. For example, a solvable fourth-root equation has two solutions, not four. The squarefree coordinate structure is responsible for this behavior.","\\section{The quotient formula}\n\\begin{theorem}[All integer quotients]\n\\label{thm:quotientformula}\nFor $m\\ge1$, the coordinate map\n\\[\n\\pi_m:G_A\\longrightarrow\\prod_{p\\mid m}C_p\n\\]\nis surjective with kernel $mG_A$. Thus\n\\[\nG_A/mG_A\\cong\\prod_{p\\mid m}C_p\\cong C_{\\rad(m)}.\n\\]\nFor $m\\mid n$, the transition map is the corresponding coordinate projection.\n\\end{theorem}\n\\begin{proof}\nA vector supported on the prime divisors of $m$ realizes any tuple in the target, proving surjectivity. The root criterion identifies the kernel. The Chinese remainder theorem identifies the finite product with the squarefree cyclic group. The transition assertion follows by taking the same coordinates before and after quotienting.\n\\end{proof}","mula}\n\\begin{theorem}[All integer quotients]\n\\label{thm:quotientformula}\nFor $m\\ge1$, the coordinate map\n\\[\n\\pi_m:G_A\\longrightarrow\\prod_{p\\mid m}C_p\n\\]\nis surjective with kernel $mG_A$. Thus\n\\[\nG_A/mG_A\\cong\\prod_{p\\mid m}C_p\\cong C_{\\rad(m)}.\n\\]\nFor $m\\mid n$, the transition map is the corresponding coordinate projection.\n\\end{theorem}\n\\begin{proof}\nA vector supported on the prime divisors of $m$ realizes any tuple in the target, proving surjectivity. The root criterion identifies the kernel. The Chinese remainder theorem identifies the finite product with the squarefree cyclic group. The transition assertion follows by taking the same coordinates before and after quotienting.\n\\end{proof}","\\begin{center}\n\\begin{tabular}{@{}rrrr@{}}\n\\toprule\n$m$&$\\rad(m)$&$G_A/mG_A$&Root-fiber size when nonempty","midrule\n1&1&$0$&1","4&2&$C_2$&2","8&2&$C_2$&2","12&6&$C_6$&6","18&6&$C_6$&6","30&30&$C_{30}$&30","60&30&$C_{30}$&30","The kernel itself depends only on the distinct prime divisors:\n\\[\nmG_A=\\rad(m)G_A.\n\\]\nThis is stronger than equality of the quotient orders. It is equality of the subgroups inside the fixed source group.","\\section{Why these are all finite quotients}\nLet $f:G_A\\to F$ be any homomorphism to a finite group. Its image is an abelian subgroup of $F$, and every element of the image is annihilated by $\\lvert F\\rvert$. Therefore $f$ kills $\\lvert F\\rvert G_A$ and factors through the quotient in Theorem~\\ref{thm:quotientformula}.","\\begin{corollary}[Common profinite completion]\n\\label{thm:GAcompletion}\nFor every $A\\subseteq\\N$, the inclusion $G_A\\subseteq P$ extends to a canonical isomorphism\n\\[\n\\widehat{G_A}\\cong P.\n\\]\nThe induced topology on $G_A$ is its full profinite topology.\n\\end{corollary}\n\\begin{proof}\nThe coordinate kernels are cofinal among finite-index normal subgroups by the preceding factorization argument. Their inverse limit is $P$. Neighborhoods defined by finitely many coordinate zeros coincide with the kernels $mG_A$, so the induced and profinite topologies agree.\n\\end{proof}\nThe product $P$ has a computable profinite presentation: list the primes, form the finite addition tables of $A_N$, and give the coordinate projections. This is the standard inverse-system meaning of a computable profinite","\\begin{corollary}[Common profinite completion]\n\\label{thm:GAcompletion}\nFor every $A\\subseteq\\N$, the inclusion $G_A\\subseteq P$ extends to a canonical isomorphism\n\\[\n\\widehat{G_A}\\cong P.\n\\]\nThe induced topology on $G_A$ is its full profinite topology.\n\\end{corollary}\n\\begin{proof}\nThe coordinate kernels are cofinal among finite-index normal subgroups by the preceding factorization argument. Their inverse limit is $P$. Neighborhoods defined by finitely many coordinate zeros coincide with the kernels $mG_A$, so the induced and profinite topologies agree.\n\\end{proof}\nThe product $P$ has a computable profinite presentation: list the primes, form the finite addition tables of $A_N$, and give the coordinate projections. This is the standard inverse-system meaning of a computable profinite presentation \\cite[Definition 3","\\begin{corollary}[Common profinite completion]\n\\label{thm:GAcompletion}\nFor every $A\\subseteq\\N$, the inclusion $G_A\\subseteq P$ extends to a canonical isomorphism\n\\[\n\\widehat{G_A}\\cong P.\n\\]\nThe induced topology on $G_A$ is its full profinite topology.\n\\end{corollary}\n\\begin{proof}\nThe coordinate kernels are cofinal among finite-index normal subgroups by the preceding factorization argument. Their inverse limit is $P$. Neighborhoods defined by finitely many coordinate zeros coincide with the kernels $mG_A$, so the induced and profinite topologies agree.\n\\end{proof}\nThe product $P$ has a computable profinite presentation: list the primes, form the finite addition tables of $A_N$, and give the coordinate projections. This is the standard inverse-system meaning of a computable profinite presentation \\cite[Definition 3.3]{Lupini}. It is independent of $A$. The uncountable product","\\begin{corollary}[Common profinite completion]\n\\label{thm:GAcompletion}\nFor every $A\\subseteq\\N$, the inclusion $G_A\\subseteq P$ extends to a canonical isomorphism\n\\[\n\\widehat{G_A}\\cong P.\n\\]\nThe induced topology on $G_A$ is its full profinite topology.\n\\end{corollary}\n\\begin{proof}\nThe coordinate kernels are cofinal among finite-index normal subgroups by the preceding factorization argument. Their inverse limit is $P$. Neighborhoods defined by finitely many coordinate zeros coincide with the kernels $mG_A$, so the induced and profinite topologies agree.\n\\end{proof}\nThe product $P$ has a computable profinite presentation: list the primes, form the finite addition tables of $A_N$, and give the coordinate projections. This is the standard inverse-system meaning of a computable profinite presentation \\cite[Definition 3.3]{Lupini}. It is independent of $A$. The uncountable product is presented by an effective i","mmon profinite completion]\n\\label{thm:GAcompletion}\nFor every $A\\subseteq\\N$, the inclusion $G_A\\subseteq P$ extends to a canonical isomorphism\n\\[\n\\widehat{G_A}\\cong P.\n\\]\nThe induced topology on $G_A$ is its full profinite topology.\n\\end{corollary}\n\\begin{proof}\nThe coordinate kernels are cofinal among finite-index normal subgroups by the preceding factorization argument. Their inverse limit is $P$. Neighborhoods defined by finitely many coordinate zeros coincide with the kernels $mG_A$, so the induced and profinite topologies agree.\n\\end{proof}\nThe product $P$ has a computable profinite presentation: list the primes, form the finite addition tables of $A_N$, and give the coordinate projections. This is the standard inverse-system meaning of a computable profinite presentation \\cite[Definition 3.3]{Lupini}. It is independent of $A$. The uncountable product is presented by an effective inverse system, not by a computable listing of all its elements. This does not turn an arbitrary $G_A$ into a computable discrete group.","sion $G_A\\subseteq P$ extends to a canonical isomorphism\n\\[\n\\widehat{G_A}\\cong P.\n\\]\nThe induced topology on $G_A$ is its full profinite topology.\n\\end{corollary}\n\\begin{proof}\nThe coordinate kernels are cofinal among finite-index normal subgroups by the preceding factorization argument. Their inverse limit is $P$. Neighborhoods defined by finitely many coordinate zeros coincide with the kernels $mG_A$, so the induced and profinite topologies agree.\n\\end{proof}\nThe product $P$ has a computable profinite presentation: list the primes, form the finite addition tables of $A_N$, and give the coordinate projections. This is the standard inverse-system meaning of a computable profinite presentation \\cite[Definition 3.3]{Lupini}. It is independent of $A$. The uncountable product is presented by an effective inverse system, not by a computable listing of all its elements. This does not turn an arbitrary $G_A$ into a computable discrete group.","P.\n\\]\nThe induced topology on $G_A$ is its full profinite topology.\n\\end{corollary}\n\\begin{proof}\nThe coordinate kernels are cofinal among finite-index normal subgroups by the preceding factorization argument. Their inverse limit is $P$. Neighborhoods defined by finitely many coordinate zeros coincide with the kernels $mG_A$, so the induced and profinite topologies agree.\n\\end{proof}\nThe product $P$ has a computable profinite presentation: list the primes, form the finite addition tables of $A_N$, and give the coordinate projections. This is the standard inverse-system meaning of a computable profinite presentation \\cite[Definition 3.3]{Lupini}. It is independent of $A$. The uncountable product is presented by an effective inverse system, not by a computable listing of all its elements. This does not turn an arbitrary $G_A$ into a computable discrete group.","The coordinate kernels are cofinal among finite-index normal subgroups by the preceding factorization argument. Their inverse limit is $P$. Neighborhoods defined by finitely many coordinate zeros coincide with the kernels $mG_A$, so the induced and profinite topologies agree.\n\\end{proof}\nThe product $P$ has a computable profinite presentation: list the primes, form the finite addition tables of $A_N$, and give the coordinate projections. This is the standard inverse-system meaning of a computable profinite presentation \\cite[Definition 3.3]{Lupini}. It is independent of $A$. The uncountable product is presented by an effective inverse system, not by a computable listing of all its elements. This does not turn an arbitrary $G_A$ into a computable discrete group.","coordinate zeros coincide with the kernels $mG_A$, so the induced and profinite topologies agree.\n\\end{proof}\nThe product $P$ has a computable profinite presentation: list the primes, form the finite addition tables of $A_N$, and give the coordinate projections. This is the standard inverse-system meaning of a computable profinite presentation \\cite[Definition 3.3]{Lupini}. It is independent of $A$. The uncountable product is presented by an effective inverse system, not by a computable listing of all its elements. This does not turn an arbitrary $G_A$ into a computable discrete group.","resentation: list the primes, form the finite addition tables of $A_N$, and give the coordinate projections. This is the standard inverse-system meaning of a computable profinite presentation \\cite[Definition 3.3]{Lupini}. It is independent of $A$. The uncountable product is presented by an effective inverse system, not by a computable listing of all its elements. This does not turn an arbitrary $G_A$ into a computable discrete group.","\\section{The information retained by a marked element}\nFor the distinguished element $u_A=s_A(1)$, the coordinate criterion simplifies to\n\\begin{equation}\n\\label{eq:markroot}\np\\mid_{G_A}u_A\\quad\\Longleftrightarrow\\quad p\\notin S_A.\n\\end{equation}\nConsequently, for each $n$,\n\\begin{align}\n p_{2n}\\mid_{G_A}u_A&\\quad\\Longleftrightarrow\\quad n\\in A,\\label{eq:evenroot}","\\section{The information retained by a marked element}\nFor the distinguished element $u_A=s_A(1)$, the coordinate criterion simplifies to\n\\begin{equation}\n\\label{eq:markroot}\np\\mid_{G_A}u_A\\quad\\Longleftrightarrow\\quad p\\notin S_A.\n\\end{equation}\nConsequently, for each $n$,\n\\begin{align}\n p_{2n}\\mid_{G_A}u_A&\\quad\\Longleftrightarrow\\quad n\\in A,\\label{eq:evenroot}\\\\\n p_{2n+1}\\mid_{G_A}u_A&\\quad\\Longleftrightarrow\\quad n\\notin A.\\label{eq:oddroot}\n\\end{align}\nExactly one root equation is solvable. This binary pattern is the selector written in intrinsic group language.","\\section{The signature of an infinite-order element}\nThe rational coordinate of a normal form detects torsion: $s_A(r)+d$ has infinite order exactly when $r\\ne0$.","\\begin{lemma}[Tail signature]\n\\label{thm:tailsignature}\nIf $g=s_A(r)+d$ has infinite order, then for all but finitely many primes,\n\\[\np\\mid_{G_A}g\\quad\\Longleftrightarrow\\quad p\\notin S_A.\n\\]\nAn explicit exception set consists of the primes dividing the nonzero numerator or denominator of $r$, together with $\\supp(d)$.\n\\end{lemma}\n\\begin{proof}\nOutside that set, the reduction of $r$ is defined and nonzero. The coordinate of $g$ is therefore nonzero exactly at the selected primes. Apply Theorem~\\ref{thm:rootcertificate} with multiplier $p$.\n\\end{proof}\nThe full signature of the distinguished element is exact. The signature of an arbitrary infinite-order element is exact only in the tail. This finite discrepancy is what determines the correct unmarked classifica","\\begin{lemma}[Tail signature]\n\\label{thm:tailsignature}\nIf $g=s_A(r)+d$ has infinite order, then for all but finitely many primes,\n\\[\np\\mid_{G_A}g\\quad\\Longleftrightarrow\\quad p\\notin S_A.\n\\]\nAn explicit exception set consists of the primes dividing the nonzero numerator or denominator of $r$, together with $\\supp(d)$.\n\\end{lemma}\n\\begin{proof}\nOutside that set, the reduction of $r$ is defined and nonzero. The coordinate of $g$ is therefore nonzero exactly at the selected primes. Apply Theorem~\\ref{thm:rootcertificate} with multiplier $p$.\n\\end{proof}\nThe full signature of the distinguished element is exact. The signature of an arbitrary infinite-order element is exact only in the tail. This finite discrepancy is what determines the correct unmarked classification.","\\label{thm:tailsignature}\nIf $g=s_A(r)+d$ has infinite order, then for all but finitely many primes,\n\\[\np\\mid_{G_A}g\\quad\\Longleftrightarrow\\quad p\\notin S_A.\n\\]\nAn explicit exception set consists of the primes dividing the nonzero numerator or denominator of $r$, together with $\\supp(d)$.\n\\end{lemma}\n\\begin{proof}\nOutside that set, the reduction of $r$ is defined and nonzero. The coordinate of $g$ is therefore nonzero exactly at the selected primes. Apply Theorem~\\ref{thm:rootcertificate} with multiplier $p$.\n\\end{proof}\nThe full signature of the distinguished element is exact. The signature of an arbitrary infinite-order element is exact only in the tail. This finite discrepancy is what determines the correct unmarked classification.","_A}g\\quad\\Longleftrightarrow\\quad p\\notin S_A.\n\\]\nAn explicit exception set consists of the primes dividing the nonzero numerator or denominator of $r$, together with $\\supp(d)$.\n\\end{lemma}\n\\begin{proof}\nOutside that set, the reduction of $r$ is defined and nonzero. The coordinate of $g$ is therefore nonzero exactly at the selected primes. Apply Theorem~\\ref{thm:rootcertificate} with multiplier $p$.\n\\end{proof}\nThe full signature of the distinguished element is exact. The signature of an arbitrary infinite-order element is exact only in the tail. This finite discrepancy is what determines the correct unmarked classification.","\\section{Isomorphism means eventual agreement}\n\\begin{theorem}[Unmarked isomorphism classification]\n\\label{thm:isoclass}\nFor all $A,B\\subseteq\\N$,\n\\[\nG_A\\cong G_B\\quad\\Longleftrightarrow\\quad A\\eqfin B.\n\\]\nWhen $A\\eqfin B$, the two groups are equal as subgroups of the fixed product $P$.\n\\end{theorem}\n\\begin{proof}\nSuppose $f:G_A\\to G_B$ is an isomorphism. The image of $u_A$ has infinite order. An isomorphism preserves the existence of a $p$th root for every prime $p$. Combining the exact signature \\eqref{eq:markroot} with Lemma~\\ref{thm:tailsignature} for $f(u_A)$ gives $S_A\\eqfin S_B$. Since each set chooses exactly one prime from each pair, the corresponding sets of bits differ finitely.","ent}\n\\begin{theorem}[Unmarked isomorphism classification]\n\\label{thm:isoclass}\nFor all $A,B\\subseteq\\N$,\n\\[\nG_A\\cong G_B\\quad\\Longleftrightarrow\\quad A\\eqfin B.\n\\]\nWhen $A\\eqfin B$, the two groups are equal as subgroups of the fixed product $P$.\n\\end{theorem}\n\\begin{proof}\nSuppose $f:G_A\\to G_B$ is an isomorphism. The image of $u_A$ has infinite order. An isomorphism preserves the existence of a $p$th root for every prime $p$. Combining the exact signature \\eqref{eq:markroot} with Lemma~\\ref{thm:tailsignature} for $f(u_A)$ gives $S_A\\eqfin S_B$. Since each set chooses exactly one prime from each pair, the corresponding sets of bits differ finitely.","ed isomorphism classification]\n\\label{thm:isoclass}\nFor all $A,B\\subseteq\\N$,\n\\[\nG_A\\cong G_B\\quad\\Longleftrightarrow\\quad A\\eqfin B.\n\\]\nWhen $A\\eqfin B$, the two groups are equal as subgroups of the fixed product $P$.\n\\end{theorem}\n\\begin{proof}\nSuppose $f:G_A\\to G_B$ is an isomorphism. The image of $u_A$ has infinite order. An isomorphism preserves the existence of a $p$th root for every prime $p$. Combining the exact signature \\eqref{eq:markroot} with Lemma~\\ref{thm:tailsignature} for $f(u_A)$ gives $S_A\\eqfin S_B$. Since each set chooses exactly one prime from each pair, the corresponding sets of bits differ finitely.","quad\\Longleftrightarrow\\quad A\\eqfin B.\n\\]\nWhen $A\\eqfin B$, the two groups are equal as subgroups of the fixed product $P$.\n\\end{theorem}\n\\begin{proof}\nSuppose $f:G_A\\to G_B$ is an isomorphism. The image of $u_A$ has infinite order. An isomorphism preserves the existence of a $p$th root for every prime $p$. Combining the exact signature \\eqref{eq:markroot} with Lemma~\\ref{thm:tailsignature} for $f(u_A)$ gives $S_A\\eqfin S_B$. Since each set chooses exactly one prime from each pair, the corresponding sets of bits differ finitely.","ghtarrow\\quad A\\eqfin B.\n\\]\nWhen $A\\eqfin B$, the two groups are equal as subgroups of the fixed product $P$.\n\\end{theorem}\n\\begin{proof}\nSuppose $f:G_A\\to G_B$ is an isomorphism. The image of $u_A$ has infinite order. An isomorphism preserves the existence of a $p$th root for every prime $p$. Combining the exact signature \\eqref{eq:markroot} with Lemma~\\ref{thm:tailsignature} for $f(u_A)$ gives $S_A\\eqfin S_B$. Since each set chooses exactly one prime from each pair, the corresponding sets of bits differ finitely.","Conversely, suppose $A\\eqfin B$. For every rational $r$, the vectors $s_A(r)$ and $s_B(r)$ differ only on the finitely many prime pairs where the selectors differ. Thus their difference belongs to $D$. The descriptions $G_A=s_A(\\Q)+D$ and $G_B=s_B(\\Q)+D$ define the same subset of $P$.\n\\end{proof}\nThe normal forms need not be identical. Their coordinate change is\n\\begin{equation}\n\\label{eq:transport}\n(r,d)\\longmapsto\n\\bigl(r,d+s_A(r)-s_B(r)\\bigr).\n\\end{equation}\nIt fixes the represented ambient element. The correction is finite exactly because the selectors differ only finitely.","\\begin{corollary}[Marked classification]\n\\label{thm:markedclass}\nAn isomorphism $(G_A,u_A)\\cong(G_B,u_B)$ exists if and only if $A=B$.\n\\end{corollary}\n\\begin{proof}\nA marked isomorphism preserves the exact two-prime root test \\eqref{eq:evenroot}--\\eqref{eq:oddroot} at every index. Therefore every bit agrees. The converse is the identity map.\n\\end{proof}","\\begin{example}[The same group with a different marked point]\nLet $A=\\varnothing$ and $B=\\{0\\}$. Only the first prime pair changes. Then $G_A=G_B$ inside $P$, but $u_A-u_B=e_2-e_3$. The distinguished element is not preserved. In particular, the statement $G_A\\cong G_B\\iff A=B$ would be false for unmarked groups.\n\\end{example}","\\section{All homomorphisms are diagonal}\nThe product $P$ is also a commutative ring under coordinatewise multiplication. Multiplication by any $\\lambda\\in P$ preserves $D$, so it induces an additive endomorphism $\\overline M_\\lambda$ of $V$. Because division in $V$ is unique, this induced endomorphism is $\\Q$-linear.","\\begin{theorem}[Complete homomorphism description]\n\\label{thm:allhom}\nEvery homomorphism $f:G_A\\to G_B$ has the form\n\\[\nf(g)_p=\\lambda_p g_p\\qquad(p\\text{ prime})\n\\]\nfor a unique $\\lambda\\in P$. Conversely, coordinate multiplication by $\\lambda$ defines such a homomorphism exactly when\n\\begin{equation}\n\\label{eq:homcondition}\nq(\\lambda u_A)\\in\\Q q(u_B).\n\\end{equation}\nEquivalently, there is a rational $r$ for which $\\lambda u_A-s_B(r)$ has finite support.\n\\end{theorem}\n\\begin{proof}\nFor a homomorphism $f$, the composite $G_A\\to G_B\\to C_p$ kills $pG_A$. The quotient $G_A/pG_A$ is canonically $C_p$. Every endomorphism of this cyclic additive group is multiplication by a unique scalar $\\lambda_p$. Hence the displayed coordinate rule holds. Uniqueness also follows by evaluating $f(e_p)$.","{theorem}[Complete homomorphism description]\n\\label{thm:allhom}\nEvery homomorphism $f:G_A\\to G_B$ has the form\n\\[\nf(g)_p=\\lambda_p g_p\\qquad(p\\text{ prime})\n\\]\nfor a unique $\\lambda\\in P$. Conversely, coordinate multiplication by $\\lambda$ defines such a homomorphism exactly when\n\\begin{equation}\n\\label{eq:homcondition}\nq(\\lambda u_A)\\in\\Q q(u_B).\n\\end{equation}\nEquivalently, there is a rational $r$ for which $\\lambda u_A-s_B(r)$ has finite support.\n\\end{theorem}\n\\begin{proof}\nFor a homomorphism $f$, the composite $G_A\\to G_B\\to C_p$ kills $pG_A$. The quotient $G_A/pG_A$ is canonically $C_p$. Every endomorphism of this cyclic additive group is multiplication by a unique scalar $\\lambda_p$. Hence the displayed coordinate rule holds. Uniqueness also follows by evaluating $f(e_p)$.","lete homomorphism description]\n\\label{thm:allhom}\nEvery homomorphism $f:G_A\\to G_B$ has the form\n\\[\nf(g)_p=\\lambda_p g_p\\qquad(p\\text{ prime})\n\\]\nfor a unique $\\lambda\\in P$. Conversely, coordinate multiplication by $\\lambda$ defines such a homomorphism exactly when\n\\begin{equation}\n\\label{eq:homcondition}\nq(\\lambda u_A)\\in\\Q q(u_B).\n\\end{equation}\nEquivalently, there is a rational $r$ for which $\\lambda u_A-s_B(r)$ has finite support.\n\\end{theorem}\n\\begin{proof}\nFor a homomorphism $f$, the composite $G_A\\to G_B\\to C_p$ kills $pG_A$. The quotient $G_A/pG_A$ is canonically $C_p$. Every endomorphism of this cyclic additive group is multiplication by a unique scalar $\\lambda_p$. Hence the displayed coordinate rule holds. Uniqueness also follows by evaluating $f(e_p)$.","the form\n\\[\nf(g)_p=\\lambda_p g_p\\qquad(p\\text{ prime})\n\\]\nfor a unique $\\lambda\\in P$. Conversely, coordinate multiplication by $\\lambda$ defines such a homomorphism exactly when\n\\begin{equation}\n\\label{eq:homcondition}\nq(\\lambda u_A)\\in\\Q q(u_B).\n\\end{equation}\nEquivalently, there is a rational $r$ for which $\\lambda u_A-s_B(r)$ has finite support.\n\\end{theorem}\n\\begin{proof}\nFor a homomorphism $f$, the composite $G_A\\to G_B\\to C_p$ kills $pG_A$. The quotient $G_A/pG_A$ is canonically $C_p$. Every endomorphism of this cyclic additive group is multiplication by a unique scalar $\\lambda_p$. Hence the displayed coordinate rule holds. Uniqueness also follows by evaluating $f(e_p)$.","\\lambda_p g_p\\qquad(p\\text{ prime})\n\\]\nfor a unique $\\lambda\\in P$. Conversely, coordinate multiplication by $\\lambda$ defines such a homomorphism exactly when\n\\begin{equation}\n\\label{eq:homcondition}\nq(\\lambda u_A)\\in\\Q q(u_B).\n\\end{equation}\nEquivalently, there is a rational $r$ for which $\\lambda u_A-s_B(r)$ has finite support.\n\\end{theorem}\n\\begin{proof}\nFor a homomorphism $f$, the composite $G_A\\to G_B\\to C_p$ kills $pG_A$. The quotient $G_A/pG_A$ is canonically $C_p$. Every endomorphism of this cyclic additive group is multiplication by a unique scalar $\\lambda_p$. Hence the displayed coordinate rule holds. Uniqueness also follows by evaluating $f(e_p)$.","{ prime})\n\\]\nfor a unique $\\lambda\\in P$. Conversely, coordinate multiplication by $\\lambda$ defines such a homomorphism exactly when\n\\begin{equation}\n\\label{eq:homcondition}\nq(\\lambda u_A)\\in\\Q q(u_B).\n\\end{equation}\nEquivalently, there is a rational $r$ for which $\\lambda u_A-s_B(r)$ has finite support.\n\\end{theorem}\n\\begin{proof}\nFor a homomorphism $f$, the composite $G_A\\to G_B\\to C_p$ kills $pG_A$. The quotient $G_A/pG_A$ is canonically $C_p$. Every endomorphism of this cyclic additive group is multiplication by a unique scalar $\\lambda_p$. Hence the displayed coordinate rule holds. Uniqueness also follows by evaluating $f(e_p)$.","ique $\\lambda\\in P$. Conversely, coordinate multiplication by $\\lambda$ defines such a homomorphism exactly when\n\\begin{equation}\n\\label{eq:homcondition}\nq(\\lambda u_A)\\in\\Q q(u_B).\n\\end{equation}\nEquivalently, there is a rational $r$ for which $\\lambda u_A-s_B(r)$ has finite support.\n\\end{theorem}\n\\begin{proof}\nFor a homomorphism $f$, the composite $G_A\\to G_B\\to C_p$ kills $pG_A$. The quotient $G_A/pG_A$ is canonically $C_p$. Every endomorphism of this cyclic additive group is multiplication by a unique scalar $\\lambda_p$. Hence the displayed coordinate rule holds. Uniqueness also follows by evaluating $f(e_p)$.","rsely, coordinate multiplication by $\\lambda$ defines such a homomorphism exactly when\n\\begin{equation}\n\\label{eq:homcondition}\nq(\\lambda u_A)\\in\\Q q(u_B).\n\\end{equation}\nEquivalently, there is a rational $r$ for which $\\lambda u_A-s_B(r)$ has finite support.\n\\end{theorem}\n\\begin{proof}\nFor a homomorphism $f$, the composite $G_A\\to G_B\\to C_p$ kills $pG_A$. The quotient $G_A/pG_A$ is canonically $C_p$. Every endomorphism of this cyclic additive group is multiplication by a unique scalar $\\lambda_p$. Hence the displayed coordinate rule holds. Uniqueness also follows by evaluating $f(e_p)$.","{equation}\n\\label{eq:homcondition}\nq(\\lambda u_A)\\in\\Q q(u_B).\n\\end{equation}\nEquivalently, there is a rational $r$ for which $\\lambda u_A-s_B(r)$ has finite support.\n\\end{theorem}\n\\begin{proof}\nFor a homomorphism $f$, the composite $G_A\\to G_B\\to C_p$ kills $pG_A$. The quotient $G_A/pG_A$ is canonically $C_p$. Every endomorphism of this cyclic additive group is multiplication by a unique scalar $\\lambda_p$. Hence the displayed coordinate rule holds. Uniqueness also follows by evaluating $f(e_p)$.","ere is a rational $r$ for which $\\lambda u_A-s_B(r)$ has finite support.\n\\end{theorem}\n\\begin{proof}\nFor a homomorphism $f$, the composite $G_A\\to G_B\\to C_p$ kills $pG_A$. The quotient $G_A/pG_A$ is canonically $C_p$. Every endomorphism of this cyclic additive group is multiplication by a unique scalar $\\lambda_p$. Hence the displayed coordinate rule holds. Uniqueness also follows by evaluating $f(e_p)$.","Conversely, multiplication by any $\\lambda$ is a group endomorphism of $P$ and preserves $D$. Since its induced map on $V$ is rational-linear, it carries $\\Q q(u_A)$ into $\\Q q(u_B)$ exactly when it carries the generating vector into that line. This is \\eqref{eq:homcondition}. Taking preimages proves that the restriction maps $G_A$ into $G_B$. The equivalent finite-support condition follows from the definition of the quotient.\n\\end{proof}\nNo continuity assumption was imposed. Every resulting homomorphism is nevertheless continuous for the induced profinite topologies, because each output coordinate depends only on the corresponding input coordinate. Its extension to the completion is the same coordinatewise scalar map on $P$.","$ into $G_B$. The equivalent finite-support condition follows from the definition of the quotient.\n\\end{proof}\nNo continuity assumption was imposed. Every resulting homomorphism is nevertheless continuous for the induced profinite topologies, because each output coordinate depends only on the corresponding input coordinate. Its extension to the completion is the same coordinatewise scalar map on $P$.","\\begin{theorem}[Torsion-image obstruction]\n\\label{thm:homobstruction}\nIf $A\\neqfin B$, every homomorphism $G_A\\to G_B$ has image contained in $D$. Consequently neither group embeds in the other, and neither is a quotient of the other.\n\\end{theorem}\n\\begin{proof}\nA homomorphism preserves torsion and induces a map $\\Q\\to\\Q$ on the quotients, necessarily multiplication by a rational $r$. If $r=0$, its image lies in $D$.","orsion-image obstruction]\n\\label{thm:homobstruction}\nIf $A\\neqfin B$, every homomorphism $G_A\\to G_B$ has image contained in $D$. Consequently neither group embeds in the other, and neither is a quotient of the other.\n\\end{theorem}\n\\begin{proof}\nA homomorphism preserves torsion and induces a map $\\Q\\to\\Q$ on the quotients, necessarily multiplication by a rational $r$. If $r=0$, its image lies in $D$.","Assume $r\\ne0$. Then $f(u_A)$ has the tail signature of $S_B$. For every $p\\notin S_A$, the element $u_A$ has a $p$th root, and its image has a $p$th root as well. Lemma~\\ref{thm:tailsignature} implies that $S_B\\setminus S_A$ is finite. Each differing prime pair contributes one element to this difference, so $A\\eqfin B$, contrary to the assumption.","An embedding cannot send an infinite-order element to torsion. A surjection onto $G_B$ cannot have torsion image, since $G_B/D\\cong\\Q\\ne0$.\n\\end{proof}\nA useful precise consequence of Theorem~\\ref{thm:allhom} is that, when $A\\neqfin B$, the possible diagonal sequences are exactly those for which $\\lambda_p=0$ at all but finitely many primes of $S_A$. Their values on the complementary primes are unrestricted. These maps can be nonzero, and their torsion images need not be finite.","\\section{The endomorphism ring and its units}\nWrite $\\overline r_p$ for reduction of a rational $r$ at a prime not dividing its denominator. The phrase ``for almost every selected prime'' below means outside a finite set, which may include those denominator primes.","\\begin{corollary}[Endomorphism ring]\n\\label{thm:endGA}\nUnder coordinate multiplication,\n\\[\n\\End(G_A)\\cong\nE_A=\\{\\lambda\\in P:\\exists r\\in\\Q,\\ \n\\lambda_p=\\overline r_p\\text{ for almost every }p\\in S_A\\}.\n\\]\nThe rational $r$ is unique. It is the scalar induced on $G_A/D$, and the map $E_A\\to\\Q$ sending $\\lambda$ to $r$ is a surjective ring homomorphism.\n\\end{corollary}\n\\begin{proof}\nApply Theorem~\\ref{thm:allhom} with $B=A$. At unselected coordinates, both $\\lambda u_A$ and $s_A(r)$ vanish. At selected coordinates the finite-support condition is exactly the displayed eventual equality.","omorphism ring]\n\\label{thm:endGA}\nUnder coordinate multiplication,\n\\[\n\\End(G_A)\\cong\nE_A=\\{\\lambda\\in P:\\exists r\\in\\Q,\\ \n\\lambda_p=\\overline r_p\\text{ for almost every }p\\in S_A\\}.\n\\]\nThe rational $r$ is unique. It is the scalar induced on $G_A/D$, and the map $E_A\\to\\Q$ sending $\\lambda$ to $r$ is a surjective ring homomorphism.\n\\end{corollary}\n\\begin{proof}\nApply Theorem~\\ref{thm:allhom} with $B=A$. At unselected coordinates, both $\\lambda u_A$ and $s_A(r)$ vanish. At selected coordinates the finite-support condition is exactly the displayed eventual equality.","coordinate multiplication,\n\\[\n\\End(G_A)\\cong\nE_A=\\{\\lambda\\in P:\\exists r\\in\\Q,\\ \n\\lambda_p=\\overline r_p\\text{ for almost every }p\\in S_A\\}.\n\\]\nThe rational $r$ is unique. It is the scalar induced on $G_A/D$, and the map $E_A\\to\\Q$ sending $\\lambda$ to $r$ is a surjective ring homomorphism.\n\\end{corollary}\n\\begin{proof}\nApply Theorem~\\ref{thm:allhom} with $B=A$. At unselected coordinates, both $\\lambda u_A$ and $s_A(r)$ vanish. At selected coordinates the finite-support condition is exactly the displayed eventual equality.","lication,\n\\[\n\\End(G_A)\\cong\nE_A=\\{\\lambda\\in P:\\exists r\\in\\Q,\\ \n\\lambda_p=\\overline r_p\\text{ for almost every }p\\in S_A\\}.\n\\]\nThe rational $r$ is unique. It is the scalar induced on $G_A/D$, and the map $E_A\\to\\Q$ sending $\\lambda$ to $r$ is a surjective ring homomorphism.\n\\end{corollary}\n\\begin{proof}\nApply Theorem~\\ref{thm:allhom} with $B=A$. At unselected coordinates, both $\\lambda u_A$ and $s_A(r)$ vanish. At selected coordinates the finite-support condition is exactly the displayed eventual equality.","If two different rationals agreed modulo almost every selected prime, their nonzero difference would have a numerator divisible by infinitely many primes, which is impossible. Thus $r$ is unique. Addition and composition induce addition and multiplication of the quotient scalars, giving a ring homomorphism. The diagonal sequence $s_A(r)$ realizes every rational $r$, proving surjectivity.\n\\end{proof}\nThe arbitrary coordinates outside $S_A$ show why the endomorphism ring may contain infinitely specified operators. The seed-rank obstruction below remains true even when all these operators are available.","lo almost every selected prime, their nonzero difference would have a numerator divisible by infinitely many primes, which is impossible. Thus $r$ is unique. Addition and composition induce addition and multiplication of the quotient scalars, giving a ring homomorphism. The diagonal sequence $s_A(r)$ realizes every rational $r$, proving surjectivity.\n\\end{proof}\nThe arbitrary coordinates outside $S_A$ show why the endomorphism ring may contain infinitely specified operators. The seed-rank obstruction below remains true even when all these operators are available.","\\begin{corollary}[Automorphisms]\n\\label{thm:autGA}\nAn endomorphism with diagonal $\\lambda\\in E_A$ is an automorphism if and only if every coordinate $\\lambda_p$ is nonzero. In that case its rational quotient scalar is nonzero, and its inverse has diagonal $\\lambda_p^{-1}$.\n\\end{corollary}\n\\begin{proof}\nAn automorphism restricts to an automorphism of each torsion factor, so every scalar is nonzero. Conversely, if every scalar is nonzero, the eventual rational scalar cannot be zero because $S_A$ is infinite. Outside finitely many selected primes, the inverse diagonal is reduction of its reciprocal. Thus the inverse diagonal belongs to $E_A$ and defines an inverse endomorphism.\n\\end{proof}","_A$ is an automorphism if and only if every coordinate $\\lambda_p$ is nonzero. In that case its rational quotient scalar is nonzero, and its inverse has diagonal $\\lambda_p^{-1}$.\n\\end{corollary}\n\\begin{proof}\nAn automorphism restricts to an automorphism of each torsion factor, so every scalar is nonzero. Conversely, if every scalar is nonzero, the eventual rational scalar cannot be zero because $S_A$ is infinite. Outside finitely many selected primes, the inverse diagonal is reduction of its reciprocal. Thus the inverse diagonal belongs to $E_A$ and defines an inverse endomorphism.\n\\end{proof}","has diagonal $\\lambda_p^{-1}$.\n\\end{corollary}\n\\begin{proof}\nAn automorphism restricts to an automorphism of each torsion factor, so every scalar is nonzero. Conversely, if every scalar is nonzero, the eventual rational scalar cannot be zero because $S_A$ is infinite. Outside finitely many selected primes, the inverse diagonal is reduction of its reciprocal. Thus the inverse diagonal belongs to $E_A$ and defines an inverse endomorphism.\n\\end{proof}","\\section{Infinite seed rank before completion}\n\\begin{theorem}[Seed-rank separation]\n\\label{thm:rankGA}\nFor every $A$,\n\\[\n\\srank_{\\End(G_A)}(G_A)=\\aleph_0.\n\\]\nNo finite family topologically generates $G_A$, even after arbitrary endomorphisms are admitted. Yet its completion has ordinary topological generating rank one.\n\\end{theorem}\n\\begin{proof}\nEach element $s_A(r)+d$ has only finitely many nonzero coordinates outside $S_A$. A finite family therefore omits infinitely many unselected prime coordinates. By Theorem~\\ref{thm:allhom}, every endomorphism preserves a zero coordinate as zero. Finite sums of orbit elements still vanish at any omitted prime $p$. The projection onto $C_p$ is a non-generation certificate, since $e_p\\in G_A$ has nonzero projection. Its kernel is closed, so taking closure does not re","\\section{Infinite seed rank before completion}\n\\begin{theorem}[Seed-rank separation]\n\\label{thm:rankGA}\nFor every $A$,\n\\[\n\\srank_{\\End(G_A)}(G_A)=\\aleph_0.\n\\]\nNo finite family topologically generates $G_A$, even after arbitrary endomorphisms are admitted. Yet its completion has ordinary topological generating rank one.\n\\end{theorem}\n\\begin{proof}\nEach element $s_A(r)+d$ has only finitely many nonzero coordinates outside $S_A$. A finite family therefore omits infinitely many unselected prime coordinates. By Theorem~\\ref{thm:allhom}, every endomorphism preserves a zero coordinate as zero. Finite sums of orbit elements still vanish at any omitted prime $p$. The projection onto $C_p$ is a non-generation certificate, since $e_p\\in G_A$ has nonzero projection. Its kernel is closed, so taking closure does not repair the failure.","nk before completion}\n\\begin{theorem}[Seed-rank separation]\n\\label{thm:rankGA}\nFor every $A$,\n\\[\n\\srank_{\\End(G_A)}(G_A)=\\aleph_0.\n\\]\nNo finite family topologically generates $G_A$, even after arbitrary endomorphisms are admitted. Yet its completion has ordinary topological generating rank one.\n\\end{theorem}\n\\begin{proof}\nEach element $s_A(r)+d$ has only finitely many nonzero coordinates outside $S_A$. A finite family therefore omits infinitely many unselected prime coordinates. By Theorem~\\ref{thm:allhom}, every endomorphism preserves a zero coordinate as zero. Finite sums of orbit elements still vanish at any omitted prime $p$. The projection onto $C_p$ is a non-generation certificate, since $e_p\\in G_A$ has nonzero projection. Its kernel is closed, so taking closure does not repair the failure.","its completion has ordinary topological generating rank one.\n\\end{theorem}\n\\begin{proof}\nEach element $s_A(r)+d$ has only finitely many nonzero coordinates outside $S_A$. A finite family therefore omits infinitely many unselected prime coordinates. By Theorem~\\ref{thm:allhom}, every endomorphism preserves a zero coordinate as zero. Finite sums of orbit elements still vanish at any omitted prime $p$. The projection onto $C_p$ is a non-generation certificate, since $e_p\\in G_A$ has nonzero projection. Its kernel is closed, so taking closure does not repair the failure.","ary topological generating rank one.\n\\end{theorem}\n\\begin{proof}\nEach element $s_A(r)+d$ has only finitely many nonzero coordinates outside $S_A$. A finite family therefore omits infinitely many unselected prime coordinates. By Theorem~\\ref{thm:allhom}, every endomorphism preserves a zero coordinate as zero. Finite sums of orbit elements still vanish at any omitted prime $p$. The projection onto $C_p$ is a non-generation certificate, since $e_p\\in G_A$ has nonzero projection. Its kernel is closed, so taking closure does not repair the failure.","orem}\n\\begin{proof}\nEach element $s_A(r)+d$ has only finitely many nonzero coordinates outside $S_A$. A finite family therefore omits infinitely many unselected prime coordinates. By Theorem~\\ref{thm:allhom}, every endomorphism preserves a zero coordinate as zero. Finite sums of orbit elements still vanish at any omitted prime $p$. The projection onto $C_p$ is a non-generation certificate, since $e_p\\in G_A$ has nonzero projection. Its kernel is closed, so taking closure does not repair the failure.","Countably many seeds suffice because $G_A$ is countable. For ordinary topological generation, the coordinate vectors already generate the dense subgroup $D$. Finally, the completion is $P$, whose all-one vector is a topological generator by Theorem~\\ref{thm:procyclic}.\n\\end{proof}\nThe same infinite seed rank occurs for every selector. It does not measure the varying effective complexity. Algebraic generation, finite description, and computable presentation remain separate invariants.","\\section{Oracles, copies, and degrees}\nAn oracle for a set $X\\subseteq\\N$ answers membership questions about $X$. The notation $A\\le_T X$ means that a terminating oracle procedure decides membership in $A$. Two sets have the same Turing degree when each is computable from the other.","\\section{A presentation from the selector}\nReduced rationals have canonical integer codes. Finite-support vectors in $D$ have canonical codes given by increasing prime labels and nonzero residues. Their Cartesian product is a computably named countable domain with decidable equality and a computable bijection to $\\N$.","\\begin{proposition}[Uniform construction relative to the selector]\n\\label{thm:computeupper}\nIf $A\\le_T X$, then the normal-form construction gives an $X$-computable copy of $G_A$. In its canonical normal-form presentation, the group operation has Turing degree exactly $\\deg_T(A)$.\n\\end{proposition}\n\\begin{proof}\nTo add $(r,d)$ and $(t,e)$, compute the prime divisors of $\\den(r)\\den(t)$. At each of these finitely many primes, query the corresponding selector bit and evaluate the three section residues. Lemma~\\ref{thm:cocyclesupport} guarantees that every other correction coordinate is zero. Formula~\\eqref{eq:grouplaw} then gives the sum. Negation and equality are computable without querying $A$.","\\begin{proof}\nTo add $(r,d)$ and $(t,e)$, compute the prime divisors of $\\den(r)\\den(t)$. At each of these finitely many primes, query the corresponding selector bit and evaluate the three section residues. Lemma~\\ref{thm:cocyclesupport} guarantees that every other correction coordinate is zero. Formula~\\eqref{eq:grouplaw} then gives the sum. Negation and equality are computable without querying $A$.","For the lower bound on this particular presentation, compute the sum of the coded elements $(1/p,0)$ and $((p-1)/p,0)$. Its finite torsion part is the value in Proposition~\\ref{thm:cocyclebit}. For $p=p_{2n}$, that value is zero exactly when $n\\in A$. Thus the operation computes every selector bit.\n\\end{proof}\nThe number of oracle questions used by one addition is finite and bounded by the number of distinct prime divisors of the denominator product. This is a locality statement about the operation, not a claim of a polynomial-time implementation. Factoring integers and finding their prime indices contribute ordinary computational costs.","\\section{Decoding a marked copy}\n\\begin{proposition}[Uniform marked decoding]\n\\label{thm:markeddecode}\nGiven an $X$-computable copy of $G_A$ together with the name of $u_A$, one can uniformly compute $A$ in $X$.\n\\end{proposition}\n\\begin{proof}\nFor input $n$, search the domain simultaneously for solutions of\n\\[\np_{2n}y=u_A,\\qquad p_{2n+1}z=u_A.\n\\]\nEach candidate test is decidable in $X$. Exactly one equation has a solution by \\eqref{eq:evenroot}--\\eqref{eq:oddroot}. Dovetailing the searches therefore terminates. A solution to the first equation means $n\\in A$, and a solution to the second means $n\\notin A$.\n\\end{proof}\nThe procedure does not decide a negative existential statement directly. It uses two complementary positive searches. The paired selector is what converts enumerable root information into","\\section{Decoding a marked copy}\n\\begin{proposition}[Uniform marked decoding]\n\\label{thm:markeddecode}\nGiven an $X$-computable copy of $G_A$ together with the name of $u_A$, one can uniformly compute $A$ in $X$.\n\\end{proposition}\n\\begin{proof}\nFor input $n$, search the domain simultaneously for solutions of\n\\[\np_{2n}y=u_A,\\qquad p_{2n+1}z=u_A.\n\\]\nEach candidate test is decidable in $X$. Exactly one equation has a solution by \\eqref{eq:evenroot}--\\eqref{eq:oddroot}. Dovetailing the searches therefore terminates. A solution to the first equation means $n\\in A$, and a solution to the second means $n\\notin A$.\n\\end{proof}\nThe procedure does not decide a negative existential statement directly. It uses two complementary positive searches. The paired selector is what converts enumerable root information into a total decision procedur","\\section{Decoding a marked copy}\n\\begin{proposition}[Uniform marked decoding]\n\\label{thm:markeddecode}\nGiven an $X$-computable copy of $G_A$ together with the name of $u_A$, one can uniformly compute $A$ in $X$.\n\\end{proposition}\n\\begin{proof}\nFor input $n$, search the domain simultaneously for solutions of\n\\[\np_{2n}y=u_A,\\qquad p_{2n+1}z=u_A.\n\\]\nEach candidate test is decidable in $X$. Exactly one equation has a solution by \\eqref{eq:evenroot}--\\eqref{eq:oddroot}. Dovetailing the searches therefore terminates. A solution to the first equation means $n\\in A$, and a solution to the second means $n\\notin A$.\n\\end{proof}\nThe procedure does not decide a negative existential statement directly. It uses two complementary positive searches. The paired selector is what converts enumerable root information into a total decision procedure.","e name of $u_A$, one can uniformly compute $A$ in $X$.\n\\end{proposition}\n\\begin{proof}\nFor input $n$, search the domain simultaneously for solutions of\n\\[\np_{2n}y=u_A,\\qquad p_{2n+1}z=u_A.\n\\]\nEach candidate test is decidable in $X$. Exactly one equation has a solution by \\eqref{eq:evenroot}--\\eqref{eq:oddroot}. Dovetailing the searches therefore terminates. A solution to the first equation means $n\\in A$, and a solution to the second means $n\\notin A$.\n\\end{proof}\nThe procedure does not decide a negative existential statement directly. It uses two complementary positive searches. The paired selector is what converts enumerable root information into a total decision procedure.","$ in $X$.\n\\end{proposition}\n\\begin{proof}\nFor input $n$, search the domain simultaneously for solutions of\n\\[\np_{2n}y=u_A,\\qquad p_{2n+1}z=u_A.\n\\]\nEach candidate test is decidable in $X$. Exactly one equation has a solution by \\eqref{eq:evenroot}--\\eqref{eq:oddroot}. Dovetailing the searches therefore terminates. A solution to the first equation means $n\\in A$, and a solution to the second means $n\\notin A$.\n\\end{proof}\nThe procedure does not decide a negative existential statement directly. It uses two complementary positive searches. The paired selector is what converts enumerable root information into a total decision procedure.","\\section{Decoding an unmarked copy}\n\\begin{theorem}[Exact oracle criterion]\n\\label{thm:degree}\nFor every $A,X\\subseteq\\N$,\n\\[\nG_A\\text{ has an }X\\text{-computable copy}\n\\quad\\Longleftrightarrow\\quad A\\le_T X.\n\\]\nIn particular, the isomorphism type of $G_A$ has least presentation degree $\\deg_T(A)$.\n\\end{theorem}\n\\begin{proof}\nThe reverse implication is Proposition~\\ref{thm:computeupper}. For the forward implication, let $H$ be an $X$-computable copy. Choose the name $h$ of an infinite-order element. Such an element exists because $H$ has quotient $\\Q$ by its torsion subgroup.","rked copy}\n\\begin{theorem}[Exact oracle criterion]\n\\label{thm:degree}\nFor every $A,X\\subseteq\\N$,\n\\[\nG_A\\text{ has an }X\\text{-computable copy}\n\\quad\\Longleftrightarrow\\quad A\\le_T X.\n\\]\nIn particular, the isomorphism type of $G_A$ has least presentation degree $\\deg_T(A)$.\n\\end{theorem}\n\\begin{proof}\nThe reverse implication is Proposition~\\ref{thm:computeupper}. For the forward implication, let $H$ be an $X$-computable copy. Choose the name $h$ of an infinite-order element. Such an element exists because $H$ has quotient $\\Q$ by its torsion subgroup.","\\begin{theorem}[Exact oracle criterion]\n\\label{thm:degree}\nFor every $A,X\\subseteq\\N$,\n\\[\nG_A\\text{ has an }X\\text{-computable copy}\n\\quad\\Longleftrightarrow\\quad A\\le_T X.\n\\]\nIn particular, the isomorphism type of $G_A$ has least presentation degree $\\deg_T(A)$.\n\\end{theorem}\n\\begin{proof}\nThe reverse implication is Proposition~\\ref{thm:computeupper}. For the forward implication, let $H$ be an $X$-computable copy. Choose the name $h$ of an infinite-order element. Such an element exists because $H$ has quotient $\\Q$ by its torsion subgroup.","on]\n\\label{thm:degree}\nFor every $A,X\\subseteq\\N$,\n\\[\nG_A\\text{ has an }X\\text{-computable copy}\n\\quad\\Longleftrightarrow\\quad A\\le_T X.\n\\]\nIn particular, the isomorphism type of $G_A$ has least presentation degree $\\deg_T(A)$.\n\\end{theorem}\n\\begin{proof}\nThe reverse implication is Proposition~\\ref{thm:computeupper}. For the forward implication, let $H$ be an $X$-computable copy. Choose the name $h$ of an infinite-order element. Such an element exists because $H$ has quotient $\\Q$ by its torsion subgroup.","as an }X\\text{-computable copy}\n\\quad\\Longleftrightarrow\\quad A\\le_T X.\n\\]\nIn particular, the isomorphism type of $G_A$ has least presentation degree $\\deg_T(A)$.\n\\end{theorem}\n\\begin{proof}\nThe reverse implication is Proposition~\\ref{thm:computeupper}. For the forward implication, let $H$ be an $X$-computable copy. Choose the name $h$ of an infinite-order element. Such an element exists because $H$ has quotient $\\Q$ by its torsion subgroup.","mputable copy}\n\\quad\\Longleftrightarrow\\quad A\\le_T X.\n\\]\nIn particular, the isomorphism type of $G_A$ has least presentation degree $\\deg_T(A)$.\n\\end{theorem}\n\\begin{proof}\nThe reverse implication is Proposition~\\ref{thm:computeupper}. For the forward implication, let $H$ be an $X$-computable copy. Choose the name $h$ of an infinite-order element. Such an element exists because $H$ has quotient $\\Q$ by its torsion subgroup.","Under an isomorphism to $G_A$, the element $h$ has a normal form with nonzero rational part. Lemma~\\ref{thm:tailsignature} gives a finite exception set. Choose $N$ so that every prime in the pairs indexed by $n\\ge N$ avoids that set. For these $n$, exactly one of\n\\[\np_{2n}y=h,\\qquad p_{2n+1}z=h\n\\]\nhas a solution, and the first is solvable exactly when $n\\in A$.","Dovetailing the two searches decides every bit beyond $N$. Hard-code the name of $h$, the number $N$, and the finitely many bits below $N$ into the oracle procedure. These are finite constants, so the result is a Turing reduction of $A$ to $X$.","The canonical normal-form copy has exact degree $\\deg_T(A)$ by Proposition~\\ref{thm:computeupper}, while every copy computes $A$ by the forward implication. Hence this degree is the least one.\n\\end{proof}\nThere is no claim of a uniform algorithm finding an infinite-order element or its exception bound in every unmarked copy. The theorem is an existence statement about Turing reductions, and a finite list of constants is legitimate in such a reduction. For marked copies, Proposition~\\ref{thm:markeddecode} gives the genuinely uniform version.","The distinction also matches the algebra. If $A\\eqfin B$, then the unmarked groups are identical inside $P$ and $A$ and $B$ have the same Turing degree. Recovering a literal selector uniformly from the unmarked group would demand more than the isomorphism type remembers.","\\section{Every degree over one computable completion}\n\\begin{corollary}[Degree variation with fixed observations]\n\\label{thm:alldegrees}\nEvery Turing degree occurs as the least presentation degree of a countable, reduced, residually finite abelian group whose profinite completion is $P$, equipped with the fixed computable inverse-system presentation of Chapter~\\ref{ch:roots}.\n\\end{corollary}\n\\begin{proof}\nChoose any set $A$ of the desired degree and apply Theorem~\\ref{thm:degree}. All the structural and completion properties were proved in Chapters~\\ref{ch:purity}--\\ref{ch:roots}.\n\\end{proof}\nFor $A=\\varnothing$, the copy is computable. For $A=K$, the group has no computable copy and its least degree is the halting degree $\\mathbf 0'$. Iterated jumps yield further examples. An arbitrary selector supplies any desired degree without changing the finite quotient tables or their transition maps.","r one computable completion}\n\\begin{corollary}[Degree variation with fixed observations]\n\\label{thm:alldegrees}\nEvery Turing degree occurs as the least presentation degree of a countable, reduced, residually finite abelian group whose profinite completion is $P$, equipped with the fixed computable inverse-system presentation of Chapter~\\ref{ch:roots}.\n\\end{corollary}\n\\begin{proof}\nChoose any set $A$ of the desired degree and apply Theorem~\\ref{thm:degree}. All the structural and completion properties were proved in Chapters~\\ref{ch:purity}--\\ref{ch:roots}.\n\\end{proof}\nFor $A=\\varnothing$, the copy is computable. For $A=K$, the group has no computable copy and its least degree is the halting degree $\\mathbf 0'$. Iterated jumps yield further examples. An arbitrary selector supplies any desired degree without changing the finite quotient tables or their transition maps.","ntable, reduced, residually finite abelian group whose profinite completion is $P$, equipped with the fixed computable inverse-system presentation of Chapter~\\ref{ch:roots}.\n\\end{corollary}\n\\begin{proof}\nChoose any set $A$ of the desired degree and apply Theorem~\\ref{thm:degree}. All the structural and completion properties were proved in Chapters~\\ref{ch:purity}--\\ref{ch:roots}.\n\\end{proof}\nFor $A=\\varnothing$, the copy is computable. For $A=K$, the group has no computable copy and its least degree is the halting degree $\\mathbf 0'$. Iterated jumps yield further examples. An arbitrary selector supplies any desired degree without changing the finite quotient tables or their transition maps.","oup whose profinite completion is $P$, equipped with the fixed computable inverse-system presentation of Chapter~\\ref{ch:roots}.\n\\end{corollary}\n\\begin{proof}\nChoose any set $A$ of the desired degree and apply Theorem~\\ref{thm:degree}. All the structural and completion properties were proved in Chapters~\\ref{ch:purity}--\\ref{ch:roots}.\n\\end{proof}\nFor $A=\\varnothing$, the copy is computable. For $A=K$, the group has no computable copy and its least degree is the halting degree $\\mathbf 0'$. Iterated jumps yield further examples. An arbitrary selector supplies any desired degree without changing the finite quotient tables or their transition maps.","ystem presentation of Chapter~\\ref{ch:roots}.\n\\end{corollary}\n\\begin{proof}\nChoose any set $A$ of the desired degree and apply Theorem~\\ref{thm:degree}. All the structural and completion properties were proved in Chapters~\\ref{ch:purity}--\\ref{ch:roots}.\n\\end{proof}\nFor $A=\\varnothing$, the copy is computable. For $A=K$, the group has no computable copy and its least degree is the halting degree $\\mathbf 0'$. Iterated jumps yield further examples. An arbitrary selector supplies any desired degree without changing the finite quotient tables or their transition maps.","The group $G_K$ still has a finite mathematical definition. Its noncomputability is not a proof that it cannot be named in a finite semantic language. The theorem concerns effective copies with computable operations and equality.","\\section{A sharper separation than prime-spectrum coding}\nFor the initial groups $D_B$, an enumeration of the selected prime set suffices. For $G_A$, the paired complementary tests decide the selector. Meanwhile, every prime torsion factor is present in every $G_A$, so the information is no longer stored in which finite-order elements exist.","\\section{The observation contract}\nLet $\\Obs(G)$ denote the unmarked inverse system of finite quotient groups and their transition maps, considered up to isomorphism as a pro-object. The source group $G$ is not supplied again as part of the input. Nor do we retain evaluations of named source elements. Supplying the source itself would make a reconstruction problem vacuous, and supplying coherent marked elements is a stronger contract.","For the family of this book, there is no ambiguity about the comparison of pro-objects. The same concrete finite products\n\\[\n\\prod_{p\\mid m}C_p\n\\]\nwith the same coordinate projections form a cofinal model for every $\\Obs(G_A)$. Proposition~\\ref{thm:finitehom} also identifies the homomorphism sets into every finite target by a natural bijection induced by restriction to $D$.","\\begin{theorem}[No decoder from the unmarked finite-quotient system]\n\\label{thm:nodecoder}\nThere is no isomorphism-invariant assignment $R$ on these observation systems such that\n\\[\nR(\\Obs(G_A))\\cong G_A\n\\qquad\\text{for every }A\\subseteq\\N.\n\\]\nThis remains impossible without any computability restriction on $R$.\n\\end{theorem}\n\\begin{proof}\nTake $A=\\varnothing$ and $B=\\N$. Their observation systems are isomorphic by Theorem~\\ref{thm:quotientformula}. Their selectors differ infinitely, so their groups are not isomorphic by Theorem~\\ref{thm:isoclass}. An isomorphism-invariant assignment has one output isomorphism type on that input observation type. It cannot recover both nonisomorphic source types.\n\\end{proof}\nThe proof uses two objects, not a counting argument. The full family shows that the ambiguity is much larger: one observ","\\begin{theorem}[No decoder from the unmarked finite-quotient system]\n\\label{thm:nodecoder}\nThere is no isomorphism-invariant assignment $R$ on these observation systems such that\n\\[\nR(\\Obs(G_A))\\cong G_A\n\\qquad\\text{for every }A\\subseteq\\N.\n\\]\nThis remains impossible without any computability restriction on $R$.\n\\end{theorem}\n\\begin{proof}\nTake $A=\\varnothing$ and $B=\\N$. Their observation systems are isomorphic by Theorem~\\ref{thm:quotientformula}. Their selectors differ infinitely, so their groups are not isomorphic by Theorem~\\ref{thm:isoclass}. An isomorphism-invariant assignment has one output isomorphism type on that input observation type. It cannot recover both nonisomorphic source types.\n\\end{proof}\nThe proof uses two objects, not a counting argument. The full family shows that the ambiguity is much larger: one observation system supports continuum many isomorphism types, including examples with every possible least pr","\\begin{theorem}[No decoder from the unmarked finite-quotient system]\n\\label{thm:nodecoder}\nThere is no isomorphism-invariant assignment $R$ on these observation systems such that\n\\[\nR(\\Obs(G_A))\\cong G_A\n\\qquad\\text{for every }A\\subseteq\\N.\n\\]\nThis remains impossible without any computability restriction on $R$.\n\\end{theorem}\n\\begin{proof}\nTake $A=\\varnothing$ and $B=\\N$. Their observation systems are isomorphic by Theorem~\\ref{thm:quotientformula}. Their selectors differ infinitely, so their groups are not isomorphic by Theorem~\\ref{thm:isoclass}. An isomorphism-invariant assignment has one output isomorphism type on that input observation type. It cannot recover both nonisomorphic source types.\n\\end{proof}\nThe proof uses two objects, not a counting argument. The full family shows that the ambiguity is much larger: one observation system supports continuum many isomorphism types, including examples with every possible least presentation degree.","\\section{Element separation does not identify the source}\nEvery $G_A$ is residually finite. Its coordinate maps distinguish any two distinct elements. They give an injective map\n\\[\nG_A\\hookrightarrow\\varprojlim_N A_N=P.\n\\]\nThere is no conflict with Theorem~\\ref{thm:nodecoder}. The injective map identifies each element once the domain is known. The unmarked inverse system does not identify the image of that domain among all compatible tuples.","The complete product is uncountable. Each $G_A$ is countable and dense, hence proper and noncomplete. In particular, a theorem asserting that a complete compatible realization equals its inverse limit has not failed. These examples deliberately retain a different admissible subspace before completion.","\\begin{proposition}[A proper map invisible after completion]\n\\label{thm:propercompletion}\nThe inclusion $D\\hookrightarrow G_A$ is not an isomorphism, but it induces an isomorphism of profinite completions.\n\\end{proposition}\n\\begin{proof}\nThe quotient $G_A/D\\cong\\Q$ is nonzero, so the inclusion is proper. Both completions are canonically $P$, and the coordinate inclusions agree, so the induced completion map is the identity on $P$.\n\\end{proof}\nThis supplies a map-level failure: completion need not reflect isomorphisms. It is stronger than merely exhibiting two abstract groups with isomorphic completions.","p invisible after completion]\n\\label{thm:propercompletion}\nThe inclusion $D\\hookrightarrow G_A$ is not an isomorphism, but it induces an isomorphism of profinite completions.\n\\end{proposition}\n\\begin{proof}\nThe quotient $G_A/D\\cong\\Q$ is nonzero, so the inclusion is proper. Both completions are canonically $P$, and the coordinate inclusions agree, so the induced completion map is the identity on $P$.\n\\end{proof}\nThis supplies a map-level failure: completion need not reflect isomorphisms. It is stronger than merely exhibiting two abstract groups with isomorphic completions.","\\begin{proposition}[Faithfulness and failure of fullness]\n\\label{thm:faithful}\nProfinite completion is faithful on the category of residually finite groups and group homomorphisms. Its restriction to the full subcategory with objects $G_A$ is not full. On the larger full subcategory with objects $D$ and all $G_A$, it does not reflect isomorphisms.\n\\end{proposition}\n\\begin{proof}\nLet $f,g:H\\to K$ be distinct homomorphisms between residually finite groups. Choose $h$ with $f(h)\\ne g(h)$. Some finite quotient of $K$ distinguishes those elements. Therefore the induced completion maps differ on the image of $h$, proving faithfulness.","m:faithful}\nProfinite completion is faithful on the category of residually finite groups and group homomorphisms. Its restriction to the full subcategory with objects $G_A$ is not full. On the larger full subcategory with objects $D$ and all $G_A$, it does not reflect isomorphisms.\n\\end{proposition}\n\\begin{proof}\nLet $f,g:H\\to K$ be distinct homomorphisms between residually finite groups. Choose $h$ with $f(h)\\ne g(h)$. Some finite quotient of $K$ distinguishes those elements. Therefore the induced completion maps differ on the image of $h$, proving faithfulness.","l}\nProfinite completion is faithful on the category of residually finite groups and group homomorphisms. Its restriction to the full subcategory with objects $G_A$ is not full. On the larger full subcategory with objects $D$ and all $G_A$, it does not reflect isomorphisms.\n\\end{proposition}\n\\begin{proof}\nLet $f,g:H\\to K$ be distinct homomorphisms between residually finite groups. Choose $h$ with $f(h)\\ne g(h)$. Some finite quotient of $K$ distinguishes those elements. Therefore the induced completion maps differ on the image of $h$, proving faithfulness.","he category of residually finite groups and group homomorphisms. Its restriction to the full subcategory with objects $G_A$ is not full. On the larger full subcategory with objects $D$ and all $G_A$, it does not reflect isomorphisms.\n\\end{proposition}\n\\begin{proof}\nLet $f,g:H\\to K$ be distinct homomorphisms between residually finite groups. Choose $h$ with $f(h)\\ne g(h)$. Some finite quotient of $K$ distinguishes those elements. Therefore the induced completion maps differ on the image of $h$, proving faithfulness.","bcategory with objects $G_A$ is not full. On the larger full subcategory with objects $D$ and all $G_A$, it does not reflect isomorphisms.\n\\end{proposition}\n\\begin{proof}\nLet $f,g:H\\to K$ be distinct homomorphisms between residually finite groups. Choose $h$ with $f(h)\\ne g(h)$. Some finite quotient of $K$ distinguishes those elements. Therefore the induced completion maps differ on the image of $h$, proving faithfulness.","For failure of fullness, choose $A\\neqfin B$. Under the canonical identifications of the completions with $P$, consider the identity map $P\\to P$. If it were induced by a homomorphism $G_A\\to G_B$, that homomorphism would be the identity on every coordinate and hence would include $G_A$ into $G_B$. Theorem~\\ref{thm:homobstruction} excludes such an embedding. After adjoining $D$ to the source category, the proper inclusion of Proposition~\\ref{thm:propercompletion} proves failure to reflect isomorphisms.\n\\end{proof}\nThe functorial extension of a homomorphism to profinite completions is standard \\cite[Proposition 3.1.2]{Wilkes}. The distinction between the two source categories is essential. Within the paired-prime family alone, a suppl","For failure of fullness, choose $A\\neqfin B$. Under the canonical identifications of the completions with $P$, consider the identity map $P\\to P$. If it were induced by a homomorphism $G_A\\to G_B$, that homomorphism would be the identity on every coordinate and hence would include $G_A$ into $G_B$. Theorem~\\ref{thm:homobstruction} excludes such an embedding. After adjoining $D$ to the source category, the proper inclusion of Proposition~\\ref{thm:propercompletion} proves failure to reflect isomorphisms.\n\\end{proof}\nThe functorial extension of a homomorphism to profinite completions is standard \\cite[Proposition 3.1.2]{Wilkes}. The distinction between the two source categories is essential. Within the paired-prime family alone, a supplied map can be recovered more strongly.","\\begin{proposition}[Reflection within the paired-prime family]\n\\label{thm:rankonereflection}\nFor a homomorphism $f:G_A\\to G_B$, the induced completion map $\\widehat f$ is an isomorphism if and only if $f$ is an isomorphism.\n\\end{proposition}\n\\begin{proof}\nOne implication follows from functoriality. Conversely, suppose $\\widehat f$ is an isomorphism. By Theorem~\\ref{thm:allhom}, its coordinate action is multiplication by a diagonal sequence $\\lambda$. Each $\\lambda_p$ must be nonzero, so $f$ is injective. If $A\\neqfin B$, Theorem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{t","\\begin{proposition}[Reflection within the paired-prime family]\n\\label{thm:rankonereflection}\nFor a homomorphism $f:G_A\\to G_B$, the induced completion map $\\widehat f$ is an isomorphism if and only if $f$ is an isomorphism.\n\\end{proposition}\n\\begin{proof}\nOne implication follows from functoriality. Conversely, suppose $\\widehat f$ is an isomorphism. By Theorem~\\ref{thm:allhom}, its coordinate action is multiplication by a diagonal sequence $\\lambda$. Each $\\lambda_p$ must be nonzero, so $f$ is injective. If $A\\neqfin B$, Theorem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{thm:autGA} makes it an automorphism.\n\\end{","\\begin{proposition}[Reflection within the paired-prime family]\n\\label{thm:rankonereflection}\nFor a homomorphism $f:G_A\\to G_B$, the induced completion map $\\widehat f$ is an isomorphism if and only if $f$ is an isomorphism.\n\\end{proposition}\n\\begin{proof}\nOne implication follows from functoriality. Conversely, suppose $\\widehat f$ is an isomorphism. By Theorem~\\ref{thm:allhom}, its coordinate action is multiplication by a diagonal sequence $\\lambda$. Each $\\lambda_p$ must be nonzero, so $f$ is injective. If $A\\neqfin B$, Theorem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{thm:autGA} makes it an automorphism.\n\\end{proof}\nThus completion forgets distin","\\begin{proposition}[Reflection within the paired-prime family]\n\\label{thm:rankonereflection}\nFor a homomorphism $f:G_A\\to G_B$, the induced completion map $\\widehat f$ is an isomorphism if and only if $f$ is an isomorphism.\n\\end{proposition}\n\\begin{proof}\nOne implication follows from functoriality. Conversely, suppose $\\widehat f$ is an isomorphism. By Theorem~\\ref{thm:allhom}, its coordinate action is multiplication by a diagonal sequence $\\lambda$. Each $\\lambda_p$ must be nonzero, so $f$ is injective. If $A\\neqfin B$, Theorem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{thm:autGA} makes it an automorphism.\n\\end{proof}\nThus completion forgets distinctions between the objects $G_A$, and some maps between their completions have no lift. But when a homomorphism between two memb","red-prime family]\n\\label{thm:rankonereflection}\nFor a homomorphism $f:G_A\\to G_B$, the induced completion map $\\widehat f$ is an isomorphism if and only if $f$ is an isomorphism.\n\\end{proposition}\n\\begin{proof}\nOne implication follows from functoriality. Conversely, suppose $\\widehat f$ is an isomorphism. By Theorem~\\ref{thm:allhom}, its coordinate action is multiplication by a diagonal sequence $\\lambda$. Each $\\lambda_p$ must be nonzero, so $f$ is injective. If $A\\neqfin B$, Theorem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{thm:autGA} makes it an automorphism.\n\\end{proof}\nThus completion forgets distinctions between the objects $G_A$, and some maps between their completions have no lift. But when a homomorphism between two members of this family is already supplied, an isomorphism after completion certifies that the supplied map was an iso…","]\n\\label{thm:rankonereflection}\nFor a homomorphism $f:G_A\\to G_B$, the induced completion map $\\widehat f$ is an isomorphism if and only if $f$ is an isomorphism.\n\\end{proposition}\n\\begin{proof}\nOne implication follows from functoriality. Conversely, suppose $\\widehat f$ is an isomorphism. By Theorem~\\ref{thm:allhom}, its coordinate action is multiplication by a diagonal sequence $\\lambda$. Each $\\lambda_p$ must be nonzero, so $f$ is injective. If $A\\neqfin B$, Theorem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{thm:autGA} makes it an automorphism.\n\\end{proof}\nThus completion forgets distinctions between the objects $G_A$, and some maps between their completions have no lift. But when a homomorphism between two members of this family is already supplied, an isomorphism after completion certifies that the supplied map was an isomorphism. This d…","For a homomorphism $f:G_A\\to G_B$, the induced completion map $\\widehat f$ is an isomorphism if and only if $f$ is an isomorphism.\n\\end{proposition}\n\\begin{proof}\nOne implication follows from functoriality. Conversely, suppose $\\widehat f$ is an isomorphism. By Theorem~\\ref{thm:allhom}, its coordinate action is multiplication by a diagonal sequence $\\lambda$. Each $\\lambda_p$ must be nonzero, so $f$ is injective. If $A\\neqfin B$, Theorem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{thm:autGA} makes it an automorphism.\n\\end{proof}\nThus completion forgets distinctions between the objects $G_A$, and some maps between their completions have no lift. But when a homomorphism between two members of this family is already supplied, an isomorphism after completion certifies that the supplied map was an isomorphism. This does not reconstruct an unknown s…",":G_A\\to G_B$, the induced completion map $\\widehat f$ is an isomorphism if and only if $f$ is an isomorphism.\n\\end{proposition}\n\\begin{proof}\nOne implication follows from functoriality. Conversely, suppose $\\widehat f$ is an isomorphism. By Theorem~\\ref{thm:allhom}, its coordinate action is multiplication by a diagonal sequence $\\lambda$. Each $\\lambda_p$ must be nonzero, so $f$ is injective. If $A\\neqfin B$, Theorem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{thm:autGA} makes it an automorphism.\n\\end{proof}\nThus completion forgets distinctions between the objects $G_A$, and some maps between their completions have no lift. But when a homomorphism between two members of this family is already supplied, an isomorphism after completion certifies that the supplied map was an isomorphism. This does not reconstruct an unknown source from its comple…","implication follows from functoriality. Conversely, suppose $\\widehat f$ is an isomorphism. By Theorem~\\ref{thm:allhom}, its coordinate action is multiplication by a diagonal sequence $\\lambda$. Each $\\lambda_p$ must be nonzero, so $f$ is injective. If $A\\neqfin B$, Theorem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{thm:autGA} makes it an automorphism.\n\\end{proof}\nThus completion forgets distinctions between the objects $G_A$, and some maps between their completions have no lift. But when a homomorphism between two members of this family is already supplied, an isomorphism after completion certifies that the supplied map was an isomorphism. This does not reconstruct an unknown source from its completion alone.","ehat f$ is an isomorphism. By Theorem~\\ref{thm:allhom}, its coordinate action is multiplication by a diagonal sequence $\\lambda$. Each $\\lambda_p$ must be nonzero, so $f$ is injective. If $A\\neqfin B$, Theorem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{thm:autGA} makes it an automorphism.\n\\end{proof}\nThus completion forgets distinctions between the objects $G_A$, and some maps between their completions have no lift. But when a homomorphism between two members of this family is already supplied, an isomorphism after completion certifies that the supplied map was an isomorphism. This does not reconstruct an unknown source from its completion alone.","rem~\\ref{thm:homobstruction} would force its image to be torsion, contradicting injectivity on an infinite-order element. Hence $A\\eqfin B$, and the source and target are equal as subgroups of $P$. The map is now an endomorphism of that group with every diagonal entry nonzero. Corollary~\\ref{thm:autGA} makes it an automorphism.\n\\end{proof}\nThus completion forgets distinctions between the objects $G_A$, and some maps between their completions have no lift. But when a homomorphism between two members of this family is already supplied, an isomorphism after completion certifies that the supplied map was an isomorphism. This does not reconstruct an unknown source from its completion alone.","\\section{What a marked coherent point restores}\nThe coherent image of $u_A$ in the finite quotients records, at prime $p$, whether its coordinate is zero or one. Within the declared family, these zero patterns recover $S_A$ and therefore $A$.","One point is sufficient to identify our family because the construction contract says that the rational line through its tail class is the admissible quotient. Without that contract, a point in $P$ would not identify an arbitrary dense subgroup containing it. The realization rule and its parameter must be retained together.","\\section{Two different inverse-limit assertions}\nFor the subgroup family $D_B$, a complete coherent sequence of finite membership decisions determines $B$ exactly. For the family $G_A$, the unmarked finite quotient system is independent of $A$. These are different observations, even though both involve finite structures and inverse limits.","\\section{Fix the semantics before diagonalizing}\nLet $\\mathfrak G$ be a grammar with countably many finite, parameter-free descriptions. Each valid description has one group isomorphism type as its semantics. Enumerate the strings as\n\\[\nd_0,d_1,d_2,\\ldots\n\\]\nand write $R_j$ for the semantic output. Invalid strings may be assigned a fixed dummy group. No decidability assumption is made about validity or evaluation.","For each $j$ whose output lies in the family, choose $B_j\\subseteq\\N$ such that $R_j\\cong G_{B_j}$. For other $j$, set $B_j=\\varnothing$. This is a set-theoretic choice of representatives. It is not a procedure for deciding whether an arbitrary description denotes a group in the family.","The isomorphism criterion says that a selector must avoid the whole eventual-equivalence class of $B_j$, not merely disagree at one coordinate. A naive Cantor diagonal with one disagreement for each $j$ would not establish that requirement.","\\section{Repeated diagonal coding}\nFix a computable pairing bijection $\\langle j,k\\rangle:\\N^2\\to\\N$. For a freely chosen set $C\\subseteq\\N$, define a new selector $A_C$ by\n\\begin{align}\n\\one_{A_C}(2\\langle j,k\\rangle)\n &=1-\\one_{B_j}(2\\langle j,k\\rangle),\\label{eq:diageven}","\\section{Repeated diagonal coding}\nFix a computable pairing bijection $\\langle j,k\\rangle:\\N^2\\to\\N$. For a freely chosen set $C\\subseteq\\N$, define a new selector $A_C$ by\n\\begin{align}\n\\one_{A_C}(2\\langle j,k\\rangle)\n &=1-\\one_{B_j}(2\\langle j,k\\rangle),\\label{eq:diageven}\\\\\n\\one_{A_C}(2\\langle j,k\\rangle+1)\n &=\\one_C(j).\\label{eq:diagodd}\n\\end{align}\nEvery natural number has exactly one of these forms, so the rules define a total subset. The even positions defeat grammar outputs. The odd positions retain an independent continuum of choices, with each chosen bit repeated infinitely often.","\\begin{theorem}[Escape within one profinite-completion class]\n\\label{thm:escape}\nFor every $C\\subseteq\\N$, the group $G_{A_C}$ is not isomorphic to any output of $\\mathfrak G$. Distinct $C$ give pairwise nonisomorphic groups, with no embeddings in either direction. Every group is countable, reduced, residually finite, and has profinite completion $P$.\n\\end{theorem}\n\\begin{proof}\nFix a description index $j$. If $R_j$ is outside the family, it cannot be isomorphic to $G_{A_C}$. Otherwise rule \\eqref{eq:diageven} makes $A_C$ disagree with $B_j$ at every index $2\\langle j,k\\rangle$. These are infinitely many distinct indices, so $A_C\\neqfin B_j$. Theorem~\\ref{thm:isoclass} excludes the isomorphism.","theorem}[Escape within one profinite-completion class]\n\\label{thm:escape}\nFor every $C\\subseteq\\N$, the group $G_{A_C}$ is not isomorphic to any output of $\\mathfrak G$. Distinct $C$ give pairwise nonisomorphic groups, with no embeddings in either direction. Every group is countable, reduced, residually finite, and has profinite completion $P$.\n\\end{theorem}\n\\begin{proof}\nFix a description index $j$. If $R_j$ is outside the family, it cannot be isomorphic to $G_{A_C}$. Otherwise rule \\eqref{eq:diageven} makes $A_C$ disagree with $B_j$ at every index $2\\langle j,k\\rangle$. These are infinitely many distinct indices, so $A_C\\neqfin B_j$. Theorem~\\ref{thm:isoclass} excludes the isomorphism.","[Escape within one profinite-completion class]\n\\label{thm:escape}\nFor every $C\\subseteq\\N$, the group $G_{A_C}$ is not isomorphic to any output of $\\mathfrak G$. Distinct $C$ give pairwise nonisomorphic groups, with no embeddings in either direction. Every group is countable, reduced, residually finite, and has profinite completion $P$.\n\\end{theorem}\n\\begin{proof}\nFix a description index $j$. If $R_j$ is outside the family, it cannot be isomorphic to $G_{A_C}$. Otherwise rule \\eqref{eq:diageven} makes $A_C$ disagree with $B_j$ at every index $2\\langle j,k\\rangle$. These are infinitely many distinct indices, so $A_C\\neqfin B_j$. Theorem~\\ref{thm:isoclass} excludes the isomorphism.","m:escape}\nFor every $C\\subseteq\\N$, the group $G_{A_C}$ is not isomorphic to any output of $\\mathfrak G$. Distinct $C$ give pairwise nonisomorphic groups, with no embeddings in either direction. Every group is countable, reduced, residually finite, and has profinite completion $P$.\n\\end{theorem}\n\\begin{proof}\nFix a description index $j$. If $R_j$ is outside the family, it cannot be isomorphic to $G_{A_C}$. Otherwise rule \\eqref{eq:diageven} makes $A_C$ disagree with $B_j$ at every index $2\\langle j,k\\rangle$. These are infinitely many distinct indices, so $A_C\\neqfin B_j$. Theorem~\\ref{thm:isoclass} excludes the isomorphism.","C}$ is not isomorphic to any output of $\\mathfrak G$. Distinct $C$ give pairwise nonisomorphic groups, with no embeddings in either direction. Every group is countable, reduced, residually finite, and has profinite completion $P$.\n\\end{theorem}\n\\begin{proof}\nFix a description index $j$. If $R_j$ is outside the family, it cannot be isomorphic to $G_{A_C}$. Otherwise rule \\eqref{eq:diageven} makes $A_C$ disagree with $B_j$ at every index $2\\langle j,k\\rangle$. These are infinitely many distinct indices, so $A_C\\neqfin B_j$. Theorem~\\ref{thm:isoclass} excludes the isomorphism.","c to any output of $\\mathfrak G$. Distinct $C$ give pairwise nonisomorphic groups, with no embeddings in either direction. Every group is countable, reduced, residually finite, and has profinite completion $P$.\n\\end{theorem}\n\\begin{proof}\nFix a description index $j$. If $R_j$ is outside the family, it cannot be isomorphic to $G_{A_C}$. Otherwise rule \\eqref{eq:diageven} makes $A_C$ disagree with $B_j$ at every index $2\\langle j,k\\rangle$. These are infinitely many distinct indices, so $A_C\\neqfin B_j$. Theorem~\\ref{thm:isoclass} excludes the isomorphism.","mathfrak G$. Distinct $C$ give pairwise nonisomorphic groups, with no embeddings in either direction. Every group is countable, reduced, residually finite, and has profinite completion $P$.\n\\end{theorem}\n\\begin{proof}\nFix a description index $j$. If $R_j$ is outside the family, it cannot be isomorphic to $G_{A_C}$. Otherwise rule \\eqref{eq:diageven} makes $A_C$ disagree with $B_j$ at every index $2\\langle j,k\\rangle$. These are infinitely many distinct indices, so $A_C\\neqfin B_j$. Theorem~\\ref{thm:isoclass} excludes the isomorphism.","embeddings in either direction. Every group is countable, reduced, residually finite, and has profinite completion $P$.\n\\end{theorem}\n\\begin{proof}\nFix a description index $j$. If $R_j$ is outside the family, it cannot be isomorphic to $G_{A_C}$. Otherwise rule \\eqref{eq:diageven} makes $A_C$ disagree with $B_j$ at every index $2\\langle j,k\\rangle$. These are infinitely many distinct indices, so $A_C\\neqfin B_j$. Theorem~\\ref{thm:isoclass} excludes the isomorphism.","If $C\\ne C'$, choose $j$ at which they differ. Rule \\eqref{eq:diagodd} makes $A_C$ and $A_{C'}$ differ at infinitely many odd indices. Theorems~\\ref{thm:isoclass} and \\ref{thm:homobstruction} then give nonisomorphism and nonembeddability. The remaining properties hold for every paired-prime group.\n\\end{proof}","\\begin{figure}[tbp]\n\\centering\\includegraphics[width=\\textwidth]{\\figfile{06_repeated_diagonal.pdf}}\n\\caption{Two infinite coordinate blocks are allocated to every $j$. On the even block the selector opposes the representative $B_j$ forever. On the odd block it repeats the free bit $C(j)$ forever. Both repetitions are necessary to survive the quotient by finite changes.}\n\\label{fig:diagonal}\n\\end{figure}","\\section{The complexity consequence}\nThe theorem implies\n\\[\nK_{\\mathfrak G}(G_{A_C})=+\\infty\n\\]\nfor every $C$. It produces continuum many distinct isomorphism types outside the grammar while preserving all the finite quotient data.","This is a statement relative to a specified language and semantics. The definition of $A_C$ may use the entire semantic table of that language in the surrounding mathematical theory. It does not yield an effective evaluator for the table. Nor does it produce a paradox by defining in one paragraph an object claimed to be beyond every possible finite definition. The object escapes the fixed internal grammar, not all larger metatheories.","\\section{Monotonicity under grammar extension}\n\\begin{proposition}[Admissible expansion]\n\\label{thm:grammarextension}\nSuppose every valid description in $\\mathfrak G$ translates to a valid description in $\\mathfrak H$ with the same semantic group. Then every $\\mathfrak G$-realizable group is $\\mathfrak H$-realizable. If every translated description has length at most $f(\\lvert d\\rvert)$, then\n\\[\nK_{\\mathfrak H}(G)\\le f(K_{\\mathfrak G}(G))\n\\]\nwhenever $K_{\\mathfrak G}(G)$ is finite.\n\\end{proposition}\n\\begin{proof}\nTranslate a shortest valid $\\mathfrak G$ description of $G$. Its semantic output remains $G$, and its length has the stated bound.\n\\end{proof}\nAn obstruction to a small grammar can therefore disappear after an admissible expansion. The power-of-two selector from Chapter~\\ref{ch:spectrum} illustrates this concretely: a unary finite-state selecto","\\section{Monotonicity under grammar extension}\n\\begin{proposition}[Admissible expansion]\n\\label{thm:grammarextension}\nSuppose every valid description in $\\mathfrak G$ translates to a valid description in $\\mathfrak H$ with the same semantic group. Then every $\\mathfrak G$-realizable group is $\\mathfrak H$-realizable. If every translated description has length at most $f(\\lvert d\\rvert)$, then\n\\[\nK_{\\mathfrak H}(G)\\le f(K_{\\mathfrak G}(G))\n\\]\nwhenever $K_{\\mathfrak G}(G)$ is finite.\n\\end{proposition}\n\\begin{proof}\nTranslate a shortest valid $\\mathfrak G$ description of $G$. Its semantic output remains $G$, and its length has the stated bound.\n\\end{proof}\nAn obstruction to a small grammar can therefore disappear after an admissible expansion. The power-of-two selector from Chapter~\\ref{ch:spectrum} illustrates this concretely: a unary finite-state selector fails, while a binary one succeeds. Such a result","\\section{Monotonicity under grammar extension}\n\\begin{proposition}[Admissible expansion]\n\\label{thm:grammarextension}\nSuppose every valid description in $\\mathfrak G$ translates to a valid description in $\\mathfrak H$ with the same semantic group. Then every $\\mathfrak G$-realizable group is $\\mathfrak H$-realizable. If every translated description has length at most $f(\\lvert d\\rvert)$, then\n\\[\nK_{\\mathfrak H}(G)\\le f(K_{\\mathfrak G}(G))\n\\]\nwhenever $K_{\\mathfrak G}(G)$ is finite.\n\\end{proposition}\n\\begin{proof}\nTranslate a shortest valid $\\mathfrak G$ description of $G$. Its semantic output remains $G$, and its length has the stated bound.\n\\end{proof}\nAn obstruction to a small grammar can therefore disappear after an admissible expansion. The power-of-two selector from Chapter~\\ref{ch:spectrum} illustrates this concretely: a unary finite-state selector fails, while a binary one succeeds. Such a result characterizes a boundary rather than discrediting the","\\section{Monotonicity under grammar extension}\n\\begin{proposition}[Admissible expansion]\n\\label{thm:grammarextension}\nSuppose every valid description in $\\mathfrak G$ translates to a valid description in $\\mathfrak H$ with the same semantic group. Then every $\\mathfrak G$-realizable group is $\\mathfrak H$-realizable. If every translated description has length at most $f(\\lvert d\\rvert)$, then\n\\[\nK_{\\mathfrak H}(G)\\le f(K_{\\mathfrak G}(G))\n\\]\nwhenever $K_{\\mathfrak G}(G)$ is finite.\n\\end{proposition}\n\\begin{proof}\nTranslate a shortest valid $\\mathfrak G$ description of $G$. Its semantic output remains $G$, and its length has the stated bound.\n\\end{proof}\nAn obstruction to a small grammar can therefore disappear after an admissible expansion. The power-of-two selector from Chapter~\\ref{ch:spectrum} illustrates this concretely: a unary finite-state selector fails, while a binary one succeeds. Such a result characterizes a boundary rather than discrediting the larger realization program.","oposition}[Admissible expansion]\n\\label{thm:grammarextension}\nSuppose every valid description in $\\mathfrak G$ translates to a valid description in $\\mathfrak H$ with the same semantic group. Then every $\\mathfrak G$-realizable group is $\\mathfrak H$-realizable. If every translated description has length at most $f(\\lvert d\\rvert)$, then\n\\[\nK_{\\mathfrak H}(G)\\le f(K_{\\mathfrak G}(G))\n\\]\nwhenever $K_{\\mathfrak G}(G)$ is finite.\n\\end{proposition}\n\\begin{proof}\nTranslate a shortest valid $\\mathfrak G$ description of $G$. Its semantic output remains $G$, and its length has the stated bound.\n\\end{proof}\nAn obstruction to a small grammar can therefore disappear after an admissible expansion. The power-of-two selector from Chapter~\\ref{ch:spectrum} illustrates this concretely: a unary finite-state selector fails, while a binary one succeeds. Such a result characterizes a boundary rather than discrediting the larger realization program.","on in $\\mathfrak G$ translates to a valid description in $\\mathfrak H$ with the same semantic group. Then every $\\mathfrak G$-realizable group is $\\mathfrak H$-realizable. If every translated description has length at most $f(\\lvert d\\rvert)$, then\n\\[\nK_{\\mathfrak H}(G)\\le f(K_{\\mathfrak G}(G))\n\\]\nwhenever $K_{\\mathfrak G}(G)$ is finite.\n\\end{proposition}\n\\begin{proof}\nTranslate a shortest valid $\\mathfrak G$ description of $G$. Its semantic output remains $G$, and its length has the stated bound.\n\\end{proof}\nAn obstruction to a small grammar can therefore disappear after an admissible expansion. The power-of-two selector from Chapter~\\ref{ch:spectrum} illustrates this concretely: a unary finite-state selector fails, while a binary one succeeds. Such a result characterizes a boundary rather than discrediting the larger realization program.","s to a valid description in $\\mathfrak H$ with the same semantic group. Then every $\\mathfrak G$-realizable group is $\\mathfrak H$-realizable. If every translated description has length at most $f(\\lvert d\\rvert)$, then\n\\[\nK_{\\mathfrak H}(G)\\le f(K_{\\mathfrak G}(G))\n\\]\nwhenever $K_{\\mathfrak G}(G)$ is finite.\n\\end{proposition}\n\\begin{proof}\nTranslate a shortest valid $\\mathfrak G$ description of $G$. Its semantic output remains $G$, and its length has the stated bound.\n\\end{proof}\nAn obstruction to a small grammar can therefore disappear after an admissible expansion. The power-of-two selector from Chapter~\\ref{ch:spectrum} illustrates this concretely: a unary finite-state selector fails, while a binary one succeeds. Such a result characterizes a boundary rather than discrediting the larger realization program.","hfrak H$ with the same semantic group. Then every $\\mathfrak G$-realizable group is $\\mathfrak H$-realizable. If every translated description has length at most $f(\\lvert d\\rvert)$, then\n\\[\nK_{\\mathfrak H}(G)\\le f(K_{\\mathfrak G}(G))\n\\]\nwhenever $K_{\\mathfrak G}(G)$ is finite.\n\\end{proposition}\n\\begin{proof}\nTranslate a shortest valid $\\mathfrak G$ description of $G$. Its semantic output remains $G$, and its length has the stated bound.\n\\end{proof}\nAn obstruction to a small grammar can therefore disappear after an admissible expansion. The power-of-two selector from Chapter~\\ref{ch:spectrum} illustrates this concretely: a unary finite-state selector fails, while a binary one succeeds. Such a result characterizes a boundary rather than discrediting the larger realization program.","\\section{Uniform families and individual instances}\nThe expression $A\\mapsto G_A$ is one uniform mathematical construction. It is not a finite description of each arbitrary instance unless the selector is also described. A single family can contain uncountably many values even though a parameter-free finite syntax has only countably many strings.","For computable selectors, a program for $A$ provides a finite effective input to the normal-form construction. For noncomputable but finitely definable selectors, one obtains a finite semantic description without an effective copy. For selectors outside a chosen finite-description language, the family formula still proves existence relative to that selector, but does not internalize it into that language.","\\section{The empty selector}\nTake $A=\\varnothing$. The selected primes are\n\\[\nS_A=\\{2,5,11,17,23,31,41,47,\\ldots\\},\n\\]\nnamely the even-indexed primes. The first coordinates of three rational lifts are\n\\begin{center}\n\\begin{tabular}{@{}crrrrrr@{}}\n\\toprule\n$p$&2&3&5&7&11&13","midrule\n$u_A=s_A(1)$&1&0&1&0&1&0","$s_A(1/2)$&0&0&3&0&6&0","$s_A(1/3)$&1&0&2&0&4&0","bottomrule\n\\end{tabular}\n\\end{center}\nEach residue is exact. At $p=2$, the lift of $1/2$ is set to zero because the denominator is not invertible. At $p=5$, its value is $3$ because $2\\cdot3=1$ in $\\F_5$.","Doubling the half-lift agrees with $u_A$ at every prime except $2$. Therefore\n\\[\n2(1/2,0)=(1,e_2).\n\\]\nThe element $u_A$ has no half in the group: there is no $y$ with $2y=u_A$, because its $2$-coordinate is nonzero. The apparent candidate $(1/2,0)$ misses by precisely the certificate coordinate.","At the unselected prime $3$, the coordinate of $u_A$ is zero. In this case\n\\[\n3(1/3,0)=(1,0).\n\\]\nThe three roots are\n\\[\n(1/3,0),\\qquad(1/3,e_3),\\qquad(1/3,2e_3).\n\\]\nTheir differences form $G_A[3]=C_3$, as predicted by Proposition~\\ref{thm:rootfiber}.","\\section{A successful composite-root calculation}\nLet\n\\[\ng=s_A(1)+e_2=(1,e_2).\n\\]\nIts coordinate at $2$ is $1+1=0$, and its coordinate at $3$ is zero. Thus the multiplier $6$ has no obstructing coordinate.","The rational candidate is $t=1/6$. At a selected prime other than $2$, multiplication by $6$ recovers the coordinate of $u_A$ because the denominator is invertible, except possibly at $3$, which is unselected. At $2$, both sides are zero. Hence\n\\[\n6(1/6,0)=(1,e_2)=g.\n\\]\nEvery other root is obtained by adding an arbitrary element of $C_2\\oplus C_3$. There are exactly six roots. This calculation exhibits a positive certificate without searching an unbounded domain.","Now contrast $u_A$ itself. The same multiplier $6$ fails at the prime $2$. No change at the $3$-coordinate and no choice of rational representative can repair that obstruction while keeping the target fixed.","\\section{A torsion correction that cannot be discarded}\nTake the purely torsion target $e_3$ and multiplier $2$. The root criterion succeeds because the target has zero $2$-coordinate. Its rational part is zero, but the answer is not the zero element:\n\\[\n2(0,2e_3)=(0,e_3).\n\\]\nIn $\\F_3$, multiplication by $2$ sends $2$ to $1$. This is the simplest example showing why the finite correction in the root algorithm is necessary even when the rational coordinate causes no difficulty.","\\section{Finite quotient representatives}\nFor any target $g$ and positive multiplier $m$, define the finite vector\n\\[\nd_m(g)=\\sum_{p\\mid m}g_p e_p.\n\\]\nThen $g-d_m(g)$ has zero coordinates at every prime dividing $m$. The root theorem provides a witness that it belongs to $mG_A$. Thus $d_m(g)$ is an exact finite representative of the coset of $g$.","For $m=12$, only coordinates $2$ and $3$ matter. The six representatives are\n\\[\n0,\\ e_2,\\ e_3,\\ 2e_3,\\ e_2+e_3,\\ e_2+2e_3.\n\\]\nThe vector $e_2+e_3$ generates this quotient, which is $C_6$, not $C_{12}$. Repeated prime powers in the multiplier do not create additional coordinates.","\\begin{figure}[tbp]\n\\centering\\includegraphics[width=\\textwidth]{\\figfile{07_squarefree_quotients.pdf}}\n\\caption{The exact order of $G_A/mG_A$ for $1\\le m\\le30$ is $\\rad(m)$. The values are independent of the selector. This plot evaluates the theorem's arithmetic formula, rather than estimating the size of an unknown quotient.}\n\\label{fig:radical}\n\\end{figure}","\\section{Changing one selector bit}\nLet $B=\\{0\\}$, so the first pair changes from selecting $2$ to selecting $3$. The groups $G_A$ and $G_B$ are equal as subgroups of $P$. Their half-lift coordinates differ, however:\n\\[\ns_A(1/2)-s_B(1/2)=e_3.\n\\]\nAt $p=2$, both half-lifts are zero. At $p=3$, the $B$ lift is $2$, so its negative is $1$. Therefore the same ambient element has normal forms\n\\[\n(1/2,0)_A=(1/2,e_3)_B.\n\\]\nDoubling the right-hand side in the $B$ group gives\n\\[\n(1,2e_3)_B,\n\\]\nwhich is the transported form of $(1,e_2)_A$. The correction in \\eqref{eq:transport} is required to compare the two group laws. Merely reusing the same pair $(r,d)$ under a new selector changes the represented element.","he first pair changes from selecting $2$ to selecting $3$. The groups $G_A$ and $G_B$ are equal as subgroups of $P$. Their half-lift coordinates differ, however:\n\\[\ns_A(1/2)-s_B(1/2)=e_3.\n\\]\nAt $p=2$, both half-lifts are zero. At $p=3$, the $B$ lift is $2$, so its negative is $1$. Therefore the same ambient element has normal forms\n\\[\n(1/2,0)_A=(1/2,e_3)_B.\n\\]\nDoubling the right-hand side in the $B$ group gives\n\\[\n(1,2e_3)_B,\n\\]\nwhich is the transported form of $(1,e_2)_A$. The correction in \\eqref{eq:transport} is required to compare the two group laws. Merely reusing the same pair $(r,d)$ under a new selector changes the represented element.","nges from selecting $2$ to selecting $3$. The groups $G_A$ and $G_B$ are equal as subgroups of $P$. Their half-lift coordinates differ, however:\n\\[\ns_A(1/2)-s_B(1/2)=e_3.\n\\]\nAt $p=2$, both half-lifts are zero. At $p=3$, the $B$ lift is $2$, so its negative is $1$. Therefore the same ambient element has normal forms\n\\[\n(1/2,0)_A=(1/2,e_3)_B.\n\\]\nDoubling the right-hand side in the $B$ group gives\n\\[\n(1,2e_3)_B,\n\\]\nwhich is the transported form of $(1,e_2)_A$. The correction in \\eqref{eq:transport} is required to compare the two group laws. Merely reusing the same pair $(r,d)$ under a new selector changes the represented element.","ps $G_A$ and $G_B$ are equal as subgroups of $P$. Their half-lift coordinates differ, however:\n\\[\ns_A(1/2)-s_B(1/2)=e_3.\n\\]\nAt $p=2$, both half-lifts are zero. At $p=3$, the $B$ lift is $2$, so its negative is $1$. Therefore the same ambient element has normal forms\n\\[\n(1/2,0)_A=(1/2,e_3)_B.\n\\]\nDoubling the right-hand side in the $B$ group gives\n\\[\n(1,2e_3)_B,\n\\]\nwhich is the transported form of $(1,e_2)_A$. The correction in \\eqref{eq:transport} is required to compare the two group laws. Merely reusing the same pair $(r,d)$ under a new selector changes the represented element.","At $p=2$, both half-lifts are zero. At $p=3$, the $B$ lift is $2$, so its negative is $1$. Therefore the same ambient element has normal forms\n\\[\n(1/2,0)_A=(1/2,e_3)_B.\n\\]\nDoubling the right-hand side in the $B$ group gives\n\\[\n(1,2e_3)_B,\n\\]\nwhich is the transported form of $(1,e_2)_A$. The correction in \\eqref{eq:transport} is required to compare the two group laws. Merely reusing the same pair $(r,d)$ under a new selector changes the represented element.","\\section{A finite tail exception}\nConsider an infinite-order element\n\\[\nx=s_A(5/6)+2e_7.\n\\]\nAn exception set for its signature is $\\{2,3,5,7\\}$. Outside these primes, the denominator is invertible, the numerator is nonzero, and the torsion correction vanishes. Thus the root pattern of $x$ agrees with that of $u_A$ in every later prime pair.","\\section{The ambient object and its admissible realizations}\nThe chain\n\\[\nD\\subseteq G_A\\subseteq P\n\\]\ncontains three different mathematical roles. The direct sum provides the finite-support corrections. The product supplies every compatible prime-coordinate tuple. The intermediate group is the intended admissible realization, determined by a rational line in the quotient.","This arrangement gives a concrete model for a central question of Komorebi Space. Once the compatible ambient object exists, what additional data identifies the realization under study? Here the answer is exact: a rational subspace of $P/D$, or in the paired family its selector class. Existence of the ambient product does not eliminate that choice.","The nonsplitting result supplies an extension obstruction. A rational quotient exists, but no homomorphic section realizes it inside the reduced product. In cocycle coordinates, the obstruction is the impossibility of removing the defect by a $D$-valued correction.","\\section{A realization profile rather than a single score}\nOne scalar cannot capture all the boundaries exhibited here. The following profile retains the distinctions:\n\\[\n\\bigl(\\text{seed rank},\\ \\text{grammar complexity},\\\n\\text{least presentation degree},\\ \\text{observation type}\\bigr).\n\\]\nFor every $G_A$, the seed rank is $\\aleph_0$ and the observation type is the same. The least presentation degree varies with $A$. Grammar complexity depends on which selectors the chosen semantic language can name.","ion profile rather than a single score}\nOne scalar cannot capture all the boundaries exhibited here. The following profile retains the distinctions:\n\\[\n\\bigl(\\text{seed rank},\\ \\text{grammar complexity},\\\n\\text{least presentation degree},\\ \\text{observation type}\\bigr).\n\\]\nFor every $G_A$, the seed rank is $\\aleph_0$ and the observation type is the same. The least presentation degree varies with $A$. Grammar complexity depends on which selectors the chosen semantic language can name.","midrule\nAmbient system&$D\\subseteq P$ and the quotient $q:P\\to P/D$.","Parameter&The selector $A$, with its declared oracle or finite semantic description.","Admissible image&$q^{-1}(\\Q q(u_A))$.","\\section{Further questions with exact starting points}\nThe general correspondence of Chapter~\\ref{ch:purity} extends the ambient construction to higher-dimensional rational subspaces $W\\le P/D$. Its finite quotient theorem is already complete. What remains is to classify those subspaces under the diagonal multipliers that induce group homomorphisms. The paired rank-one family has an exact answer, but the book does not claim that its binary signature classifies higher rank.","A second question concerns effective selection. For $G_A$, the normal-form operation and the paired root tests give a matching upper and lower bound. For a more general countable $W$, one must specify how its rational dependencies are presented and how lifts are computed. A list of candidate spanning vectors is not enough if their dependence relations are unavailable.","\\section{Conclusion}\nThe principal boundary is now concrete. There is a fixed ambient completion, presented by a computable inverse system, and a family of countable, residually finite groups densely realized inside it. Their finite-quotient systems are canonically identified, and restriction to $D$ naturally identifies their homomorphism sets into every finite target. Yet the groups have distinct algebraic types and require arbitrarily different computational information.","The selector is supplied as a callable taking a nonnegative integer $n$ to the Boolean value of $\\one_A(n)$. Integer bits zero and one are also accepted. Other return values are rejected. The group object converts a prime label to its index, determines which pair it belongs to, and queries the corresponding selector bit. This is an oracle interface. A finite Python function implements only those selectors for which it can return the correct value. No program in the package claims to compute a noncomputable selector.","\\texttt{root(x, m)}&For $m\\ge1$, return a root or an obstructing prime.","\\section{Root construction as an algorithm}\nThe proof of Theorem~\\ref{thm:rootcertificate} gives the following procedure.\n\\begin{enumerate}\n\\item Factor the positive integer $m$ and inspect the target coordinates at its prime divisors. If any is nonzero, return that prime as the obstruction.\n\\item Divide the rational coordinate by $m$ to obtain a candidate section lift.\n\\item Compute the finite residual between the target and $m$ times that lift. Its support is confined to the input torsion support and the denominator primes stated in the proof.\n\\item At primes not dividing $m$, divide the residual coordinate by $m$ in the finite field. At primes dividing $m$, the residual is zero. Add this finite correction to the candidate lift.\n\\end{enumerate}\nThe procedure returns a complete normal-form element. Its certificate is checked by the independently implemented integ","\\section{Root construction as an algorithm}\nThe proof of Theorem~\\ref{thm:rootcertificate} gives the following procedure.\n\\begin{enumerate}\n\\item Factor the positive integer $m$ and inspect the target coordinates at its prime divisors. If any is nonzero, return that prime as the obstruction.\n\\item Divide the rational coordinate by $m$ to obtain a candidate section lift.\n\\item Compute the finite residual between the target and $m$ times that lift. Its support is confined to the input torsion support and the denominator primes stated in the proof.\n\\item At primes not dividing $m$, divide the residual coordinate by $m$ in the finite field. At primes dividing $m$, the residual is zero. Add this finite correction to the candidate lift.\n\\end{enumerate}\nThe procedure returns a complete normal-form element. Its certificate is checked by the independently implemented integer multiplication routine. The negative certificate is checked by verifying both $p\\mid m$ and a nonzero target coordinate.","n algorithm}\nThe proof of Theorem~\\ref{thm:rootcertificate} gives the following procedure.\n\\begin{enumerate}\n\\item Factor the positive integer $m$ and inspect the target coordinates at its prime divisors. If any is nonzero, return that prime as the obstruction.\n\\item Divide the rational coordinate by $m$ to obtain a candidate section lift.\n\\item Compute the finite residual between the target and $m$ times that lift. Its support is confined to the input torsion support and the denominator primes stated in the proof.\n\\item At primes not dividing $m$, divide the residual coordinate by $m$ in the finite field. At primes dividing $m$, the residual is zero. Add this finite correction to the candidate lift.\n\\end{enumerate}\nThe procedure returns a complete normal-form element. Its certificate is checked by the independently implemented integer multiplication routine. The negative certificate is checked by verifying both $p\\mid m$ and a nonzero target coordinate.","inspect the target coordinates at its prime divisors. If any is nonzero, return that prime as the obstruction.\n\\item Divide the rational coordinate by $m$ to obtain a candidate section lift.\n\\item Compute the finite residual between the target and $m$ times that lift. Its support is confined to the input torsion support and the denominator primes stated in the proof.\n\\item At primes not dividing $m$, divide the residual coordinate by $m$ in the finite field. At primes dividing $m$, the residual is zero. Add this finite correction to the candidate lift.\n\\end{enumerate}\nThe procedure returns a complete normal-form element. Its certificate is checked by the independently implemented integer multiplication routine. The negative certificate is checked by verifying both $p\\mid m$ and a nonzero target coordinate.","rime divisors. If any is nonzero, return that prime as the obstruction.\n\\item Divide the rational coordinate by $m$ to obtain a candidate section lift.\n\\item Compute the finite residual between the target and $m$ times that lift. Its support is confined to the input torsion support and the denominator primes stated in the proof.\n\\item At primes not dividing $m$, divide the residual coordinate by $m$ in the finite field. At primes dividing $m$, the residual is zero. Add this finite correction to the candidate lift.\n\\end{enumerate}\nThe procedure returns a complete normal-form element. Its certificate is checked by the independently implemented integer multiplication routine. The negative certificate is checked by verifying both $p\\mid m$ and a nonzero target coordinate.","t prime as the obstruction.\n\\item Divide the rational coordinate by $m$ to obtain a candidate section lift.\n\\item Compute the finite residual between the target and $m$ times that lift. Its support is confined to the input torsion support and the denominator primes stated in the proof.\n\\item At primes not dividing $m$, divide the residual coordinate by $m$ in the finite field. At primes dividing $m$, the residual is zero. Add this finite correction to the candidate lift.\n\\end{enumerate}\nThe procedure returns a complete normal-form element. Its certificate is checked by the independently implemented integer multiplication routine. The negative certificate is checked by verifying both $p\\mid m$ and a nonzero target coordinate.","At primes not dividing $m$, divide the residual coordinate by $m$ in the finite field. At primes dividing $m$, the residual is zero. Add this finite correction to the candidate lift.\n\\end{enumerate}\nThe procedure returns a complete normal-form element. Its certificate is checked by the independently implemented integer multiplication routine. The negative certificate is checked by verifying both $p\\mid m$ and a nonzero target coordinate.","midrule\nOmit the cocycle&Twice the half-lift can differ from $u_A$ at prime $2$.","Use $C_m$ for every integer quotient&Prime powers collapse to $C_{\\rad(m)}$.","\\ref{thm:torP}&$\\Tor(P)=D$, and $D$ is dense.&Distinct prime coordinates.","\\ref{thm:endD}&All endomorphisms of $D$ are diagonal.&Prime annihilators.","\\ref{thm:rankD}&$D$ has infinite endomorphism-seed rank.&Finite support.","\\ref{thm:procyclic}&$P$ is topologically cyclic.&Chinese remainder theorem.","\\ref{thm:Dcompletion}&$\\widehat D=P$.&Cofinal finite coordinate kernels.","\\ref{thm:Vrational}&$P/D$ is a rational vector space.&Unique integer division modulo $D$.","\\ref{thm:subD}&$\\Sub(D)\\cong\\mathcal P(\\N)$.&Coordinate isolation by CRT.","\\ref{thm:spectrumce}&Computable copies of $D_B$ correspond to enumerable selectors.&Prime-order witness searches.","\\ref{thm:generalquotients}&All preimages $H_W$ have completion $P$.&Purity and finite-target exponent.","\\ref{thm:normalform}&$G_A$ has unique $(r,d)$ names.&Nonzero rational line.","\\ref{thm:cocyclelaw}&The normal forms form the actual group $G_A$.&Cocycle identity.","\\ref{thm:rootfiber}&A nonempty root fiber has $\\rad(m)$ elements.&The prime-coordinate kernel.","\\ref{thm:quotientformula}&$G_A/mG_A=C_{\\rad(m)}$.&Root kernel and finite surjectivity.","\\ref{thm:isoclass}&$G_A\\cong G_B$ iff $A\\eqfin B$.&Intrinsic divisibility and finite transport.","\\ref{thm:allhom}&Every homomorphism is a diagonal multiplier.&The quotients $G_A/pG_A$.","\\ref{thm:rankGA}&Every $G_A$ has infinite seed rank.&Infinitely many omitted unselected coordinates.","\\ref{thm:degree}&An oracle computes a copy iff it computes $A$.&Cocycle evaluation and complementary root searches.","The traced implementation erases to the executable solver. If $B$ is its finite correction bound, it uses at most $m+1+2B$ selector queries and at most $m+1+B$ loop candidates. These are explicit oracle and iteration counts. They are not bounds on the bit cost of rational arithmetic, primality testing, division, or modular inversion. The Lean proofs certify the Lean implementation and its mathematical interpretation. They do not claim kernel verification of the separate Python runtime.","\\item[Cocycle] Here, a symmetric map $c:\\Q\\times\\Q\\to D$ satisfying the identity that makes the twisted addition on $\\Q\\times D$ associative.\n\\Needspace{5\\baselineskip}\\item[Cofinal quotient system] A family of finite quotients through which every finite quotient factors. Its inverse limit computes the full profinite completion.\n\\Needspace{5\\baselineskip}\\item[Computable copy] A countable realization with effective names, decidable equality, and computable operations. An oracle version permits the stated oracle.\n\\Needspace{5\\baselineskip}\\item[Compu","\\item[Cocycle] Here, a symmetric map $c:\\Q\\times\\Q\\to D$ satisfying the identity that makes the twisted addition on $\\Q\\times D$ associative.\n\\Needspace{5\\baselineskip}\\item[Cofinal quotient system] A family of finite quotients through which every finite quotient factors. Its inverse limit computes the full profinite completion.\n\\Needspace{5\\baselineskip}\\item[Computable copy] A countable realization with effective names, decidable equality, and computable operations. An oracle version permits the stated oracle.\n\\Needspace{5\\baselineskip}\\item[Computable profinite presentation] An effective inverse system of finite grou","\\item[Divisible abelian group] An abelian group in which every equation $my=x$ with $m\\ge1$ is solvable. A divisible quotient need not split off as a subgroup.\n\\Needspace{5\\baselineskip}\\item[Endomorphism-seed rank] The minimum number of elements generating a group as a module over its full endomorphism ring.\n\\Needspace{5\\baselineskip}\\item[Eventual equality] Equality modulo finitely many coordinates, written $A\\eqfin B$. This is the unmarked selector invariant of $G_A$.\n\\Needspace{5\\baselineskip}\\item[Finite description] A valid finite code in a specified language with","\\item[Divisible abelian group] An abelian group in which every equation $my=x$ with $m\\ge1$ is solvable. A divisible quotient need not split off as a subgroup.\n\\Needspace{5\\baselineskip}\\item[Endomorphism-seed rank] The minimum number of elements generating a group as a module over its full endomorphism ring.\n\\Needspace{5\\baselineskip}\\item[Eventual equality] Equality modulo finitely many coordinates, written $A\\eqfin B$. This is the unmarked selector invariant of $G_A$.\n\\Needspace{5\\baselineskip}\\item[Finite description] A valid finite code in a specified language with declared sema","\\item[Eventual equality] Equality modulo finitely many coordinates, written $A\\eqfin B$. This is the unmarked selector invariant of $G_A$.\n\\Needspace{5\\baselineskip}\\item[Finite description] A valid finite code in a specified language with declared semantics. It need not provide computable operations.\n\\Needspace{5\\baselineskip}\\item[Finite support] A vector has only finitely many nonzero coordinates. Such vectors form $D$ inside $P$.\n\\Needspace{5\\baselineskip}\\item[Fully invariant subgroup] A subgroup preserved by every endomorphism of its ambient group. Every subgroup of the squ","\\item[Eventual equality] Equality modulo finitely many coordinates, written $A\\eqfin B$. This is the unmarked selector invariant of $G_A$.\n\\Needspace{5\\baselineskip}\\item[Finite description] A valid finite code in a specified language with declared semantics. It need not provide computable operations.\n\\Needspace{5\\baselineskip}\\item[Finite support] A vector has only finitely many nonzero coordinates. Such vectors form $D$ inside $P$.\n\\Needspace{5\\baselineskip}\\item[Fully invariant subgroup] A subgroup preserved by every endomorphism of its ambient group. Every subgroup of the squarefree direct sum has this property.\n\\Needspace{5","\\item[Finite support] A vector has only finitely many nonzero coordinates. Such vectors form $D$ inside $P$.\n\\Needspace{5\\baselineskip}\\item[Fully invariant subgroup] A subgroup preserved by every endomorphism of its ambient group. Every subgroup of the squarefree direct sum has this property.\n\\Needspace{5\\baselineskip}\\item[Grammar-relative nonrealizability] Failure to have an admissible description in one fixed grammar. It is not absolute mathematical nonexistence or indescribability.\n\\Needspace{5\\baselineskip}\\item[Marked group] A group together with named elements that isomorphisms mu","\\item[Finite support] A vector has only finitely many nonzero coordinates. Such vectors form $D$ inside $P$.\n\\Needspace{5\\baselineskip}\\item[Fully invariant subgroup] A subgroup preserved by every endomorphism of its ambient group. Every subgroup of the squarefree direct sum has this property.\n\\Needspace{5\\baselineskip}\\item[Grammar-relative nonrealizability] Failure to have an admissible description in one fixed grammar. It is not absolute mathematical nonexistence or indescribability.\n\\Needspace{5\\baselineskip}\\item[Marked group] A group together with named elements that isomorphisms must preserve","\\item[Marked group] A group together with named elements that isomorphisms must preserve. The mark $u_A$ retains the literal selector.\n\\Needspace{5\\baselineskip}\\item[Normal form] The unique pair $(r,d)$ representing $s_A(r)+d$. Equality of these names is simpler than the selector-dependent addition law.\n\\Needspace{5\\baselineskip}\\item[Oracle] An external set-membership resource available to a computation. Its information is charged to the effective realization contract.\n\\Needspace{5\\baselineskip}\\item[Prime spectrum] The prime orders of nonzero torsion elements. It classifies the groups $D_B$ bu","\\item[Normal form] The unique pair $(r,d)$ representing $s_A(r)+d$. Equality of these names is simpler than the selector-dependent addition law.\n\\Needspace{5\\baselineskip}\\item[Oracle] An external set-membership resource available to a computation. Its information is charged to the effective realization contract.\n\\Needspace{5\\baselineskip}\\item[Prime spectrum] The prime orders of nonzero torsion elements. It classifies the groups $D_B$ but is constant across all $G_A$.\n\\Needspace{5\\baselineskip}\\item[Profinite completion] The inverse l","\\item[Normal form] The unique pair $(r,d)$ representing $s_A(r)+d$. Equality of these names is simpler than the selector-dependent addition law.\n\\Needspace{5\\baselineskip}\\item[Oracle] An external set-membership resource available to a computation. Its information is charged to the effective realization contract.\n\\Needspace{5\\baselineskip}\\item[Prime spectrum] The prime orders of nonzero torsion elements. It classifies the groups $D_B$ but is constant across all $G_A$.\n\\Needspace{5\\baselineskip}\\item[Profinite completion] The inverse limit of all finite quoti","\\item[Prime spectrum] The prime orders of nonzero torsion elements. It classifies the groups $D_B$ but is constant across all $G_A$.\n\\Needspace{5\\baselineskip}\\item[Profinite completion] The inverse limit of all finite quotient groups, denoted $\\widehat G$. It is not the Pontryagin dual.\n\\Needspace{5\\baselineskip}\\item[Procyclic group] A topological group with a dense cyclic subgroup. The product $P$ is procyclic.\n\\Needspace{5\\baselineskip}\\item[Pure subgroup] A subgroup $H\\le M$ for which $H\\cap mM=mH$ for all positive integers $m$.\n\\Needspace{5\\baselineskip}\\item[Rational line] A one-dime","\\item[Prime spectrum] The prime orders of nonzero torsion elements. It classifies the groups $D_B$ but is constant across all $G_A$.\n\\Needspace{5\\baselineskip}\\item[Profinite completion] The inverse limit of all finite quotient groups, denoted $\\widehat G$. It is not the Pontryagin dual.\n\\Needspace{5\\baselineskip}\\item[Procyclic group] A topological group with a dense cyclic subgroup. The product $P$ is procyclic.\n\\Needspace{5\\baselineskip}\\item[Pure subgroup] A subgroup $H\\le M$ for which $H\\cap mM=mH$ for all positive integers $m$.\n\\Needspace{5\\baselineskip}\\item[Rational line] A one-dimensional subspace over $\\Q$. The p","\\item[Profinite completion] The inverse limit of all finite quotient groups, denoted $\\widehat G$. It is not the Pontryagin dual.\n\\Needspace{5\\baselineskip}\\item[Procyclic group] A topological group with a dense cyclic subgroup. The product $P$ is procyclic.\n\\Needspace{5\\baselineskip}\\item[Pure subgroup] A subgroup $H\\le M$ for which $H\\cap mM=mH$ for all positive integers $m$.\n\\Needspace{5\\baselineskip}\\item[Rational line] A one-dimensional subspace over $\\Q$. The preimage of $\\Q q(u_A)$ is $G_A$.\n\\Needspace{5\\baselineskip}\\item[Reduced group] A group with no nonzero divisible subgroup. T","\\item[Procyclic group] A topological group with a dense cyclic subgroup. The product $P$ is procyclic.\n\\Needspace{5\\baselineskip}\\item[Pure subgroup] A subgroup $H\\le M$ for which $H\\cap mM=mH$ for all positive integers $m$.\n\\Needspace{5\\baselineskip}\\item[Rational line] A one-dimensional subspace over $\\Q$. The preimage of $\\Q q(u_A)$ is $G_A$.\n\\Needspace{5\\baselineskip}\\item[Reduced group] A group with no nonzero divisible subgroup. This property does not exclude divisible quotients.\n\\Needspace{5\\baselineskip}\\item[Residual finiteness] Every nonzero element is detected by some f","\\item[Pure subgroup] A subgroup $H\\le M$ for which $H\\cap mM=mH$ for all positive integers $m$.\n\\Needspace{5\\baselineskip}\\item[Rational line] A one-dimensional subspace over $\\Q$. The preimage of $\\Q q(u_A)$ is $G_A$.\n\\Needspace{5\\baselineskip}\\item[Reduced group] A group with no nonzero divisible subgroup. This property does not exclude divisible quotients.\n\\Needspace{5\\baselineskip}\\item[Residual finiteness] Every nonzero element is detected by some finite quotient. It does not imply source reconstruction from the unmarked quotien","\\item[Pure subgroup] A subgroup $H\\le M$ for which $H\\cap mM=mH$ for all positive integers $m$.\n\\Needspace{5\\baselineskip}\\item[Rational line] A one-dimensional subspace over $\\Q$. The preimage of $\\Q q(u_A)$ is $G_A$.\n\\Needspace{5\\baselineskip}\\item[Reduced group] A group with no nonzero divisible subgroup. This property does not exclude divisible quotients.\n\\Needspace{5\\baselineskip}\\item[Residual finiteness] Every nonzero element is detected by some finite quotient. It does not imply source reconstruction from the unmarked quotient system.\n\\Needspace{5\\b","\\item[Pure subgroup] A subgroup $H\\le M$ for which $H\\cap mM=mH$ for all positive integers $m$.\n\\Needspace{5\\baselineskip}\\item[Rational line] A one-dimensional subspace over $\\Q$. The preimage of $\\Q q(u_A)$ is $G_A$.\n\\Needspace{5\\baselineskip}\\item[Reduced group] A group with no nonzero divisible subgroup. This property does not exclude divisible quotients.\n\\Needspace{5\\baselineskip}\\item[Residual finiteness] Every nonzero element is detected by some finite quotient. It does not imply source reconstruction from the unmarked quotient system.\n\\Needspace{5\\baselineskip}\\item[Root certifi","\\item[Rational line] A one-dimensional subspace over $\\Q$. The preimage of $\\Q q(u_A)$ is $G_A$.\n\\Needspace{5\\baselineskip}\\item[Reduced group] A group with no nonzero divisible subgroup. This property does not exclude divisible quotients.\n\\Needspace{5\\baselineskip}\\item[Residual finiteness] Every nonzero element is detected by some finite quotient. It does not imply source reconstruction from the unmarked quotient system.\n\\Needspace{5\\baselineskip}\\item[Root certificate] Either an exact element $y$ with $my=g$, or a prime divisor of $m$ at which the","\\item[Rational line] A one-dimensional subspace over $\\Q$. The preimage of $\\Q q(u_A)$ is $G_A$.\n\\Needspace{5\\baselineskip}\\item[Reduced group] A group with no nonzero divisible subgroup. This property does not exclude divisible quotients.\n\\Needspace{5\\baselineskip}\\item[Residual finiteness] Every nonzero element is detected by some finite quotient. It does not imply source reconstruction from the unmarked quotient system.\n\\Needspace{5\\baselineskip}\\item[Root certificate] Either an exact element $y$ with $my=g$, or a prime divisor of $m$ at which the target has a nonzero coordin","\\item[Rational line] A one-dimensional subspace over $\\Q$. The preimage of $\\Q q(u_A)$ is $G_A$.\n\\Needspace{5\\baselineskip}\\item[Reduced group] A group with no nonzero divisible subgroup. This property does not exclude divisible quotients.\n\\Needspace{5\\baselineskip}\\item[Residual finiteness] Every nonzero element is detected by some finite quotient. It does not imply source reconstruction from the unmarked quotient system.\n\\Needspace{5\\baselineskip}\\item[Root certificate] Either an exact element $y$ with $my=g$, or a prime divisor of $m$ at which the target has a nonzero coordinate.\n\\Nee","\\item[Root certificate] Either an exact element $y$ with $my=g$, or a prime divisor of $m$ at which the target has a nonzero coordinate.\n\\Needspace{5\\baselineskip}\\item[Selector] A subset $A\\subseteq\\N$ whose bits choose one prime from each consecutive pair. It determines the admissible rational line.\n\\Needspace{5\\baselineskip}\\item[Squarefree radical] The integer $\\rad(m)$ formed by retaining each prime divisor of $m$ once.\n\\Needspace{5\\baselineskip}\\item[Tail signature] The pattern of prime-root solvability outside a finite exception set. Eve","\\item[Root certificate] Either an exact element $y$ with $my=g$, or a prime divisor of $m$ at which the target has a nonzero coordinate.\n\\Needspace{5\\baselineskip}\\item[Selector] A subset $A\\subseteq\\N$ whose bits choose one prime from each consecutive pair. It determines the admissible rational line.\n\\Needspace{5\\baselineskip}\\item[Squarefree radical] The integer $\\rad(m)$ formed by retaining each prime divisor of $m$ once.\n\\Needspace{5\\baselineskip}\\item[Tail signature] The pattern of prime-root solvability outside a finite exception set. Every infinite-","\\item[Root certificate] Either an exact element $y$ with $my=g$, or a prime divisor of $m$ at which the target has a nonzero coordinate.\n\\Needspace{5\\baselineskip}\\item[Selector] A subset $A\\subseteq\\N$ whose bits choose one prime from each consecutive pair. It determines the admissible rational line.\n\\Needspace{5\\baselineskip}\\item[Squarefree radical] The integer $\\rad(m)$ formed by retaining each prime divisor of $m$ once.\n\\Needspace{5\\baselineskip}\\item[Tail signature] The pattern of prime-root solvability outside a finite exception set. Every infinite-order element of $G_A$ has","\\item[Selector] A subset $A\\subseteq\\N$ whose bits choose one prime from each consecutive pair. It determines the admissible rational line.\n\\Needspace{5\\baselineskip}\\item[Squarefree radical] The integer $\\rad(m)$ formed by retaining each prime divisor of $m$ once.\n\\Needspace{5\\baselineskip}\\item[Tail signature] The pattern of prime-root solvability outside a finite exception set. Every infinite-order element of $G_A$ has the selector's tail signature.\n\\Needspace{5\\baselineskip}\\item[Torsion] Elements of finite order. In the product","\\item[Squarefree radical] The integer $\\rad(m)$ formed by retaining each prime divisor of $m$ once.\n\\Needspace{5\\baselineskip}\\item[Tail signature] The pattern of prime-root solvability outside a finite exception set. Every infinite-order element of $G_A$ has the selector's tail signature.\n\\Needspace{5\\baselineskip}\\item[Torsion] Elements of finite order. In the product and in every $G_A$, the torsion subgroup is exactly $D$.\n\\Needspace{5\\baselineskip}\\item[Turing degree] A measure of relative computability under mutual oracle computation.","\\item[Squarefree radical] The integer $\\rad(m)$ formed by retaining each prime divisor of $m$ once.\n\\Needspace{5\\baselineskip}\\item[Tail signature] The pattern of prime-root solvability outside a finite exception set. Every infinite-order element of $G_A$ has the selector's tail signature.\n\\Needspace{5\\baselineskip}\\item[Torsion] Elements of finite order. In the product and in every $G_A$, the torsion subgroup is exactly $D$.\n\\Needspace{5\\baselineskip}\\item[Turing degree] A measure of relative computability under mutual oracle computation. The least presentation degree of $G_A$ is the","\\item[Tail signature] The pattern of prime-root solvability outside a finite exception set. Every infinite-order element of $G_A$ has the selector's tail signature.\n\\Needspace{5\\baselineskip}\\item[Torsion] Elements of finite order. In the product and in every $G_A$, the torsion subgroup is exactly $D$.\n\\Needspace{5\\baselineskip}\\item[Turing degree] A measure of relative computability under mutual oracle computation. The least presentation degree of $G_A$ is the degree of $A$.\n\\Needspace{5\\baselineskip}\\item[Unmarked observations] Finite quotient objects and transition maps without a named coherent source element or the o","\\item[Torsion] Elements of finite order. In the product and in every $G_A$, the torsion subgroup is exactly $D$.\n\\Needspace{5\\baselineskip}\\item[Turing degree] A measure of relative computability under mutual oracle computation. The least presentation degree of $G_A$ is the degree of $A$.\n\\Needspace{5\\baselineskip}\\item[Unmarked observations] Finite quotient objects and transition maps without a named coherent source element or the original admissible image.\n\\end{description}","\\item[Turing degree] A measure of relative computability under mutual oracle computation. The least presentation degree of $G_A$ is the degree of $A$.\n\\Needspace{5\\baselineskip}\\item[Unmarked observations] Finite quotient objects and transition maps without a named coherent source element or the original admissible image.\n\\end{description}"]}