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1 \chap{The additive obstruction and the space of extensions}\label{sec:obstruction}
2 The strict definition gives a system of equations. In an additive setting, that system has an actual differential, and the failure to extend prescribed data has a quotient valued obstruction.
3 
4 \subsection{The compatibility differential}
5 Assume each fiber \(A_v\) is an abelian group and each restriction is additive. Set
6 \[
7  E=\prod_v A_v,\qquad R=\prod_{f:v\to w}A_v,
8  \qquad (\delta x)_f=f^*(x_w)-x_v.
9 \]
10 The group \(R\) retains an entry for every arrow. It is not silently identified with a cohomology group.
11 
12 \begin{proposition}[Differential identities]\label{KS08}
13 The map \(\delta:E\to R\) is additive, \(\Gamma(A)=\ker\delta\), and
14 \[
15  (\delta x)_{1_v}=0,\qquad
16  (\delta x)_{g\circ f}=f^*((\delta x)_g)+(\delta x)_f.
17 \]
18 \end{proposition}
19 \begin{proof}
20 Additivity follows coordinate by coordinate. Vanishing is exactly the defining coherence equation. For composition, add and subtract \(f^*x_w\):
21 \[
22  f^*g^*x_z-x_v
23  =f^*(g^*x_z-x_w)+(f^*x_w-x_v).
24 \]
25 The identity equation uses \((1_v)^*=\id\).
26 \end{proof}
27 \source{KS08}
28 This 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.
29 
30 \subsection{The admissible corrections}
31 Fix \(U\subseteq\Ob\V\). 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
32 \[
33  B_U=\delta(\ker p_U)\le R.
34 \]
35 For a partial section \(s\), choose any raw extension \(\widetilde s\), for example zero off \(U\), and define
36 \begin{equation}\label{eq:obstruction}
37  Q_U=R/B_U,\qquad
38  \mathfrak o_U(s)=[\delta\widetilde s]\in Q_U.
39 \end{equation}
40 If 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.
41 
42 \begin{theorem}[Extension criterion]\label{KS09}
43 For every coherent partial section,
44 \[
45  \mathfrak o_U(s)=0\quad\Longleftrightarrow\quad\Ext_U(s)\ne\varnothing.
46 \]
47 \end{theorem}
48 \begin{proof}
49 Let $\pi_U:R\to R/B_U$ be the quotient homomorphism. By the definition of a quotient by a subgroup,
50 \[
51  \pi_U(r)=0\quad\Longleftrightarrow\quad r\in B_U.
52 \]
53 Here $B_U$ 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.
54 
55 If such $h$ exists, set $x=\widetilde s-h$. Additivity gives the two equations separately:
56 \[
57  \delta x=\delta\widetilde s-\delta h=0,
58  \qquad p_Ux=p_U\widetilde s-p_Uh=s.
59 \]
60 The first equation puts $x$ in $\Gamma(A)$, and the second puts that coherent family in $\Ext_U(s)$.
61 
62 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.
63 
64 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
65 \[
66  \delta\widetilde s'-\delta\widetilde s=\delta k\in B_U.
67 \]
68 Thus 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.
69 \end{proof}
70 \source{KS09}
71 The 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.
72 
73 \plate{02_extension}{The raw family is repaired only along directions invisible on the prescribed views. The obstruction records whether its defect lies in the image of those admissible directions. Once an extension exists, a different kernel measures the remaining freedom.}{fig:extension}
74 
75 \subsection{Existence is not uniqueness}
76 Define the ambiguity group
77 \[
78  H_U=\ker\delta\cap\ker p_U.
79 \]
80 It consists of coherent changes that are invisible on the prescribed domain.
81 
82 \begin{theorem}[Classification of a nonempty extension fiber]\label{KS10}
83 If \(x_0\in\Ext_U(s)\), translation gives an equivalence
84 \[
85  H_U\simeq\Ext_U(s),\qquad h\longmapsto x_0+h.
86 \]
87 Consequently the extension is unique exactly when \(\mathfrak o_U(s)=0\) and \(H_U=0\).
88 \end{theorem}
89 \begin{proof}
90 Define $\Theta:H_U\to\Ext_U(s)$ by $\Theta(h)=x_0+h$. It lands in the extension set because
91 \[
92  \delta(x_0+h)=\delta x_0+\delta h=0,
93  \qquad p_U(x_0+h)=s+0=s.
94 \]
95 Define $\Psi:\Ext_U(s)\to H_U$ 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
96 \[
97  \Psi\Theta(h)=x_0+h-x_0=h,\qquad
98  \Theta\Psi(x)=x_0+(x-x_0)=x.
99 \]
100 This proves the equivalence rather than only a comparison of cardinalities.
101 
102 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.
103 
104 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.
105 \end{proof}
106 \source{KS10}
107 An 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.
108 
109 \begin{example}[Parity and an independent ambiguity]
110 For 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.
111 \end{example}
112 \source{KS11}
113 
114 \subsection{The exact sequence behind the criterion}
115 The additive structures make restriction a homomorphism
116 \(r_U:\Gamma(A)\to\Gamma_U(A)\). Using the zero extension, or lift independence, \(\mathfrak o_U\) is also a homomorphism.
117 
118 \begin{proposition}[Exactness at partial sections]\label{KS12}
119 The sequence
120 \[
121  \Gamma(A)\xrightarrow{r_U}\Gamma_U(A)
122  \xrightarrow{\mathfrak o_U}Q_U
123 \]
124 is 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.
125 \end{proposition}
126 \begin{proof}
127 First 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
128 \[
129  \mathfrak o_U(s+t)
130  =[\delta(\widetilde s+\widetilde t)]
131  =[\delta\widetilde s]+[\delta\widetilde t]
132  =\mathfrak o_U(s)+\mathfrak o_U(t).
133 \]
134 The zero lift gives $\mathfrak o_U(0)=0$. Restriction $r_U$ is a homomorphism because addition of global sections is coordinatewise.
135 
136 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.
137 
138 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 $Q_U$.
139 \end{proof}
140 \source{KS12}
141 No 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.
142 
143 \subsection{Extension in stages}
144 For \(U\subseteq W\subseteq\Ob\V\), an extension first to \(W\) and then globally must retain the intermediate choice. There is an explicit equivalence
145 \begin{equation}\label{eq:stages}
146  \Ext_U(s)\simeq
147  \coprod_{\substack{t\in\Gamma_W(A)\\t|_U=s}}\Ext_W(t).
148 \end{equation}
149 It 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.
150 \source{KS13}
151 
152 \chap{Protected correction as a general principle}\label{sec:protected}
153 The 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.
154 
155 \begin{theorem}[Semantics preserving correction]\label{KS14}
156 For \(x\in E\), the following conditions are equivalent:
157 \[
158 \begin{aligned}
159  &[dx]=0\text{ in }R/d(\ker p),\\
160  &\exists h\in E:\ p(h)=0,\quad d(h)=d(x),\\
161  &\exists y\in E:\ d(y)=0,\quad p(y)=p(x).
162 \end{aligned}
163 \]
164 The class of \(dx\) is unchanged when an admissible correction is subtracted from \(x\).
165 \end{theorem}
166 \begin{proof}
167 The image $d(\ker p)$ is a subgroup of $R$. The defining property of the quotient says
168 \[
169  [dx]=0\quad\Longleftrightarrow\quad
170  dx\in d(\ker p)
171  \quad\Longleftrightarrow\quad
172  \exists h\in E:\ ph=0,\ dh=dx.
173 \]
174 This is the first equivalence with its witness made explicit.
175 
176 For the second, a permitted $h$ gives $y=x-h$. Then
177 \[
178  dy=dx-dh=0,\qquad py=px-ph=px.
179 \]
180 Conversely 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.
181 
182 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.
183 \end{proof}
184 \source{KS14}
185 Here \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.
186 
187 \begin{proposition}[A certificate of impossibility]\label{KS15}
188 Let \(L:R\to D\) be additive. If \(L\) vanishes on \(d(\ker p)\) but \(L(dx)\ne0\), no exact \(y\) can preserve \(p(x)\).
189 \end{proposition}
190 \begin{proof}
191 Suppose, for contradiction, that $dy=0$ and $py=px$. Then $h=x-y$ satisfies $ph=0$ and $dh=dx$. By the annihilation hypothesis,
192 \[
193  L(dx)=L(dh)=0,
194 \]
195 contrary to the assumed nonzero value. This already proves impossibility directly.
196 
197 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.
198 \end{proof}
199 \source{KS15}
200 This 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.
201 
202 \subsection{Why unrestricted cancellation is empty information}
203 If 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.
204 
205 \begin{example}[A rigid obstruction]
206 Take \(E=R=S=\Z\), \(d=p=\id\), 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.
207 \end{example}
208 
209 \subsection{What an obstruction does not yet provide}
210 An 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.
211 
212 \begin{insight}
213 A 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.
214 \end{insight}

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