Compact Support Sheaf Calculus

Finite shrinking. Let $Y$ be a normal space and let $(A_1,\dots ,A_m)$ be a finite open cover of $Y$. Then there are open sets $B_1,\dots ,B_m$ with $\overline{B_i}\subseteq A_i$ for every $i$ and $Y=B_1\cup \cdots \cup B_m$. Prove this by recursion on $i=1,\dots ,m$, maintaining the invariant $Y=\left({\bigcup }_{j<i}{B}_j\right)\cup \left({\bigcup }_{j\ge i}{A}_j\right)\mathrm{.}$ The invariant holds at $i=1$ because the $A_j$ cover $Y$. Given the invariant for some $i\le m$, the set $F_i:=Y\setminus ({\bigcup }_{j<i}{B}_j\cup {\bigcup }_{j>i}{A}_j)$ is closed, and it is contained in $A_i$: if $x\in F_i$ and $x\mathrm{\notin }{A}_i$, then the invariant puts $x$ in ${\bigcup }_{j<i}{B}_j\cup {\bigcup }_{j\ge i}{A}_j$; the first of these two sets is excluded by $x\in F_i$, while membership in the second together with $x\mathrm{\notin }{A}_i$ forces $x\in A_j$ for some $j>i$, also excluded by $x\in F_i$. Now $F_i$ and $Y\setminus A_i$ are disjoint closed sets, so by normality there are disjoint open sets $G\supseteq F_i$ and $H\supseteq Y\setminus A_i$; set $B_i:=G$. Then $\overline{B_i}\subseteq Y\setminus H\subseteq A_i$ by [F3], and the invariant for $i+1$ holds because ${\bigcup }_{j\le i}{B}_j\cup {\bigcup }_{j>i}{A}_j$ contains ${\bigcup }_{j<i}{B}_j\cup F_i\cup {\bigcup }_{j>i}{A}_j=Y$. At $i=m+1$ the invariant reads $Y=B_1\cup \cdots \cup B_m$.

The base case $p=0$. Suppose $p=0$, so that for every $i\in I$ a family $(W_{i,k}{)}_{k\in K_i}$ of open subsets of $U_i$ covering $U_i$ is given. The family of all $W_{i,k}$ with $i\in I$ and $k\in K_i$ is an open cover of $X$: a point $x$ lies in some $U_i$, hence in some $W_{i,k}$. By compactness [F1] there are finitely many pairs $(i_1,k_1),\dots ,(i_m,k_m)$ with $X=W_{i_1,k_1}\cup \cdots \cup W_{i_m,k_m}$. Take $J:=\{1,\dots ,m\}$, $V_t:=W_{i_t,k_t}$ and $\alpha (t):=i_t$. Then $V_t\subseteq U_{\alpha (t)}$ and the one-element tuple $(t)$ is contained in the member $W_{\alpha (t),k_t}$ of the family indexed by the tuple $(\alpha (t))$. Hence the statement holds for $p=0$.

Induction hypothesis and the auxiliary covers. Assume now $p\ge 1$ and assume the statement is proved for $p-1$ with the same space $X$: for every finite open cover of $X$ and every family of covers indexed by the $(p)$-tuples, a finite refinement $(V_j,\alpha )$ as in the statement exists. For each $p$-tuple ${\underset{‾}{i}}^{\mathrm{\prime }}=(i_0^{\mathrm{\prime }},\dots ,i_{p-1}^{\mathrm{\prime }})$ of elements of $I$ define a cover of the open set $U_{{\underset{‾}{i}}^{\mathrm{\prime }}}$ as follows. Let $S:=\{i_0^{\mathrm{\prime }},\dots ,i_{p-1}^{\mathrm{\prime }}\}$, the set of entries of ${\underset{‾}{i}}^{\mathrm{\prime }}$; it is a nonempty finite set of size at most $p$. Let $\mathrm{\Phi }$ be the set of $(p+1)$-tuples $\underset{‾}{i}\in I^{p+1}$ whose set of entries is exactly $S$; the set $\mathrm{\Phi }$ is finite and nonempty, and for every $\underset{‾}{i}\in \mathrm{\Phi }$ one has $U_{\underset{‾}{i}}={\bigcap }_{i\in S}{U}_i=U_{{\underset{‾}{i}}^{\mathrm{\prime }}}$. Let the new family consist of all intersections $W_{{\underset{‾}{i}}^{\mathrm{\prime }},(k_{\underset{‾}{i}}{)}_{\underset{‾}{i}\in \mathrm{\Phi }}}^{\mathrm{\prime }}:={\bigcap }_{\underset{‾}{i}\in \mathrm{\Phi }}{W}_{\underset{‾}{i},k_{\underset{‾}{i}}}(k_{\underset{‾}{i}}\in K_{\underset{‾}{i}}\text{ for each }\underset{‾}{i}\in \mathrm{\Phi }),$ indexed by the tuples of choices $(k_{\underset{‾}{i}}{)}_{\underset{‾}{i}\in \mathrm{\Phi }}$. Each of these intersections is an open subset of $U_{{\underset{‾}{i}}^{\mathrm{\prime }}}$, and the family covers $U_{{\underset{‾}{i}}^{\mathrm{\prime }}}$: a point $x\in U_{{\underset{‾}{i}}^{\mathrm{\prime }}}$ lies in $U_{\underset{‾}{i}}$ for every $\underset{‾}{i}\in \mathrm{\Phi }$, hence in some member $W_{\underset{‾}{i},k_{\underset{‾}{i}}}$ of each of the finitely many covers, and therefore in the corresponding intersection.

The intermediate refinement. Apply the induction hypothesis of step 1.3 to the cover $(U_i{)}_{i\in I}$ and these covers indexed by $p$-tuples. This gives a finite set $J$, a map $\alpha :J\to I$ and an open cover $(V_j{)}_{j\in J}$ of $X$ with $V_j\subseteq U_{\alpha (j)}$ such that for every $p$-tuple $(j_0,\dots ,j_{p-1})$ of elements of $J$ with nonempty intersection some member of the family $W_{\alpha (j_0)\dots \alpha (j_{p-1})}^{\mathrm{\prime }}$ contains $V_{j_0}\cap \cdots \cap V_{j_{p-1}}$. Apply the shrinking of step 1.1 to this finite cover (the space $X$ is normal by [F2]) and replace every $V_j$ by its shrinking; rename the result $(V_j{)}_{j\in J}$. All containments of intersections in members of the families $W^{\mathrm{\prime }}$ persist, because each new $V_j$ is a subset of the old one. The outcome is a finite open cover with $V_j\subseteq \overline{V_j}\subseteq U_{\alpha (j)}(j\in J)$ and with the $p$-tuple property just recalled.

Tuples with a repetition. Call a tuple compatible if its intersection is empty or is contained in a member of the prescribed cover. Suppose $(j_0,\dots ,j_p)$ has nonempty intersection and repeated $\alpha$-values. Drop a position whose value repeats; the remaining $p$-tuple has the same set $S$ of $\alpha$-values and nonempty intersection. Step 2.1 puts that intersection in a member of the meet cover of step 1.3. Such a member is contained in a $W_{\underset{‾}{i},k}$ for each value tuple $\underset{‾}{i}$ with entry set $S$, in particular $\underset{‾}{i}=(\alpha (j_0),\dots ,\alpha (j_p))$. The full intersection is smaller, hence compatible.

The fix-up operation. Fix a pairwise-distinct value tuple $\underset{‾}{i}=(i_0,\dots ,i_p)$ and an incompatible tuple $(j_0,\dots ,j_p)$ with $\alpha (j_t)=i_t$. Put $C:={\bigcap }_{t=1}^p\left({\bigcup }_{\alpha (j)=i_t}\overline{V_j}\right),\mathrm{\Sigma }:=\overline{V_{j_0}}\cap C\mathrm{.}$ The unions are finite, so $C$ and $\mathrm{\Sigma }$ are closed. Each union in the intersection is contained in $U_{i_t}$ by step 2.1; hence $\mathrm{\Sigma }\subseteq U_{\underset{‾}{i}}$ and $\mathrm{\Sigma }$ is compact [F4]. The given incompatible intersection is nonempty and contained in $\mathrm{\Sigma }$, so $\mathrm{\Sigma }\mathrm{\ne }\mathrm{\varnothing }$. Choose a finite nonempty set $E\subseteq K_{\underset{‾}{i}}$ with $\mathrm{\Sigma }\subseteq {\bigcup }_{k\in E}{W}_{\underset{‾}{i},k}$ by [F5]: the $W_{\underset{‾}{i},k}$ are open in $X$ because $U_{\underset{‾}{i}}$ is open in $X$. Replace $V_{j_0}$ by $H:=V_{j_0}\setminus C,B_k:=V_{j_0}\cap W_{\underset{‾}{i},k}(k\in E),$ using disjoint new index labels, each mapped to $i_0$. These opens cover $V_{j_0}$: a point in $C\cap V_{j_0}$ belongs to $\mathrm{\Sigma }$ and hence to some $B_k$, and a point outside $C$ belongs to $H$. They are subsets of $V_{j_0}$, so their closures remain in $U_{i_0}$. The other members are unchanged; the resulting family is a finite open refinement with the same closure property.

The counts. For a pairwise-distinct value tuple $\underset{‾}{i}$, let $N(\underset{‾}{i})$ be the number of incompatible tuples of indices whose $\alpha$-values are $\underset{‾}{i}$. This is finite. Let $M$ be the number of value tuples with $N(\underset{‾}{i})>0$; this is finite because $I$ is finite. Compatibility for repeated values was proved in step 3.1, so $M=0$ suffices for the conclusion.

The chosen count decreases. After step 3.2, a tuple with value tuple $\underset{‾}{i}$ using $B_k$ is contained in $W_{\underset{‾}{i},k}$. A tuple using $H$ has empty intersection: its remaining members, with values $i_1,\dots ,i_p$, have intersection contained in $C$. Tuples using no new index are unchanged. Every tuple using the deleted $j_0$ has disappeared, including the chosen incompatible tuple, and all its replacements are compatible. Thus $N(\underset{‾}{i})$ strictly decreases.

Zero counts remain zero. Map every new index to $j_0$ and keep the other indices fixed, obtaining $\beta :J^{\ast }\to J$. The new opens satisfy $V_{j^{\ast }}^{\ast }\subseteq V_{\beta (j^{\ast })}$ and ${\alpha }^{\ast }=\alpha \circ \beta$. Any tuple whose old $\beta$-image was compatible remains compatible, because its intersection is smaller; an empty old intersection stays empty. Therefore every zero count remains zero, and compatibility for repeated values persists. Positive counts for other value tuples may increase, since $\beta$ need not be injective on tuples; no bound on those counts is asserted.

Finite descent. If $M>0$, fix one value tuple $\underset{‾}{i}$ with $N(\underset{‾}{i})>0$ and repeatedly perform step 3.2 for this same value tuple. Step 5.1 strictly decreases its finite count at each operation, so after finitely many operations its count is zero. By step 5.2 all previously zero counts remain zero, so this lowers $M$. Repeating for at most the original $M$ value tuples produces $M=0$, while compatibility for repeated values persists. Each inner and outer descent is finite, so it requires only finite selections.

Conclusion. The finite open family produced by step 6.1 refines $(U_i)$ and is compatible for every tuple, by steps 3.1 and 4.1. This proves the induction step from $p-1$ to $p$; together with step 1.2 it proves the result for all $p\ge 0$. Empty intersections require no witness, and for $X=\mathrm{\varnothing }$ take $J=\mathrm{\varnothing }$.

Reduction to the presheaf value at $K$. The source of the canonical map of the statement is $P(K)$ and the map is the component ${\eta }_{P,K}$, by [F1]; the structure maps of the colimits and the sheafification map are additive (and $k$-linear in the vector-space case), so ${\eta }_{P,K}$ is additive. If $K=\mathrm{\varnothing }$, the empty open set is a terminal stage of the neighbourhood diagram, and both $P(K)$ and $(i^{-1}\mathcal{F})(K)$ are zero by the empty-cover sheaf axiom; hence the assertion holds. For surjectivity below assume $K\mathrm{\ne }\mathrm{\varnothing }$. It therefore suffices to prove that ${\eta }_{P,K}$ is injective and surjective.

Classes and germs. For later use, record the descriptions of [F2]: elements of $P(K)$ are classes $[V,a]$ with $V\supseteq K$ open and $a\in \mathcal{F}(V)$, equality being agreement after restriction to an open $W$ with $K\subseteq W\subseteq V\cap V^{\mathrm{\prime }}$; the stalk $P_x$ at $x\in K$ is the filtered colimit of the $P(U)$ and is identified with ${\mathcal{F}}_x$, and equality of germs in $P_x$ is agreement on a smaller open neighbourhood.

${\eta }_{P,K}$ is injective. Let $p,q\in P(K)$ with ${\eta }_{P,K}(p)={\eta }_{P,K}(q)$; write $p=[V,a]$ and $q=[V^{\mathrm{\prime }},a^{\mathrm{\prime }}]$ with $K\subseteq V,V^{\mathrm{\prime }}$ open and $a\in \mathcal{F}(V)$, $a^{\mathrm{\prime }}\in \mathcal{F}(V^{\mathrm{\prime }})$ by F2. Fix $z\in K$: the germ of ${\eta }_{P,K}(p)$ at $z$ is ${\eta }_{P,z}(p_z)$ and that of ${\eta }_{P,K}(q)$ is ${\eta }_{P,z}(q_z)$ by [F3], and these are equal, so $p_z=q_z$ in $P_z$ because ${\eta }_{P,z}$ is injective; by F2 this means that $a$ and $a^{\mathrm{\prime }}$ agree on an open neighbourhood of $z$. Let $\mathcal{O}$ be the family of all open subsets $O\subseteq V\cap V^{\mathrm{\prime }}$ with $a{\mathrm{\mid }}_O=a^{\mathrm{\prime }}{\mathrm{\mid }}_O$: it is cut out by a formula, and the previous sentence shows that it covers $K$, so by compactness of $K$ ([F5]) there are finitely many $O_1,\dots ,O_m\in \mathcal{O}$ with $K\subseteq O_1\cup \cdots \cup O_m$. Put $W:=O_1\cup \cdots \cup O_m$, an open set with $K\subseteq W\subseteq V\cap V^{\mathrm{\prime }}$; the restrictions $a{\mathrm{\mid }}_W$ and $a^{\mathrm{\prime }}{\mathrm{\mid }}_W$ agree by locality of the sheaf $\mathcal{F}$ ([F4]), since they agree on each $O_t$. Hence $p=[V,a]=[W,a{\mathrm{\mid }}_W]=[W,a^{\mathrm{\prime }}{\mathrm{\mid }}_W]=[V^{\mathrm{\prime }},a^{\mathrm{\prime }}]=q$ by F2, and ${\eta }_{P,K}$ is injective.

Local representatives of a section. Let $\sigma \in (i^{-1}\mathcal{F})(K)$. For $x\in K$ the map ${\eta }_{P,x}:P_x\to (i^{-1}\mathcal{F}{)}_x$ is onto by [F3], so there is $\xi \in P_x$ with ${\eta }_{P,x}(\xi )={\sigma }_x$; by the description of the stalk $P_x$ as a filtered colimit, F2, the element $\xi$ is the germ at $x$ of an element $b\in P(U)$ for some open $U\subseteq K$ containing $x$, and by F2 we can write $b=[V,a]$ with $V\supseteq U$ open in $X$ and $a\in \mathcal{F}(V)$. Then the section ${\zeta }_V(a){\mathrm{\mid }}_U={\eta }_{P,U}(b)$ of $i^{-1}\mathcal{F}$ over $U$ has at $x$ the same germ as $\sigma$ by [F3], and two sections of the sheaf $i^{-1}\mathcal{F}$ with the same germ at $x$ agree on some open neighbourhood $U^{\mathrm{\prime }}\subseteq U$ of $x$ by the description of equality in the stalk (F2 applied to $i^{-1}\mathcal{F}$, [F4]). Form the family $\mathcal{V}$ of all open subsets $U\subseteq K$ for which there are an open $V\subseteq X$ with $U\subseteq V$ and a section $a\in \mathcal{F}(V)$ satisfying ${\zeta }_V(a){\mathrm{\mid }}_U=\sigma {\mathrm{\mid }}_U$: the family is cut out by a formula, and the argument above shows that every point of $K$ lies in a member of $\mathcal{V}$.

Finitely many pieces with compact cores. By compactness of $K$ ([F5]) there are finitely many members $U_0,\dots ,U_m\in \mathcal{V}$ with $K=U_0\cup \cdots \cup U_m$, and for each $i$ we select witnesses: an open $V_i\subseteq X$ with $U_i\subseteq V_i$ and a section $a_i\in \mathcal{F}(V_i)$ with ${\zeta }_{V_i}(a_i){\mathrm{\mid }}_{U_i}=\sigma {\mathrm{\mid }}_{U_i}$; these are finitely many selections ([F5], [F3]). The compact Hausdorff space $K$ is normal ([F5]), so its finite open cover $(U_0,\dots ,U_m)$ can be shrunk: recursively for $i=0,\dots ,m$, maintaining $K=({\bigcup }_{j<i}{B}_j)\cup ({\bigcup }_{j\ge i}{U}_j)$, the set $F_i:=K\setminus ({\bigcup }_{j<i}{B}_j\cup {\bigcup }_{j>i}{U}_j)$ is closed in $K$ and contained in $U_i$, and normality supplies an open $B_i\subseteq K$ with $F_i\subseteq B_i\subseteq \overline{B_i}\subseteq U_i$, the closure taken in $K$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$; the closure is the smallest closed superset and equals $A$ together with its derived set); the invariant is maintained and at the end $K=B_0\cup \cdots \cup B_m$. Put $C_i:=\overline{B_i}$, a compact subset of $K$ with $C_i\subseteq U_i$ ([F5]); then $C_0,\dots ,C_m$ are compact subsets of $K$ with $K=C_0\cup \cdots \cup C_m$ and $C_i\subseteq U_i\subseteq V_i$ for every $i$.

Neighbourhoods of the pairwise intersections. For all indices $i,j$ let ${\mathcal{O}}_{ij}$ be the family of all open subsets $O\subseteq V_i\cap V_j$ with $a_i{\mathrm{\mid }}_O=a_j{\mathrm{\mid }}_O$. It covers $C_i\cap C_j$: for $z\in C_i\cap C_j\subseteq U_i\cap U_j$ the sections ${\zeta }_{V_i}(a_i)=\sigma {\mathrm{\mid }}_{U_i}$ and ${\zeta }_{V_j}(a_j)=\sigma {\mathrm{\mid }}_{U_j}$ have the same germ ${\sigma }_z$ at $z$, so $a_i$ and $a_j$ have the same germ at $z$ by [F3], and by F2 they agree on an open neighbourhood of $z$. Since $C_i\cap C_j$ is a closed subset of the compact set $C_i$, hence compact ([F5]), finitely many members of ${\mathcal{O}}_{ij}$ cover it; their union is an open set $W_{ij}$ with $C_i\cap C_j\subseteq W_{ij}\subseteq V_i\cap V_j$ and, by locality of $\mathcal{F}$ ([F4]), $a_i{\mathrm{\mid }}_{W_{ij}}=a_j{\mathrm{\mid }}_{W_{ij}}$.

Fusion of two pieces. Let $C_1,C_2\subseteq X$ be compact, let $V_1,V_2\subseteq X$ be open with $C_k\subseteq V_k$ and let $s_k\in \mathcal{F}(V_k)$ for $k=1,2$; suppose that $W\subseteq V_1\cap V_2$ is open with $C_1\cap C_2\subseteq W$ and $s_1{\mathrm{\mid }}_W=s_2{\mathrm{\mid }}_W$. Then there are an open $V^{\mathrm{\prime }}\subseteq X$ with $C_1\cup C_2\subseteq V^{\mathrm{\prime }}$ and a section $s^{\mathrm{\prime }}\in \mathcal{F}(V^{\mathrm{\prime }})$, together with open $V_k^{\mathrm{\prime }}\subseteq V_k$ with $C_k\subseteq V_k^{\mathrm{\prime }}$ and $V^{\mathrm{\prime }}=V_1^{\mathrm{\prime }}\cup V_2^{\mathrm{\prime }}$, such that $s^{\mathrm{\prime }}{\mathrm{\mid }}_{V_k^{\mathrm{\prime }}}=s_k{\mathrm{\mid }}_{V_k^{\mathrm{\prime }}}$ for $k=1,2$. Indeed, $A:=C_1\setminus W$ and $B:=C_2\setminus W$ are closed subsets of the compact sets $C_1$ and $C_2$, hence compact, and they are disjoint because $C_1\cap C_2\subseteq W$; as $X$ is Hausdorff there are disjoint open sets $G_1\supseteq A$ and $G_2\supseteq B$ ([F5]). Set $V_1^{\mathrm{\prime }}:=V_1\cap (G_1\cup W)$ and $V_2^{\mathrm{\prime }}:=V_2\cap (G_2\cup W)$: these are open, $C_1=(C_1\cap W)\cup A\subseteq V_1^{\mathrm{\prime }}$ and $C_2=(C_2\cap W)\cup B\subseteq V_2^{\mathrm{\prime }}$, and $V_1^{\mathrm{\prime }}\cap V_2^{\mathrm{\prime }}\subseteq (G_1\cup W)\cap (G_2\cup W)=W\cup (G_1\cap W)\cup (G_2\cap W)\cup (G_1\cap G_2)\subseteq W$, since $G_1\cap G_2=\mathrm{\varnothing }$. Hence $s_1{\mathrm{\mid }}_{V_1^{\mathrm{\prime }}}$ and $s_2{\mathrm{\mid }}_{V_2^{\mathrm{\prime }}}$ agree on $V_1^{\mathrm{\prime }}\cap V_2^{\mathrm{\prime }}$ and glue, by the sheaf axiom for $\mathcal{F}$ ([F4]), to a section $s^{\mathrm{\prime }}$ of $\mathcal{F}$ over $V^{\mathrm{\prime }}=V_1^{\mathrm{\prime }}\cup V_2^{\mathrm{\prime }}\supseteq C_1\cup C_2$ with $s^{\mathrm{\prime }}{\mathrm{\mid }}_{V_k^{\mathrm{\prime }}}=s_k{\mathrm{\mid }}_{V_k^{\mathrm{\prime }}}$ for $k=1,2$.

Iterated fusion. Fuse the pieces of step 1.5 one after another: for each $k\in \{0,\dots ,m\}$ there are an open $V^{(k)}\supseteq C_0\cup \cdots \cup C_k$ and a section $a^{(k)}\in \mathcal{F}(V^{(k)})$ such that ${\zeta }_{V^{(k)}}(a^{(k)})$ and $\sigma$ agree on an open neighbourhood of $D_k:=C_0\cup \cdots \cup C_k$ in $K$. For $k=0$ this is step 1.5. For the inductive step, $D_k$ is compact ([F5]) and on the compact set $D_k\cap C_{k+1}$ the sections ${\zeta }_{V^{(k)}}(a^{(k)})$ and ${\zeta }_{V_{k+1}}(a_{k+1})$ both agree with $\sigma$ on open neighbourhoods, so their germs agree everywhere on $D_k\cap C_{k+1}$ and [F3] gives that $a^{(k)}$ and $a_{k+1}$ have equal germs at every point of it; the compactness-and-locality argument of step 1.6 applied to these two sections produces an open $W\subseteq V^{(k)}\cap V_{k+1}$ with $D_k\cap C_{k+1}\subseteq W$ and $a^{(k)}{\mathrm{\mid }}_W=a_{k+1}{\mathrm{\mid }}_W$. Apply the fusion of step 1.7 with $(C_1,V_1,s_1):=(D_k,V^{(k)},a^{(k)})$ and $(C_2,V_2,s_2):=(C_{k+1},V_{k+1},a_{k+1})$: it gives open $V^{\mathrm{\prime }\mathrm{\prime }}=V_1^{\mathrm{\prime }}\cup V_2^{\mathrm{\prime }}$ and $a^{\mathrm{\prime }\mathrm{\prime }}\in \mathcal{F}(V^{\mathrm{\prime }\mathrm{\prime }})$ with $a^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\mid }}_{V_1^{\mathrm{\prime }}}=a^{(k)}{\mathrm{\mid }}_{V_1^{\mathrm{\prime }}}$, $a^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\mid }}_{V_2^{\mathrm{\prime }}}=a_{k+1}{\mathrm{\mid }}_{V_2^{\mathrm{\prime }}}$, where $D_k\subseteq V_1^{\mathrm{\prime }}$ and $C_{k+1}\subseteq V_2^{\mathrm{\prime }}$. Let $E_k\subseteq V^{(k)}\cap K$ be an open neighbourhood of $D_k$ on which ${\zeta }_{V^{(k)}}(a^{(k)})=\sigma$. By restriction compatibility [F3], ${\zeta }_{V^{\mathrm{\prime }\mathrm{\prime }}}(a^{\mathrm{\prime }\mathrm{\prime }})$ agrees with $\sigma$ on $(V_1^{\mathrm{\prime }}\cap E_k)\cup (V_2^{\mathrm{\prime }}\cap U_{k+1})$, an open neighbourhood of $D_{k+1}$ in $K$; locality [F4] gives agreement on this union. This proves the claim for $k+1$.