[Expose V bis, 4.1.3, SGA4]

Lemma 20.16.3. Let $X$ be a topological space. Let $Z \subset X$ be a quasi-compact subset such that any two points of $Z$ have disjoint open neighbourhoods in $X$. For every abelian sheaf $\mathcal{F}$ on $X$ the canonical map

$\mathop{\mathrm{colim}}\nolimits H^ p(U, \mathcal{F}) \longrightarrow H^ p(Z, \mathcal{F}|_ Z)$

where the colimit is over open neighbourhoods $U$ of $Z$ in $X$ is an isomorphism.

Proof. We first prove this for $p = 0$. Injectivity follows from the definition of $\mathcal{F}|_ Z$ and holds in general (for any subset of any topological space $X$). Next, suppose that $s \in H^0(Z, \mathcal{F}|_ Z)$. Then we can find opens $U_ i \subset X$ such that $Z \subset \bigcup U_ i$ and such that $s|_{Z \cap U_ i}$ comes from $s_ i \in \mathcal{F}(U_ i)$. It follows that there exist opens $W_{ij} \subset U_ i \cap U_ j$ with $W_{ij} \cap Z = U_ i \cap U_ j \cap Z$ such that $s_ i|_{W_{ij}} = s_ j|_{W_{ij}}$. Applying Topology, Lemma 5.13.7 we find opens $V_ i$ of $X$ such that $V_ i \subset U_ i$ and such that $V_ i \cap V_ j \subset W_{ij}$. Hence we see that $s_ i|_{V_ i}$ glue to a section of $\mathcal{F}$ over the open neighbourhood $\bigcup V_ i$ of $Z$.

To finish the proof, it suffices to show that if $\mathcal{I}$ is an injective abelian sheaf on $X$, then $H^ p(Z, \mathcal{I}|_ Z) = 0$ for $p > 0$. This follows using short exact sequences and dimension shifting; details omitted. Thus, suppose $\overline{\xi }$ is an element of $H^ p(Z, \mathcal{I}|_ Z)$ for some $p > 0$. By Lemma 20.16.2 the element $\overline{\xi }$ comes from $\check{H}^ p(\mathcal{V}, \mathcal{I}|_ Z)$ for some open covering $\mathcal{V} : Z = \bigcup V_ i$ of $Z$. Say $\overline{\xi }$ is the image of the class of a cocycle $\xi = (\xi _{i_0 \ldots i_ p})$ in $\check{\mathcal{C}}^ p(\mathcal{V}, \mathcal{I}|_ Z)$.

Let $\mathcal{I}' \subset \mathcal{I}|_ Z$ be the subpresheaf defined by the rule

$\mathcal{I}'(V) = \{ s \in \mathcal{I}|_ Z(V) \mid \exists (U, t),\ U \subset X\text{ open}, \ t \in \mathcal{I}(U),\ V = Z \cap U,\ s = t|_{Z \cap U} \}$

Then $\mathcal{I}|_ Z$ is the sheafification of $\mathcal{I}'$. Thus for every $(p + 1)$-tuple $i_0 \ldots i_ p$ we can find an open covering $V_{i_0 \ldots i_ p} = \bigcup W_{i_0 \ldots i_ p, k}$ such that $\xi _{i_0 \ldots i_ p}|_{W_{i_0 \ldots i_ p, k}}$ is a section of $\mathcal{I}'$. Applying Topology, Lemma 5.13.5 we may after refining $\mathcal{V}$ assume that each $\xi _{i_0 \ldots i_ p}$ is a section of the presheaf $\mathcal{I}'$.

Write $V_ i = Z \cap U_ i$ for some opens $U_ i \subset X$. Since $\mathcal{I}$ is flasque (Lemma 20.12.2) and since $\xi _{i_0 \ldots i_ p}$ is a section of $\mathcal{I}'$ for every $(p + 1)$-tuple $i_0 \ldots i_ p$ we can choose a section $s_{i_0 \ldots i_ p} \in \mathcal{I}(U_{i_0 \ldots i_ p})$ which restricts to $\xi _{i_0 \ldots i_ p}$ on $V_{i_0 \ldots i_ p} = Z \cap U_{i_0 \ldots i_ p}$. (This appeal to injectives being flasque can be avoided by an additional application of Topology, Lemma 5.13.7.) Let $s = (s_{i_0 \ldots i_ p})$ be the corresponding cochain for the open covering $U = \bigcup U_ i$. Since $\text{d}(\xi ) = 0$ we see that the sections $\text{d}(s)_{i_0 \ldots i_{p + 1}}$ restrict to zero on $Z \cap U_{i_0 \ldots i_{p + 1}}$. Hence, by the initial remarks of the proof, there exists open subsets $W_{i_0 \ldots i_{p + 1}} \subset U_{i_0 \ldots i_{p + 1}}$ with $Z \cap W_{i_0 \ldots i_{p + 1}} = Z \cap U_{i_0 \ldots i_{p + 1}}$ such that $\text{d}(s)_{i_0 \ldots i_{p + 1}}|_{W_{i_0 \ldots i_{p + 1}}} = 0$. By Topology, Lemma 5.13.7 we can find $U'_ i \subset U_ i$ such that $Z \subset \bigcup U'_ i$ and such that $U'_{i_0 \ldots i_{p + 1}} \subset W_{i_0 \ldots i_{p + 1}}$. Then $s' = (s'_{i_0 \ldots i_ p})$ with $s'_{i_0 \ldots i_ p} = s_{i_0 \ldots i_ p}|_{U'_{i_0 \ldots i_ p}}$ is a cocycle for $\mathcal{I}$ for the open covering $U' = \bigcup U'_ i$ of an open neighbourhood of $Z$. Since $\mathcal{I}$ has trivial higher Čech cohomology groups (Lemma 20.11.1) we conclude that $s'$ is a coboundary. It follows that the image of $\xi$ in the Čech complex for the open covering $Z = \bigcup Z \cap U'_ i$ is a coboundary and we are done. $\square$

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