Lemma 63.2.1. Let $X$ be a scheme. Let $\mathcal{F}$ be an abelian sheaf on $X_{\acute{e}tale}$. Let $\varphi : U' \to U$ be a morphism of $X_{\acute{e}tale}$. Let $Z' \subset U'$ be a closed subscheme such that $Z' \to U' \to U$ is a closed immersion with image $Z \subset U$. Then there is a canonical bijection

\[ \{ s \in \mathcal{F}(U) \mid \text{Supp}(s) \subset Z\} = \{ s' \in \mathcal{F}(U') \mid \text{Supp}(s') \subset Z'\} \]

which is given by restriction if $\varphi ^{-1}(Z) = Z'$.

**Proof.**
Consider the closed subscheme $Z'' = \varphi ^{-1}(Z)$ of $U'$. Then $Z' \subset Z''$ is closed because $Z'$ is closed in $U'$. On the other hand, $Z' \to Z''$ is an étale morphism (as a morphism between schemes étale over $Z$) and hence open. Thus $Z'' = Z' \amalg T$ for some closed subset $T$. The open covering $U' = (U' \setminus T) \cup (U' \setminus Z')$ shows that

\[ \{ s' \in \mathcal{F}(U') \mid \text{Supp}(s') \subset Z'\} = \{ s' \in \mathcal{F}(U' \setminus T) \mid \text{Supp}(s') \subset Z'\} \]

and the étale covering $\{ U' \setminus T \to U, U \setminus Z \to U\} $ shows that

\[ \{ s \in \mathcal{F}(U) \mid \text{Supp}(s) \subset Z\} = \{ s' \in \mathcal{F}(U' \setminus T) \mid \text{Supp}(s') \subset Z'\} \]

This finishes the proof.
$\square$

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