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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