Lemma 66.20.2 (04K9)—The Stacks project

Lemma 66.20.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $\mathcal{F}$ be an abelian sheaf on $X_{spaces, {\acute{e}tale}}$. Let $\sigma \in \mathcal{F}(U)$ be a local section. There exists an open subspace $W \subset U$ such that

$W \subset U$ is the largest open subspace of $U$ such that $\sigma |_ W = 0$,
for every $\varphi : V \to U$ in $X_{spaces, {\acute{e}tale}}$ we have

\[ \sigma |_ V = 0 \Leftrightarrow \varphi (V) \subset W, \]
for every geometric point $\overline{u}$ of $U$ we have

\[ (U, \overline{u}, \sigma ) = 0\text{ in }\mathcal{F}_{\overline{x}} \Leftrightarrow \overline{u} \in W \]

where $\overline{x} = (U \to X) \circ \overline{u}$.

Proof. Since $\mathcal{F}$ is a sheaf in the étale topology the restriction of $\mathcal{F}$ to $U_{Zar}$ is a sheaf on $U$ in the Zariski topology. Hence there exists a Zariski open $W$ having property (1), see Modules, Lemma 17.5.2. Let $\varphi : V \to U$ be an arrow of $X_{spaces, {\acute{e}tale}}$. Note that $\varphi (V) \subset U$ is an open subspace (Lemma 66.16.7) and that $\{ V \to \varphi (V)\} $ is an étale covering. Hence if $\sigma |_ V = 0$, then by the sheaf condition for $\mathcal{F}$ we see that $\sigma |_{\varphi (V)} = 0$. This proves (2). To prove (3) we have to show that if $(U, \overline{u}, \sigma )$ defines the zero element of $\mathcal{F}_{\overline{x}}$, then $\overline{u} \in W$. This is true because the assumption means there exists a morphism of étale neighbourhoods $(V, \overline{v}) \to (U, \overline{u})$ such that $\sigma |_ V = 0$. Hence by (2) we see that $V \to U$ maps into $W$, and hence $\overline{u} \in W$. $\square$

Comments (2)

Comment #8434 by ZL on May 28, 2023 at 13:16

I'm a little confused about the statement of the lemma and its proof. In the statement, we see $U$ is an algebraic space étale over $X$ . But in the proof, after citing lemma 17.5.2, we see that $U$ , having an underlying topological space, is a scheme.

I guess that previous lemma 66.20.1 is used to prove the space case. Since the $\mathcal{F}\vert_U \in U_{\mathrm{space,étale}}$ , we may assume $U=X$ . Then consider the sheaf $\mathcal{G}$ assigning $V\rightarrow X$ to a singleton if $\sigma\vert_V=0$ and assigning $V$ to $\emptyset$ otherwise. Then using 66.20.1 we get the open subspace $W$ .

There is also a typo for the conclusion $(3)$ : where $\bar{s}=(U\rightarrow X)\circ \bar{u}$ .

Comment #9058 by Stacks project on May 06, 2024 at 11:39

OK, the proof was correct as written but it was confusing because of the mix up of the different sites used in the statement and proof. The key is just Lemma 66.16.7 and the fact that in the topology on $X_{spaces, etale}$ the surjective etale morphisms are coverings. See changes here.

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