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

  1. $W \subset U$ is the largest open subspace of $U$ such that $\sigma |_ W = 0$,

  2. for every $\varphi : V \to U$ in $X_{\acute{e}tale}$ we have

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

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

    where $\overline{s} = (U \to S) \circ \overline{u}$.

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 $X_{\acute{e}tale}$. 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 $\mathcal{F}_{\overline{s}}$, 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 (1)

Comment #8434 by ZL on

I'm a little confused about the statement of the lemma and its proof. In the statement, we see is an algebraic space étale over . But in the proof, after citing lemma 17.5.2, we see that , 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 , we may assume . Then consider the sheaf assigning to a singleton if and assigning to otherwise. Then using 66.20.1 we get the open subspace .

There is also a typo for the conclusion : where .


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