Lemma 59.75.5. Let $S$ be a scheme such that every quasi-compact open of $S$ has finite number of irreducible components (for example if $S$ has a Noetherian underlying topological space, or if $S$ is locally Noetherian). Let $\mathcal{F}$ be a sheaf of sets on $S_{\acute{e}tale}$. The following are equivalent

$\mathcal{F}$ is finite locally constant, and

all stalks of $\mathcal{F}$ are finite sets and all specialization maps $sp : \mathcal{F}_{\overline{s}} \to \mathcal{F}_{\overline{t}}$ are bijective.

**Proof.**
Assume (2). Let $\overline{s}$ be a geometric point of $S$ lying over $s \in S$. In order to prove (1) we have to find an étale neighbourhood $(U, \overline{u})$ of $(S, \overline{s})$ such that $\mathcal{F}|_ U$ is constant. We may and do assume $S$ is affine.

Since $\mathcal{F}_{\overline{s}}$ is finite, we can choose $(U, \overline{u})$, $n \geq 0$, and pairwise distinct elements $\sigma _1, \ldots , \sigma _ n \in \mathcal{F}(U)$ such that $\{ \sigma _1, \ldots , \sigma _ n\} \subset \mathcal{F}(U)$ maps bijectively to $\mathcal{F}_{\overline{s}}$ via the map $\mathcal{F}(U) \to \mathcal{F}_{\overline{s}}$. Consider the map

\[ \varphi : \underline{\{ 1, \ldots , n\} } \longrightarrow \mathcal{F}|_ U \]

on $U_{\acute{e}tale}$ defined by $\sigma _1, \ldots , \sigma _ n$. This map is a bijection on stalks at $\overline{u}$ by construction. Let us consider the subset

\[ E = \{ u' \in U \mid \varphi _{\overline{u}'}\text{ is bijective}\} \subset U \]

Here $\overline{u}'$ is any geometric point of $U$ lying over $u'$ (the condition is independent of the choice by Remark 59.29.8). The image $u \in U$ of $\overline{u}$ is in $E$. By our assumption on the specialization maps for $\mathcal{F}$, by Remark 59.75.3, and by Lemma 59.75.4 we see that $E$ is closed under specializations and generalizations in the topological space $U$.

After shrinking $U$ we may assume $U$ is affine too. By Descent, Lemma 35.16.3 we see that $U$ has a finite number of irreducible components. After removing the irreducible components which do not pass through $u$, we may assume every irreducible component of $U$ passes through $u$. Since $U$ is a sober topological space it follows that $E = U$ and we conclude that $\varphi $ is an isomorphism by Theorem 59.29.10. Thus (1) follows.

We omit the proof that (1) implies (2).
$\square$

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