Lemma 58.5.5. Let $X$ be a connected scheme. Let $\overline{x}$ be a geometric point. The functor

$F_{\overline{x}} : \textit{FÉt}_ X \longrightarrow \textit{Sets},\quad Y \longmapsto |Y_{\overline{x}}|$

defines a Galois category (Definition 58.3.6).

Proof. After identifying $\textit{FÉt}_{\overline{x}}$ with the category of finite sets (Example 58.5.1) we see that our functor $F_{\overline{x}}$ is nothing but the base change functor for the morphism $\overline{x} \to X$. Thus we see that $\textit{FÉt}_ X$ has finite limits and finite colimits and that $F_{\overline{x}}$ is exact by Lemma 58.5.2. We will also use that finite limits in $\textit{FÉt}_ X$ agree with the corresponding finite limits in the category of schemes over $X$, see Remark 58.5.3.

If $Y' \to Y$ is a monomorphism in $\textit{FÉt}_ X$ then we see that $Y' \to Y' \times _ Y Y'$ is an isomorphism, and hence $Y' \to Y$ is a monomorphism of schemes. It follows that $Y' \to Y$ is an open immersion (Étale Morphisms, Theorem 41.14.1). Since $Y'$ is finite over $X$ and $Y$ separated over $X$, the morphism $Y' \to Y$ is finite (Morphisms, Lemma 29.44.14), hence closed (Morphisms, Lemma 29.44.11), hence it is the inclusion of an open and closed subscheme of $Y$. It follows that $Y$ is a connected objects of the category $\textit{FÉt}_ X$ (as in Definition 58.3.6) if and only if $Y$ is connected as a scheme. Then it follows from Topology, Lemma 5.7.7 that $Y$ is a finite coproduct of its connected components both as a scheme and in the sense of Definition 58.3.6.

Let $Y \to Z$ be a morphism in $\textit{FÉt}_ X$ which induces a bijection $F_{\overline{x}}(Y) \to F_{\overline{x}}(Z)$. We have to show that $Y \to Z$ is an isomorphism. By the above we may assume $Z$ is connected. Since $Y \to Z$ is finite étale and hence finite locally free it suffices to show that $Y \to Z$ is finite locally free of degree $1$. This is true in a neighbourhood of any point of $Z$ lying over $\overline{x}$ and since $Z$ is connected and the degree is locally constant we conclude. $\square$

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