Proposition 58.3.10. Let $(\mathcal{C}, F)$ be a Galois category. Let $G = \text{Aut}(F)$ be as in Example 58.3.5. The functor $F : \mathcal{C} \to \textit{Finite-}G\textit{-Sets}$ (58.3.5.1) an equivalence.

This is a weak version of [ExposÃ© V, SGA1]. The proof is borrowed from [Theorem 7.2.5, BS].

**Proof.**
We will use the results of Lemma 58.3.7 without further mention. In particular we know the functor is faithful. By Lemma 58.3.9 we know that for any connected $X$ the action of $G$ on $F(X)$ is transitive. Hence $F$ preserves the decomposition into connected components (existence of which is an axiom of a Galois category). Let $X$ and $Y$ be objects and let $s : F(X) \to F(Y)$ be a map. Then the graph $\Gamma _ s \subset F(X) \times F(Y)$ of $s$ is a union of connected components. Hence there exists a union of connected components $Z$ of $X \times Y$, which comes equipped with a monomorphism $Z \to X \times Y$, with $F(Z) = \Gamma _ s$. Since $F(Z) \to F(X)$ is bijective we see that $Z \to X$ is an isomorphism and we conclude that $s = F(f)$ where $f : X \cong Z \to Y$ is the composition. Hence $F$ is fully faithful.

To finish the proof we show that $F$ is essentially surjective. It suffices to show that $G/H$ is in the essential image for any open subgroup $H \subset G$ of finite index. By definition of the topology on $G$ there exists a finite collection of objects $X_ i$ such that

is contained in $H$. We may assume $X_ i$ is connected for all $i$. We can choose a Galois object $Y$ mapping to a connected component of $\prod X_ i$ using Lemma 58.3.8. Choose an isomorphism $F(Y) = G/U$ in $G\textit{-sets}$ for some open subgroup $U \subset G$. As $Y$ is Galois, the group $\text{Aut}(Y) = \text{Aut}_{G\textit{-Sets}}(G/U)$ acts transitively on $F(Y) = G/U$. This implies that $U$ is normal. Since $F(Y)$ surjects onto $F(X_ i)$ for each $i$ we see that $U \subset H$. Let $M \subset \text{Aut}(Y)$ be the finite subgroup corresponding to

Set $X = Y/M$, i.e., $X$ is the coequalizer of the arrows $m : Y \to Y$, $m \in M$. Since $F$ is exact we see that $F(X) = G/H$ and the proof is complete. $\square$

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