The Stacks project

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. $\square$