Lemma 58.3.3. Let $G$ be a topological group. The automorphism group of the functor (58.3.2.1) endowed with its profinite topology from Lemma 58.3.1 is the profinite completion of $G$.

Proof. Denote $F_ G$ the functor (58.3.2.1). Any morphism $X \to Y$ in $\textit{Finite-}G\textit{-Sets}$ commutes with the action of $G$. Thus any $g \in G$ defines an automorphism of $F_ G$ and we obtain a canonical homomorphism $G \to \text{Aut}(F_ G)$ of groups. Observe that any finite $G$-set $X$ is a finite disjoint union of $G$-sets of the form $G/H_ i$ with canonical $G$-action where $H_ i \subset G$ is an open subgroup of finite index. Then $U_ i = \bigcap gH_ ig^{-1}$ is open, normal, and has finite index. Moreover $U_ i$ acts trivially on $G/H_ i$ hence $U = \bigcap U_ i$ acts trivially on $F_ G(X)$. Hence the action $G \times F_ G(X) \to F_ G(X)$ is continuous. By the universal property of the topology on $\text{Aut}(F_ G)$ the map $G \to \text{Aut}(F_ G)$ is continuous. By Lemma 58.3.1 and the universal property of profinite completion there is an induced continuous group homomorphism

$G^\wedge \longrightarrow \text{Aut}(F_ G)$

Moreover, since $G/U$ acts faithfully on $G/U$ this map is injective. If the image is dense, then the map is surjective and hence a homeomorphism by Topology, Lemma 5.17.8.

Let $\gamma \in \text{Aut}(F_ G)$ and let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. We will show there is a $g \in G$ such that $\gamma$ and $g$ induce the same action on $F_ G(X)$. This will finish the proof. As before we see that $X$ is a finite disjoint union of $G/H_ i$. With $U_ i$ and $U$ as above, the finite $G$-set $Y = G/U$ surjects onto $G/H_ i$ for all $i$ and hence it suffices to find $g \in G$ such that $\gamma$ and $g$ induce the same action on $F_ G(G/U) = G/U$. Let $e \in G$ be the neutral element and say that $\gamma (eU) = g_0U$ for some $g_0 \in G$. For any $g_1 \in G$ the morphism

$R_{g_1} : G/U \longrightarrow G/U,\quad gU \longmapsto gg_1U$

of $\textit{Finite-}G\textit{-Sets}$ commutes with the action of $\gamma$. Hence

$\gamma (g_1U) = \gamma (R_{g_1}(eU)) = R_{g_1}(\gamma (eU)) = R_{g_1}(g_0U) = g_0g_1U$

Thus we see that $g = g_0$ works. $\square$

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