Lemma 58.3.4. Let $G$ be a topological group. Let $F : \textit{Finite-}G\textit{-Sets} \to \textit{Sets}$ be an exact functor with $F(X)$ finite for all $X$. Then $F$ is isomorphic to the functor (58.3.2.1).

**Proof.**
Let $X$ be a nonempty object of $\textit{Finite-}G\textit{-Sets}$. The diagram

is cocartesian. Hence we conclude that $F(X)$ is nonempty. Let $U \subset G$ be an open, normal subgroup with finite index. Observe that

where the summand corresponding to $gU$ corresponds to the orbit of $(eU, gU)$ on the left hand side. Then we see that

Hence $|F(G/U)| = |G/U|$ as $F(G/U)$ is nonempty. Thus we see that

is nonempty (Categories, Lemma 4.21.7). Pick $\gamma = (\gamma _ U)$ an element in this limit. Denote $F_ G$ the functor (58.3.2.1). We can identify $F_ G$ with the functor

where $f : G/U \to X$ corresponds to $f(eU) \in X = F_ G(X)$ (details omitted). Hence the element $\gamma $ determines a well defined map

Namely, given $x \in X$ choose $U$ and $f : G/U \to X$ sending $eU$ to $x$ and then set $t_ X(x) = F(f)(\gamma _ U)$. We will show that $t$ induces a bijective map $t_{G/U} : F_ G(G/U) \to F(G/U)$ for any $U$. This implies in a straightforward manner that $t$ is an isomorphism (details omitted). Since $|F_ G(G/U)| = |F(G/U)|$ it suffices to show that $t_{G/U}$ is surjective. The image contains at least one element, namely $t_{G/U}(eU) = F(\text{id}_{G/U})(\gamma _ U) = \gamma _ U$. For $g \in G$ denote $R_ g : G/U \to G/U$ right multiplication. Then set of fixed points of $F(R_ g) : F(G/U) \to F(G/U)$ is equal to $F(\emptyset ) = \emptyset $ if $g \not\in U$ because $F$ commutes with equalizers. It follows that if $g_1, \ldots , g_{|G/U|}$ is a system of representatives for $G/U$, then the elements $F(R_{g_ i})(\gamma _ U)$ are pairwise distinct and hence fill out $F(G/U)$. Then

and the proof is complete. $\square$

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