More precisely, let $$G$$ be a finite group, $$J$$ be a cofiltered category and $$(X_j)_{j \in J}$$ a $$J$$-indexed system of $$G$$-sets, i.e., a functor $$X : J \to \text{Set}$$. Then the canonical map $$(\lim_{j\in J} X_j)/G \to \lim_{j \in J}(X_j/G)$$ is a bijection. I learned this fact from the appendix to André Joyal’s lovely Foncteurs analytiques et espèces de structures, and since I couldn’t find a written proof anywhere, I decided to put one here.

## Warm up: $$J$$ is just a sequence.

Let’s do the case when $$J$$ is just the natural numbers ordered by $$\ge$$. We’ll think of $$\lim X_j$$ concretely as a subset of $$\prod X_j$$.

The canonical map is injective. Let $$(x_j)_j$$ and $$(y_j)_j$$ be such that for each $$j$$, $$y_j = g_j x_j$$ for some $$g_j \in G$$. We need to prove a single $$g \in G$$ works for all $$j$$. But whenever $$g$$ works1 for a particular $$j$$ it also works for all $$i \le j$$. Since $$G$$ is finite, at least one $$g$$ occurs infinitely often among the $$g_j$$, and this $$g$$ will therefore satisfy $$y_j =g x_j$$ for all $$j$$.

The canonical map is surjective. Let $$([x_j])_j$$ be an arbitrary element of $$\lim(X_j/G)$$ –where $$[x_j]$$ denotes the orbit of $$x_j \in X_j$$. Then there exist $$g_j$$ such that $$X(j \to j-1) g_j x_j = x_{j-1}$$, and we can set $$y_j = g_1 g_2 \cdots g_j x_j$$ to get a sequence in $$\lim X_j$$, i.e., satisfying $$X(j \to j-1) y_j = y_{j-1}$$.

## General cofiltered $$J$$

Injectivity. Again let $$(x_j)_j$$ and $$(y_j)_j$$ be such that $$[x_j] = [y_j]$$ for all $$j$$. Define $$A_j = \{ g \in G : g x_j = y_j \}$$. We need to prove that $$\bigcap A_j$$ is non-empty assuming each $$A_j$$ is non-empty. Note that if there is a morphism $$j \to k$$ in $$J$$, $$A_j \subset A_k$$. If for each of the finitely many $$g \in G$$ there is a $$j(g)$$ such that $$g \notin A_{j(g)}$$, we get a contradiction by taking any $$j$$ such that there are morphisms $$j \to j(g)$$ for all $$g$$: any element of $$g_0 \in A_j$$ will then belong to $$A_{j(g_0)}$$

Surjectivity. Again, take an arbitrary element $$([y_j])_j$$ of $$\lim(X_j/G)$$. Let $$T = \{(x_j)_j \in \prod X_j : [x_j]=[y_j] \} = \prod [y_j]$$; give each $$[y_j]$$ the discrete topology, and $$T$$ the product topology. Since each $$[y_j]$$ is finite, $$T$$ is compact. For each morphism $$f : i \to j$$, let $$T_f = \{ (x_j)_j \in T : X(f) x_i = x_j \}$$. To establish surjectivity we need to show the intersection of all $$T_f$$ is non-empty. Notice that each $$T_f$$ is a closed subset of the compact space $$T$$, so to show they have a non-empty intersection it is enough to show any finite collection $$\{T_f : f \in F\}$$ ($$F$$ some finite collection of morphisms in $$J$$) has a non-empty intersection. Because $$J$$ is cofiltered, there is a cone over the diagram formed by $$F$$, say the cone is $$(h_j : j_0 \to j)_j$$ where $$j$$ ranges over the domains and codomains of the morphisms in $$F$$. Using this cone we can construct an element of $$\cap_{f \in F} T_f$$ : let $$z_j = X(h_j) x_{j_0}$$ if $$j$$ is a domain or codomain of a morphism in $$F$$; and choose all other $$z_j \in [y_j]$$ arbitrarily. Then $$(z_j)_j \in \cap_{f \in F} T_f$$ because if $$f : i \to j$$ is in $$F$$, we have $$X(f) z_i = X(f) X(h_i) x_{j_0} = X(f h_i) x_{j_0} = X(h_j) x_{j_0} = z_j$$.

1. By “$$g$$ works for $$j$$” I just mean $$g x_j = y_j$$.