Taking orbits of a finite group action commutes with inverse limits

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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 \(X(f) z_i = X(f) X(h_i) x_{j_0} = X(f h_i) x_{j_0} = X(h_j) x_{j_0} = z_j\).


By “\(g\) works for \(j\)” I just mean \(g x_j = y_j\).

Omar Antolín Camarena