# Taking orbits of a finite group action commutes with inverse limits

## Table of Contents

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
\(g\) works^{1} 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 \(g x_j = y_j\).