Show that if a group \(G\) acts freely and properly discontinuously on a Hausdorff space \(X\), then the action is a covering space action. (Here “properly discontinuously” means that each \(x \in X\) has a neighborhood \(U\) such that \(\{g \in G \mid U \cap g(U) \neq \emptyset\}\) is finite.) In particular, a free action of a finite group on a Hausdorff space is a covering space action.
Solution
Let \[ q:X\to X/G \] be the orbit map.
1. Fix \(x\in X\).By proper discontinuity there is an open neighborhood \(U\) of \(x\) such that\[ F=\{g\in G:U\cap gU\ne\varnothing\} \]is finite.
This is exactly the stated proper-discontinuity hypothesis.
2. There is an open neighborhood \(V\subseteq U\) of \(x\) such that\[ V\cap gV=\varnothing \qquad\text{for every }g\ne e. \]
For each nonidentity \(g\in F\), freeness gives \[ gx\ne x. \] Since \(X\) is Hausdorff, choose disjoint open neighborhoods \[ A_g\ni x, \qquad B_g\ni gx. \] Replace \(A_g\) by \[ A_g\cap g^{-1}(B_g), \] so that \[ gA_g\subseteq B_g \] and therefore \[ A_g\cap gA_g=\varnothing. \] Because \(F\setminus\{e\}\) is finite, the intersection \[ V=U\cap\bigcap_{g\in F\setminus\{e\}}A_g \] is an open neighborhood of \(x\).
If \(g\in F\setminus\{e\}\), then \[ V\subseteq A_g, \qquad gV\subseteq gA_g, \] so \(V\cap gV=\varnothing\). If \(g\notin F\), then already \(U\cap gU=\varnothing\), and since \(V\subseteq U\) we again have \[ V\cap gV=\varnothing. \] Thus the asserted disjointness holds for every nonidentity \(g\).
3. The translates \(gV\) are pairwise disjoint.
If \[ gV\cap hV\ne\varnothing, \] then applying \(g^{-1}\) gives \[ V\cap g^{-1}hV\ne\varnothing. \] By 2 this forces \(g^{-1}h=e\), hence \(g=h\).
4. The orbit map restricts to a homeomorphism\[ q|_V:V\longrightarrow q(V). \]Moreover,\[ q^{-1}(q(V))=\coprod_{g\in G}gV. \]
The second equality follows from the definition of an orbit. By 3 the union is disjoint. The map \(q|_V\) is injective because two points of \(V\) in the same orbit would lie in \(V\cap gV\) for some \(g\), forcing \(g=e\). It is surjective onto \(q(V)\) by definition. The quotient map \(q\) is open because \[ q^{-1}(q(W))=\bigcup_{g\in G}gW \] is open for every open \(W\subseteq X\). Hence \(q|_V\) is an open continuous bijection, thus a homeomorphism.
5. Therefore \(q:X\to X/G\) is a covering map, so the action is a covering space action.
The neighborhood \(q(V)\) of \(q(x)\) is evenly covered by the pairwise disjoint sheets \(gV\), each mapped homeomorphically onto \(q(V)\) by \(q\). Since \(x\) was arbitrary, this holds at every orbit.
6. If \(G\) is finite and acts freely on a Hausdorff space \(X\), then the action is a covering space action.
For every neighborhood \(U\) of any point, \[ \{g\in G:U\cap gU\ne\varnothing\}\subseteq G \] is finite because \(G\) itself is finite. Thus the action is properly discontinuous in the stated sense, and 5 applies.