Proper and discontinuous action of a group

Question

I came across the following problem of algebraic topology that I couldn't solve.

Let $\Gamma$ be a group which acts properly and discontinuously in a topological Haussdorf space $X$. Let $H \triangleleft \Gamma$. Prove that $\Gamma / H$ acts properly and discontinuously in $X/H$.

For an action to be proper and discontinuous it must verify two properties:

• $g \neq e, x\in X \implies g \cdot x \neq x$
• $\forall ~ x,y \in X ~~ \exists ~U^x, V^y ~\text{such that} ~ \{ g \in \Gamma | gU^x \cap V^y \neq \emptyset \} ~~\text{is finite}$

I've been able to prove the first condition but I couldn't prove the second one.

I thought that it could be possible to take the union of neighbourhoods of the points of the orbit of x (the class of x under H) in the space $X$ which satisfy the second condition, intersect them (knowing beforehand that every intersection is finite) and then take the union of them. But this would not work in the general case of an infinite group.

How could I prove the second condition? Thanks.

Answer 1

I finally got the answer to this exercise.

First let's start by proving some equivalences that will make everything easier later. We want to see that $\forall ~ \bar{x}, \bar{y} \in X/H, \quad \exists \bar{U}^{\bar{x}}, \bar{V}^{\bar{y}} \subset X/H$ such that:

$$\#(\{gH, \quad gH\bar{U}^{\bar{x}}\cap \bar{V}^{\bar{y}} \neq \emptyset \})< \infty$$

We will use that:

$$gH\bar{U} \cap \bar{V} \neq \emptyset \iff \bar{gU} \cap \bar{V} \neq \emptyset \iff \pi^{-1}(\bar{gU}) \cap \pi^{-1}(\bar{V}) \neq \emptyset \iff$$

$$ \iff (\cup_{h \in H} hgU) \cap (\cup_{h' \in H} h'V) \neq \emptyset$$

On the other hand, we will also use that $\exists ~ h,h' \in H, \quad hgU \cap h'V \neq \emptyset \iff h''gU \cap V \neq \emptyset$

Taking now into account that the action of $\Gamma$ over X is proper and discontinuous we know that $\exists ~ U^x, V^y, \quad \{ k \in \Gamma, \quad kU \cap V \neq \emptyset \} = \{ k1, \dots, k_m \}$. Considering $\bar{U}=\pi(U)$ and $\bar{V}=\pi(V)$, if $gH\bar{U}\cap \bar{V} \neq \emptyset$, then $h''g = k1, \dots, k_m$ which implies $\{ gH, \quad gH \bar{U} \cap \bar{V} \neq \emptyset \} \subset \{k_1H, \dots, k_mH \}$.

Answer 2

Shortest proof using covering space theory:

$\Gamma$ acts properly discontinous on $X \Leftrightarrow$ $X\to X/G$ is a covering. Now the statement follows by passing to an intermediate cover $X \to X/H \to X/\Gamma$.