Prove that $G \cong \pi_{1} (X/G)$.

Question

This is a question from an exam.

Let $X$ be a topological space which is simply-connected, and let $G$ be a group of homeomorphisms of $X$ which acts properly discontinuously, meaning

$$\forall \ x\in X,\ \exists\text{ open neighbourhood } U \text{ such that }\ \forall \ g \in G \setminus {e},\ U \cap g(U) = \emptyset. $$

We need to show that $G \cong \pi_{1} (X/G)$.

I would appreciate any hints or clarification for the question.

Thanks in advance.

Answer 1

The key to this problem is noticing that the condition on the action of $G$ implies that the quotient map $q\colon X \to X/G$ is a covering space. (Hint: for each $x\in X$ we can choose a neighbourhood $U$ satisfying the given condition, and then $q(U)\subset X/G$ will be evenly covered $q^{-1}(q(U)) = \cup_{g\in G}g\cdot U$.)

Then since $X$ is simply connected it is actually the universal covering space of $X/G$, and from here you can prove $\pi_1(X/G) \cong G$ in a number of ways. Here are some suggestions:

1. Prove that $G$ is the automorphism group (aka group of Deck transformations) of $q$.
2. For each basepoint $x_0\in X$ the fibre over $q(x_0)$ is the $G$-orbit $\{g\cdot x_0\mid g\in G\}$, which is in bijection with $G$. Use path/homotopy lifting to construct a map $\pi_1(X/G, q(x_0)) \to G$ and show it is a bijective homomorphism.
3. Use the long exact sequence of a fibration (if you know it).

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Since OP requested it in a comment, I will try to elaborate on the third suggestion.

A covering space $E\to B$ with fibre $F$ satisfies the homotopy lifting property with respect to any space, or in other words it is a fibration (specifically a Hurewicz fibration). In particular a covering spaces satisfies the homotopy-lifting property for CW complexes (aka a Serre fibration) so it's possible to derive a long exact sequence of homotopy groups $$ \dots \to \pi_n(F) \to \pi_n(E) \to \pi_n(B) \to \pi_{n-1}(F) \to $$ $$ \dots \to \pi_1(B) \to \pi_0(F) \to \pi_0(E)$$ where for $\pi_0$ "exactness" is in terms of pointed sets.

Since $F$ is discrete $\pi_n(F) = 0$ for all $n>0$ (in particular $\pi_n(E) \cong \pi_n(B)$ for $n \geq 2$) and if $E$ is simply connected we have $\pi_1(B) \cong \pi_0(F) \cong F$, as pointed sets.

In our case we actually have $F = G$, a group, and $B = E/G$, but it remains to show that the bijection $\pi_1(B) \to \pi_0(G)\cong G$ is a homomorphism of groups, and I haven't managed to come up with an argument for this yet.