When is Quotient Map a Covering Map

Group $G$ acts on topological space $X$. Also, $x,x'\in X$ not in the same orbit of $G$ have open $U$, $U'$ such that $g(U)\cap U'=\varnothing$ for all $g\in G$. I have shown that $X/G$ is Hausdorff. The question asks whether the quotient map $q : X \rightarrow X/G$ a covering map.

I've worked at this for several days, but I feel like my understanding is shaky to begin with (especially since I am a beginner in both Abstract Algebra and Topology). I want to say "no", since the quotient map is not, in general, injective so it itself can't be a homeomorphism. However, a covering map has a union of disjoint open sets each mapped homeomorphically.

I can think of a (possible) example: $X=[0,1)\times[0,1]$ with the relative topology from $\mathbb{R}^2$. $G$ is the group $[0,1)\times\{0\}$ with the operation of componentwise addition modulo $1$. Then, every $[0,1)\times\{y\} \subset X$ maps to a point in $G/X$. Yet, $[0,1)\times\{y\}$ isn't a union of open sets each homeomorphic to a point.

Solution 1:

The following theorem is given with proof here:

Let $X$ be a topological space that is path-connected and locally path-connected. Let $G < \text{Hom}(X)$ be a group of homeomorphisms on $X$. Then the projection map $q: X \to X/G$ is a covering map if and only if $G$ acts properly discontinuously on $X$.

Properly discontinuously means that for every $x \in X$ there is a neighbourhood $U$ s.t. $gU \cap U = \varnothing$ for all $g \in G - \{e\}$.

The condition you give in the question is almost the condition that $G$ acts p.d. on $X$ so I suspect the question boils down to whether your $X$ is path-connected and locally path-connected.

Hope this helps.