$\mathbf{Question}$. Let us assume that $M^n$ is a topological manifold of dimension n, with a group action $\Gamma$, which acts discontinuously and freely on $M^n$. Is the orbit space $M^n / \Gamma$ still a manifold of dimension n?


"Discontinuously" here means the orbit of a certain point does not have a limit point. "Freely" here means a member of the group $\Gamma$ has a fixed point only if it is the identity.


Here is what I know about the problem.


$\mathbf{Proposition \ 1}.$ We know that if $M^n$ carries a metric $d$, from which the topology of $M^n$ derives, then the statement above holds simply.


Denote the orbit map by $\pi : M^n \to M^n / \Gamma$. Consider the sub-metric $d'$ defined by
$d'( \pi A, \pi B )
= \text{min}\{ d( A', B' ) \ | \ A' \in \Gamma A, B' \in \Gamma B \}$.


Now, fix a point $A \in M^n$, and let
$\epsilon = \text{ min } \{ d( A', A ) \ | \ A' \in \Gamma A, A' \neq A \}$.
It is easy to show that $\pi$ restricted on an open disk surrounding $A$ with radii $\epsilon / 3$ is a homeomorphism to a neighborhood of $\pi A$. Hence the proposition holds.


$\mathbf{Proposition \ 2}$. In addition to the criterion that $\Gamma$ should be discontinuous and free, we restrict further that $\Gamma$ acts properly on $M^n$, i.e., assign a discrete topology to $\Gamma$, then the map
$\Gamma \times M^n \to M^n \times M^n$, $( \gamma, A ) \mapsto ( \gamma A, A )$ is proper. Now the orbit space $M^n / \Gamma$ is a manifold of dimension n.


I found this fact here. <https://math.stackexchange.com/questions/496571/under-what-conditions-the-quotient-space-of-a-manifold-is-a-manifold>


Proposition 1 let us (at least me) believe strongly that our question holds true. However, in proposition 2 we see an extra prerequisite on $\Gamma$, that it acts properly, and maybe the prerequisite is substaintial.


I will appreciate it if anyone can answer this question, either by providing a proof or a counter example.



When is the orbit space of a manifold still a manifold of the same dimension?
No, here is the standard counterexample. Let $M$ be the complement of the origin in $\mathbb{R}^2$, let $T$ be the linear transformation $T(x,y) = (2x,y/2)$, and let $\Gamma$ be the cyclic group generated by $T$. Then $\Gamma$ acts discontinuously and freely on $M$, but the orbit space $M/\Gamma$ is not Hausdorff, so it is certainly not a manifold. (The $\Gamma$-orbit of every neighborhood of $(0,1)$ intersects every neighborhood of $(1,0)$.)


