TITLE: Can the torus and the Klein bottle be thought as $\mathbb{R}^2/G$ with $G$ a finite group acting freely
QUESTION [8 upvotes]: I have recently proved the following exercise,

Let $G$ be a finite group acting freely on a (compact) topological manifold of dimension $n$. Then, $X/G$ is a (compact) manifold of dimension $n$.

I have done this in the context of an introductory course on topology: thus, we have not talked about anything much more profound than the definition of manifold itself. More so, the aforementioned exercise was guided. 
In the same spirit, there is an excersice which asks to prove that the torus and the Klein bottle are $2$-dimensional manifolds. I presume this can be proved 'manually' i.e. by looking at an explicit construction of the quotient of $[0,1]^2$. However, I want to know whether there is a proof involving an free action from a finite group acting on $\mathbb{R}^2$. Any hints would be much appreciated.

REPLY [5 votes]: While not being a finite group, $\Bbb{Z}^2$ acts on $\Bbb{R}^2$ by translation (concretely $(n,m) \cdot (a,b) :=(a+n,b+m)$) and the torus arises as the quotient space for this action.
Similarly, let $f,g:\Bbb{R}^2\to \Bbb{R}^2$ be the bijections given by $(a,b)\mapsto (a,b+1)$ and $(a,b)\mapsto (a+1, 1-b)$ respectively. If $G$ denotes the (infinite!) group generated by $f$ and $g$, the quotient $\Bbb{R}^2/G$ is the Klein bottle. You may check this by inspecting how the points in $[0,1]^2$ are identified by the action.
In both cases, the action is properly discontinuous: this ensures that each point in the quotient has a locally euclidean neighborhood (since the quotient map is a local homeomorphism). The fact that $\Bbb{R}^2/G$ is Hausdorff is not automatic, but is implied by the hypothesis

for each $x,y\in\Bbb{R}^2$ with disjoint orbits, there exist neighborhoods $U\ni x$, $V\ni y$ such that $gU\cap V$ is empty for each $g\in G$,

which is easily verified in both examples (and is in fact automatic provided $G$ is finite).