Let $\Gamma=\mathbb{Z}\times\mathbb{Z}$ with the usual group operation
$$(m,n)+(m',n')=(m+m',n+n').$$
I already showed that $\phi_{m,n}(x,y):=(x+m,y+n)$ defines an action of $\Gamma$ on $\mathbb{R}^2$, namely
$$\phi_{p,q}(\phi{m,n}(x,y))=\phi_{p,q}(x+m,y+n)=(x+m+p,y+n+q)=\phi_{m+p,n+q}.$$
Right?
Now I have to show that $\mathbb{R}^2/\Gamma$ is homeomorphic to the torus. My definition of the torus is $S^1\times S^1$ and I already did prove once that this is homeomorphic to the parametrization given below. This is my attempt, but I'm not sure this is totally correct. Who can give me tips to prove it correctly?
My attempt
Use the following corollary:
Assume that $(Y,\pi)$ is a quotient of the topological space $X$ modulo $R$. Then, for any topological space $Z$, there is a 1-1 correspondence between continuous maps $f: Y\rightarrow Z$ and continuous maps $\tilde{f}:X\rightarrow Z$ such that $\tilde{f}(x)=\tilde{f}(x')$ whenever $(x,x')\in R$. This correspondence is characterized by $\tilde{f}=f\circ \pi$.
Using this we can see that the map $\tilde{f}: \mathbb{R}^2 \rightarrow S^1\times S^1, (a,b)\mapsto (R+r\cos(a))\cos(b),(R+r\cos(a))\sin(b), r\sin(a))$ with $a,b\in[0,2\pi]$ induces a continuous bijection $f: \mathbb{R}^2/\mathbb{Z}^2 \rightarrow S^1\times S^1$. Now use the fact that $\tilde{f}$ is an open map (but I really do not know how to prove this, please help me with this!), and then we can conclude $f$ is a homeomorphism (continuous, bijection, inverse is continuous).
So, what I ask you is the following: Is my proof correct? And how do I prove $f$ is an open map?
Thanks a lot!
Best Answer
Modulo showing $\tilde{f}$ is open and the minor corrections given by David in the comments, your proof looks good to me. You could prove that $\tilde{f}$ is open, but in this case there's actually a trick that gives you that $f^{-1}$ is continuous for free without having to think about $\tilde{f}$. Namely, if you can show that $\mathbb{R}^2/\mathbb{Z}^2$ is compact, then $f$ is a continuous bijection from a compact space to a Hausdorff space, so it is automatically a homeomorphism. And to show that $\mathbb{R}^2/\mathbb{Z}^2$ is compact, you can just note that it is the continuous image of the compact space $[0,2\pi]^2$ under the continuous map $\pi$.