The integral lattice $\Bbb Z^n$ is a discrete subgroup of the Lie group $\Bbb R^n$. Therefore, it acts freely and properly discontinuously on $\Bbb R^n$ and the orbit space $\Bbb R^n/\Bbb Z^n$ has a smooth manifold structure. I wanted to show that $\Bbb R^n/\Bbb Z^n$ is diffeomorphic to the toral group $T^n = S^1\times\dots\times S^1$. I realized that I can show something more general:
Let $G$ be a Lie group and let $F:G \to H$ be a surjective Lie group homomorphim. If $\Gamma=\ker F$ is a discrete subgroup, then the orbit space $G/\Gamma$ is diffeomorphic to $H$.
Partial Proof: Let $\pi:G \to G/\Gamma$ be the quotient map. Define $\tilde{F}: G/\Gamma \to H$ by $\tilde{F}(\Gamma x) = F(x)$. This is a well defined bijection which is also a homeomorphism. Now since $\pi$ is a covering map, for each $p \in G/\Gamma$ there exists a connected neighbourhood $U$ of $p$ and a connected neighbourhood $\tilde{U}$ in $G$ such that $\pi|\tilde{U}: \tilde{U} \to U$ is a diffeomorphism. Therefore $\tilde{F}|U = F\circ\pi^{-1}$ and hence is a smooth map. Therefore $\tilde{F}$ is smooth because it's locally smooth.
This proof isn't complete since I haven't shown that $\tilde{F}^{-1}$ is smooth. This is where I'm stuck. It'll be helpful if someone gives a hint as to how to show that $\tilde{F}$ is a diffeomorphism.
Best Answer
One way to see this is as follows : you proved that $G/\Gamma \to H$ was a homeomorphism. $\Gamma$ is discrete so this implies that $G$ and $H$ have the same dimension.
Also, $G\to H$ is a submersion (you have to prove that), therefore on tangent spaces it is surjective, so by a dimension argument it is an isomorphism on tangent spaces, therefore $G\to H$ is a local diffeomorphism (by the local inversion theorem).
This should be enough to conclude
(By the way, you probably know this but of course the $T^n \cong \mathbb{R^n/Z^n}$ case is completely elementary and easier than the general case)