Quotient of $\mathbb{R}$ by $2\pi \mathbb{Z}$ I'd like help for the problem:
Let the additive group $2\pi \mathbb{Z}$ act on $\mathbb{R}$ on the right by $x · 2\pi n = x+2\pi n$, where $n$ is an integer. Show that the orbit space $\frac{\mathbb{R}}{2\pi n\mathbb{Z}} $  is a smooth manifold.
I proved that $\frac{\mathbb{R}}{2\pi n\mathbb{Z}} $ is hausdorff and second countable, but I don't know how to find an atlas, I was thinking about $ \psi([x])=e^{ix}$ but I don't know how to show that this is a homeomorphism.
 A: This is easier to do in general than in the special case, so I'll just generalize.
Most of the claims below require proof, by the way!
Suppose $M$ is a smooth manifold and that $G$ is a group which acts on $M$ smoothly and properly discontinuously. Let $N=M/G$ be the quotient topological space and $\pi:M\to N$ the quotient map. Since the action is properly discontinuous, $N$ is Hausdorff and the map $\pi$ is open —in particular, the second-countability of $M$ implies that of $N$.
Let us now construct an atlas. 
Let $p\in M$. Proper-discontinuity implies there is an open neighborhood $U\subseteq M$ of $p$ such that $U\cap gU=\emptyset$ for all $g\in G\setminus\{1\}$. The restriction $\pi|_U:U\to N$ is an homeomorphism onto its image, which is open in $N$. By replacing $U$ by a smaller open neighborhood of $p$, we can suppose that there is a chart $\phi:V\to\mathbb R^n$ in the (maximal) atlas of $M$ with $U\subseteq V$. We define $\psi_p=\phi\circ(\pi|_U)^{-1}:\pi(U)\to\mathbb R^n$.
Now the set $\mathcal A=\{\psi_p:p\in M\}$ is an atlas on $N$.
