Quotient of $R $ by $2πZ$ Let the additive group $2πZ$ act on $R $ on the right by $x · 2πn = x +2πn$, where $n$ is an integer.
Show that the orbit space $R/2πZ$ is a smooth manifold.
 A: To simplify notations, we prove that $\mathbb{T} = \mathbb{R}/\mathbb{Z}$ is a smooth manifold.
It can be proved similarly that $\mathbb{R}/2\pi\mathbb{Z}$ is also a smooth manifold.
Since $\mathbb{Z}$ is a closed subgroup of $\mathbb{R}$, $\mathbb{T}$ is a Hausdorff topological group.
Let $p\colon \mathbb{R} \rightarrow \mathbb{T}$ be the canonical map.
Let $V$ be an open subset of $\mathbb{R}$.
It is easy to see that $p^{-1}p(V) = V + \mathbb{Z}$.
Since $V + \mathbb{Z}$ is open, $p$ is an open map.
Let $U = \{x\in \mathbb{R} | |x| < 1/2\}$.
Suppose $p(x) = p(y)$ for $x, y \in U$.
Then $x - y \in \mathbb{Z}$.
Since $|x - y| \le |x| + |y| < 1$, $x - y$ must be $0$.
Hence $p|U$ is injective.
Since $p$ is an open continuous map, $p|U$ is a homeomorphism onto $p(U)$.
Since $p$ is open, $p(U) + p(x)$ is an open neighborhood of $p(x)$ for $x \in \mathbb{R}$.
Let $f_x\colon U \rightarrow p(U) + p(x)$ be the map defined by $f_x(t) = p(t) + p(x)$.
Clearly $f_x$ is a homeomorphism.
Hence $\mathbb{T}$ is a topological manifold.
Let $p(x), p(y)$ be any two points of $\mathbb{T}$.
Suppose $p(z) \in (p(U) + p(x)) \cap (p(U) + p(y))$.
There exist unique $s, t \in U$ such that
$p(z) = p(s) + p(x) = p(t) + p(y)$.
Note that $s$ and $t$ are local coordinates of $p(z)$ in $p(U) + p(x)$ and $p(U) + p(y)$ respectively.
Since $s + x \equiv t + y$ mod $\mathbb{Z}$, $t = s + x - y + n$ for some $n \in \mathbb{Z}$.
Since $t$ is a continuous function of $s$, $n$ is constant in a neighborhood of $s$. 
Thus $t$ is a smooth function of $s$.
Hence $\mathbb{T}$ is a smooth manifold.
