How can I prove that $\mathbb{R}/2\pi\mathbb{Z}$ is a smooth manifold?

Question

First, does $\mathbb{R}/2\pi\mathbb{Z}$ equals to $[0,2\pi)$? Second how can I prove $\mathbb{R}/2\pi\mathbb{Z}$ is a manifold?

Answer 1

What do you mean by "equals"? They are different sets, but there is a bijection beetween them (can you give one?).

They are not homeomorphic. The topological space $\mathbf R/2\pi\mathbf Z$ is compact (why?) but $[0,2\pi)$ is not.

In fact $\mathbf R/2\pi\mathbf Z$ is homeomorphic to a well known space, can you find it?

What do you mean by manifold? (topological, smooth...)