Q:

Real projective space $\mathbb{R}P^1$ is diffeomorphic to $S^1$?

I am reading Lee, Manifolds and Differential Geometry, and I am a bit confused. It is seen that the real projective space $\mathbb{R}P^n$ is homeomorphic to $S^n/ \sim$, where $p \sim q$ iff $p = \pm q$. 
However, in one of the exercises (1.48) one is asked to prove that $\mathbb{R}P^1$ is diffeomorphic to $S^1$. 
Is this really true (is there a simple proof of this) and does this imply something for the relation between $S^1$ and $S^1/\sim$? 
Does this also mean that $\mathbb{R}P^1$ is homeomorphic to $S^1$ (since diffeomorphism is a stronger condition)? 

A:

The unique 1-dimensional compact connected manifold is $S^1$. 
To see that  $RP^1$ is $S^1$, you view $S^1=\{ e^{it}, t=\in [0,2\pi])\}$.
The identification $p\simeq q$ here can be described like so. Consider the interval $I=\{e^{it}, t\in [0,\pi]\}$. If $t\in (0,\pi)$ there exists a unique $t'\in (\pi,2\pi)$ such that $e^{it}\simeq e^{it'}$ and $e^{i0}=1\simeq e^{i\pi}=-1$. Thus the quotient space is obtained by identifying $0$ and $-1$ in $I$. That is you identify, the endpoints of an interval. This is the circle.

