TITLE: Real projective space, the quotient map and covering projection
QUESTION [33 upvotes]: As I understand it, real projective space $\mathbb R \mathbb P^n$ is defined to be the quotient $S^n / \sim$, where $x \sim y$ iff $x = \pm y$. In other words, elements of $\mathbb R \mathbb P^n$ are pairs of antipodal points on $S^n$. We have the associated quotient map $q : S^n \to \mathbb R \mathbb P^n$, which maps points to their equivalence class. 
I have heard that $q$ is a covering projection of $\mathbb R \mathbb P^n$, and I'm trying to see why. I don't know how to 'visualise' real projective space, which I suppose is my problem. This is highlighted by the fact that I fail to see why $\mathbb R \mathbb P^1 \cong S^1$. I guess my real questions are:

Why is $\mathbb R \mathbb P^1 \cong S^1 $?
Why is $q$ a covering projection? If we're given an open set $U$ in $\mathbb R \mathbb P^n$, then we know necessarily that $q^{-1}(U)$ is open in $S^n$ (by definition of the quotient topology). Why is $q^{-1}(U)$ a disjoint union of homeomorphic copies of $U$? I can see that it might be, by considering $q^{-1}(u)$ for some equivalence class $u$. I don't know how to think of open sets in a non-metric sense. 
Why does the quotient $q: S^1 \to \mathbb R \mathbb P^1 $ induce multiplication by $2$ on $\Pi_1(\mathbb R \mathbb P^1)$? (in case this isn't clear, I mean the homomorphism the map induces)
In fact, come to think of it, why is real projective space path connected? Why is it locally path connected?

Answers to any of the above and more would be greatly appreciated.

REPLY [9 votes]: Why is $\mathbb R \mathbb P^1 \cong S^1$?

$f(e^{i\theta})=e^{i2\theta}$ maps $S^1$ onto $S^1$, it's continuous and $f(e^{i\theta_1})=f(e^{i\theta_1})$ if and only if $|\theta_1 - \theta_2| \in \{0, \pi\}$, i.e. when $e^{i\theta_1}$ and $e^{i\theta_1}$ are antipodal points of $S^1$. By universal property the quotient map $\tilde{f}:S^1/f=\mathbb{R}\mathbb P^1 \to S^1$ is continuos and it's now bijective. Being a continuous map  between a compact space (quotient of a compact space) to an Hausdorff space (a subspace of an Hausdorff) it's closed, thus it's a homeomorphism.

Why is $q$ a covering projection?

A standard way to build coverings is to act with a group of homeomorphism on a space with some reasonable hypotesis and see the quotient space as result of a "folding" of the original space on itself. You can prove as an exercise that if $X$ is locally path connected and $G<Hom(X)$ is finite and acts freely on $X$, then the projection to the quotient $\pi:X\to X/G$ is a covering. In our case you can see $\mathbb{R}\mathbb{P}^n = S^n/G$ where $G=<a>$ is the group of order two generated by the antipodal map $a(x)=-x$  of $S^n$, that is a homeomorphism.

In fact, come to think of it, why is real projective space path connected?

The quotient of a path connected space is path connected, in fact if $\pi :X\to Y$
 is the projection map and you take $y_1, y_2 \in Y$, than you can choose a path $\alpha$ in $X$ between $x_1 \in \pi^{-1}(y_1)$ and $x_2 \in \pi^{-1}(y_2)$, then $\pi\alpha$ gives a path between $y_1$ and $y_2$. If you know how to prove that $S^n$ is path connected for $n>0$ you are ok.