Real projective space, the quotient map and covering projection

Solution 1:

Let's think about what the equivalence relation we put on points of $S^n$ is. We identify antipodal points, which means that the fiber of a point in $\mathbb{R}P^n$ under the quotient map is just two opposite points. Similarly, if you look at the lift of a neighborhood in $\mathbb{R}P^n$, as long as it is small enough, it's just going to be two copies of that neighborhood on opposite ends of the sphere.

In dimensions one and two, we can quickly come up with a model for $\mathbb{R}P^n$ that we can put down on paper and stare at. For $n=1$, we start with the circle $S^1$. Our equivalence relation is to identify opposite points, so let's find a set of representative points on $S^1$ that are in bijection with points in $\mathbb{R}P^1$. Any line through the center of the circle will intersect at least one point in the upper half of the circle. Furthermore, the only line that intersects it in two places is the horizontal line. So this means that a set of representatives can be taken to be those $(x,y)$ on the circle with $y > 0$, along with one of the points $(1,0)$ and $(-1,0)$; it doesn't matter which.

So now we have $\mathbb{R}P^1$ as a set, and we want to make sure it has the right topology. A neighborhood of our exceptional point $(1,0)$ is "missing" some neighbors: it is also close to points of the form $(x,y)$ with $x > 0$ and $y< 0$ small. But these points get identified with $(-x,-y)$, which are just points on the other end of the upper half-circle. So this tells us that we need to "glue" together $(1,0)$ and $(-1,0)$. In our construction of $\mathbb{R}P^1$, we started with a half-open line segment, and then glued the ends together, and this is nothing more than a circle.

As for your question about fundamental groups (incidentally, the standard notation is to use a lowercase $\pi$), we only need to look at what happens to a generator of $\pi_1(S^1)$, the loop that goes around once. Since we built $\mathbb{R}P^1$ from $S^1$ essentially by shrinking our circle by a factor of two, we see that the loop that goes around the big circle $S^1$ once will go around the smaller circle $\mathbb{R}P^1$ twice.

You should think about what $\mathbb{R}P^2$ "looks like", as you can do an analogous construction to get a model for $\mathbb{R}P^2$ from $S^2$ like I did for $S^1$. What happens at the boundary gets more interesting, but it's something to think about.

Solution 2:

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.