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:
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Why is $\mathbb R \mathbb P^1 \cong S^1 $?
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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.
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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)
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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.
Best Answer
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.