Why is the projective space $\Bbb RP^{2n}$ not the total space of a nontrivial covering map? I've heard this in class but I can't see why it holds.
The projective space $\Bbb RP^{2n}$ cannot be the total space of a nontrivial covering map
algebraic-topologycovering-spacesprojective-space
Related Solutions
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.
As Jim pointed out, it's not enough to show that $f_*$ induces the trivial map on fundamental groups. To finish it, we can proceed as follows. Let $p\colon \mathbb R\to S^1$ be the covering projection. Since $f_*(\pi_1(P^2))=0$ (as Fredrik showed), it's clearly contained in $p_*(\pi_1(\mathbb R))=0$. So, this means there is a map $\tilde f\colon P^2\to\mathbb R$ such that $f=p\tilde f$ (Proposition 1.33 in Hatcher). Now, $\mathbb R$ is contractible so we have a homotopy $H\colon P^2\times I\to \mathbb R$ such that $H_0=\tilde f$ and $H_1$ is constant at some $x_0$. But then $pH$ is a homotopy from $p\tilde f=f$ to $p(x_0)$ meaning that $f$ is null-homotopic.
Best Answer
You've already worked out the answer in the comments, but just so that there is an official answer I will write it down.
Recall that the Euler characteristic is multiplicative for covering spaces, specifically if $E \to B$ is a covering space with $n$-sheets where $B$ is of finite type (by which I mean homology is finitely-generated in all degrees and non-zero in finitely many degrees so that the Euler characteristic is defined) then $\chi(E) = n\cdot \chi(B)$.
Note that $\chi(\mathbb{RP}^{2n}) = 1$. If $\mathbb{RP}^{2n} \to B$ is a covering space then by compactness it has finite fibres (say of size $n$) and $B$ is compact and hence of finite type, therefore
$$ 1 = n\cdot \chi(B).$$
The only possibility is that $n = 1$ and it is a trivial covering.
It's worth noting that $\mathbb{RP}^{2n+1}$ is the total space of non-trivial coverings, as it covers Lens spaces. See for example Covering $\Bbb RP^\text{odd}\longrightarrow X$, what can be said about $X$?