Q:

The projective space $\Bbb RP^{2n}$ cannot be the total space of a nontrivial covering map

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.

A:

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 $X$?

