TITLE: Can $\mathbb RP^2$ cover $\mathbb S^2$?
QUESTION [10 upvotes]: I know $\mathbb{S}^2$ is the universal cover of $\mathbb{R}P^2$, but can $\mathbb{R}P^2$ be a covering space (at all) of $\mathbb{S}^2$?
Attempt at solution
It's clear that for a umramified covering with degree $n$ between surfaces, the euler characteristic of the covering space must be $n$ times the Euler characteristic of the image space. $\chi(\mathbb{R}P^2)=1$ and $\chi(\mathbb{S}^2)=2$, and clearly $1\neq n\cdot 2$ for any $n$. So the cover cannot be $n$-fold, but can the fibers be countable or the cover non-ramified?

REPLY [7 votes]: Note that even the following is true:
suppose $\mathbb{R}P^2 \to X$ is a covering. Then the map is a homeomorphism.
A nice argument involves multiplicativity of the euler characteristic as you adressed it.
Hence $\chi(\mathbb{R}P^2) = 1$ and hence we get $k\cdot \chi(X) = 1$ where $k$ is the number of leaves of the covering (which must be finite since $\mathbb{R}P^2$ is compact). Hence $k = \chi(X) = 1$ and the claim follows.
So not only does $\mathbb{R}P^2$ not cover $S^2$, it acutally only covers itself.