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?