A Covering Map $\mathbb{R}P^2\longrightarrow X$ is a homeomorphism

Question

I came across the following problem: Any covering map $\mathbb{R}P^2\longrightarrow X$ is a homeomorphism. To solve the problem you can look at the composition of covering maps $$ S^2\longrightarrow \mathbb{R}P^2\longrightarrow X $$ and examine the deck transformations to show that the covering $S^2\longrightarrow X$ only has the identity and antipodal maps as deck transformations.

I've seen these types of problems solved by showing that the covering is one-sheeted. Is there a solution to the problem along those lines?

EDIT: Even if there isn't a way to do it by showing it is one-sheeted, are there other ways?

Answer 1

What about using Euler characteristic? Euler characteristic is multiplicative for a covering map: If $E\to B$ is an $n$-sheeted covering space and $E$ is compact, then $\chi(E)=n\chi(B)$. Since $\chi(\mathbb RP^2)=1$, we're done.

Answer 2

1-Prove that $X$ has to be a compact topological surface;

2-Prove that such a covering has to be finite-sheeted;

3-Deduce from 2 and from $\pi_1(\mathbb{R}P^2)$ that $\pi_1(X)$ is finite;

4- Since the map induced by the covering projection on $\pi_1$ is injective you get $\mathbb{Z}/2\mathbb{Z}< \pi_1(X)$;

5-Conclude using the classification of compact topological surfaces.