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

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.
