$\mathbb{R} \mathbb{P}^2$ as the total space of a covering Suppose there exists a covering $\xi :\mathbb{R} \mathbb{P}^2 \rightarrow X$.
How can I show that $\xi$ is a homeomorphism? Thanks!
 A: The basic theory of covering spaces says if you have such a cover, $\pi_1 \mathbb RP^2$ would be a subgroup of $\pi_1 X$ and the index would be the number of sheets of the covering map, which is a homeomorphism if and only if that number is $1$.  
So consider the ways $\mathbb Z_2$ sits inside the fundamental groups of surfaces. 
A: If you have such a covering, composing with the usual map $S^2\to P^2$ will also be a covering. Then $X$ can be obtained as a properly discontinuous quotient of $S^2$. But a group of homeomorphisms acting on $S^2$ properyl discontinuously is finite and, with a little more work, of order $2$, with the non/trivial element acting by central inversion. 
Can you finish?
A: There is a nice argument of this fact using the euler characteristic.
Suppose $\mathbb{R}P^2$ covers $X$. Then $\chi(\mathbb{R}P^2) = k \cdot \chi(X)$ where $k$ denotes the number of leaves (which must be finite since $\mathbb{R}P^2$ is compact).
But since $\chi(\mathbb{R}P^2) = 1$ it follows that $k = 1 = \chi(X)$. Hence the number of leaves is $1$ which proves the claim.
