# $\mathbb{R} \mathbb{P}^2$ 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!

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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?

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Thanks to both of you, but I can only choose one answer and this seems to be closer to my current skillset. I have figured out that $S^2$ is in fact the universal covering of $P^2$, while the deck transformations are the identity and the antipodal map. I only know that $S^2$ is then also the universal covering of X (because it is simply connected), but I am not sure how this yields homeomorphy between $P^2$ and X.. –  jules Nov 22 '11 at 17:20

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.

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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.

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