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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?

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What other examples od "these types of problems" which are solved by doing that have you seen? – Mariano Suárez-Alvarez Jun 27 '13 at 20:50
Show that $X$ is manifold like $\mathbb{R}P^2$ and then use fact that $\pi_1(X)$ have subgroup isomorphic with $\pi_1(\mathbb{R}P^2) \cong \mathbb{Z}_2$ and classification of surfaces (like Ted proposed). – Cortizol Jun 27 '13 at 20:50
up vote 9 down vote accepted

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.

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Why is it that the fibre having cardinality $1$ implies that $f$ is a homeomorphism? Is $f$ assumed to be surjective (Hatcher does not)? – user38268 Jun 28 '13 at 4:30
So far as I am concerned, covering maps are surjective. I don't remember Hatcher's rationale for being non-standard. – Ted Shifrin Jun 28 '13 at 4:33
Thanks Ted. So your solution has shown that any such $f$ is bijective we already know $f$ is a local homeomorphism so $f$ is a homeomorphism. – user38268 Jun 28 '13 at 4:44
Perhaps you should point out that by compactness it can't be an infinite-sheeted cover either. – Grumpy Parsnip Jun 28 '13 at 12:53
The surjectivity is a consequence of the definition: each point of the covered space has an evenly covered neighborhood. – Edoardo Lanari Jun 28 '13 at 13:45

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.

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Exactly what I meant. (+1) – Cortizol Jun 28 '13 at 17:23

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