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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
    

3 Answers 3

up vote 8 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
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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. –  Lano 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

Hint: $\pi_1(X)$ acts freely on the cover $S^2$, which should tell you what $\pi_1(X)$ is. Then use the classification of covers in terms of subgroups of the fundamental group, and note that $S^2\to \mathbb{R}P^2$ is a 2-fold cover.

This is the style you were after.

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