Let $\tilde{X}$ be a two-sheeted cover of a connected topological space X. If $\tilde{X}$ is disconnected then this has exactly two components. Further each component is homeomorphic to X by the covering projection. Can anyone give the argument for this statement ?
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What are your thoughts about the problem? – Moishe Kohan Jul 12 '14 at 07:07
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I was just applying the definition of covering space. But I don't have any clue, how this comes from the definition ?. – user93620 Jul 12 '14 at 07:10
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1If you would have a section $s:X \to \bar X$, you would be happy since you know the projection is a local homeomorphism. This is easy when you consider path-connected spaces, but gives you the right idea. – Daniel Valenzuela Jul 12 '14 at 07:12
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1Hint: Verify that the restriction of a covering map to a connected component is still a covering map to its image. – Moishe Kohan Jul 12 '14 at 07:16
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Sorry, here what is mean by section? – user93620 Jul 12 '14 at 07:17
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studiosus ! thanks, but how this followes from this hint? – user93620 Jul 12 '14 at 07:28
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Try to think a bit more: This is a very good problem to try on your own. what would you call a covering map which has multiplicity (degree) 1? – Moishe Kohan Jul 13 '14 at 09:10