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Prove that any homeomorphism is a covering map.

My thought:

Let $p:X\to Y$ be a homeomorphism. Choose $y\in Y$. Then $Y$ is a open neighbourhood of $y$. Since $p$ is a homeomorphism, $p^{-1}(Y)=X$ is an open set and also homeomorphic to $Y$.
Am I correct? Please give me your valuable suggestion. Thanks.

Stefan Hamcke
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poton
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  • Yes, it seems correct, and also $X$ is open because it is the whole space, and of course you then have $f_| X \rightarrow Y$ is a homeomorphism, and the same argument holds for any {$y$}.. – DBFdalwayse Oct 31 '13 at 03:03
  • The definition of covering space involve a statement like, "for each point, there exists a neighborhood around it such that ........". For a homeomorphism, the condition will be satisfied with the full space being the required open set. –  Oct 31 '13 at 16:46

1 Answers1

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Yes, your proof is correct: You have shown that each $y\in Y$ has an open neighborhood $Y$ whose preimage is a disjoint union of open sets (here $X$) each of which is mapped homeomorphically onto $Y$ via $p$.

Stefan Hamcke
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