Any homeomorphism is a covering map 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.

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 $p$.
