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This problem just came to my mind. It looks easy, but i can't do it. Is there any counterexample of the problem given in the title?

Brmbrm
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – CrSb0001 Jun 07 '23 at 00:39
  • Are the two spaces path connected? – Charlie Frohman Jun 07 '23 at 00:53
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    Hint: There is a counterexample with $X$ a wedge of two circles. – Jason DeVito - on hiatus Jun 07 '23 at 02:10
  • @CharlieFrohman You can assume that they are path connected – Brmbrm Jun 07 '23 at 09:55
  • @JasonDeVito can you say a bit more? The question, whether the covering maps are mutually inverse, obviously has to be answered to the negative, any two nontrivial endo-coverings $\Bbb S^1 \rightarrow \Bbb S^1$ will do. But any space is homeomorphic to itself. I read the question as "can covering maps detect isomorphism classes?". So we need two non-homeomorphic mutually covering spaces. If you allow non-connected spaces the disjoint union of a space with itself should provide a counterexample. But I am not so sure about the connected case. – Jonas Linssen Jun 07 '23 at 11:48
  • @JonasLinssen: I was originally thinking of $X = S^1\vee S^1$ and $Y = S^1 \vee S^1 \vee S^1$. Because $\pi_1(X)$ (which is isomorphic to the free group on two generators) contains a copy of $\pi_1(Y)$, $X$ has a covering with the homotopy type of $Y$. I originally was thinking that this covering was actually homeomorphic to $Y$, but now I'm not so sure. – Jason DeVito - on hiatus Jun 07 '23 at 16:12

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