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I want to prove A path connected, locally path connected has a universal covering space implies that the space is semi-locally simply connected. Looking forward to some hints.

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  • $\begingroup$ Would you be satisfied with a counterexample rather than a hint? The Hawaiian earring is path connected and locally path connected, and it has many covering spaces. In fact it is its own covering space, because the identity map to itself is a covering map. And there are many other covering spaces of the Hawaiian earring besides itself. Nonetheless, the Hawaiian earring is not semi-locally simply connected. en.wikipedia.org/wiki/Hawaiian_earring $\endgroup$
    – Lee Mosher
    Commented Apr 18, 2018 at 21:41
  • $\begingroup$ That helps, thanks. But I was looking for some hints to prove this. $\endgroup$
    – Arindam
    Commented Apr 18, 2018 at 21:44
  • $\begingroup$ Rule #1: Never try to prove a false theorem. And this theorem is false, because I gave you a counterexample. $\endgroup$
    – Lee Mosher
    Commented Apr 18, 2018 at 21:55
  • $\begingroup$ Sorry I edited the question, I wanted to ask about the universal cover. $\endgroup$
    – Arindam
    Commented Apr 18, 2018 at 22:01

1 Answer 1

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If $f:E\rightarrow X$ is a covering with $E$ simply-connected, then for every point $x\in X$, it has a neighborhood $U$ which can be mapped homeomorphically to $\widetilde{U}\subset E$, so for any loop $\gamma$ in $\pi_1(U)$, we have a corresponding loop $\widetilde{\gamma}$ in $\widetilde{U}$, which is null-homotopic in $E$, so $\gamma$ is null-homotopic in X.

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