# The simply connected coverings of two homotopy equivalent spaces are homotopy equivalent

This is exercise 1.3.8 in Hatcher:

Let $\tilde{X}$ and $\tilde{Y}$ be simply-connected covering spaces of path connected, locally path-connected spaces $X$ and $Y$. Show that if $X\simeq Y$ then $\tilde{X}\simeq \tilde{Y}$.

I tried applying the lifting criterion, but I seem to be hitting a dead end. Any help would be much appreciated.

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If $f : X \to Y$ is a homotopy equivalence, pre-composing with the covering map $\widetilde{X} \to X$ you get a map $\widetilde{X} \to Y$ and by the lifting property of maps into covering spaces, there's a lift $\widetilde{X} \to \widetilde{Y}$. The idea is to try to argue this is a homotopy-equivalence. I suggest doing the same with the homotopy-inverse to $f$, and then try to check if the lifts are homotopy-inverses to each other (or not). –  Ryan Budney Aug 22 '11 at 4:53
Note that $p\times id_I: \tilde{X}\times I \to X \times I$ is a covering if $p: \tilde{X} \to X$ is a covering. Now $\tilde{X}\times I$ is also simply connected and locally path-connected, so you can lift the homotopy. –  Alexander Thumm Aug 22 '11 at 11:34
Thank you! That seemed to do the trick. –  John Aug 23 '11 at 3:53
@RyanBudney Could you give more details please? I could not show that the lifts are homotopy inverses to each other. –  James Apr 17 '12 at 23:04
@Karatug: you might be interested in having a look at this question –  t.b. Apr 24 '12 at 20:13