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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).2011-08-22
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    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.2011-08-22
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    Thank you! That seemed to do the trick.2011-08-23
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    @RyanBudney Could you give more details please? I could not show that the lifts are homotopy inverses to each other.2012-04-17
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    @Karatug: you might be interested in having a look at [this question](http://math.stackexchange.com/q/136405/5363)2012-04-24

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