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Suppose that $X,Y$ are arc-connected and locally arc-connected spaces and that $p:\widetilde{X}\to X$ and $q:\widetilde{Y}\to Y$ are universal covering of $X$ and $Y$ respectively. Show that if $X$ is homotopically equivalent to $Y$, then $\widetilde{X}$ is homotopically equivalent to $\widetilde{Y}$

I have to find a homotopic equivalence from $\widetilde{X}$ to $\widetilde{Y}$, I know that there is a homotopic equivalence $f:X\to Y$ and therefore $f_*:\pi_1(X)\to \pi_1(Y)$ is an isomorphism, besides $\widetilde{X}$ and $\widetilde{Y}$ are the universal covers then $\pi_1(\widetilde{X})=\{1\}$ and $\pi_1(\widetilde{Y})=\{1\}$ and I have the next diagram but I do not know what else to do. Could someone help me please?, could I take $g=q^{-1}\circ f\circ p$ ? Thank you very much.

enter image description here

user402543
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    The composite function $f\circ p$ maps the simply connected space $\tilde X$ into $Y$. Because $q$ is a covering map, $f\circ q$ factors through $q$, and that gives you a map $\tilde X\to\tilde Y$. Similarly, get a map $\tilde Y\to\tilde X$. If you've been careful about base points when doing the factoring, these ought to be homotopy inverses. – Andreas Blass May 01 '18 at 00:58
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    @AndreasBlass in the second sentence, $f \circ p$? – Andres Mejia May 01 '18 at 01:03
  • @AndreasBlass Why $f\circ q$ factors through $q$? – user402543 May 01 '18 at 16:51
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    Sorry about the typo. I meant $f\circ p$ factors through $q$. Since $\tilde X$ is simply connected, any map from it to $Y$ factors through any covering space of $Y$. – Andreas Blass May 02 '18 at 02:16
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    @AndresMejia Yes, sorry about the typo. – Andreas Blass May 02 '18 at 02:17
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    It is a theorem that for a covering $p:C \to Y$, and a map $f:X \to Y$, then $f$ lifts to a map in $c$ if and only if the induced homomorphisms $f_(\pi_1(X)) \subset f_(p(\pi_1(C))$. This is trivially satisfied in the above since a universal cover is assumed to be simply connected – Andres Mejia May 02 '18 at 02:20
  • See page 3 here – Andres Mejia May 02 '18 at 02:24

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Hint: Use the following description of the universal cover of $X$: Let $x_0\in X$, consider $\hat X$ the set of paths:$c:[0,1]\rightarrow X$ such that $c(0)=x_0$. Consider the relation $cRc'$ of $\hat X$ if $c(1)=c'(1)$ and they are homotopically equivaelent. The quotient of $\hat X$ by this relation is the universal cover $\tilde X$. It $f:X\rightarrow Y$ and $g:\rightarrow X$ define an homotopy equivalence, you can define $\hat f:\hat X\rightarrow \hat Y$ and $\tilde f:\tilde X\rightarrow\tilde Y$. Show that $(\tilde f,\tilde g)$ define an homotopy equivalence between $X$ and $Y$.

Tsemo Aristide
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  • Why does $\bar{X}=\widehat{X}/\sim$ define a topological space?, why this is a universal covering of $X$? How is $\widehat{f}: \widehat{X}\to \widehat{Y}$ defined? To construct $\bar{f}:\bar{X}\to \bar{Y}$, can the universal property of the quotient topology be used? – user402543 May 01 '18 at 16:48