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.
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 $Y$.