As part of am exam question (Q21F here), I'm trying to prove that if $X$ and $Y$ are path-connected, locally path-connected spaces with universal covers $\widetilde{X}$ and $\widetilde{Y}$, respectively, then if $X \simeq Y$ then $\widetilde{X} \simeq \widetilde{Y}$.
My attempt might be correct, but it seems too complicated and amongst the jumble I may have made some incorrect assumptions. So what I'm looking for is (a) verification or correction; and (b) ideas as to how I could simplify my argument.
My argument:
Let $p:\widetilde{X} \to X$ and $q:\widetilde{Y} \to Y$ be the covering maps, and let $X \overset{f}{\underset{g}{\leftrightarrows}} Y$ be a homotopy equivalence.
Fix a point $\widetilde{x} \in \widetilde{X}$, let $x=p(\widetilde{x}) \in X$ and $y=f(x) \in Y$, and fix $\widetilde{y} \in q^{-1}(\{y\}) \subseteq \widetilde{Y}$. Also let $x' = g(y) \in X$ and fix $\widetilde{x'} \in p^{-1}(\{x'\}) \subseteq \widetilde{X}$.
Define a map $\widetilde{f} : \widetilde{X} \to \widetilde{Y}$ as follows. For $\widetilde{z} \in \widetilde{X}$ let $\widetilde{u}:[0,1] \to \widetilde{X}$ be a path from $\widetilde{x}$ to $\widetilde{z}$. Let $z=p(z)$ so that $u=p\widetilde{u}$ is a path in $X$ from $x$ to $z$. Let $v=fu$, so that $v$ is a path in $Y$ from $y$ to $f(z)$. Lift $v$ to a path $\widetilde{v}$ in $\widetilde{Y}$ with $\widetilde{v}(0) = \widetilde{y}$. Define $\widetilde{f}(\widetilde{z}) = \widetilde{v}(1)$.
Notice that $q\widetilde{v}=v$ so that $q\widetilde{f} = fp$.
Define $\widetilde{g}:Y \to X$ analogously: For $\widetilde{z} \in \widetilde{Y}$ let $\widetilde{a}:[0,1] \to \widetilde{Y}$ be a path from $\widetilde{y}$ to $\widetilde{z}$. Let $z=q(z)$ so that $a=q\widetilde{a}$ is a path in $Y$ from $y$ to $z$. Let $b=fa$, so that $b$ is a path in $X$ from $x'$ to $g(z)$. Lift $b$ to a path $\widetilde{b}$ in $\widetilde{X}$ with $\widetilde{b}(0) = \widetilde{x'}$. Define $\widetilde{g}(\widetilde{z}) = \widetilde{b}(1)$.
Likewise, notice that $p\widetilde{g}=gq$.
Claim: $\widetilde{X} \overset{\widetilde{f}}{\underset{\widetilde{g}}{\leftrightarrows}} \widetilde{Y}$ is a homotopy equivalence.
We have $p\widetilde{g}\widetilde{f} = gq\widetilde{f}=gfp \simeq p$ and $q\widetilde{f}\widetilde{g} = fp\widetilde{g} = fgq \simeq q$.
This shows that $\widetilde{g}\widetilde{f}$ and $\widetilde{f}\widetilde{g}$ are covering translations (deck transformations) and are therefore homotopic to the respective identity maps. So we have a homotopy equivalence.
Any comments would be appreciated.
Best Answer
This is more of a long comment than an answer. Your way to go is certainly correct but a little more needs to be justified. First, I should say that there is no need to redefine (IMO in an exam) your $\tilde{f} : \tilde{X} \to \tilde{Y}$. This is already given by the lifting criterion (Hatcher Proposition 1.33):
Since $\tilde{X}$ and $\tilde{Y}$ are simply connected the lifts will always exist. Of course your definition of how the lift is defined is exactly that given in Hatcher. However you are not exactly done yet because you have only proven that $p\tilde{g}\tilde{f} \simeq p$ and $q\tilde{f}\tilde{g} \simeq q$.
To finish the problem I think you can use the homotopy lifting property to tell you the following. Suppose that the homotopy between $q \tilde{f}\tilde{g}$ and $q$ is given by some $F :\tilde{Y} \times I \to Y$. Now the restriction of $F$ to $\tilde{Y} \times \{0\}$ lifts to some map $\tilde{Y} \times \{0\} \to \tilde{Y}$. By uniqueness of lifts, we deduce that this map must be $\tilde{f}\tilde{g}$. Then we know that there exists a unique homotopy $\tilde{F} : \tilde{Y} \times I \to \tilde{Y}$ such that $\tilde F|_{\tilde{Y} \times \{0\}} = \tilde{f}\tilde{g}$ and
$$q \circ \tilde{F} = F.$$
Now we have that $\tilde{F}(y,0) = \tilde{f}\tilde{g}$, $q \tilde{F}(y,1) = F(y,1) = q$. However we now have by uniqueness of lifts that $\tilde{F}(y,1) = \textrm{id}_\tilde{Y}$, and so $\tilde{F}$ is a homotopy between $\tilde{f}\tilde{g}$ and $\textrm{id}_{\tilde{Y}}$. The proof that $\tilde{g}\tilde{f}$ is homotopic to $\textrm{id}_{\tilde{X}}$ is similar and we are done.