Homotopy equivalent spaces have homotopy equivalent universal covers

Question

A problem in section 1.3 of Hatcher's Algebraic Topology is

Let $\tilde{X}$ and $\tilde{Y}$ be simply-connected covering spaces of the path-connected, locally path-connected spaces $X$ and $Y$. Show that if $X \simeq Y$ then $\tilde{X} \simeq \tilde{Y}$. [Exercise 11 in Chapter 0 may be helpful.]

Exercise 11 in chapter 0 says

Show that $f:X \to Y$ is a homotopy equivalence if there exist maps $g, h:Y \to X$ such that $fg \simeq 1$ and $hf \simeq 1$. More generally, ,show that $f$ is a homotopy equivalence if $fg$ and $hf$ are homotopy equivalences.

These two questions on Stack ask about this problem, but one is unanswered and the solution to the other is unclear to me (and may be wrong).

What I have so far: Given universal covering maps $p: \tilde{X} \to X$ and $q: \tilde{Y} \to Y$, and homotopy inverses $f:X \to Y$ and $g: Y \to X$, we can find a lift $\tilde{f}: \tilde{X} \to \tilde{Y}$ such that $q\tilde{f} = fp$. (In fact, I think there are as many such $\tilde{f}$ as there are elements of $q^{-1}(y)$ for any basepoint $y \in Y$). Similarly we can find $\tilde{g}$ such that $gq = p\tilde{g}$.

From the homotopy $gf \simeq 1$ we have a unique lift of a homotopy $p\tilde{g}\tilde{f} = gfp \Rightarrow p$ starting at $\tilde{g}\tilde{f}$, but how do we know it ends at $1_{\tilde{X}}$?

I notice I haven't used exercise 11...

I'm guessing these are unbased homotopies, since that's generally how Hatcher uses the term.

Answer 1

You are right that the lift of the homotopy $p\tilde{g}\tilde{f}$ starting at $ \tilde{g} \tilde{f}$ may not end at the identity. However, it does end at a lift of the identity, that is, a deck transformation $\phi$ of $\widetilde{X}$. Now, this implies that $\tilde{g}\tilde{f} \simeq \phi$ so $\phi^{-1} \tilde{g}\tilde{f} \simeq id_{\widetilde{X}}$. Similarly, there is a deck transformation $\psi$ of $\widetilde{Y}$ such that $\tilde{f}\tilde{g} \psi^{-1} \simeq id_{\widetilde{Y}}$.

Now, apply exercise 11 in chapter 0 to conclude that $\tilde{f}$ is a homotopy equivalence.

Answer 2

$p:(\widetilde{X},\tilde{x}_0)\to (X,x_0)$ and $q:(\widetilde{Y},\tilde{y}_0)\to (Y,y_0)$ be covering maps. $f:(X,x_0)\to(Y,y_0)$ be a homotopy equivalence, with inverse $g$. Then, $fg\simeq1_Y$ and $gf\simeq1_X$

Lift $fp:\widetilde{X}\to Y$ to $F=\widetilde{fp}:\widetilde{X}\to \widetilde{Y}$. Since $\widetilde{X}$ is simply connected, lift exists. Again by lifting, $G=\widetilde{gq}:\widetilde{Y}\to\widetilde{X}$

Now, $qFG=q\widetilde{fp}\widetilde{gq}=fp\widetilde{gq}=fgq$. So $FG$ is the (unique) lift of $fgq:\widetilde{Y}\to Y$. But, $fgq\simeq1_Yq=q$ and by homotopy lifting, $FG\simeq \tilde{q}=1_\widetilde{Y}$

Similarly, $GF\simeq 1_\tilde{X}$. Thus, $F:\widetilde{X}\to\widetilde{Y}$ is a homotopy equivalence.

Answer 3

While Andrew Hanlon gives in his answer what seems to me to be the correct idea (the lift is determined up to a deck transformation), here is an amusing (probably circular) alternative, just for fun.

We know that there is a lift $\tilde{f} : \tilde{X} \to \tilde{Y}$. So $\tilde{f}$ is a continuous map between two spaces, and since the map $f$ induced isomorphisms on all higher homotopy groups, being a homotopy equivalence, and since a covering map induces isomorphisms on the higher homotopy groups, it follows from the simply-connectedness of universal covers that $\tilde{f}$ induces isomorphisms on all homotopy groups.

So in the case that $X$ and $Y$ are connected CW-complexes, it follows from Whiteheads theorem that $f$ is a homotopy equivalence.