# Homotopy equivalent spaces have homotopy equivalent universal covers

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.

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.

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.

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.

$p:(\widetilde{X},\tilde{x}_0)\to (X,x_0)$ 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.

• Ah! Somehow I assumed $X$, $Y$ are simply connected. – ChesterX Dec 24 '15 at 15:46
• @EricAuld, I've edited my answer. – ChesterX Dec 24 '15 at 19:44
• If you fix base-points, then the lifts are always unique. – ChesterX Dec 25 '15 at 7:36
• I agree that $1_{\tilde{Y}}$ is the unique lift of $q$ taking $\tilde{y}_0 \mapsto \tilde{y}_0$. Now when you lift the homotopy taking $fgq$ to $q$, how do you know that the final map is a map taking $\tilde{y}_0$ to $\tilde{y}_0$? – Eric Auld Dec 25 '15 at 19:52