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.
Best Answer
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.