# Universal covering spaces of homotopy equivalent spaces are homotopy equivalent

I'm working on an exercise from Hatcher's Algebraic Topology (exercise 1.3.8):

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 $\tilde X \simeq \tilde Y$.

The exercise recommends using the following result from earlier in the textbook:

A map $f: X \to Y$ is a homotopy equivalence if there exist maps $g, h: Y \to X$ such that $fg \simeq \mathbb{1}$ and $hf \simeq \mathbb{1}$.

This problem has been solved here and also here, but both proofs include a step that seems to be a bit of a jump. It's possible I'm missing a very simple argument, but this is the one piece I'm having trouble with.

Here's what I have so far: Let $p : \tilde X \to X$ and $q: \tilde Y \to Y$ be the respective covering maps, and let $f : X \to Y$ and $g : Y \to X$ be homotopy inverses (so $gf \simeq \mathbb{1}_X$ and $fg \simeq \mathbb{1}_Y$). Consider the map $fp : \tilde X \to Y$, and the induced homomorphism $(fp)_*: \pi_1(\tilde X) \to \pi_1(Y)$. Since $\tilde X$ and $\tilde Y$ are simply-connected, $(fp)_*(\pi_1(\tilde X)) = 0 = q_*(\pi_1(\tilde Y))$, so by the lifting criterion (prop. 1.33 in Hatcher), we have a lift $\tilde f : \tilde X \to \tilde Y$ of $fp : \tilde X \to Y$. We similarly have a lift $\tilde g : \tilde Y \to \tilde X$ of $gq: \tilde Y \to X$.

This gives us that $p\tilde g \tilde f = gq\tilde f = gfp \simeq p$. Call this homotopy $h_t : \tilde X \to X$, so that $h_0 = p\tilde g \tilde f$ and $h_1 = p$. By the homotopy lifting property, since $\tilde g \tilde f : \tilde X \to \tilde X$ lifts $p \tilde g \tilde f$, there is a homotopy $\tilde h_t$ between $\tilde g \tilde f$ and a lift $\tilde p$ of the covering map $p : \tilde X \to X$. In particular, $p\tilde p = p$.

Here's my problem: In the first of the links I shared, they claim that $\tilde p$ is a deck transformation of $\tilde X$ (ie a homeomorphism with the property that $p \tilde p = p$). In the second of the links, they claim something stronger: that by uniqueness of lifts, in fact $\tilde p = \mathbb{1}_{\tilde X}$. The trouble with the first of these answers is that it's not clear to me that $\tilde p : \tilde X \to \tilde X$ is a homeomorphism. The trouble with the second is that uniqueness of lifts only applies if we can prove that $\tilde p$ fixes a point of $\tilde X$, which is also not clear. (Plus $\tilde p = \mathbb{1}_\tilde X$ seems like too much to ask for because then one could similarly show that $\tilde q = \mathbb{1}_{\tilde Y}$, which gives that $\tilde g \tilde f \simeq \mathbb{1}_\tilde{X}$ and $\tilde f \tilde g \simeq \mathbb{1}_\tilde{Y}$, and we are done without using the recommended hint.)

The proof can easily be completed after this is taken care of, but I'm having trouble justifying this. Is there a good reason $\tilde p : \tilde X \to \tilde X$ is a homeomorphism, or better, the identity map of $\tilde X$? Thanks!

You'll be able to understand this more clearly if you keep track of base points. Once that's done, they key is to apply, over and over, the uniqueness properties of lifting theory, which you use to tell you that this map is equal to that map.

Let's fix base points $x \in X$ and a lifted base point $\tilde x \in X$. Let $\tilde x' = \tilde p(\tilde x)$, also a lift of $x$.

The map $\tilde p : \tilde X \to \tilde X$ is the unique map (by covering theory) which is a lift of $p : \tilde X \to X$ such that $\tilde p(x)=x'$.

There also exists (by covering theory) a unique map $\tilde p_1 : \tilde X \to \tilde X$ which the lift of the map $p : \tilde X \to \tilde X$ such that $\tilde p_1(x')=x$.

The composition $\tilde p_1 \tilde p : \tilde X \to \tilde X$ is a lift of the map $p$ such that $\tilde p_1 \tilde p(x)=x$. The identity map $\mathbb{1}_{\tilde X}$ satisfies the same property: it is a lift of the map $p$ such that $\mathbb{1}_{\tilde X}(x)=x$. The map characterized by this property is unique (again by covering theory) and so $\tilde p_1 \tilde p = \mathbb{1}_{\tilde X}$.

A similar argument shows that $\tilde p \tilde p_1 = \mathbb{1}_{\tilde X}$.

Thus, $\tilde p$ and $\tilde p_1$ are inverses of each other, hence each is a homeomorphism.

Now, as to the issue of showing that $\tilde p = \mathbb{1}_{\tilde X}$, you've not included enough information for that purpose. The lifting lemma guarantees uniqueness of a lift for a given point and a given image of that point. And, as you have reported the proof, that information is not given. If you could structure the proof with specifically named points and specifically named images of those points, and if you can do this so that $\tilde p$ fixes some point of $\tilde X$, then uniqueness of lifting would guarantee that $\tilde p = \mathbb{1}_{\tilde X}$.

• Thank you! This solves my problem perfectly. Yes, I was skeptical of the argument that $\tilde p = \mathbb{1}_{\tilde X}$ for that same reason. Perhaps it could be proven if we assume that, if $x_0 \in X$ and $y_0 \in Y$ are basepoints, then $f(x_0) = y_0$ and $g(y_0) = x_0$. I'd need to think about it some more. – D Ford Dec 23 '16 at 19:52

So perhaps there are some technical considerations I'm overlooking, but here's how I would think about this:

Let $x\in X$ be any point. For any connected cover $X'\rightarrow X$, if one chooses a point in the fiber $x'\in p^{-1}(x)$, the map $p$ induces a homomorphism: $$p_* : \pi_1(X',x')\rightarrow \pi_1(X,x)$$ By the homotopy lifting property of covering spaces, this map is an injection, and hence (after checking a few more details) we obtain a bijection between isomorphism classes of covering spaces of $X$ and conjugacy classes of subgroups of $\pi_1(X,x)$ (where different choices of base points $x'\in p^{-1}(x)$ yield conjugate subgroups). This is just Theorem 1.38 of Hatcher.

Now, if $X' = \tilde{X}$ is simply connected, then for any choice of basepoint $\tilde{x}\in\tilde{X}$, $\pi_1(\tilde{X},\tilde{x}) = 1$, and hence via the above correspondence, it corresponds to the trivial subgroup of $\pi_1(X,x)$. Since the trivial subgroup is unique, the above correspondence implies that any two simply connected covering spaces of $X$ are isomorphic.

Now, if $f : X\rightarrow Y$ is a covering map (for example, an isomorphism), then we obtain two simply connected covering spaces of $Y$, namely $$\tilde{X}\rightarrow X\stackrel{f}{\rightarrow}Y,\qquad\text{and}\qquad \tilde{Y}\rightarrow Y$$ Since $\tilde{X},\tilde{Y}$ are both simply connected, the above discussion implies that they are isomorphic as coverings of $Y$, and hence a fortiori isomorphic as topological spaces.

Hope that helps.