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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 $z=p(\tilde{z})$ 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.

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    $\begingroup$ This looks like the way to go to me... Note that at the very beginning of the two paragraphs starting with "Define (a map) $\widetilde{f}$/$\widetilde{g}$ ..." the maps should go between $\widetilde{X}$ and $\widetilde{Y}$ not $X$ and $Y$. You could add a word why those maps are well-defined (not dependent on the choices of paths). $\endgroup$
    – t.b.
    Apr 24, 2012 at 20:10
  • $\begingroup$ There was this question covering the same grounds and Ryan's hint amounts to what you're doing. $\endgroup$
    – t.b.
    Apr 24, 2012 at 20:13
  • $\begingroup$ @t.b.: Thanks, and sorry - I didn't realise the same question had been asked! (I'd searched for it but never found it.) $\endgroup$ Apr 24, 2012 at 21:50
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    $\begingroup$ @CliveNewstead Why are deck transformations homotopic to identity maps? $\endgroup$ Jun 21, 2014 at 1:45

1 Answer 1

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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):

Given a covering space $p : (\tilde{X},\tilde{x_0}) \to (X,x_0)$ and a map $f : (Y,y_0) \to (X,x_0)$ with $Y$ path connected and locally path-connected. Then a lift $\tilde{f} : (Y,y_0) \to (\tilde{X},\tilde{x_0})$ of $f$ exists iff $f_\ast(\pi_1(Y,y_0)) \subset p_\ast (\pi_1(\tilde{X},\tilde{x_0}))$.

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.

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  • $\begingroup$ This is now quite an old question (thanks for your reply nevertheless)! I would have been required to define $\tilde f$ though; an annoying thing about exams at my institution is that you can only assume results that are in the lecture notes and don't trivialise the proof. This result doesn't trivialise the proof, but it isn't in the lecture notes! $\endgroup$ Nov 8, 2012 at 0:26
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    $\begingroup$ ...I have since completed the problem, though. I'll post the answer when I get the time to. $\endgroup$ Nov 8, 2012 at 0:26
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    $\begingroup$ A pony would be nice. Thanks for your answer. $\endgroup$ Nov 8, 2012 at 1:36
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    $\begingroup$ This is an old question, but I wanted to add that I don't think the answer is correct. Uniqueness of lifts only applies when two lifts agree on at least one point. But I don't believe there is any requirement that $\tilde{F}(y,1)=y$ for any $y$. $\endgroup$
    – Anonymous
    Dec 19, 2019 at 2:45
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    $\begingroup$ Anonymous is correct, this proof has a non-trivial hole (unless I'm missing something). There is a correct solution on math.stackexchange.com/questions/3176802. It also puts light on why Hatcher suggests exercise 0.11 as a hint. $\endgroup$
    – Qi Zhu
    Dec 9, 2021 at 18:57

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