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I was trying to solve the following exercise in Hatcher (1.3.8). Let $p:(\tilde{X},\tilde{x})\to(X,x)$ and $q:(\tilde{Y},\tilde{y})\to(Y,y)$ simply-connected covering spaces. Assume $X,Y$ path-connected and locally path-connected spaces such that $X\simeq Y$. Then $\tilde{X}\simeq \tilde{Y}$. My thoughts:

Let $f:X\to Y$ be a homotopy equivalence, $x_0\in X$. Define $y_0:=f(x_0)$. We get the following diagram. enter image description here

There exists a unique lift of $f$ if $f_*(\pi_1(X))\leq p_*(\pi_1(Y))=\{*\}$ called $F:(X,x_0)\to (\tilde{Y},\tilde{y_0})$ with $f=q\circ F$. Define $\tilde{f}:\tilde{X}\to\tilde{Y}$ by $\tilde{f}=F\circ p$. I want to show that $\tilde{f}$ is a homotopoy equivalence. I don't see why $f_*(\pi_1(X))=\{*\}$. Any thoughts?


marked as duplicate by Paul Frost, Lee Mosher, Leucippus, Juniven, Lord Shark the Unknown Apr 7 at 3:12

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  • $\begingroup$ It's not the case that $f_*\pi_1(X) = \{*\}$. In fact, $f$ being a homotopy equivalence implies that $f_*$ is an isomorphism on $\pi_1$. You want to look at $(f\circ p)_*$ $\endgroup$ – Max Apr 6 at 9:28
  • $\begingroup$ But if you want to lift $fp$, why bother about (locally)-path connectedness of $X$, we only need $\tilde{X}$ to be locally path connected $\endgroup$ – Lucas Smits Apr 8 at 6:12
  • $\begingroup$ Local things are the same in $\tilde{X}$ and $X$ $\endgroup$ – Max Apr 8 at 6:16
  • $\begingroup$ Yeah so we just need it for locally path connectedness of $\tilde{X}$ $\endgroup$ – Lucas Smits Apr 8 at 6:20
  • $\begingroup$ And why did we need $X$ to be path connected?🤔 $\endgroup$ – Lucas Smits Apr 8 at 6:29

Since $\tilde{X}, \tilde{Y}$ are simply connected, you get lifts $\tilde{f} : \tilde{X} \to \tilde{Y}$ of $fp$ and $\tilde{g} : \tilde{Y} \to \tilde{X}$ of $gq$. We show that there exists $\phi : \tilde{Y} \to \tilde{X}$ such that $\phi \tilde{f} \simeq id$. The existence of $\psi : \tilde{X} \to \tilde{Y}$ such that $\tilde{f} \psi \simeq id$ can be shown similarly, and by exercise 0.11 (which is referenced by Hatcher) we shall be done.

Choose a homtopy $H : g f \simeq id$. Since $\tilde{g} \tilde{f}$ is a lift of $g f p$, we find a unique lift of $H(p \times id_I)$ to a homotopy $\tilde{H}$ such that $\tilde{H}_0 = \tilde{g} \tilde{f}$. Clearly $h = \tilde{H}_1$ is a lift of $p$, i.e $hp = p$. Let $\xi \in \tilde{X}$. We have $p(h(\xi)) = p(\xi)$, i.e. $\xi, h(\xi)$ are in the same fiber of $p$. Since a simply connected covering space is a normal covering space, we find a covering transformation $u : \tilde{X} \to \tilde{X}$ such that $u(h(\xi)) = \xi$. Because $u h$ and $id$ are lifts of $p$ which agree at $\xi$, we conclude $u h = id$. Define $\phi = u \tilde{g}$. Then $\phi \tilde{f} = u \tilde{g} \tilde{f} \simeq u h = id$.

  • $\begingroup$ Why do we assume $X$ to be path connected and locally path connected, we should assume $\tilde{X}$ to be locally path connected to fullfil the lifting criterion? $\endgroup$ – Lucas Smits Apr 8 at 6:09
  • $\begingroup$ @LucasSmits I see your point. The lifting theorem for maps $f : Z \to X$ makes assumptions on $Z$ and not on $X$. However, we apply it to maps defined on $\tilde{X}$ and $\tilde{Y}$. Covering projections are local homeomorphisms, therefore local path connectivity of the covering space $\tilde{X}$ is equivalent to local path connectivity of the base space $X$. But in fact the assumption that $X$ is path connected is unneccesary. If we assume that $\tilde{X}$ is simply connected, then $X$ must automatically be path connected. $\endgroup$ – Paul Frost Apr 8 at 9:04
  • $\begingroup$ Thank you, this is a very clear explanation $\endgroup$ – Lucas Smits Apr 8 at 10:30

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