# Homotopy equivalence of covering spaces [duplicate]

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.

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 UnknownApr 7 at 3:12

• 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 $(f\circ p)_*$ – Max Apr 6 at 9:28
• 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 – Lucas Smits Apr 8 at 6:12
• Local things are the same in $\tilde{X}$ and $X$ – Max Apr 8 at 6:16
• Yeah so we just need it for locally path connectedness of $\tilde{X}$ – Lucas Smits Apr 8 at 6:20
• And why did we need $X$ to be path connected?🤔 – 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 $$\phi \tilde{f} = u \tilde{g} \tilde{f} \simeq u h = id$$.
• 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? – Lucas Smits Apr 8 at 6:09
• @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. – Paul Frost Apr 8 at 9:04