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Let $p: E \rightarrow B$ be a covering map with $E$ path-connected. Show that if $B$ is simply-connected, then $p$ is a homeomorphism.

I'm checking to see if my solution is flawed.

Since $p$ is a covering map it is a continuous, surjective and open map. That means that all I need to do to show that $p$ is a homeomorphism is to show that it is injective.

So for $p(a) = p(b)$ for $a,b \in E$, we want to show that $a = b$.

Now, since $E$ is path-connected there exists a path from $a$ to $b$ denote it by $\psi$.

Then $p\circ\psi$ is a loop in $B$ since $p(a) = p(b)$.

But since $B$ is simply-connected $p\circ\psi$ is homotopic to a point.

So $\psi$ must be homotopic to a point when we lift it and therefore $a = b$.

So $p$ is injective and thus a homeomorphism.

Stefan Hamcke
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EgoKilla
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1 Answers1

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Your proof works fine, but you should say that $p\psi$ is homotopic rel endpoints or path homotopic to a constant path and, since a path homotopy lifts to a path homotopy, $ψ$ is path homotopic to a constant map, too, which implies $a=b$.
With little more effort, we can show that if a loop $\phi$ at $b_0$ in $B$ is in the image $p_*(\pi_1(E,e_0))$ (which is a subgroup of $\pi_1(B,b_0)$), then $ϕ$ lifts to a loop in $E$. For if $[ϕ]\in p_*(\pi_1(E,e_0))$, then there is a loop $\lambda$ at $e_0$ such that $pλ$ is path homotopic to $ϕ$, and this homotopy lifts to a path homotopy $λ\simeqψ$, where $ψ$ is the lift of $ϕ$ at $e_0$, which is thus a loop.
In particular this implies that a null-homotopic $ϕ=pψ$ lifts to a loop $ψ$ since $[ϕ]=0$ is always in $p_*(\pi_1(E,e_0))$. Therefore $ψ(0)=ψ(1)$.

Stefan Hamcke
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  • "since a path homotopy lifts to a path homotopy": Can you please explain this part? I know each path in $B$ has a lifting path in $E$, but what does a homotopy being lifted mean? And why does it necessarily mean that $\psi$ is path homotopic to a constant map and not to some other arbitrary map? – Dean Gurvitz Jan 26 '19 at 18:48
  • @DeanGurvitz: Just as a path in $B$ has a lifting path in $E$, a homotopy in $B$, which is a map $H:I\times I\to B$ has a lifting $\tilde H:I\times I\to E$ such that $p\tilde H=H$. If $H$ is a homotopy $ϕ≃0$, then $H(0,t)=ϕ(t)$ and $H(1,t)=b_0$. Then the lift $\tilde H$ satisfies $\tilde H(0,t)=ψ(t)$ and $\tilde H(1,t)=a$. Note that $\tilde H(1,t)$ is constant since it lifts $H(1,t)$ which is constant. And it must be equal to $a$ because $H(s,0)=ϕ(0)=p(a)$ since $H$ is fixed at the starting point $p(a)$, so the lift $\tilde H(s,0)$ must be fixed at $a$ – Stefan Hamcke Feb 02 '19 at 17:59
  • Or were you unaware of the fact that a homotopy can be lifted as well? It is a theorem in the theory of covering spaces and can be found in textbooks about Algebraic Topology, for example Hatcher's book @DeanGurvitz – Stefan Hamcke Feb 02 '19 at 18:05
  • I was aware but didn't put 1+1 together. However, I still don't understand why $\bar{H}(1,t)$ is constant, since we only know $p\bar{H}(1,t)$ is constant, which doesn't directly imply anything about $\bar{H}$ on its own – Dean Gurvitz Feb 02 '19 at 19:55
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    @DeanGurvitz. Actually, it does. Since $H(1,t)$ is constant, there is the constant path at $a$, which lifts $H(1,t)$. But the lift is unique, so the only lift for $H(1,t)$ (starting at $a$) is the constant path. – Stefan Hamcke Feb 02 '19 at 21:49
  • The uniqueness part of the lift is what I missed, thank you. Maybe you can add it to your answer as something that's worth mentioning in a proof (or at least be a little more explicit)? – Dean Gurvitz Feb 03 '19 at 07:08