# Show that if B is simply-connected, then p is a homeomorphism.

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.

• I think this works fine! Commented Apr 15, 2014 at 20:51
• It works, but you'll want to be precise and say that $\psi$ is homotopic rel. endpoints to a constant path, for any path is freely homotopic to a constant path. Commented Apr 15, 2014 at 21:00
• There is a surjection from $\pi_1(B,b_0)$ to the fibre $p^{-1}(b_0)$, and so the fibre is a single point. Commented Jul 23, 2019 at 12:22

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 $$ψ(0)=ψ(1)$$.
• "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? Commented Jan 26, 2019 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 $a$ Commented Feb 2, 2019 at 17:59
• 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 Commented Feb 2, 2019 at 19:55
• @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. Commented Feb 2, 2019 at 21:49