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.
Best Answer
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 $ψ(0)=ψ(1)$.