Let $p : E \to B$ be a covering map with $E$ path connected and $B$
simply connected. Then $p$ is a homeomorphism.
My approach : Let $p(e_0)=b_0.$ Since $E$ is path connected, the lifting correspondence $\phi : \pi_1(B,b_0) \to p^{-1}(b_0)$ is surjective. But since $B$ is simply connected, $\pi_1(B,b_0)=\{e\}.$ Thus $|p^{-1}(b_0)|=1.$ Next, connectedness of $B$ implies that $|p^{-1}(b)|=1 \; \forall \; b \in B.$
Thus every $b \in B$ has an evenly covered neighborhood $U$ such that $p^{-1}(U)=V$ for some open neighborhood of $e_0$ and $p|_U : U \to V$ is a homeomorphism.
I think I am close to finishing the proof but I am stuck here. What is the idea to complete this proof?
Best Answer
the induced map $p_*:\pi_1(E) \to (B)$ will be injective, so $\pi_1(E)$ also needs to be trivial (making $E$ simply connected as well.)
Since $E$ is path connected, suppose that $|p^{-1}(b)|>1$, and take a path between two points: $\alpha:a \mapsto b$. $p \circ \alpha$ is a loop in $B$, $\dots$
Using the theorem in your question, you have injectivity. You should show that a local homeomorphism is open (maps open sets to open sets) and then you will have the result. See here for a hint.