A covering map is a homeomorphism

Question

Let $p : E \rightarrow X$ be a covering map. Suppose $E$ is path connected and that $\pi_1(X, b) = \{1\}$. Prove that $p$ is a homeomorphism.

I have no idea how to approach this problem.

Answer 1

As $p$ is already open, continuous and onto, we only need that $p$ is injective.

So suppose $p$ is not and we have $e_1 \neq e_2$ in $E$ with $p(e_1)=p(e_2)$. Let $x_0 = p(e_1)=p(e_2)$. Now appply 54.4 in Munkres (2nd ed p. 345) where it says that $\pi_1(X,x_0)$ has a surjection onto $p^{-1}[\{x_0\}]$ and note that we have an outright contradiction, as the former set has size $1$ by assumption and the fibre on the right has at least size $2$, containing $\{e_1,e_2\}$. So $p$ is indeed 1-1.