General Topology – Proving that a Covering Map with Certain Domain and Range is Homeomorphism

Question

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

Well I don't know exactly what can I do here, maybe I have to start with this:

A covering map is surjective and is a continuous and open map. So it remains to prove that it is injective.
If I suppose that there exist $b\in B$ that have at least two preimages $e_0,e_1$. Let $U$ be a neighborhood of $b$ that is evenly covered by $p$, let's call $ V_0,V_1$ the slices of $p^{-1} (U)$ that contains $e_0,e_1$ respectively. Since $E$ is path connected there exist a path $ f:I\to E$ such that $f(0)=e_0 , f(1)=e_1 $ . If I consider the new path $ p(f): I\to B$ note that this map is a loop based on $b$, and since $B$ is simply connected $p(f)$ is homotopic to the constant map $ e_b :I\to B$ given by $e_b(t)=b \forall t\in I$.

I don't know what else I can do :S

Answer 1

The facts you need are:

1. For every covering map $p\colon E \to B$ and point $e \in E$ the map $p^\ast\colon\pi_1(E, e) \to \pi_1(B, p(e))$ induced on fundamental groups is injective.
2. If $p\colon E \to B$ is a covering map and $E$ is simply connected then each fiber of $p$ has the same cardinality as the fundamental group of $B$.
3. Every injective covering map is an isomorphism.