proving that a covering map with certain domain and range is homeomorphism

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

• Hint: you can also lift the homotopy, right? What does this mean when $t = 1$? Better hint: have you learned yet what covering spaces of a space correspond to? – Paul VanKoughnett Feb 22 '13 at 23:09

You are almost done. Note that since $p(f)$ is nullhomotopic, the class $[p(f)]$ is in the trivial image $p_*(\pi_1(E,e_0))$. The elements in the image of $p_*$ are classes of loops whose lifts at $e_0$ in $E$ are also loops, i.e. the lift $f$ of $p(f)$ is a loop.

• I don't understand what are you trying to say :S sorry for being so stupid – Miguel Feb 23 '13 at 3:22
• There is a lemma which says that the subgroup $p_*(\pi_1(E,e_0))$ of $\pi_1(B,b)$ consists of homotopy classes of loops whose lifts are also loops. More precisly, if $c$ is in $p_*(\pi_1(E,e_0))$ and the loop $\gamma$ is in $c$, i.e. $[\gamma]=c$, then the lift of $\gamma$ must also be a loop. In your case this implies that $f$ is a loop. – Stefan Hamcke Feb 23 '13 at 12:59
• Thanks for the answer, but I want to know , where I can see all that results? I'm studying the book topology Munkres, and I did not read nothing about that. – Miguel Feb 23 '13 at 16:26
• @Miguel. In Hatcher's Algebraic Topology, chapter 1.3 you will find a good treatment about covering spaces. – Stefan Hamcke Feb 23 '13 at 17:20
• @Miguel I understand your concern. Let's do this step by step: You assume that $e_0$ and $e_1$ have the same image $b$ and $f:e_0\rightsquigarrow e_1$. We could write $[f]$, the class of all paths homotopic to $f$ via a homotopy fixing the endpoints, but this $[f]$ is not in $\pi_1(E,e_0)$ if $e_0\neq e_1$. But we can compose $f$ with $p$ to get a path in $B$, and this path is a loop at $b$, so $[p(f)]\in\pi_1(B,b)$. But $\pi_1(B,b)=p_*(\pi_1(E,e_0))=0$, so $[p(f)]$ is in the image of $p_*$. The lemma then says that $f$ must be a loop because $f$ is a lift for $p(f)$. – Stefan Hamcke Feb 23 '13 at 20:42

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.

Well, you know that the number of sheets exactly equal index of subgroup $p_*(\pi(E))$ in $\pi(B)$, since $B$ is simply connected, it implies that the index is $1$, so $p$ is one-to-one.