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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

Miguel
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    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

3 Answers3

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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.
Jim
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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.

Xiang Yu
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Hoang Nguyen
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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.

Stefan Hamcke
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  • I don't understand what are you trying to say :S sorry for being so stupid – Miguel Feb 23 '13 at 03:22
  • 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
  • @Miguel Hey, where did your comment go?! – Stefan Hamcke Feb 23 '13 at 20:47
  • Sorry I deleted because I noticed what you were trying to say , but thanks for the new answer :D – Miguel Feb 23 '13 at 21:17