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I'm aware of this and this answer but I still don't seem to understand whether my proof is correct. Please point me to any errors or jumps in reasoning.

Pick $e_0 \in E$ and let $b_0 = p^{-1}(e_0)$. Since $E$ is path connected, $\pi_1(E, e_0) \simeq \pi_1(E, e)$ $\forall e\neq e_0$.

Consider the induced homomorphism $p_{*}: \pi_1(E, e_0) \to \pi_1(B, b_0)$. Let $[f]$,$[g]$$\in \pi_1(E, e_0)$ be such that $p_{*}([f]) = p_{*}([g])$. WTS $[f]=[g]$.

Since $p\circ f$ and $p\circ g$ are loops with the same start and end points in the simply connected space $B$, there exists a homotopy $H$ between $p\circ f$ and $p\circ g$. Then $G = p^{-1}\circ H$ is a homotopy between $f$ and $g$. Therefore $[f] = [g]$.

Since the induced $p_{*}$ is injective, $p$ is injective. $p$ is a covering map so it is also surjective and thus bijective and continuous. Since $p$ is also an open map, $p^{-1}$ is continuous. Therefore $p$ is a homeomorphism.

  • Was I correct in defining $G$ to be the path homotopy between $f$ and $g$?
  • My argument that $p$ is injective because $p_{*}$ is injective stems from $p_{*}([f]) = [p\circ f]$. Is this true?
  • Did I need to say that $\pi_1(E, e_0) \simeq \pi_1(E, e)$ $\forall e\neq e_0$ and therefore the choice of base point is irrelevant?

Edit: Thanks to everyone comments, I can see where I was going wrong.. Here is my corrected version of the proof

$p$ is a covering map so it is continuous and surjective. We need to show it is injective. The continuous inverse follows because an open bijection has a continuous inverse.

If $p$ is injective, then $\forall b \in B$, $|p^{-1}(b)|=1$. Suppose $\exists b_0$ such that $e_0,e_1 \in p^{-1}(b_0)$. Since $E$ is path connected, there exists a path $\gamma$ between $e_0$ and $e_1$.Then $p\circ \gamma$ is a loop in B starting and ending at $b_0$.

Since $B$ is simply connected, this loop is homotopic to the constant map at $b_0$. Call this homotopy $H$, then $H(0, 0) = b_0$. Then there exists a unique lifting of $H$, to a homotopy in E, call it $\tilde{H}$ such that $\tilde{H}(0,t)=e_0$. But then $\tilde{H}(1, t) = e_1$ which is a contradiction.

Amin Sammara
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  • It's convenient to explain independence of (isomorphism class of) the fundamental group because of path-connectedness, but you never bothered to use it later, so it's probably extraneous. It turns out that for any covering map, the induced map on fundamental groups is always injective, which your argument showed. But not every covering map is injective, so you'll need more work to get injectivity of $p$ (the second answer you linked should help for that). – kamills Dec 12 '19 at 17:43
  • Since when does injectivity of $p_*$ imply injectivity of $p$ ? – Maxime Ramzi Dec 12 '19 at 18:30
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    Thank you for the tip. I was able to see why the injectivity of $p_$ does not imply the injectivity of $p$. Take any path $\gamma$ in a contractible space, this path is homotopic to a single point. But since $\pi_1({I})$ is trivial, $\gamma_{}$ is injective while $\gamma$ is most certainly not. – Amin Sammara Dec 13 '19 at 17:08

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Other basepoints than $e_0$ are irrelevant. Here are two problems in your proof.

  1. You say that $G = p^{-1} \circ H$ is a homotopy. But there is no map $p^{-1} : B \to E$ unless you know that $p$ is a homeomorphism. That is what you want to show, so you cannot use it here.

  2. You say that $p_*$ injective implies $p$ injective. This is not true. $p_*$ is always injective, but no covering map which is not a homeomorphism is injective.

Edited: Your recent proof is essentially correct, but I think it should be more detailed. Here is a suggestion:

You start with a path $\gamma : [0,1] \to E$ such that $p(0) = e_0, p(1) = e_1$. Then $p \circ \gamma$ is a loop in $B$ beginning and ending at $b_0$. Since $B$ is simply connected, there exists a homotopy $H : [0,1] \times [0,1] \to B$ such that $H(x,0) = (p \circ \gamma)(x)$, $H(x,1) = b_0$ for all $x$ and $H(i,t) = b_0$ for $i=0,1$ and all $t$. Thus with $R = [0,1] \times \{1\} \cup \{0,1\} \times [0,1]$ we have $H(R) = \{b_0\}$.

There exists a unique lift $\overline{H} : [0,1] \times [0,1] \to E$ such that $\overline{H}(x,0) = \gamma(x)$ for all $x$. Then $\overline{H}(R) \subset p^{-1}(b_0)$. The fiber $p^{-1}(b_0)$ is discrete and $R$ is connected (in fact, $R \approx [0,1]$), therefore $\overline{H}(R) = \{e\}$ for some $e \in p^{-1}(b_0)$. But now $e_i = \gamma(i) = \overline{H}(i,0) = e$, i.e. $e_0 = e_1$.

Paul Frost
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