Show that if $p: E \to B$ is a covering map and $E$ is path connected while $B$ is simply connected then $p$ is a homeomorphism. 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. 



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


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*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.

*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. $e_0 = e_1$.
