I have a post about this here Sufficient condition for subgroup being normal but it has no answer, I would like to know how to solve this problem.
I have thought that I can use the following theorem:
Taking into account the theorem, if we assume that $p_*(\pi_1(E,e_0))$ is a normal subgroup of $\pi_1(B, b_0)$, it is sufficient to show that given $e_1, e_2\in p^{-1}(b_0)$ then $p_*(\pi_1(E,e_1))=p_*(\pi_1(E,e_2))$, to do this let's take $[\alpha]\in p_*(\pi_1(E,e_1))$, let's see that $[\alpha]\in p_*(\pi_1(E,e_2))$, as $[\alpha]\in p_*(\pi_1(E,e_1))$ then $[p\circ \gamma]=[\alpha]$ for some loop $\gamma$ in $E$ based on $e_1$ , I want to prove that there is a loop $\delta$ in $E$ based on $e_2$ such that $[p\circ \delta]=[\alpha]$ but I do not know how to do this, could someone help me please? I've thought about solving this problem by taking a path $\beta$ in $E$ from $e_2$ to $e_1$ since $E$ is arc-connected, so $[\beta*\gamma*\bar{\beta}]$ is a loop in $E$ based on $e_2$, but I do not know what else to do.
How can I do the other direction? Thank you very much.