I recently encountered the following:
Let $p:(E, e_0) \to (B, b_0)$ be a covering map. Assume that $p_∗(\pi_1(E, e_0)) \subseteq \pi_1(B, b_0)$ is a normal subgroup. If $e_1\in p^{−1}(\{b_0\})$, then $p_∗(\pi_1(E, e_1)) = p_∗(\pi_1(E, e_0))$. Additionally, the group of covering transformations of the covering space $p : E \to B$ acts transitively on $p^{−1}(b_0)$.
I am trying to prove this. Is it true that changing $e_0$ to $e_1$ corresponds to conjugating $p_∗(\pi_1(E, e_0))$ by an element in $\pi_1(B, b_0)$ (a loop that lifts to take $e_0$ the $e_1$)? If this is true, then $p_∗(\pi_1(E, e_1)) = p_∗(\pi_1(E, e_0))$. But I am having trouble figuring out if this is true or not. Any hints would be appreciated.