Normal subgroup of $\pi_{1}(B,b_{0})$

Question

Let $p : E \to B$ be a covering map ; let $p(e_{0})=b_{0}$ . Show that $p_{*}(\pi_{1}(E,e_{0}))$ is a normal subgroup of $\pi_{1}(B,b_{0})$ if and only if every pair of point $e_{1},e_{2}$ of $p^{-1}(b_{0})$ there is an equivalence $h : E \to E$ with $h(e_{1})=e_{2}$

In this post , the symbol $f_{*}$ mean , given an continuous map $f : X \to Y$ then $f_{*} : \pi(X) \to \pi(Y)$ and $f_{*}([h]) = [f \cdot h]$ ( $(f \cdot h)(x) = f(h(x))$ ). We call two covering map :

$$p : E \to B$$ $$p' : E' \to B$$

is equivalence if exist a homeomorphism lifitng $h$ such that :

$$h : E' \to E$$

$$p \cdot h = p'$$

Answer 1

Hint 1: Given a covering space $(E,e_0) \to (B, b_0)$ and a map $f: (Y, y_0) \to (B, b_0)$ with $Y$ path connected and locally path connected, there exists a lift $\tilde f: (Y, y_0) \to (E, e_0)$ with $p \circ \tilde f = f$ iff $f_\star (\pi_1 (Y, y_0)) \subset p_\star (\pi_1 (E, e_0))$. [Hatcher pages 61-62]

Hint 2: Let $\gamma$ be any path in $E$ between your two chosen basepoints $e_1, e_2$. Let $g = p\circ \gamma$ be the projection of this path down to $B$; this $g$ can be thought of as a loop in $\pi_1(B, b_0)$. Then $p_\star(\pi_1(E,e_1)) = g^{-1} p_\star (\pi_1(E,e_2)) g$. [Hatcher, pages 67-68]

Does this help?