Lemma 81.1 in Munkres In the proof of Munkres, which is given in the picture below, I don't understand why we need to show that $h,\ h(e_0)=e_1$ exists iff $[\alpha]\in N(H_0)$. So I tried to organize the proof in the way I see it, but I got stuck. Here is my attempt.
The statement we need to prove is $Im(\Psi)=\Phi(N(H_0)/H_0)$. So we need to show 2 inclusions.
First, suppose $e_1\in Im(\Psi)$. Then there is a covering transformation $h:E\to E$ with $h(e_0)=e_1$. We need to show that there exists $[\beta]\in N(H_0)$ such that $\Phi([\beta]H_0)=e_1$. By the definition of $\Phi$, the LHS of the last equation is the endpoint of the unique lift of $\beta$ starting at $e_0$. So at least we must find a loop in $B$ based at $b_0$ whose lift starting at $e_0$ ends at $e_1$ (and then we could try to prove that the homotopy class of such loop lies in $N(H_0)$). I'm not sure how to apply Lemma 74.1. Should I apply it with $Y=I$? But then the inclusion stated there always holds because $\pi_1$ of $I$ is trivial. How to finish this part?
Conversely, suppose $e_1\in \Phi(N(H_0)/H_0)$. This means that there is $[\beta]\in N(H_0)$ with $\Phi([\beta]H_0)=e_1$, i.e., there is a loop $\beta$ based at $b_0$ in $B $ s.t. its lift starting at $e_0$ ends at $e_1$ and its homotopy class lies in $N(H_0)$. We need to show that there exists a covering transformation $h: E\to E$ s.t. $h(e_0)=e_1$. By Theorem 79.2, such a map exists iff $p_\ast \pi_1(E,e_0)=p_\ast\pi_1 (E,e_1)$. Why is this so?

 A: Let me try and attempt your questions in order.
Given $e_0, e_1 \in F$, there is always a loop $[\beta] \in \pi_1(B,b)$ with a lift $\tilde \beta$ with $\tilde \beta(0) = e_0$ and $\tilde \beta(1) = e_1$. This is because $E$ is path-connected, so such a $\tilde \beta$ exists, and then we may set $\beta = p\circ \tilde\beta$. The issue is to determine in what subgroup $[\beta]$ might lie in if there is a deck transformation $h$ taking $e_0$ to $e_1$. We don't need Lemma 79.1 for this direction: if $h$ exists, then $h$ is a lift of the identity map on $B$, i.e. $p \circ h = p$. But this implies that
$$ p_\ast (\pi_1(E, e_0)) = (p\circ h)_\ast (\pi_1(E,e_0)) = p_\ast (\pi_1(E,e_1)).$$ 
But since $\tilde \beta$ is a path from $e_0$ to $e_1$, we have
$$ \pi_1(E,e_0) = \tilde\beta^{-1} \ast \pi_1(E,e_1) 
 \ast\tilde\beta$$
and thus 
$$p_\ast(\pi_1(E,e_0))= [\beta]^{-1} \ast p_\ast(\pi_1(E,e_0)) \ast [\beta]^{-1},$$
i.e. $[\beta]$ normalizes $p_\ast(\pi_1(E,e_0))$, as desired.
Lemma 79.1 tells us that everything above goes through backwards: if $[\beta]$ normalizes $p_\ast(\pi_1(E,e_0))$, then if lifts to endpoint $e_1$, we have $
p_\ast(\pi_1(E,e_1)) = p_\ast(\pi_1(E,e_0))$ and so Lemma 79.1 gives that $p: E \to B$ can be lifted to a map $h: E \to E$ such that $h(e_0) = e_1$. Such a lift is a homeomorphism, since a lift $k: E \to E$ of $p: E \to B$ such that $k(e_1) = e_0$ must be an inverse by the uniqueness of Lemma 79.1. As $p \circ h = p$ by definition, $h$ is a deck transformation.
All of this is hidden inside Munkres' proofs of Lemma 79.1, Theorem 79.2, and Lemma 79.3.
