(2.48) We will write down a surjective homomorphism $\phi\colon N_H\to Deck(Y,p)$ whose kernel is $H$. The first isomorphism theorem will then imply that $Deck(Y,p)\cong N_H/H$.

(3.57) We define $\phi$ as follows. Given an element $\beta\in N_H$ we have

$$\beta p_*\pi_1(Y,y)\beta^{-1}=p_*\pi_1(Y,y),$$ and we also have $\beta p_*\pi_1(Y,y)\beta^{-1}=p_*\pi_1(Y,\sigma_\beta(y))$, so by the existence and uniqueness results for deck transformations, there exists a unique deck transformation $F_\beta\colon Y\to Y$ such that $F_\beta(y)=\sigma_\beta(y)$. We define $\phi(\beta)=F_\beta$.

(6.00) We need to check that $\phi$ is a homomorphism, that $\phi$ is surjective, and that $\ker\phi=H$.

To show that $\phi$ is a homomorphism:

$$\begin{align*} F_{\beta_1}(F_{\beta_2}(y))&=F_{\beta_1}(\sigma_{\beta_2}(y))\\ &=\sigma_{\beta_1}\sigma_{\beta_2}(y)\\ &=\sigma_{\beta_1\cdot\beta_2}(y)\\ &=F_{\beta_1\cdot\beta_2}(y), \end{align*}$$ where we used that the monodromy $\beta\mapsto\sigma_\beta$ is a homomorphism.

(7.46) To see that $\phi$ is surjective, given a deck transformation $F$ we want to find $\beta\in N_H$ such that $F=F_{\beta}$. Let $\alpha$ be a path in $Y$ from $y$ to $F(y)$. Let $\beta=p\circ\alpha$. This $\beta$ is a loop in $X$ because $p(y)=p(F(y))=x$ and $F(y)=\sigma_\beta(y)$ by definition of monodromy. Therefore $F=F_\beta$, because these covering transformations agree at $y$.

(10.38) To see that $\ker\phi=H$, suppose that $\beta\in\ker\phi$ (so that $F_\beta=id_Y$). Therefore $y=F_\beta(y)=\sigma_\beta(y)$, so $y$ is fixed by the monodromy around $\beta$. This means that the unique lift of $\beta$ starting at $y$ is a loop $\tilde{\beta}$, so $[\beta]=p_*[\tilde{\beta}]\in p_*\pi_1(Y,y)$. This shows that $\ker\phi\subset p_*\pi_1(Y,y)$. The inclusion $p_*\pi_1(Y,y)\subset\ker\phi$ is an exercise.