Proof:

Let $\tilde x_1$ and $\tilde x_2$ be two arbitrary points of $p^{-1}(x_0)$. The action of Deck $(\widetilde X, X)$ is transitive on $p^{-1}(x_0)$ if and only if there is a $\phi \in$Deck $(\widetilde X, X)$ carrying $\tilde x_1$ to $\tilde x_2$, which is the case if and only if (19.1) holds. This in turn implies that the conjugacy class

$$\Big\{ p_{*}\pi_1(\widetilde X, \tilde x_0) : \tilde x_0 \in p^{-1}(x_0) \Big\}$$

reduces to a singleton and conversely, in other words, if and only if the covering is regular. $\square$

We now relate the (perhaps intransitive) action of Deck $(\widetilde X, X)$ on $p^{-1}(x_0)$ with the transitive action of $\pi_1(X, x_0)$ on $p^{-1}(x_0)$. Pick $\phi\in\;$Deck $(\widetilde X, X)$ and $\phi(\tilde x_1) = \tilde x_2$. Then on the one hand (19.1) must hold while since $p{*}\pi_1(\widetilde X, \tilde x_1) = \;$stab $\;\tilde x_1$ (for the action of $\pi_1(X, x_0)$), we have on the other hand

   stab$$\;\tilde x_1 =$$   stab$$\;\tilde x_2 = g($$stab$$\;\tilde x_1)g^{-1}, \eqno(19.2)$$

for some $g \in \pi_1(X, x_0)$. In fact (19.2) states that $g$ belongs to the normalizer

$$N($$stab$$\;\tilde x_1) = N(p{*}(\pi_1(\widetilde X, \tilde x_1)) \subset \pi_1(X, x_0).$$

This suggests that we must relate $\phi$ to the element $g\in N(p{*}(\pi_1(\widetilde X, \tilde x_1))$. However since there may be several such elements $g$ it is expedient to define the map in the opposite direction.

Let $g \in N(p{*}(\pi_1(\widetilde X, \tilde x_1)) \subset \pi_1(X, x_0)$ and $\tilde x_1 \cdot g = \tilde x_2$. Then (19.1) holds since $g$ is in the normalizer of stab $\;\tilde x_1$. There is a unique $\phi_g\in\;$Deck $(\widetilde X, X)$ such that $\phi_g(\tilde x_1) = \tilde x_2 = \tilde x_1 \cdot g.$ The map

$$\psi \;: \;N(p{*}(\pi_1(\widetilde X, \tilde x_1))\longrightarrow$$   Deck$$(\widetilde X, X),\quad g\mapsto \phi_g \eqno(19.3)$$

is a homomorphism. To see that it is surjective, let $\phi\in\;$Deck $(\widetilde X, X)$. There is a $g \in \pi_1(X, x_0)$ such that

$$\tilde x_1 \cdot g = \phi(\tilde x_1)$$

then stab $\;\tilde x_1$ and stab $\;\phi(\tilde x_1)$ are conjugate by $g$ but they are also equal by (iii) of Theorem (19.1), whereby we conclude $g$ is in the normalizer $N(p{*}(\pi_1(\widetilde X, \tilde x_1))$ and $\phi = \phi_g$. To determine the kernel of $\psi$, observe that $\phi_g =$   id if and only if

$$\phi_g(\tilde x_1) = \tilde x_1 \cdot g$$

that is, if and only if $g\in\;$stab $\;\tilde x_1$. But stab $\;\tilde x_1 = p{*}(\pi_1(\widetilde X, \tilde x_1))$. Summarizing these observations,