Proof:

We first show that the action is well-defined. That is to say if $\gamma_1$ and $\gamma_2$ are homotopic loops based at $x_0$ then for ${\tilde x}\in p^{-1}(x_0)$

$${\tilde \gamma}_1(1) = {\tilde \gamma}_2(1),$$

where ${\tilde \gamma}_1$ and ${\tilde \gamma}_2$ are lifts of $\gamma_1$ and $\gamma_2$ starting at ${\tilde x}$. Well, if $F$ is the homotopy between $\gamma_1$ and $\gamma_2$ then $F$ has a unique lift ${\tilde F}$ satisfying ${\tilde F}(0, 0) = {\tilde x}$. In other words, ${\tilde F}: [0, 1]\times [0, 1]\longrightarrow {\tilde X}$ is the unique continuous map such that

$$p\circ {\tilde F} = F,\quad {\tilde F}(0, 0) = {\tilde x}$$

In particular the image set $\{{\tilde F}(s, 1)\}$ as $s$ runs through $[0, 1]$, must be a connected subset of ${\tilde X}$. But since $F$ is a homotopy of loops based at $x_0$,

$$F(s, 1) = p\circ {\tilde F}(s, 1) = x_0,$$    for all $$s \in [0, 1].$$

Hence $\{{\tilde F}(s, 1)/ s\in [0, 1]\} \subset p^{-1}(x_0)$ which means $\{{\tilde F}(s, 1)/ s\in [0, 1]\}$ is a singleton since $p^{-1}(x_0)$ is discrete. In particular,

$${\tilde F}(0, 1) = {\tilde F}(1, 1),$$    that is, $$\quad {\tilde \gamma}_1(1) = {\tilde \gamma}_2(1).$$

Next, we show that (15.1) defines a right group action. First let us note that if ${\tilde x}_1, {\tilde x}_2$ and ${\tilde x}_3$ are three points in $p^{-1}(x_0)$ and ${\tilde \gamma}_1$ and ${\tilde \gamma}_2$ is a pair of paths joining ${\tilde x}_1$ to ${\tilde x}_2$ and ${\tilde x}_2$ to ${\tilde x}_3$ respectively then

$$p\circ({\tilde \gamma}_1*{\tilde \gamma}_2) = (p\circ{\tilde \gamma}_1) * (p\circ{\tilde \gamma}_2).$$

Now let $\gamma_1$ and $\gamma_2$ be two loops in $X$ based at $x_0$. Assume that ${\tilde \gamma}_1$ is the unique lift of $\gamma_1$ starting at ${\tilde x}_1$ and ${\tilde \gamma}_2$ is the unique lift of $\gamma_2$ starting at the point ${\tilde x}_2 = {\tilde \gamma}_1(1)$ then the juxtaposition ${\tilde \gamma}_1*{\tilde \gamma}_2$ is defined and is the unique lift of $\gamma_1 * \gamma_2$ starting at ${\tilde x}_1$. Thus,

$${\tilde x}_1\cdot([\gamma_1][\gamma_2]) = {\tilde x}_1\cdot[\gamma_1*\gamma_2] = {\tilde \gamma_1}*{\tilde \gamma}_2(1) = {\tilde \gamma}_2(1)$$

On the other hand,

$${\tilde \gamma}_2(1) = {\tilde x}_2\cdot[\gamma_2] = ({\tilde x}_1\cdot [\gamma_1])\cdot [\gamma_2].$$

Note that if we had tried to operate from the left we would instead get an anti-action. This is one of the instances where it is important to have the book-keeping done correctly from the very outset.

Finally the constant loop $\varepsilon_{x_0}$ at $x_0$ lifts as the constant loop starting at ${\tilde x}_1\in p^{-1}(x_0)$ and so (17.1) implies

$${\tilde x}_1\cdot [{\varepsilon}_{x_0}] = {\tilde x}_1,\quad {\tilde x}_1\in p^{-1}(x_0).$$

We now examine the issues related to this group action namely, its transitivity and the stabilizer subgroups of various points of $p^{-1}(x_0)$.