Proposition $4.3$ of Chapter $8$ of Riemannian Geometry by Do Carmo I never did a course on algebraic topology, so I would like to a reference to understand why $p$ is a regular covering and why this implies $p(\tilde{x}) = p(\tilde{y})$ if and only if $\Gamma \tilde{x} = \Gamma \tilde {y}$ in the proof below

Thanks in advance!
 A: I understood why $p$ is a regular covering and why this implies $p(\tilde{x}) = p(\tilde{y})$ if and only if $\Gamma \tilde{x} = \Gamma \tilde {y}$.
I will answer considering that the people that will see this answer didn't a course in algebraic topology like me.
First of all, regular covering is also known as normal covering in the literature.
Let $\mathcal{C}_p(\tilde{X})$ denote the set of all covering transformations of $\tilde{X}$ with respect to $p$.
The proposition below answer why $p(\tilde{x}) = p(\tilde{y})$ if and only if $\Gamma \tilde{x} = \Gamma \tilde {y}$ once that $p$ is a regular covering:

$\textbf{Proposition $11.29$ (Orbit Criterion).}$ Let $p: \tilde{X} \longrightarrow X$ be a covering map.
(b) $\mathcal{C}_p(\tilde{X})$ acts transitively on each fiber if and only if $p$ is a normal covering.

A proof of this result can be found in 'Introduction to Topological Manifolds' by John Lee.
Observing that $\tilde{M}$ is simply connected in the proposition $4.3$ because $\tilde{M} = S^n, \mathbb{R}^n$ or $H^n$ by hypothesis, $\mathcal{C}_p(\tilde{X}) \cong \pi_1(X,q)$, where $q = p(\tilde{q})$ (see corollary $11.32$ of John Lee's book). Then $p$ be a regular covering is a consequence from the fact that every connected manifold is path-connected and from the result below

Suppose $p: Y \longrightarrow X$ is a covering space with $X$ path-connected, and let $G = \pi_1(X, x_0)$. Then, the $G$–set $M = p^{−1}(x_0)$ is transitive if and only if $Y$ is path-connected.

See Lemma $0.57$ of this lecture notes for a proof.
