Why is a space homeomorphic to the orbit space of its universal cover by its deck transformations? Below is an excerpt from Hatcher's Algebraic Topology, page 72:

I don't understand why the part above highlighted in red is generally true. I can relate it to an example where it is true though: We can consider $S^1$ covered by $\mathbb{R}$. Then the deck transformations of $\mathbb{R}$ correspond to the action of $\mathbb{Z}$ on $\mathbb{R}$ via translation by an integer. Hence, indeed we have $\mathbb{R} / \mathbb{Z} \cong S^1$. But I don't know why the statement $\tilde{X} / G(\tilde{X}) \cong X$ is true in general.
 A: Here's the idea: given a "normal" covering space $\widetilde{X}\xrightarrow{p} X$, a deck transformation is an automorphism $\phi\in \operatorname{Aut}(\widetilde{X})$ that preserves fibres, so that $ p=p\circ \phi.$ It can be shown (and is in Hatcher) that given $x\in X$, and any pair of points $y,z\in p^{-1}(x)$ there exists a deck transformation $\phi \in G(\widetilde{X})$ with $\phi(y)=z$. That is, the action is transitive. 
So, the intuition here is that when we quotient by $G(\widetilde{X})$ action on  $\widetilde{X}$, we think of points as being "the same" if they can be related by the action of an element of $G(\widetilde{X})$. Over each $x$ there is a fibre $p^{-1}(x)$ which  corresponds to a single orbit $\mathcal{O}_x$ under the $G(\widetilde{X})-$action. After quotienting, we are left with the space $\{\mathcal{O}_x:x\in X\}$. You can check that its quotient topology of $X/G(\widetilde{X})$ is precisely the topology of $X$ so  that $\widetilde{X}/G(\widetilde{X})\cong X$ by the obvious map $\mathcal{O}_x\mapsto x$. (This isn't too hard to see if you leverage the local triviality of the covering space.)
