If I have a group $G$ acting on a topological space $X$, and this action is free (no fixed points), what can I say about the orbit space $X/G$?
I am aware that freeness plus proper discontinuity leads to a result regarding quotient maps and covering spaces. This is not what my question is about.
My particular inquiry has to do with arXiv:1701.02293. The set of points of $M$ that belong to a flow trajectory of $−$grad$(f)$ that join $p$ to $q$, $M(f ; p, q)$, admits a natural action of $\mathbb{R}$ (by translations). On p.6-7 Pedroza says "The action is in fact free and the orbit space of this action is denoted by $\hat{M}(f ; p, q)$. Hence $\hat{M}(f ; p, q)$ is identified as the space of trajectories that joint $p$ to $q$."
But this has generated confusion for me. It seems that Pedroza is saying that freeness implies that we can make some sort of identification, but I do not see what this identification is or why we can say that.
Is it true in general that when an action is free, $X$ can be identified (homotopy equivalent, homeomorphic?) with $X/G$? I doubt this is true.
If not, what is a sufficient condition for $X$ to be, say, homotopy equivalent with its orbit space under some action?
In the notation of the article, the quotient space $\hat{M}(f;p,q)$ is the space of trajectories in the sense that a point of it corresponds to a unique trajectory in $M(f;p,q)\subset M$, concretely , its pre-image under the quotient map.