Path lifting property for $\pi:M\rightarrow M/G$ for $G$ compact Lie acting smoothly and freely Let $M$ be a smooth manifold and let $G$ be a compact Lie group acting smoothly and freely over $M$. Let $\pi:M\rightarrow M/G$ be the canonical projection, and endow $M/G$ with the unique differentiable structure such that it is a smooth manifold and $\pi$ is a surjective submersion.
I would like to prove that, in the previous situation, the map $\pi$ satisfies the path lifting property, in the sense that for each smooth path $\gamma:I\rightarrow M/G$ defined in a compact interval $I$, there exists a smooth lift $\tilde\gamma:I\rightarrow M$ with $\pi\circ\tilde\gamma=\gamma$.
This is an exercise from the book Mathematical Gauge Theory from Mark J. D. Hamilton. It appears in a chapter on group actions, preceding those about fiber bundles, connections, curvature, etc. So apparently this can be proved attending only to the basic topological properties of the objects at hand.
In this SE post the same question is asked; however, the answer given is not complete. At first, this is what I thought the answer might look like; since surjective submersions admit local sections, we can first take an open cover of the trace of $\gamma$ with open sets for which we can find sections of $\pi$, then we take a finite subcover due to the compactness of the trace, and then we patch toghether the images of the pieces of the curve via the aforamentiones sections, using the fact that, by construction, $G$ acts transitively on the fibers of $\pi$. However, this approach is not quite perfect, for there might be problems with differentiability at the points where we patch the pieces toghether. Unfortunately, it doesn't look like this problem can be remedied.
I am out of ideas, so I would really appreciate any kind of help at this point.
 A: Bredon [Bre72,Theorem II.6.2] shows that for any Hausdorff space $X$ with the action of a compact Lie group $K$, the natural map $X \to X/K$ has the path-lifting property.
This follows since there is a slice at each point.  Bredon attributes this to Montgomery and Yang (The existence of a slice. Ann. of Math. 65 (1957), 108-116).
The general philosophy is that whenever there are slices, there is path-lifting.
Addendum (after it was pointed out I misread the question):
So the above result applies to the situation when $X$ is a manifold and $K$ is acting smoothly and freely to produce a lifted path.  But that path may not be smooth a priori.
However, unless I am missing something (I often am, and so please forgive me for that), the proof (I just re-read) in Bredon can be adapted line-by-line to show the resulting path can be taken to be smooth if $K$ acts smoothly and freely on a smooth manifold $X$.
Let $f:I\to X/K$ be a smooth path.  $X/K$ is smooth since $K$ is acting freely and properly (since $K$ is compact).  Let $\pi:X\to X/K$ be the projection (which is smooth).  There are smooth slices (Koszul, 1953) and so there exists a smooth slice $\sigma:I\to f^*X$, which gives your desired smooth path through the pull-back diagram.

[Bre72] Bredon, Glen E. Introduction to compact transformation groups. Pure and Applied Mathematics, 46. Academic Press, New York, 1972.
