Let $G$ be a compact Lie group acting on a topological space $X$.
I need to show that the orbit map $p:X\rightarrow X/G$ has the path lifting property.
Here is the proof -
Let $f:I\rightarrow X/G$ be a path in $X/G$. Let $G$ act trivially on $I$.
Consider $f^*X$ , the pullback of $X$ via $f$. That is $f^*X=\{(x,y)\in X\times I : p(x)=f(y)\}$. Let $p':f^*X\rightarrow X$ be the projection to first co-ordinate.
Then $f^*X$ has a natural $G$ action and the orbit space $(f^*X)/G\cong I$. Thus we have a section $\sigma:I\rightarrow f^*X$ (a section for the projection $p:X\rightarrow X/G$ is a continuous map $\sigma:X/G\rightarrow X$ such that $p\circ \sigma =id_{X/G}$)
So the map $p'\circ\sigma:I\rightarrow X$ is a lift of $f$ and we are done.
I don't see where we have used the fact that $G$ is a compact Lie group in this proof
As I understand, we can define pullback or sections in the context of any topological group $G$ acting on a topological space $X$. Or am I wrong?
Thank you!
If $X$ is Hausdorff then the path-lifting property for the quotient map $X\to G/X$ is Theorem 6.2 in G.Bredon "Transformation Groups". The theorem itself is attributed to Montgomery and Yang. The proof is nontrivial.