Is the quotient $X/G$ homeomorphic to $\tilde{X}/G'$? Let $G$ be a Lie group (not necessarily connected) acting effectively/faithfully on a connected, locally path connected, semi-locally simply connected space $X$ (not necessarily with fixed points). Let $p:\tilde{X}\to X$ be the universal covering of $X$.
For any $g\in G$, $\theta_g:X\to X$ is the map given by $x\mapsto g\cdot x$.
Then $\theta_g$ can be covered by a homeomorphism of $\tilde{X}$ since $\tilde{X}$ is simply connected, and any two such liftings differ by a deck transformation. Clearly, all such liftings for all $g$ form a subgroup $G'$ of $\operatorname{Homeo}(\tilde{X})$.
My question : Are the quotient spaces $X/G$ and $\tilde{X}/G'$ homeomorphic?
My Attempt :
Let $q:X\to X/G$ be the quotient map. I am trying to show that $\psi=q\circ p:\tilde{X}\to X/G$ satisfies the universal property. Let $Z$ be any topological space and let $f:\tilde{X}\to Z$ be a continuous map such that for all $x,y\in\tilde{X}$ , $x\sim y\Longrightarrow f(x)=f(y)$. We need to show that there exists a unique continuous map $\phi:X/G\to Z$ such that $f=\phi\circ \psi$. Let $\bar{x}\in X/G$. Pick $x\in \psi^{-1}(\bar{x})$ and define $\phi(\bar{x})=f(x)$. If this is a well defined function then it is clear that it is continuous and satisfies $f=\phi\circ \psi$.
To check it is well defined-
Suppose $y\in \psi^{-1}(\bar{x})\Longrightarrow \psi(y)=\psi(x)$. From here I am unsure how to proceed. I have to show that $x\sim y$ that is I need to find a $g\in G'$ such that $y=g\cdot x$. Can someone help?
Thank you.

By definition, $q \circ p(x) = \psi(x)=\psi(y) = q \circ p(y) \in X/G$. So $p(x),p(y)$ are in the same orbit of the action of $G$ on $X$. Pick $g \in G$ such that $g \cdot p(x) = p(y)$. In $\tilde X$, the point $x$ is a lift of $p(x)$ and the point $y$ is a lift of $p(y)$. As shown in my answer to your previous question, there exists $g' \in G'$ which is a lift of $g$ such that $g' \cdot x = y$.
