Let $G$ be a finite group, $H\subset G$ a subgroup and $f: X\to Y$ an $H$-equivariant continuous map between topological spaces endowed with an $H$-action. Define the balanced product $$ G\times_H X := (G\times X)/H $$ to be the orbit space with respect to the group action of $H$ on $G\times X$ given by $$ h\cdot(g,x) := (gh^{-1},h\cdot x) $$ for all $h\in H$, $g\in G$ and $x\in X$. (Equivalently, we have $(gh,x)\sim(g,h\cdot x)$ for all $h\in H$, $g\in G$ and $x\in X$.) Note that $f$ induces a well-defined map $$ G\times _H f: G\times _H X\to G\times_H Y $$ given by $(G\times _H f)([g,x]) = [g,f(x)]$ for all $g\in G$ and $x\in X$. In other words, we have $(G\times_H f)\circ p = q\circ f$ where $p: G\times X\to G\times_H X$ and $q: G\times Y\to G\times_H Y$ are the canonical quotient maps.
I have the following question:
Is it true that if $f$ is a closed embedding, then also $G\times_H f$ is a closed embedding?
The map $G\times_H f$ is certainly continuous and injective, so it would be sufficient to show that it is closed.
My attempt:
Let $A\subset G\times_H X$ be a closed subspace, which means that $p^{-1}(A)$ is closed in $G\times X$. Since $f$ is closed, it follows that $f(p^{-1}(A))$ is closed in $Y$. We have $(G\times_H f)(A) = (G\times_H f)(p(p^{-1}(A)) = q(f(p^{-1}(A)))$, so to show that this is closed, it would suffice to show that $q$ is a closed map. However, I am not sure if this is really true.
Would it make a difference if we suppose that $X$ and $Y$ are compactly generated weakly Hausdorff spaces? A topological space $Z$ is weakly Hausdorff if for every compact Hausdorff space $K$ and for every continuous map $u: K\to Z$, the image $u(K)$ is closed in $Z$. A weakly Hausdorff space $Z$ is compactly generated if the following holds: A subspace $A\subset X$ is closed in $X$ if and only if for every compact Hausdorff subspace $K\subset X$, the intersection $A\cap K$ is closed in $K$. (Note that the only if direction always holds.)
Any help is appreciated! Many thanks in advance!