Let $X,Y$ two contractible spaces. Assume there is a free action of a group $G$ on both spaces.
$X$ and $Y$ are obviously homotopy equivalent. In particular, we can consider the homotopy equivalence given by the projections
$X\xleftarrow{p_X} X\times Y\xrightarrow{p_Y}Y$
Does this equivalence induce an equivalence between the quotient spaces $X/G$ and $Y/G$?
My motivation is proving that any two little disks operads are equivalent. I'm not going to define what an operad is because it is irrelevant, but we can think of a family of contractible spaces $X_n$ for $n\in\mathbb{N}$. If each $X_n$ is equipped with a free action of the pure braid group $PB_n$, then the family $X_n/PB_n$ is called a little disks operad. The way I'm trying to prove it is inspired by this paper (page 3 of the pdf).
I believe that my question can be reduced to, given two spaces $X_G$ and $Y_G$ with $\pi_1(X)=\pi_1(Y)=G$, a homotopy equivalence between the (contractible) universal coverings $\widetilde{X}_G\to \widetilde{Y}_G$ induces a homotopy equivalence $X_G\to Y_G$.
Since, in this case, the unviersal coverings are contractible, $X_G$ and $Y_G$ are both aspherical and path connected, so we get isomorphisms between all homotopy groups, but that's not enought to say they're homotopy equivalent.
Best Answer
No, this is not true without additional hypotheses. For instance, let $X$ be $G$ with the indiscrete topology; then $G$ acts on $X$ freely and $X$ is contractible but $X/G$ is a point. On the other hand, if $Y=EG$ is the universal bundle on the classifying space of $G$ then $Y/G=BG$ is not contractible if $G$ is nontrivial.
It is true if you assume that the actions are nice enough so that $X/G$ and $Y/G$ are CW-complexes and the quotient maps $X\to X/G$ and $Y\to Y/G$ are covering maps. Indeed, in that case $X/G$ and $Y/G$ are both $K(G,1)$ spaces and so are homotopy equivalent. Or, going through the product as you suggest, the projection $X\times Y\to X$ is $G$-equivariant (for the product action on $X\times Y$) and so induces a map $X\times Y/G \to X/G$. This map is a fiber bundle with fiber $Y$ (it is trivialized over any open subset of $X/G$ that is evenly covered by $X\to X/G$) and thus a homotopy equivalence since $Y$ is contractible. Similarly, the other projection gives a homotopy equivalence $X\times Y/G\to Y/G$ and combining them we get a homotopy equivalence $X/G\to Y/G$.