TITLE: When are quotients of homeomorphic spaces homeomorphic?
QUESTION [8 upvotes]: Let $X$ and $Y$ be homeomorphic topological spaces, connected by the homeomorphism $f : X \rightarrow Y$. Let $\sim$ be an equivalence relation on $X$ and $\approx$ be an equivalence relation on $Y$.
I would think that there would be some structure that I could place on the equivalence relations $\sim$ and $\approx$ that would allow me to construct, using the homeomorphism $f$, a homeomorphism $\bar{f}$ from $X/\sim$ to $Y/\approx$. I suspect that that relationship is $a \sim b \leftrightarrow f(a) \approx f(b), \forall a, b \in X$.
That is, I suspect that "equivalent" quotients of homeomorphic spaces are homeomorphic, but I don't know exactly how to formulate this or prove it.
(This result, by the way, to me, seems to hinge on the existence of a "reverse" universal quotient property: juts as functions $f : X/\sim \rightarrow Y$ correspond uniquely to a subset of functions $f : X \rightarrow Y$, I suspect that functions $f : X \rightarrow Y/\approx$ correspond uniquely to some subset of functions $f : X \rightarrow Y$, but I don't know how this works either.)

REPLY [3 votes]: For a categorical argument, a topological space equipped with an equivalence relation can be viewed as a parallel pair of arrows $R\rightrightarrows X$, where $R\subset X\times X$ is the subspace consisting of the related pairs of points. Then the claim follows from reinterpreting your condition that $f$ preserve and reflect the equivalence relation as asking that $f$ extend to a natural isomorphism between the diagrams $R\rightrightarrows X$ and $S\rightrightarrows Y$. Then your claim is a special case of the fact that naturally isomorphic diagrams have naturally isomorphic colimits, which follows from the fact that colimits are functorial. 
This perspective clarifies that this really has nothing to do with equivalence relations, nor with topological spaces. As far as the technical details of the proof, any map $f:X\to Y$ induces a natural transformation between the diagrams $X\times X\rightrightarrows X$ and $Y\times Y\rightrightarrows Y$, so one is reduced to proving that $f$ induces a bijection $R\times R\to S\times S$, which is exactly your assumption. There's no need to say anything more nitty-gritty.