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.)
Best Answer
Your suspicion is correct. If $X$ and $Y$ are topological spaces with equivalence relations $\sim$ and $\approx$ and $f:X\to Y$ is a homeomorphism such that $x\sim y$ if and only if $f(x)\approx f(y)$, then $f$ passes to the quotient as $X/_\sim \cong Y/_\approx$.
Proof goes roughly like this: define $\widetilde{f}([x]_\sim)\doteq [f(x)]_\approx$. We have that $\widetilde{f}$ is well-defined and injective because of the condition relating $\sim$ and $\approx$, while surjectivity follows from the surjectivity of $f$. Continuity of $\widetilde{f}$ and $\widetilde{f}^{-1}$ follow from the universal property of the quotient topologies applied to the relations $\widetilde{f}\circ\pi_{\sim}=\pi_{\approx}\circ f$ and $\widetilde{f}^{-1}\circ \pi_{\approx}=\pi_\sim\circ f^{-1}$ with the continuity of $f$ and $f^{-1}$.