# When are quotients of homeomorphic spaces homeomorphic?

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.)

• There is no such "reverse universal property". For instance you have many maps $S^1\to S^1$ that don't factor through the quotient map $\mathbb{R}\to S^1$ – Max Dec 15 '18 at 15:59
• Hmm, interesting. Is there a familiar subset of maps that do factor through in this fashion? I wonder if the question I'm asking here is best answered topologically or categorically. – Billy Smith Dec 15 '18 at 17:59
• In the specific case of $\mathbb{R}\to S^1$ the maps $X\to S^1$ that factor through it are precisely the nullhomotopic maps if $X$ is connected and locally path-connected. I don't know if there's a criterion for general quotient maps, but I don't think there is. – Max Dec 15 '18 at 18:04

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 $$f^{-1}$$.

I think you are correct. Here is a sketch of the argument:

$$\phi:=\pi_Y\circ f$$ is a continuous surjection: $$X\to Y/\approx$$ and it induces a bijection: $$X^*=\left \{ \phi^{-1}([y]): [y]\in Y/\approx \right \}\overset{\overline\phi}{\rightarrow} Y/\approx ,\$$ which is a homeomorphism if and only if $$\phi$$ is a quotient map.

As $$\phi$$ is indeed a quotient map, the induced map is a homeomorphism. So, $$X/\sim$$ and $$Y/\approx$$ will be homoeomorphic if and only if $$X^*$$ and $$X/\sim$$ are.

This means that the map $$[x]\mapsto f^{-1}\circ \pi^{-1}_Y([y])$$ must be a homeomorphism. That is, $$f([x])\mapsto \pi^{-1}_Y([y])$$ must be a homeomorphism.

But the above map is well-defined if and only if $$x\sim x'\Rightarrow f(x)\approx f(x'),$$ and in this case, it is obviously continuous, and it should be easy to check that it is bijective with continuous inverse.

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.