Does $A\simeq_\varphi B$ imply $A/{\sim}\,\, \simeq B/\varphi(\sim)$?

Question

Let $A,B$ be two homeomorph topological space under homeomorphism $\varphi$ and let $\sim$ be an equivalence relation on $A$. Then

Does $A\simeq_\varphi B$ imply $A/{\sim}\,\, \simeq B/\varphi(\sim)$?

Where $\varphi(\sim)$ is defined as follow $$x,y\in A,\quad x\sim y \iff \varphi(x)\,\varphi(\sim)\,\varphi(y).$$

Answer 1

You are correct. Consider the composition $A \to B \to B/ \varphi(\sim)$. Denote the resulting map $A \to B/\varphi(\sim)$ by $\sigma$. If $x \sim y$ in $A$, then $\sigma(x)= \pi( \varphi(x))= \pi(\varphi(y))= \sigma(y)$, where $\pi: B \to B/ \varphi(\sim)$ is the projection map. Hence $\sigma$ respects the equivalence relation $\sim$, so $\sigma$ induces a map $A/ \sim \to B/ \varphi(\sim)$. Since $\varphi$ is a homeomorphism, we get a similar map $\tau:B/ \varphi(\sim) \to A/ \sim$. Can you show these maps are inverse to each other?