$\mu$ is an equivalence relation on $Y$ and $X/\rho= Y/\mu.$

Question

Let $f$: $X\to Y$ be a homeomorphism and $\rho$ be an equivalence relation on $X$. For $x,y $ in $X$ let $f(x)\mu f(y)$ iff $x\rho y$. Then

Show $\mu$ is an equivalence relation on $Y$ and $X/\rho= Y/\mu.$

Answer 1

Define $\hat f\colon X/\rho \to Y/\mu$ by $\hat f ([x]_\rho) = [f(x)]_\mu$ (where $[]$ denotes the respective equivalence class). We will show that $\hat f$ is a homeomorphism.

First, $\hat f$ is well defined, as if $[x] = [x']$, we have $x\,\rho\, x'$, hence $f(x)\,\mu\,f(x')$, that is $[f(x)] = [f(x')]$.

To see that $\hat f$ is one to one, suppose $\hat f([x]) = \hat f([x'])$, then $[f(x)] = [f(x')]$, so $f(x) \,\mu\, f(x')$, hence $x\,\rho\, x'$, giving $[x] = [x']$.

$\hat f$ is onto, as $f$ is onto: If $[y] \in Y/\mu$ is given, choose $x \in X$ with $f(x) = y$, then $\hat f([x]) = [y]$.

To see that $\hat f$ is continuous, let $U \subseteq Y/\mu$ be open. Then if $\pi_\mu$ denotes the projection, $\pi_\mu^{-1}[U]$ is open in $Y$, hence $f^{-1}\circ \pi_\mu^{-1}[U] = \pi_\rho^{-1}\bigl[\hat f^{-1}[U]\bigr] $ is open in $X$. By definition of the quotient, $\hat f^{-1}[U]$ is open in $X/\rho$. Hence $\hat f$ is continous.

To see that $\hat f$ is open, argue along the same line.