Proving that two quotient spaces are homeomorphic.

Question

I am trying to solve this problem:

Let $R_1$ and $R_2$ be two equivalence relations in $X$ such that if $xR_1y \implies xR_2y$. Prove that if $R$ is a equivalence relation in $X/R_1$ such that $[x]R[y] \iff xR_2y$ then $\frac{X/R_1}{R}$ is homeomorphic to $X/R_2$ with their quotient topologies.

My attempt:

I want to apply the property (P):

(P): Given a surjective application $f:X \longrightarrow Y$ is defined $R_f$ such that $xR_fy \iff f(x)=f(y)$. Then $(X/R_f,\tau_\pi)$ is homeomorphic to $(Y,\tau_f)$.

For it I defined $f:X/R_1 \longrightarrow X/R_2$ such that $f([x]_1)= [x]_2$ where $[x]_1$ and $[x]_2$ denote the equivalence classes for the relation $R_1$ and $R_2$ respectively.

$f$ is well defined because if $[x]_1=[y]_1 \implies [x]_2=[y]_2$

$f$ is surjective because given $[x]_2 \in X/R_2$ then we can take $f([x]_1)=[x]_2$.

Up to this point we have proved that $(\frac{X/R_1}{R},\tau_{\pi})$ is homeomorphic to $(X/R_2,\tau_f)$ (where $\pi: X/R_1 \longrightarrow \frac{X/R_1}{R}$ is the natural proyection).

My problem is to prove that the quotient topology $\tau_{\pi_2}$ is equal to the final topology $\tau_f$. For this I have observed that $\pi_2=f \circ \pi_1$ and I think it follows from here.

Answer 1

Your proof is essentially correct. I have a few suggestions.

When you write

Given a surjective application $f:X⟶Y$

you should explicitly mention that $X$ is a topological space, $Y$ is a set and the word application is a synonym for function. You should also mention that $\tau_f$ denotes the final topology (aka quotient topology) induced by $f$ on the set $Y$.

Your observation that $\pi_2 = f \circ \pi_1$ is in fact the key for proving $\tau_{\pi_2} = \tau_f$. Let us consider more generally topological spaces $X =(X,\tau_X),Y = (Y,\tau_Y),Z = (Z,\tau_Z)$ and quotient maps $\varphi : X \to Y , \psi : Y \to Z$. That is, we have $\tau_Y = \tau_\varphi$ and $\tau_Z = \tau_\psi$. It is well-known (and easy to prove) that $\psi \circ \varphi : X \to Z$ is also a quotient map, i.e. $\tau_Z = \tau_{\psi \circ \varphi}$. Therefore $\tau_\psi = \tau_{\psi \circ \varphi}$.

Apply this to $\varphi = \pi_1$ and $\psi = f$. Since $\pi_2 = f \circ \pi_1$, we are done.