Let $R_1$ and $R_2$ be equivalence relations on the spaces $X$ and $Y$, respectively. Given a continuous function $f$ such that $xR_1y$ implies $f(x)R_2f(y)$, prove that the induced function $\smash[t]{\hat{f}}: X/R_1 \to Y/R_2$, defined as $\smash[t]{\hat{f}}([x]) = [f(x)]$, is well-defined and continuous.
I am trying to prove that $\hat{f}\bigl([x]\bigr) = \bigl[f(x)\bigr]$ is continuous, so here is my attempt.
Consider the following diagram: $\require{AMScd}$ \begin{CD} X @>{f}>> Y \\ @V{q_1}VV @VV{q_2}V \\ X/R_1 @>{\smash[t]{\hat{f}}}>> Y/R_2 \end{CD}
Note that $\hat{f}^{-1} = (q_2 \circ f \circ q_1^{-1})^{-1}$. We must show that $q_1(f(q_2^{-1}(V^{\text{ab}})) \subset X/R_1$, where $(V^{\text{open}}) \subseteq Y/R_2$. We will demonstrate the continuity of $\smash[t]{\hat{f}}$. Consider the following implications:
If $(V^{\text{open}}) \subseteq Y/R_2$, it follows that $(q_2^{-1}(V^{\text{open}})) \subset Y$ due to the properties of the quotient function.
Since $f$ is a continuous function, we can conclude that $f^{-1}(q_2^{-1}(V^{\text{open}})) \subset X$.
Furthermore, as $q_{1}$ is also a quotient function, we have $q_1(f^{-1}(q_2^{-1}(V^{\text{open}}))) \subset X/R_1$.
We have shown that the preimage of any open set in $Y/R_2$ is an open set in $X/R_1$, which means that $\hat{f}$ is continuous. Therefore, we have demonstrated that $\smash[t]{\hat{f}}: X/R_1 \to Y/R_2$ is a continuous function.
Any correction or help is well received.
Consider the function $q_{2}: Y \rightarrow Y/R_{2}$ given by $q_{2}: y \mapsto [y]$, which is the definition of the quotient function. Let $U^{\text{open}}\subseteq Y/R_{2}$, then $(q_{2}^{-1}(U))^{\text{ab}}\subseteq Y$, this is by the definition of the quotient function, and since $f$ is continuous, we have $$\left(f^{-1}\left(q_{2}^{-1}(U)\right)\right)^{\text{ab}}\subseteq X$$
Note also that $$f^{-1}\left(q_{2}^{-1}(U)\right)=\left((q_{2} \circ f)^{-1}(U) \right)^{\text{open}}\subseteq X$$
So $(q_{2} \circ f)$ is continuous.
Furthermore, note that as $\hat{f} \circ q_{1} = q_{2} \circ f$, then $\hat{f} \circ q_{1}$ is continuous, where $q_{1}: X \rightarrow X/R_{1}$ is the canonical map induced by the equivalence relation $R_{1}$; hence, $$\hat{f} \text{is continuous}$$