Identification Topology equals Quotient Topology on Induced Equivalence

Theorem
Let $T_1 = \left({S_1, \tau_1}\right)$ be a topological space.

Let $S_2$ be a set.

Let $f: S_1 \to S_2$ be a mapping.

Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\left({S_1, \tau_1}\right)$:
 * $\tau_2 = \left\{{V \in \mathcal P \left({S_2}\right): f^{-1} \left({V}\right) \in \tau_1}\right\}$

Let $\mathcal R_f \subseteq S_1 \times S_1$ be the equivalence on $S_1$ induced by $f$:
 * $\left({s_1, s_2}\right) \in \mathcal R_f \iff f \left({s_1}\right) = f \left({s_2}\right)$

Let $\tau_{\mathcal R_f}$ be the quotient topology on $S / \mathcal R_f$ by $q_{\mathcal R_f}$:
 * $\tau_{\mathcal R_f} := \left\{{U \subseteq S / \mathcal R_f: q_{\mathcal R_f}^{-1} \left({U}\right) \in \tau_1}\right\}$

Then:
 * $\tau_2$ is homeomorphic to $\tau_{\mathcal R_f}$.