Identification Topology equals Quotient Topology on Induced Equivalence

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ 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 $\struct {S_1, \tau_1}$:
 * $\tau_2 = \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$

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

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

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