Existence and Uniqueness of Identification Topology

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 $\struct {S_1, \tau_1}$.

Then the identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$ always exists and is unique.

Proof
Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$.

By definition:
 * $\tau_2 := \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1} \subseteq \powerset {S_2}$

where $\powerset {S_2}$ is the power set of $S_2$.

Let $V \subseteq S_2$.

Then either:
 * $f^{-1} \sqbrk V \in \tau_1 \implies V \in \tau_2$

or:
 * $f^{-1} \sqbrk V \notin \tau_1 \implies V \notin \tau_2$

In particular:
 * $f^{-1} \sqbrk {S_2} = S_1 \in \tau_1 \implies S_2 \in \tau_2$

Thus $\tau_2$ exists.

By the same coin, given any $V \subseteq S_2$, either $V \in \tau_2$ or $V \notin \tau_2$, and so the set:
 * $\tau_2 := \set {V \in \powerset {S_2}: f^{-1} \sqbrk V \in \tau_1}$

consists of a uniquely defined set of elements.