Existence and Uniqueness of Identification Topology

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 $\left({S_1, \tau_1}\right)$.

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

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

By definition:
 * $\tau_2 := \left\{{V \in \mathcal P \left({S_2}\right): f^{-1} \left({V}\right) \in \tau_1}\right\} \subseteq \mathcal P \left({S_2}\right)$

where $\mathcal P \left({S_2}\right)$ is the power set of $S_2$.

Let $V \subseteq S_2$.

Then either:
 * $f^{-1} \left({V}\right) \in \tau_1 \implies V \in \tau_2$

or:
 * $f^{-1} \left({V}\right) \notin \tau_1 \implies V \notin \tau_2$

In particular:
 * $f^{-1} \left({S_2}\right) = S_1 \in \tau_1 \implies S_2 \in \tau_2$

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:
 * $\tau_2 := \left\{{V \in \mathcal P \left({S_2}\right): f^{-1} \left({V}\right) \in \tau_1}\right\}$

consists of a uniquely defined set of elements.