Identification Topology is Finest Topology for Mapping to be Continuous

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

Let $T_2 := \left({S_2, \tau_2}\right)$ be the resulting topological space.

Then $\tau_2$ is the finest topology on $S_2$ such that $f: T_1 \to T_2$ is continuous.

Proof
It is established in Identification Mapping is Continuous that $f$ is continuous.

Let $\tau \subseteq \mathcal P \left({S_2}\right)$ be a topology on $S_2$ for which $f$ is $\left({\tau_1, \tau}\right)$-continuous.

Let $U \in \tau$.

Then by definition of $\left({\tau_1, \tau}\right)$-continuous:
 * $f^{-1} \left({U}\right) \in \tau_1$

By definition of $\tau_2$ it then follows that $U \in \tau_2$.

By definition of subset it follows that $\tau \subseteq \tau_2$.

Thus it has been shown that $\tau_2$ is finer than any topology on $S_2$ for which $f$ is $\left({\tau_1, \tau}\right)$-continuous.

Hence the result, by definition of finest topology.