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 $\tau_3$ be a topology on $S_2$ which is strictly finer than $\tau_2$.

Let $T_3 := \left({S_2, \tau_3}\right)$ be the resulting topological space.

Then by definition of strictly finer:
 * $\exists U \in \tau_3: U \notin \tau_2$

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

As $U \notin \tau_2$ it follows that:
 * $f^{-1} \left({U}\right) \notin \tau_1$

and so $f: T_1 \to T_3$ is not a continuous mapping.

Hence the result by definition of finest topology.