# Identification Topology is Finest Topology for Mapping to be Continuous

## 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 $\struct {S_1, \tau_1}$.

Let $T_2 := \struct {S_2, \tau_2}$ 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 $\tau \subseteq \powerset {S_2}$ be a topology on $S_2$ for which $f$ is $\struct {\tau_1, \tau}$-continuous.

Let $U \in \tau$.

Then by definition of $\struct {\tau_1, \tau}$-continuous:

$f^{-1} \sqbrk U \in \tau_1$

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 $\struct {\tau_1, \tau}$-continuous.

Hence the result, by definition of finest topology.

$\blacksquare$