Continuity of Composite with Inclusion/Uniqueness of Induced Topology

Theorem
Let $T = \left({A, \tau}\right)$ and $T' = \left({A', \tau'}\right)$ be topological spaces.

Let $H \subseteq A$.

Let $T_H = \left({H, \tau_H}\right)$ be a topological subspace of $T$.

Let $i: H \to A$ be the inclusion mapping.

Let $g: A' \to H$ be a mapping. The induced topology $\tau_H$ is the only topology on $H$ satisfying Continuity of Composite with Inclusion: Inclusion on Mapping for all possible $g$.

Proof
Suppose $\tau''$ is a topology on $H$ such that:
 * $(1) \quad$ For any topological space $T' = \left({A', \tau'}\right)$, and
 * $(2) \quad$ For any mapping $g: A' \to H$:

$g$ is $\left({\tau', \tau''}\right)$-continuous iff $i \circ g$ is $\left({\tau', \tau}\right)$-continuous.

It needs to be shown that $\tau''$ must be the same as $\tau_H$.

Let $A' = H$ and $\tau' = \tau''$.

Let $g$ be the identity mapping on $H$.

From Identity Mapping is Continuous, $g$ is $\left({\tau, \tau}\right)$-continuous

Thus from Continuity of Composite Mapping, $i \circ g$ is $\left({\tau'', \tau}\right)$-continuous.

Hence for any $U \in \tau$:
 * $\left({i \circ g}\right)^{-1} \left({U}\right) \in \tau''$

But:
 * $\left({i \circ g}\right)^{-1} \left({U}\right) = i^{-1} \left({U}\right) = U \cap H$

Hence $\tau_H \subseteq \tau''$.

Next, take take $A' = H$ and $\tau' = \tau_H$.

Let $g$ be the identity mapping on $H$.

We have that $i \circ g = i$ is $\left({\tau_H, \tau}\right)$-continuous,

From Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $g$ is $\left({\tau_H, \tau''}\right)$-continuous.

But by definition of continuity, this is the same as saying $\tau'' \subseteq \tau_H$.

So $\tau'' = \tau_H$, as required.