# Continuity of Composite with Inclusion/Uniqueness of Induced Topology

## Theorem

Let $T = \struct {A, \tau}$ and $T' = \struct {A', \tau'}$ be topological spaces.

Let $H \subseteq A$.

Let $T_H = \struct {H, \tau_H}$ 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 $T' = \struct {A', \tau'}$, and
- $(2) \quad$ For any mapping $g: A' \to H$:

$g$ is $\tuple {\tau', \tau''}$-continuous if and only if $i \circ g$ is $\tuple {\tau', \tau}$-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 $\tuple {\tau'', \tau''}$-continuous

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

Hence for any $U \in \tau$:

- $\map {\paren {i \circ g}^{-1} } U \in \tau''$

But:

- $\map {\paren {i \circ g}^{-1} } U = \map {i^{-1} } U = U \cap H$

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 $\tuple {\tau_H, \tau}$-continuous,

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

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

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

$\blacksquare$

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $3.4$: Subspaces: Proposition $3.4.3$