Continuity of Composite with Inclusion

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

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

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

Let $f: A \to A'$ and $g: A' \to H$ be mappings.

Then:
 * $(1): \quad f \restriction_H = f \circ i$, where $f \restriction_H$ is the restriction of $f$ to $H$
 * $(2): \quad$ If $f$ is $\left({\vartheta, \vartheta'}\right)$-continuous, then so is $f \circ i$
 * $(3): \quad g$ is $\left({\vartheta', \vartheta_H}\right)$-continuous iff $i \circ g$ is $\left({\vartheta', \vartheta}\right)$-continuous
 * $(4): \quad$ The induced topology $\vartheta_H$ is the only topology on $H$ satisfying (3) for all possible $g$.

Proof

 * $(1): \quad f f \restriction_H = f \circ i$:

Follows directly from the definition of the inclusion mapping.


 * $(2): \quad$ If $f$ is $\left({\vartheta, \vartheta'}\right)$-continuous, then so is $f \circ i$:

By Continuity of Composite Mapping, it is enough to prove that $i$ is $\left({\vartheta_H, \vartheta}\right)$-continuous.

This follows, because $U \in \vartheta \Longrightarrow i^{-1} \left({U}\right) = U \cap H \in \vartheta_H$.


 * $(3): \quad g$ is $\left({\vartheta', \vartheta_H}\right)$-continuous iff $i \circ g$ is $\left({\vartheta', \vartheta}\right)$-continuous:

Suppose $g$ is continuous. Then, using the above, so is $i \circ g$.

Suppose then that $i \circ g$ is continuous.

Let $V \in \vartheta_H$.

Then from the definition of topological subspace, $V = U \cap H$ for some $U \in \vartheta$ and $U \cap H = i^{-1} \left({U}\right)$.

So $g^{-1} \left({V}\right) = g^{-1} \left({i^{-1} \left({U}\right)}\right) = \left({i \circ g}\right)^{-1} \left({U}\right)$.

Thus $g^{-1} \left({V}\right) \in \vartheta'$ by continuity of $i \circ g$.

Hence $g$ is continuous.


 * $(4): \quad$ Uniqueness of $\vartheta_H$:

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

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

We are going to show that $\vartheta''$ must be the same as $\vartheta_H$.

First, take $A' = H$ and $\vartheta' = \vartheta''$, and $g$ the identity mapping of $H$.

Since $g$ is certainly $\left({\vartheta, \vartheta}\right)$-continuous, $i \circ g$ is $\left({\vartheta'', \vartheta}\right)$-continuous.

Hence for any $U \in \vartheta$, $\left({i \circ g}\right)^{-1} \left({U}\right) \in \vartheta''$.

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

Hence $\vartheta_H \subseteq \vartheta''$.

Next, take take $A' = H$ and $\vartheta' = \vartheta_H$, and $g$ the identity mapping of $H$.

Since $i \circ g = i$ is $\left({\vartheta_H, \vartheta}\right)$-continuous, $g$ is $\left({\vartheta_H, \vartheta''}\right)$-continuous.

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

So $\vartheta'' = \vartheta_H$, as required.