Continuity of Composite with Inclusion/Inclusion on Mapping

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.

$g$ is $\tuple {\tau', \tau_H}$-continuous $i \circ g$ is $\tuple {\tau', \tau}$-continuous.

Necessary Condition
Suppose $g$ is $\tuple {\tau', \tau_H}$-continuous.

From Inclusion Mapping is Continuous, $i$ is $\tuple {\tau_H, \tau}$-continuous.

It follows from Composite of Continuous Mappings is Continuous that $i \circ g$ is $\tuple {\tau', \tau}$-continuous.

Sufficient Condition
Suppose $i \circ g$ is $\tuple {\tau', \tau}$-continuous.

Let $V \in \tau_H$.

Then from the definition of topological subspace:
 * $V = U \cap H$ for some $U \in \tau$

and by Preimage of Subset under Inclusion Mapping:
 * $U \cap H = i^{-1} \sqbrk U$

So by Preimage of Subset under Composite Mapping:
 * $g^{-1} \sqbrk V = g^{-1} \sqbrk {i^{-1} \sqbrk U} = \paren {i \circ g}^{-1} \sqbrk U$

Thus:
 * $g^{-1} \sqbrk V \in \tau'$

by continuity of $i \circ g$.

Hence $g$ is $\tuple {\tau', \tau_H}$-continuous.