Continuity of Composite with Inclusion/Inclusion on Mapping

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.

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

Proof

Necessary Condition

Suppose $g$ is $\left({\tau', \tau_H}\right)$-continuous.

From Inclusion Mapping is Continuous, $i$ is $\left({\tau_H, \tau}\right)$-continuous.

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

Sufficient Condition

Suppose $i \circ g$ is $\left({\tau', \tau}\right)$-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:

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

So by Preimage of Subset under Composite Mapping:

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

Thus:

$g^{-1} \left({V}\right) \in \tau'$

by continuity of $i \circ g$.

Hence $g$ is $\left({\tau', \tau_H}\right)$-continuous.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3.4$: Subspaces: Proposition $3.4.2 \ \text{(b)}$