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

Proof

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.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.4$: Subspaces: Proposition $3.4.2 \ \text{(b)}$