Extension of Continuous Mapping is Continuous

Theorem

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $T_H = \struct{H, \tau_H}$ be a topological subspace of $T_2$ where $H \subseteq S_2$.

Let $f: S_1 \to H$ be a $\tuple { \tau_1 , \tau_H}$-continuous mapping.

Define a mapping $g: S_1 \to S_2$ by:

$\forall x \in S_1 : \map g x = \map f x$

Then $g$ is $\tuple { \tau_1 , \tau_2}$-continuous.

Let $i_H$ be the inclusion mapping of $H$ in $S_2$.

Then:

$g = i_H \circ f$

From Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $f$ is $\tuple { \tau_1 , \tau_H}$-continuous mapping if and only if $i_H \circ f = g$ is $\tuple { \tau_1 , \tau_2}$-continuous.

Since $f$ is continuous by assumption, $g$ is continuous.

$\blacksquare$

2011: John M. Lee: Introduction to Topological Manifolds (2nd ed.) ... (previous) ... (next): $\S 3$: New Spaces From Old: Subspaces