Restriction of Continuous Mapping is Continuous/Topological Spaces/Proof 2

Theorem

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

Let $M_1 \subseteq S_1$ be a subset of $S_1$.

Let $f: S_1 \to S_2$ be a mapping which is continuous.

Let $M_2 \subseteq S_2$ be a subset of $S_2$ such that $f \sqbrk {M_1} \subseteq M_2$.

Let $f \restriction_{M_1 \times M_2}: M_1 \to M_2$ be the restriction of $f$ to $M_1 \times M_2$.

Then $f \restriction_{M_1 \times M_2}$ is continuous, where $M_1$ and $M_2$ are equipped with the respective subspace topologies.

Proof

Consider first the restriction $f \restriction_{M_1}$.

From Composition of Mapping with Inclusion is Restriction:

$f \restriction_{M_1} = f \circ i_{M_1}$

where $i_{M_1}$ is the inclusion of $M_1$ into $S_1$.

From Continuity of Composite with Inclusion: Mapping on Inclusion, it follows that $f \circ i_{M_1} = f \restriction_{M_1}$ is continuous.

$\Box$

Consider now a mapping $\tilde f : S_1 \to M_2$ where $f \sqbrk {S_1} \subseteq M_2$

Then: $f = i_{M_2} \circ \tilde f$ where $i_{M_2}$ is the inclusion of $M_2$ into $S_2$.

From Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $\tilde f$ is continuous if and only if $i_{M_2} \circ \tilde f = f$ is continuous.

Thus $\tilde f$ is continuous, since $f$ is continuous by assumption.

$\Box$

The result follows from combining these partial results.

$\blacksquare$

Sources

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