Restriction of Continuous Mapping is Continuous/Topological Spaces

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 1

Let $V \subseteq M_2$ be an open set with respect to the subspace topology of $M_2$.

By definition of subspace topology, $V = U \cap M_2$ for an open set $U \in \tau_2$.

We have that:

$\ds \paren {f \restriction_{M_1 \times M_2} }^{-1} \sqbrk V$

$=$

$\ds \set {x \in M_1 : \map {f \restriction_{M_1 \times M_2} } x \in V}$

Definition of Preimage of Subset under Mapping

$\ds $

$=$

$\ds \set {x \in M_1 : \map {f \restriction_{M_1 \times M_2} } x \in U}$

$f \sqbrk {M_1} \subseteq M_2$

$\ds $

$=$

$\ds \set {x \in M_1 : \map f x \in U}$

Definition of Restriction of Mapping

$\ds $

$=$

$\ds \set {x \in S_1 : \map f x \in U} \cap M_1$

Definition of Set Intersection

$\ds $

$=$

$\ds f^{-1} \sqbrk U \cap M_1$

Definition of Preimage of Subset under Mapping

By definition of continuous mapping, $f^{-1} \sqbrk U$ is open in $T_1$.

By definition of subspace topology, $f^{-1} \sqbrk U \cap M_1$ is open with respect to the subspace topology of $M_1$.

As $f^{-1} \sqbrk U \cap M_1 = {f \restriction_{M_1 \times M_2} }^{-1} \sqbrk V$ it follows that ${f \restriction_{M_1 \times M_2} }^{-1} \sqbrk V$ is open with respect to the subspace topology of $M_1$.

Since $V \subseteq M_2$ was an arbitrary open set, $f$ is continuous.

$\blacksquare$

Proof 2

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.

$\blacksquare$