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

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.

$\blacksquare$