Restriction of Continuous Mapping is Continuous

Theorem

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

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

Let $f: M_1 \to M_2$ be a mapping which is continuous at a point $\alpha \in S$.

Let $f \restriction_S = g: S \to M_2$ be the restriction of $f$ to $S$.

Then $g$ is continuous at $\alpha$.

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.