Restriction of Homeomorphism is Homeomorphism

Theorem
Let $T_1 = \left({S_1, \tau_1}\right)$, $T_2 = \left({S_2, \tau_2}\right)$ be topological spaces.

Let $f: S_1 \to S_2$ be a homeomorphism between $T_1$ and $T_2$.

Let $S$ be a subset of $S_1$.

Let $f {\restriction_{S \times f\left[{S}\right]}} : S \to f \left[{S}\right]$ be the restriction of $f$ to $S \times f \left[{S}\right]$.

Let $S$ and $f \left[{S}\right]$ bear their respective subspace topologies.

Then $f {\restriction_{S \times f \left[{S}\right]}}$ is a homeomorphism.

Proof
By Restriction of Continuous Mapping is Continuous, $f {\restriction_{S \times f \left[{S}\right]}}$ is continuous.

By Restriction of Inverse is Inverse of Restriction, $\left({f {\restriction_{S \times f \left[{S}\right]}}}\right)^{-1}$ is well-defined and equal to $f^{-1} {\restriction_{f \left[{S}\right] \times S}}$.

By Restriction of Continuous Mapping is Continuous, $f^{-1} {\restriction_{f \left[{S}\right] \times S}}$ is continuous.

Since $\left({f {\restriction_{S \times f \left[{S}\right]}}}\right)^{-1} = f^{-1} {\restriction_{f \left[{S}\right] \times S}}$, $\left({f {\restriction_{S \times f \left[{S}\right]}}}\right)^{-1}$ is continuous.

Hence, $f {\restriction_{S \times f \left[{S}\right]}}$ is a homeomorphism.