Mapping between Subspaces of Homeomorphic Spaces is Homeomorphism

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

Let $f: T_1 \to T_2$ be a homeomorphism.

Let $H \subseteq S_1$.

Let $T_H = \struct {H, \tau_H}$ be the topological subspace of $T_1$ under the subspace topology $\tau_H$ induced by $\tau_1$.

Let $K = f \sqbrk H$ be the image of $H$ under $f$.

Let $T_K = \struct {K, \tau_K}$ be the topological subspace of $T_2$ under the subspace topology $\tau_K$ induced by $\tau_2$.

Let $g: H \to K$ be the restriction of $f$ to $H$.

Then $g$ is a homeomorphism.

Proof
Let $U \in \tau_K$ be open in $K$.

Then either:
 * $U \in \tau_2$

or:
 * $U = K \cap V$

where $V \in \tau_2$.

Suppose $U \in \tau_2$.

Because $f$ is continuous:
 * $f^{-1} \sqbrk U \in \tau_1$

As $U \subseteq K$ we have that $g^{-1} \sqbrk U \subseteq H$.

So by definition of subspace topology, $g^{-1} \sqbrk U$ is open in $H$.

Now suppose $U = K \cap V$ for some $V \in \tau_2$.

Then:

But as $f$ is continuous:
 * $f^{-1} \sqbrk V \in \tau_1$

As $f^{-1} \sqbrk K$ it follows that:
 * $f^{-1} \sqbrk U = H \cap f^{-1} \sqbrk U$

Hence:
 * $g^{-1} \sqbrk U \in \tau_H$

and the result follows.