Mapping between Subspaces of Homeomorphic Spaces is Homeomorphism

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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.


Corollary

Let $\overline H$ denote the complement $S_1 \setminus H$ of $H$ relative to $S_1$.

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

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

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


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


Then $h$ 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:

$f^{-1} \sqbrk U \in \tau_1$
\(\ds f^{-1} \sqbrk U\) \(\in\) \(\ds \tau_1\) Definition of Continuous Mapping
\(\ds \leadsto \ \ \) \(\ds g^{-1} \sqbrk U\) \(\in\) \(\ds \tau_1\) Definition of Restriction of Mapping

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 $H$.

$\Box$


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

Then:

\(\ds g^{-1} \sqbrk U\) \(=\) \(\ds g^{-1} \sqbrk {K \cap V}\)
\(\ds \) \(=\) \(\ds g^{-1} \sqbrk K \cap g^{-1} \sqbrk V\) Preimage of Intersection under Mapping
\(\ds \) \(=\) \(\ds f^{-1} \sqbrk K \cap f^{-1} \sqbrk V\) Definition of Restriction of Mapping

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.


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