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.
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.
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 17$