# Superspace of Homeomorphic Subspaces may not have Homeomorphism to Itself containing Subspace Homeomorphism

## Theorem

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

Let $H_1 \subseteq S_1$ and $H_2 \subseteq S_2$.

Let $H_1$ and $H_2$ be a homeomorphic.

Then it may be the case that there does not exist a homeomorphism $g: T_1 \to T_2$ such that:

$g \restriction_{H_1} = f$

where:

$g \restriction_{H_1}$ is the restriction of $g$ to $H_1$
$f: H_1 \to H_2$ is a homeomorphism.

## Proof

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.

Let $H_1 := \set 0 \cup \closedint 1 2 \cup \set 3$, where $\closedint 1 2$ denotes the closed interval from $1$ to $2$.

Let $H_2 := \closedint 0 1 \cup \set 2 \cup \set 3$.

$H_1$ and $H_2$ are homeomorphic, as can be demonstrated by the mapping $\phi: H_1 \to \H_2$ defined as:

$\forall x \in H_1: \map \phi x = \begin {cases} 2 & : x = 0 \\ 3 & : x = 3 \\ x - 1 & : x \in \closedint 1 2 \end {cases}$

which is trivially a homeomorphism.

So, let $g$ be a homeomorphism from $H_1$ and $H_2$.

Each of the singletons in $H_1$ has to map to one of the singletons in $H_2$.

As a result, $g$ is not monotone.

Let $f: \R \to \R$ be a homeomorphism such that $g$ is a restriction of $f$.

Hence $f$ cannot be a bijection.

Hence $f$ cannot be a homeomorphism.

$\blacksquare$