Characterization of Homeomorphic Topological Spaces

This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code.

Theorem

Let $T_1 = \struct{S_1, \tau_1}$ be topological space.

Let $S_2$ be a set.

Let $\tau_2$ be a subset of the powerset $\powerset {S_2}$.

Then:

$\struct{S_2, \tau_2}$ is a topological space homeomorphic to $T_1$

if and only if:

there exists a mapping $f : S_1 \to S_2$:

$(1)\quad f$ is a bijection

$(2)\quad f^\to \restriction_{\tau_1}$ is a surjection from $\tau_1$ to $\tau_2$

where

$f^\to \restriction_{\tau_1}$ denotes the restriction of $f^\to$ to $\tau_1$

$f^\to$ denotes the direct image mapping of $f$

Proof

Necessary Condition

Let $\struct{S_2, \tau_2}$ be a topological space homeomorphic to $T_1$.

Let $f: S_1 \to S_2$ be a homeomorphism.

By definition of a homeomorphism:

$f$ is a bijection

$f$ is an open mapping

$f$ is a continuous mapping

By definition of an open mapping:

$\forall U \in \tau_1 : f \sqbrk U \in \tau_2$

By definition of direct image mapping:

$\forall U \in \tau_1 : \map {f^\to} U \in \tau_2$

By definition of continuous mapping:

$\forall V \in \tau_2 : f^{-1} \sqbrk V \in \tau_1$

We have:

\(\ds \forall V \in \tau_2: \, \)

\(\ds \map {f^\to} {f^{-1} \sqbrk V}\)

\(=\)

\(\ds f \sqbrk {f^{-1} \sqbrk V}\)

Definition of Direct Image Mapping

\(\ds \)

\(=\)

\(\ds \map {\paren{f \circ f^{-1} } } V\)

Definition of Composite Mapping

\(\ds \)

\(=\)

\(\ds V\)

Image of Preimage of Subset under Surjection equals Subset

It follows that $f^\to \restriction_{\tau_1}$ is a surjection from $\tau_1$ to $\tau_2$.

$\Box$

Sufficient Condition

Let $f : S_1 \to S_2$ be a mapping such that:

$(1)\quad f$ is a bijection

$(2)\quad f^\to \restriction_{\tau_1}$ is a surjection from $\tau_1$ to $\tau_2$

From Direct Image Mapping is Bijection iff Mapping is Bijection

$f^\to$ is a bijection

From Restriction of Injection is Injection:

$f^\to \restriction_{\Sigma_L}$ is an injection

Hence $f^\to \restriction_{\Sigma_L}$ is a bijection onto $\Sigma'_L$.

$\tau_2$ satisfies Open Set Axiom $(\text O 1)$

Let $\set{V_i : i \in I} \subseteq \tau_2$ be an indexed family of sets of $\tau_2$.

By definition of a bijection:

$\forall i \in I : \exists U_i \in \tau_1 : \map {f^\to} {U_i} = V_i$

By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:

$\bigcup_{i \in I} U_i \in \tau_1$

We have:

\(\ds \bigcup_{i \in I} V_i\)

\(=\)

\(\ds \bigcup_{i \in I} \map {f^\to} {U_i}\)

\(\ds \)

\(=\)

\(\ds \map {f^\to} {\bigcup_{i \in I} U_i}\)

Image of Union under Mapping

\(\ds \)

\(\in\)

\(\ds \tau_2\)

by hypothesis $f^\to \restriction_{\tau_1}$ is a bijection from $\tau_1$ to $\tau_2$

It follows that $\tau_2$ satisfies Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets.

$\Box$

$\tau_2$ satisfies Open Set Axiom $(\text O 2)$

Let $V_1, V_2 \in \tau_2$.

By definition of a bijection:

$\forall i \in I : \exists U_1, U_2 \in \tau_1 : \map {f^\to} {U_1} = V_1, \map {f^\to} {U_2} = V_2$

By Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets:

$U_1 \cap U_2 \in \tau_1$

We have:

\(\ds V_1 \cap V_2\)

\(=\)

\(\ds \map {f^\to} {U_1} \cap \map {f^\to} {U_1}\)

\(\ds \)

\(=\)

\(\ds \map {f^\to} {U_1 \cap U_2}\)

Image of Intersection under Injection

\(\ds \)

\(\in\)

\(\ds \tau_2\)

by hypothesis $f^\to \restriction_{\tau_1}$ is a bijection from $\tau_1$ to $\tau_2$

It follows that $\tau_2$ satisfies Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets.

$\Box$

$\tau_2$ satisfies Open Set Axiom $(\text O 3)$

By Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology:

$S_1 \in \tau_1$

We have:

\(\ds S_2\)

\(=\)

\(\ds \map {f^\to} {S_1}\)

Definition of Bijection applied to $f$

\(\ds \)

\(\in\)

\(\ds \tau_2\)

by hypothesis $f^\to \restriction_{\tau_1}$ is a bijection from $\tau_1$ to $\tau_2$

It follows that $\tau_2$ satisfies Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology.

$\Box$

$\tau_2$ is a Topology

We have shown that $\tau_2$ satisfies the open set axioms.

Hence $\struct{S_2, \tau_2}$ is a topological space by definition.

$\Box$

$f$ is an Open Mapping

We have by hypothesis:

$f^\to \restriction_{\tau_1}$ is a bijection from $\tau_1$ to $\tau_2$

By definition of mapping:

$\forall U \in \tau_1 : \map {f^\to} U \in \tau_2$

By definition of direct image mapping:

$\forall U \in \tau_1 : f \sqbrk U \in \tau_2$

It follows that $f$ is an open mapping by definition.

$\Box$

$f$ is a Continuous Mapping

We have by hypothesis:

$f$ is a bijection from $S_1$ to $S_2$.

From [[Mapping is Bijection iff Direct Image Mapping is Bijection:

$f^\to$ is a bijection from $\powerset {S_1}$ to $\powerset {S_2}$

We have by hypothesis:

$f^\to \restriction_{\tau_1}$ is a bijection from $\tau_1$ to $\tau_2$

From Inverse of Bijection is Bijection:

$\paren{f^\to \restriction_{\tau_1} }^{-1}$ is a bijection from $\tau_2$ to $\tau_1$

From Restriction of Inverse is Inverse of Restriction:

$\paren{f^\to}^{-1} \restriction_{\tau_2}$ is a bijection from $\tau_2$ to $\tau_1$

From Inverse Image Mapping of Bijection is Inverse of Direct Image Mapping:

$f^\gets \restriction_{\tau_2}$ is a bijection from $\tau_2$ to $\tau_1$

By definition of mapping:

$\forall V \in \tau_2 : \map {f^\gets} V \in \tau_1$

By definition of inverse image mapping:

$\forall V \in \tau_2 : f^{-1} \sqbrk V \in \tau_1$

It follows that $f$ is a continuous mapping by definition.

$\Box$

It follows that $f$ is a homeomorphism and $\struct{S_2, \tau_2}$ is a topological space homeomorphic to $T_1$.

$\blacksquare$