Theorem: Characterization of Homeomorphic Topological Spaces/Sufficient Condition

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 $\powerset {S_2}$.
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 $\tau_1$ onto $\tau_2$
where
:$f^\to \restriction_{\tau_1}$ denotes the restriction of $f^\to$ to $\tau_1$
:$f^\to$ denotes the direct image mapping of $f$
Then:
:$\struct{S_2, \tau_2}$ is a topological space homeomorphic to $T_1$

Proof:
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|1}}:
:$\bigcup_{i \in I}  U_i \in \tau_1$
We have:
{{begin-eqn}}
{{eqn | l = \bigcup_{i \in I} V_i
      | r = \bigcup_{i \in I} \map {f^\to} {U_i}
}}
{{eqn | r = \map {f^\to} {\bigcup_{i \in I}  U_i}
      | c = Image of Union under Mapping
}}
{{eqn | o = \in
      | r = \tau_2
      | c = {{hypothesis}} $f^\to \restriction_{\tau_1}$ is a bijection from $\tau_1$ to $\tau_2$
}}
{{end-eqn}}
It follows that $\tau_2$ satisfies {{Open-set-axiom|1}}.
{{qed|lemma}}
==== $\tau_2$ 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|2}}:
:$U_1 \cap U_2 \in \tau_1$
We have:
{{begin-eqn}}
{{eqn | l = V_1 \cap V_2
      | r = \map {f^\to} {U_1} \cap \map {f^\to} {U_1} 
}}
{{eqn | r = \map {f^\to} {U_1 \cap U_2}
      | c = Image of Intersection under Injection
}}
{{eqn | o = \in
      | r = \tau_2
      | c = {{hypothesis}} $f^\to \restriction_{\tau_1}$ is a bijection from $\tau_1$ to $\tau_2$
}}
{{end-eqn}}
It follows that $\tau_2$ satisfies {{Open-set-axiom|2}}.
{{qed|lemma}}
==== $\tau_2$ satisfies Open Set Axiom $(\text O 3)$ ====
By {{Open-set-axiom|3}}:
:$S_1 \in \tau_1$
We have:
{{begin-eqn}}
{{eqn | l = S_2
      | r = \map {f^\to} {S_1}
      | c = {{Defof|Bijection}} applied to $f$
}}
{{eqn | o = \in
      | r = \tau_2
      | c = {{hypothesis}} $f^\to \restriction_{\tau_1}$ is a bijection from $\tau_1$ to $\tau_2$
}}
{{end-eqn}}
It follows that $\tau_2$ satisfies {{Open-set-axiom|3}}.
{{qed|lemma}}
==== $\tau_2$ 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.
{{qed|lemma}}
==== $f$ is an Open Mapping ====
We have {{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.
{{qed|lemma}}
==== $f$ is a Continuous Mapping ====
We have {{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 {{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.
{{qed|lemma}}
It follows that $f$ is a homeomorphism and $\struct{S_2, \tau_2}$ is a topological space homeomorphic to $T_1$.