Inverse of Open Set in Product Space is Open in Inverse Product Space

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

Let $\tau_{S_1 \times S_2}$ be the product topology on the Cartesian product $S_1 \times S_2$.

Let $\tau_{S_2 \times S_1}$ be the product topology on the Cartesian product $S_2 \times S_1$.

Let $V \subseteq S_1 \times S_2$.

Let $V^{-1} \subseteq S_2 \times S_1$ denote the inverse of $V$ viewed as a relation on $S_1 \times S_2$.

Then:
 * $V$ is open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$ $V^{-1}$ is open in $\struct{S_2 \times S_1, \tau_{S_2 \times S_1}}$

Necessary Condition
Let $V$ be open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$.

Let $\tuple{s_2, s_1} \in V^{-1}$

By definition of inverse:
 * $\tuple{s_1, s_2} \in V$

By definition of product topology:
 * $\BB_{12} = \set{U \times W : U \in \tau_1, W \in \tau_2}$ is a basis for $\tau_{S_1 \times S_2}$

By definition of basis:
 * $\exists U \in \tau_1, W \in \tau_2 : \tuple{s_1, s_2} \in U \times W \subseteq V$

By definition of inverse:
 * $\tuple{s_2, s_1} \in W \times U \subseteq V^{-1}$

By definition of product topology:
 * $\BB_{21} = \set{W \times U : W \in \tau_2, U \in \tau_1}$ is a basis for $\tau_{S_2 \times S_1}$

Since $\tuple{s_2, s_1}$ was arbitrary, we have:
 * $\forall \tuple{s_2, s_1} \in V^{-1} : \exists W \times U \in \BB_{21} : \tuple{s_2, s_1} \in W \times U \subseteq V^{-1}$

From Characterization of Set Equals Union of Sets:
 * $V^{-1} = \cup \set{W \times U \in \BB_{21} : W \times U \subseteq V^{-1}}$

By :
 * $V^{-1}$ is open in $\struct{S_2 \times S_1, \tau_{S_2 \times S_1}}$

Sufficient Condition
Let $V^{-1}$ be open in $\struct{S_2 \times S_1, \tau_{S_2 \times S_1}}$.

From Necessary Condition above:
 * $\paren{V^{-1}}^{-1}$ is open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$

From Inverse of Inverse Relation:
 * $\paren{V^{-1}}^{-1} = V$

Hence:
 * $V$ is open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$