Cartesian Product of Preimage with Image of Relation is Correspondence

Theorem
Let $\RR \subseteq S \times T$ be a relation.

Then the restriction of $\RR$ to $\Preimg \RR \times \Img \RR$ is a correspondence.

Proof
By the definition of a correspondence it will be shown that $\RR$ is both left-total and right-total.

$\RR$ is left-total :


 * $\forall x \in S: \exists y \in T: x \mathrel \RR y$

By the definition of the pre-image of $\RR$:


 * $\Preimg \RR = \set {x \in S: \exists y \in T: x \mathrel \RR y}$

Therefore $\RR$ is left-total.

$\RR$ is right-total :


 * $\forall x \in T: \exists y \in S: x \mathrel \RR y$

By the definition of the image of $\RR$:


 * $\Img \RR = \set {x \in T: \exists y \in S: x \mathrel \RR y}$

Therefore $\RR$ is right-total.

Hence $\RR$ is a correspondence.