Cartesian Product of Preimage with Image of Relation is Correspondence

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

Then the restriction of $\mathcal R$ to $\operatorname{Im}^{-1} \left ({\mathcal R}\right) \times \operatorname{Im} \left ({\mathcal R}\right)$ is a correspondence.

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

$\mathcal R$ is left-total iff:


 * $\forall x \in S: \exists y \in T: x \mathcal R y$

By the definition of the pre-image of $\mathcal R$:


 * $\operatorname{Im}^{-1} \left ({\mathcal R}\right) = \left\{{x \in S: \exists y \in T: x \mathcal R y }\right\}$

Therefore $\mathcal R$ is left-total.

$\mathcal R$ is right-total iff:


 * $\forall x \in T: \exists y \in S: x \mathcal R y$

By the definition of the image of $\mathcal R$:


 * $\operatorname{Im} \left ({\mathcal R}\right) = \left\{{x \in T: \exists y \in S: x \mathcal R y }\right\}$

Therefore $\mathcal R$ is right-total.

Hence $\mathcal R$ is a correspondence.