Cartesian Product of Preimage with Image of Relation is Correspondence
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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 if and only if:
- $\forall x \in S: \exists y \in T: x \mathrel \RR y$
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 if and only if:
- $\forall x \in T: \exists y \in S: x \mathrel \RR y$
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.
$\blacksquare$