Image of Relation is Domain of Inverse Relation

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Theorem

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

Let $\mathcal R^{-1} \subseteq T \times S$ be the inverse of $\mathcal R$.


Then:

$\Img {\mathcal R} = \Dom {\mathcal R^{-1} }$

That is, the image of a relation is the domain of its inverse.


Proof

By definition:

$\Img {\mathcal R} := \set {t \in T: \exists s \in S: \tuple {s, t} \in \mathcal R}$
$\Dom {\mathcal R^{-1} } := \set {t \in T: \exists s \in S: \tuple {t, s} \in \mathcal R^{-1} }$


\(\displaystyle x\) \(\in\) \(\displaystyle \Img {\mathcal R}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \exists s \in S: \tuple {s, x}\) \(\in\) \(\displaystyle \mathcal R\) Definition of Image of Relation
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \exists s \in S: \tuple {x, s}\) \(\in\) \(\displaystyle \mathcal R^{-1}\) Definition of Inverse Relation
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(\in\) \(\displaystyle \Dom {\mathcal R^{-1} }\) Definition of Domain of Relation

$\blacksquare$


Also see


Sources