# Image of Relation is Domain of Inverse Relation

## 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

$\blacksquare$