# 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:

$\operatorname{Im} \left ({\mathcal R}\right) = \operatorname{Dom} \left ({\mathcal R^{-1}}\right)$

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

## Proof

By definition:

$\operatorname{Im} \left({\mathcal R}\right) := \left\{{t \in T: \exists s \in S: \left({s, t}\right) \in \mathcal R}\right\}$
$\operatorname{Dom} \left({\mathcal R^{-1}}\right) := \left\{{t \in T: \exists s \in S: \left({t, s}\right) \in \mathcal R^{-1}}\right\}$

 $\displaystyle x$ $\in$ $\displaystyle \operatorname{Im} \left({\mathcal R}\right)$ $\quad$ $\quad$ $\displaystyle \iff \ \$ $\displaystyle \exists s \in S: \left({s, x}\right)$ $\in$ $\displaystyle \mathcal R$ $\quad$ Definition of Image of Relation $\quad$ $\displaystyle \iff \ \$ $\displaystyle \exists s \in S: \left({x, s}\right)$ $\in$ $\displaystyle \mathcal R^{-1}$ $\quad$ Definition of Inverse Relation $\quad$ $\displaystyle \iff \ \$ $\displaystyle x$ $\in$ $\displaystyle \operatorname{Dom} \left({\mathcal R^{-1} }\right)$ $\quad$ Definition of Domain of Relation $\quad$

$\blacksquare$