Image of Relation is Domain of Inverse Relation

Theorem

Let $\RR \subseteq S \times T$ be a relation.

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

Then:

$\Img \RR = \Dom {\RR^{-1} }$

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

Proof

By definition:

\(\ds \Img \RR\)

\(:=\)

\(\ds \set {t \in T: \exists s \in S: \tuple {s, t} \in \RR}\)

\(\ds \Dom {\RR^{-1} }\)

\(:=\)

\(\ds \set {t \in T: \exists s \in S: \tuple {t, s} \in \RR^{-1} }\)

\(\ds x\)

\(\in\)

\(\ds \Img \RR\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \exists s \in S: \, \)

\(\ds \tuple {s, x}\)

\(\in\)

\(\ds \RR\)

Definition of Image of Relation

\(\ds \leadstoandfrom \ \ \)

\(\ds \exists s \in S: \, \)

\(\ds \tuple {x, s}\)

\(\in\)

\(\ds \RR^{-1}\)

Definition of Inverse Relation

\(\ds \leadstoandfrom \ \ \)

\(\ds x\)

\(\in\)

\(\ds \Dom {\RR^{-1} }\)

Definition of Domain of Relation

$\blacksquare$

Also see

• Domain of Relation is Image of Inverse Relation

Sources

• 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Relations
• 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 10$: Inverses and Composites
• 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.11$: Relations