Inverse Relation is Left and Right Inverse iff Bijection

Theorem

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

Let:

$I_S$ be the identity mapping on $S$

$I_T$ be the identity mapping on $T$.

Let $\RR^{-1}$ be the inverse relation of $\RR$.

Then $\RR$ is a bijection if and only if:

$\RR^{-1} \circ \RR = I_S$

and

$\RR \circ \RR^{-1} = I_T$

where $\circ$ denotes composition of relations.

Proof

Necessary Condition

Let $\RR \subseteq S \times T$ be such that:

$\RR^{-1} \circ \RR = I_S$

and

$\RR \circ \RR^{-1} = I_T$.

From Left and Right Inverse Relations Implies Bijection, it follows that $\RR$ is a bijection.

$\Box$

Sufficient Condition

Suppose $\RR$ is a bijection.

From Bijective Relation has Left and Right Inverse we have that:

$\RR^{-1} \circ \RR = I_S$ and

$\RR \circ \RR^{-1} = I_T$.

$\Box$

Hence the result.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(f)}$