Inverse Relation is Left and Right Inverse iff Bijection
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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 $5.8 \ \text{(f)}$