Bijective Relation has Left and Right Inverse

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 $\RR$.

Let $\RR$ be a bijection.

Then:

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

and

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

where $\circ$ denotes composition of relations.

Proof

Suppose $\RR$ is a bijection.

Then by definition:

$(1): \quad \RR$ is a surjection and therefore right-total.

$(2): \quad \RR$ is a mapping and therefore left-total.

$(3): \quad \RR$ is a one-to-one relation and therefore also both a many-to-one relation and a one-to-many relation.

By Inverses of Right-Total and Left-Total Relations, we have that $\RR^{-1}$ is also both right-total and left-total.

By Inverse of Many-to-One Relation is One-to-Many we have that $\RR^{-1}$ is also both a many-to-one relation and a one-to-many relation.

From Condition for Composite Relation with Inverse to be Identity, it follows that:

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

and:

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

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 $5.8 \ \text{(f)}$