Inverse Relation is Left and Right Inverse iff Bijection

Theorem
Let $\mathcal R \subseteq S \times T$ 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 $\mathcal R^{-1}$ be the inverse relation of $\mathcal R$.

Then $\mathcal R$ is a bijection :
 * $\mathcal R^{-1} \circ \mathcal R = I_S$

and
 * $\mathcal R \circ \mathcal R^{-1} = I_T$

where $\circ$ denotes composition of relations.

Necessary Condition
Let $\mathcal R \subseteq S \times T$ be such that:
 * $\mathcal R^{-1} \circ \mathcal R = I_S$

and
 * $\mathcal R \circ \mathcal R^{-1} = I_T$.

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

Sufficient Condition
Suppose $\mathcal R$ is a bijection.

From Bijective Relation has Left and Right Inverse we have that:
 * $\mathcal R^{-1} \circ \mathcal R = I_S$ and
 * $\mathcal R \circ \mathcal R^{-1} = I_T$.

Hence the result.