Left and Right Inverse Relations Implies 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 $\mathcal R$.

Let $\mathcal R$ be such that: where $\circ$ denotes composition of relations.
 * $\mathcal R^{-1} \circ \mathcal R = I_S$ and
 * $\mathcal R \circ \mathcal R^{-1} = I_T$

Then $\mathcal R$ is a bijection.

Proof
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 Condition for Composite Relation with Inverse to be Identity, we have that:


 * $\mathcal R$ is many-to-one
 * $\mathcal R$ is right-total
 * $\mathcal R^{-1}$ is many-to-one
 * $\mathcal R^{-1}$ is right-total.

From Inverse of Many-to-One Relation is One-to-Many, it follows that both $\mathcal R$ and $\mathcal R^{-1}$ are by definition one-to-one.

From Inverse of Right-Total is Left-Total, it also follows that both $\mathcal R$ and $\mathcal R^{-1}$ are left-total.

By definition, an injection is a relation which is:
 * One-to-one
 * left-total.

Also by definition, a surjection is a relation which is:
 * Left-total
 * Many-to-one
 * Right-total.

It follows that $\mathcal R$ is both an injection and a surjection, and so by definition a bijection.

By the same coin, the same applies to $\mathcal R^{-1}$.