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

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 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 $$\mathcal R^{-1}$$.

Sufficient Condition
Now suppose $$\mathcal R$$ is a bijection.

As $$\mathcal R$$ is a bijection, it is a surjection and therefore right-total.

As $$\mathcal R$$ is a bijection, it is a mapping and therefore left-total.

As $$\mathcal R$$ is a bijection, it is a one-to-one relation and therefore also both a many-to-one relation and a one-to-many relation.

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

By Inverse of Many-to-One Relation is One-to-Many we have that $$\mathcal R^{-1}$$ 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:
 * $$\mathcal R^{-1} \circ \mathcal R = I_S$$ and
 * $$\mathcal R \circ \mathcal R^{-1} = I_T$$

Hence the result.