Bijective Relation has Left and Right Inverse

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 a bijection.

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

Proof
Suppose $\mathcal R$ is a bijection.

Then by definition:
 * $(1): \quad \mathcal R$ is a surjection and therefore right-total.


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


 * $(3): \quad \mathcal R$ 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.