# Left and Right Inverse Relations Implies Bijection

## 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 such that:

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

where $\circ$ denotes composition of relations.

Then $\RR$ is a bijection.

## Proof

Let $\RR \subseteq S \times T$ be such that:

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

and:

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

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

- $\RR$ is many-to-one
- $\RR$ is right-total
- $\RR^{-1}$ is many-to-one
- $\RR^{-1}$ is right-total.

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

From Inverses of Right-Total and Left-Total Relations, it also follows that both $\RR$ and $\RR^{-1}$ are left-total.

By definition, an injection is a relation which is:

Also by definition, a surjection is a relation which is:

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

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

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(f)}$