# Bijection iff Left and Right Inverse

## Theorem

Let $f: S \to T$ be a mapping.

$f$ is a bijection if and only if:

 $\text {(1)}: \quad$ $\ds \exists g_1: T \to S: \,$ $\ds g_1 \circ f$ $=$ $\ds I_S$ $\text {(2)}: \quad$ $\ds \exists g_2: T \to S: \,$ $\ds f \circ g_2$ $=$ $\ds I_T$

where:

$g_1$ and $g_2$ are mappings
$\circ$ denotes composition of mappings.

It also follows that it is necessarily the case that $g_1 = g_2$ for such to be possible.

### Corollary

Let $f: S \to T$ and $g: T \to S$ be mappings such that:

 $\ds g \circ f$ $=$ $\ds I_S$ $\ds f \circ g$ $=$ $\ds I_T$

Then both $f$ and $g$ are bijections.

## Proof

### Necessary Condition

Let $f: S \to T$ be a mapping.

Let $f$ be such that:

 $\ds \exists g_1: T \to S: \,$ $\ds g_1 \circ f$ $=$ $\ds I_S$ $\ds \exists g_2: T \to S: \,$ $\ds f \circ g_2$ $=$ $\ds I_T$

where both $g_1$ and $g_2$ are mappings.

Then from Left and Right Inverse Mappings Implies Bijection it follows that $f$ is a bijection.

From Left and Right Inverses of Mapping are Inverse Mapping it follows that:

$g_1 = g_2 = f^{-1}$

where $f^{-1}$ is the inverse of $f$.

### Sufficient Condition

Let $f: S \to T$ be a bijection.

Then from Bijection has Left and Right Inverse it follows that:

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

and:

$f \circ f^{-1} = I_T$

where $f^{-1}$ is the inverse of $f$.