# Bijection iff exists Mapping which is Left and Right Inverse

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## Theorem

Let $S$ and $T$ be sets.

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

Then $f$ is a bijection if and only if:

- there exists a mapping $g: T \to S$ such that:
- $g \circ f = I_S$
- $f \circ g = I_T$

- where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.

## Proof

### Necessary Condition

Let $f$ be a bijection.

Then for each $y \in T$ there exists one and only one $x \in S$ such that $\map f x = y$.

That is, that there exists a mapping $g: T \to S$ with the property that:

- $\forall y \in T: \exists x \in S: \map g y = x$

Let $y \in T$.

Let $x = g \map g y$.

Then:

\(\displaystyle y\) | \(=\) | \(\displaystyle \map f {\map g y}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {f \circ g} y\) |

and:

\(\displaystyle x\) | \(=\) | \(\displaystyle \map g {\map f x}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {f \circ g} x\) |

demonstrating that $g$ has the property that:

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

$\Box$

### Sufficient Condition

Suppose there exists a $g$ which satisfies the conditions on $f$.

By Condition for Composite Mapping to be Identity:

- $(1): \quad$ from $g \circ f = I_S$ it follows that $f$ is an injection.
- $(2): \quad$ from $f \circ g = I_T$ it follows that $f$ is a surjection.

Thus, by definition, $f$ is a bijection.

$\blacksquare$

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Theorem $2$