# Bijection iff Left and Right Inverse

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## Contents

## Theorem

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

$f$ is a bijection iff:

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

where both $g_1$ and $g_2$ are 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:

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

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

## Proof

### Necessary Condition

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

Let $f$ be such that:

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

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 $f$.

## Also see

- Bijection Composite with Inverse for the converse of this result.

## Sources

- George McCarty:
*Topology: An Introduction with Application to Topological Groups*(1967)... (previous)... (next): $\text{I}$: Composition of Functions - James R. Munkres:
*Topology*(2nd ed., 2000)... (previous)... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Lemma $2.1$ - James R. Munkres:
*Topology*(2nd ed., 2000)... (previous)... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.5 \ \text{(e)}$ - Paul Halmos and Steven Givant:
*Introduction to Boolean Algebras*(2008)... (previous)... (next): Appendix $\text{A}$: Set Theory: Bijections