# Bijection has Inverse Mapping

## Contents

## Theorem

Let $S$ and $T$ be sets.

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

Then there exists a mapping $g: T \to S$ which is the inverse mapping of $f$.

## Proof

Let $t \in T$.

Then as $f$ is a surjection:

- $\exists s \in S: t = f \left({s}\right)$

As $f$ is an injection, there is only one $s \in S$ such that $t = f \left({s}\right)$.

Define $g \left({t}\right) = s$.

As $t \in T$ is arbitrary, it follows that:

- $\forall t \in T: \exists s \in S: g \left({t}\right) = s$

such that $s$ is unique for a given $t$.

That is, $g: T \to S$ is a mapping.

By the definition of $g$:

- $(1): \quad \forall t \in T: f \left({g \left({t}\right)}\right) = t$

Let $s \in S$.

Let:

- $(2): \quad s' = g \left({f \left({s}\right)}\right)$.

Then:

\(\displaystyle f \left({s'}\right)\) | \(=\) | \(\displaystyle f \left({g \left({f \left({s}\right)}\right)}\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle f \left({s}\right)\) | from $(1)$ | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle s\) | \(=\) | \(\displaystyle s'\) | as $f$ is an injection | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle g \left({f \left({s}\right)}\right)\) | from $(2)$ |

Thus $f: S \to T$ and $g: T \to S$ are inverse mappings of each other.

$\blacksquare$

## Also see

## Sources

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 1.9$: Inverse Functions, Extensions, and Restrictions