# Inverse Mapping is Bijection

## Contents

## Theorem

Let $S$ and $T$ be sets.

Let $f: S \to T$ and $g: T \to S$ be inverse mappings of each other.

Then $f$ and $g$ are bijections.

## Proof

\(\displaystyle f \paren x\) | \(=\) | \(\displaystyle f \paren y\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle g \paren {f \paren x}\) | \(=\) | \(\displaystyle g \paren {f \paren y}\) | as $g$ is a mapping | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle y\) | Definition 1 of Inverse Mapping |

Thus $f$ is by definition an injection.

\(\displaystyle t\) | \(\in\) | \(\displaystyle T\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle g \paren t\) | \(\in\) | \(\displaystyle S\) | as $g$ is a mapping | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle f \paren {g \paren t}\) | \(=\) | \(\displaystyle t\) | Definition 1 of Inverse Mapping | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\, \displaystyle \exists s \in S: \, \) | \(\displaystyle f \paren s\) | \(=\) | \(\displaystyle t\) | setting $s = g \paren t$ |

Thus $f$ is by definition a surjection.

Similarly:

\(\displaystyle g \paren x\) | \(=\) | \(\displaystyle g \paren y\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle f \paren {g \paren x}\) | \(=\) | \(\displaystyle f \paren {g \paren y}\) | as $f$ is a mapping | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle y\) | Definition 1 of Inverse Mapping |

Thus $g$ is by definition an injection.

\(\displaystyle s\) | \(\in\) | \(\displaystyle S\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle f \paren s\) | \(\in\) | \(\displaystyle T\) | as $f$ is a mapping | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle g \paren {f \paren s}\) | \(=\) | \(\displaystyle s\) | Definition 1 of Inverse Mapping | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\, \displaystyle \exists t \in T: \, \) | \(\displaystyle g \paren t\) | \(=\) | \(\displaystyle s\) | setting $t = f \paren s$ |

Thus $g$ is by definition a surjection.

So $f$ and $g$ are both injections and surjections.

The result follows by definition of bijection.

$\blacksquare$

## Also see

## Sources

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