# Inverse is Mapping implies Mapping is Injection

## Theorem

Let $S$ and $T$ be sets.

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

Let the inverse $f^{-1} \subseteq T \times S$ itself be a mapping.

Then $f$ is an injection.

## Proof 1

Let $f^{-1}: T \to S$ be a mapping.

Let $\map f {x_a} = y$ and $\map f {x_b} = y$.

Then:

 $\ds \tuple {x_a, y}$ $\in$ $\ds f$ Definition of Mapping $\, \ds \land \,$ $\ds \tuple {x_b, y}$ $\in$ $\ds f$ $\ds \leadsto \ \$ $\ds \tuple {y, x_a}$ $\in$ $\ds f^{-1}$ Definition of Inverse of Mapping $\, \ds \land \,$ $\ds \tuple {y, x_b}$ $\in$ $\ds f^{-1}$ $\ds \leadsto \ \$ $\ds x_a$ $=$ $\ds x_b$ Definition 4 of Mapping: $f^{-1}$ is many-to-one

Thus, by definition, $f$ is an injection.

$\blacksquare$

## Proof 2

Let $f^{-1}: T \to S$ be a mapping.

 $\ds \map f x$ $=$ $\ds \map f y$ $\ds \leadsto \ \$ $\ds \map {f^{-1} } {\map f x}$ $=$ $\ds \map {f^{-1} } {\map f y}$ as $f^{-1}$ is a mapping $\ds \leadsto \ \$ $\ds x$ $=$ $\ds y$ Definition 1 of Inverse Mapping: as $\map {f^{-1} } {\map f x}$ and so on

Thus $f$ is by definition an injection.

$\blacksquare$