# Inverse is Mapping implies Mapping is Injection

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## 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.

$\blacksquare$