Inverse is Mapping implies Mapping is Injection

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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 $\map f {x_b} = y$.


\(\ds \tuple {x_a, y}\) \(\in\) \(\ds f\) 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.


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.


Also see