Inverse is Mapping implies Mapping is Injection/Proof 2

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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

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.

$\blacksquare$


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