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.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions