Inverse is Mapping implies Mapping is Surjection/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 a surjection.
Proof
Let $f^{-1}: T \to S$ be a mapping.
We have:
\(\ds t\) | \(\in\) | \(\ds T\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {f^{-1} } t\) | \(\in\) | \(\ds S\) | as $f^{-1}$ is a mapping | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map f {\map {f^{-1} } t}\) | \(=\) | \(\ds t\) | Definition 1 of Inverse Mapping | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists s \in S: \, \) | \(\ds \map f s\) | \(=\) | \(\ds t\) | setting $s = \map {f^{-1} } t$ |
Thus $f$ is by definition a surjection.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions