Inverse is Mapping implies Mapping is Surjection/Proof 2

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.

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.

$\blacksquare$

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