Inverse is Mapping implies Mapping is Surjection/Proof 1

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.

Aiming for a contradiction, suppose $f$ is not a surjection.

That is:

$\exists y \in T: \neg \exists x \in S: \tuple {x, y} \in f$

By definition of inverse of mapping:

$\exists y \in T: \neg \exists x \in S: \tuple {y, x} \in f^{-1}$

which would mean that $f^{-1}$ is not a mapping.

From this contradiction it follows that $f$ is a surjection.

$\blacksquare$