Inverse is Mapping implies Mapping is Injection/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 an injection.

Let $f^{-1}: T \to S$ be a mapping.

Let $\map f {x_a} = y$ and $\map f {x_b} = y$.

Then:

$\ds \tuple {x_a, y}$

$\in$

$\ds f$

Definition of Mapping

$\, \ds \land \, $

$\ds \tuple {x_b, y}$

$\in$

$\ds f$

$\ds \leadsto \ \ $

$\ds \tuple {y, x_a}$

$\in$

$\ds f^{-1}$

$\, \ds \land \, $

$\ds \tuple {y, x_b}$

$\in$

$\ds f^{-1}$

$\ds \leadsto \ \ $

$\ds x_a$

$=$

$\ds x_b$

Definition 4 of Mapping: $f^{-1}$ is many-to-one

Thus, by definition, $f$ is an injection.

$\blacksquare$