Inverse of Injective and Surjective Mapping is Mapping/Proof 1

Proof
Recall the definition of the inverse of $f$:

$f^{-1} \subseteq T \times S$ is the relation defined as:
 * $f^{-1} = \set {\tuple {t, s}: t = \map f s}$

Let $f: S \to T$ be a mapping such that:
 * $(1): \quad f$ is an injection
 * $(2): \quad f$ is a surjection.

By Inverse of Injection is Many-to-One Relation, $f^{-1}$ is many-to-one.

That is:
 * $\forall y_1, y_2 \in T: \map {f^{-1} } {y_1} \ne \map {f^{-1} } {y_2} \implies y_1 \ne y_2$

Hence the preimage $\map {f^{-1} } y$ has at most one element.

By definition of surjection:
 * $\Img f = T$

That is:
 * $\forall y \in T: \exists x \in S: \map {f^{-1} } y = x$

Hence the preimage $\map {f^{-1} } y$ has at least one element.

Thus the preimage $\map {f^{-1} } y$ has exactly one element.

Hence, by definition, $f^{-1}$ is a mapping.