Inverse of Injective and Surjective Mapping is Mapping/Proof 1

Theorem

Let $f: S \to T$ be a mapping such that:

$(1): \quad f$ is an injection

$(2): \quad f$ is a surjection.

Then the inverse $f^{-1}$ of $f$ is itself a mapping.

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.

From Inverse of Surjection is Relation both Left-Total and Right-Total $\map {f^{-1} } y$ is left-total.

Thus $f^{-1}$ is:

many-to-one

and

left-total.

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

$\blacksquare$