# Inverse of Injective and Surjective Mapping is Mapping/Proof 1

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## 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:

and

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

$\blacksquare$