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.
$\blacksquare$