Inverse of Injective and Surjective Mapping is Mapping
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 1
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.
$\blacksquare$
Proof 2
Let $f: S \to T$ be a mapping such that:
- $(1): \quad f$ is an injection
- $(2): \quad f$ is a surjection.
Let $t \in T$.
Then as $f$ is a surjection:
- $\exists s \in S: t = \map f s$
As $f$ is an injection, there is only one $s \in S$ such that $t = \map f s$.
Define $\map g t = s$.
As $t \in T$ is arbitrary, it follows that:
- $\forall t \in T: \exists s \in S: \map g t = s$
such that $s$ is unique for a given $t$.
That is, $g: T \to S$ is a mapping.
By the definition of $g$:
- $(1): \quad \forall t \in T: \map f {\map g t} = t$
Let $s \in S$.
Let:
- $(2): \quad s' = \map g {\map f s}$
Then:
\(\ds \map f {s'}\) | \(=\) | \(\ds \map f {\map g {\map f s} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map f s\) | from $(1)$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds s\) | \(=\) | \(\ds s'\) | as $f$ is an injection | ||||||||||
\(\ds \) | \(=\) | \(\ds \map g {\map f s}\) | from $(2)$ |
Thus $f: S \to T$ and $g: T \to S$ are inverse mappings of each other.
$\blacksquare$