Mapping is Injection and Surjection iff Inverse is Mapping/Proof 2
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Theorem
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
- $f: S \to T$ can be defined as a bijection in the sense that:
- $(1): \quad f$ is an injection
- $(2): \quad f$ is a surjection.
- the inverse $f^{-1}$ of $f$ is such that:
- for each $y \in T$, the preimage $\map {f^{-1} } y$ has exactly one element.
- That is, such that $f^{-1} \subseteq T \times S$ is itself a mapping.
Proof
Necessary Condition
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.
$\blacksquare$
$\Box$
Sufficient Condition
Let $f^{-1}: T \to S$ be a mapping.
By Inverse Mapping is Bijection, both $f$ and $f^{-1}$ are bijections.
Hence, in particular, $f$ is a bijection.
$\blacksquare$