Surjection from Finite Set to Itself is Permutation
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Theorem
Let $S$ be a finite set.
Let $f: S \to S$ be an surjection.
Then $f$ is a permutation.
Proof
From Surjection iff Right Inverse, $f$ has a right inverse $g: S \to S$.
From Right Inverse Mapping is Injection, $g$ is an injection.
From Injection from Finite Set to Itself is Permutation, $g$ is a permutation and so a bijection.
From Inverse of Bijection is Bijection, $f$ is also a bijection.
Thus as $f$ is a bijection to itself, it is by definition a permutation.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $3$. Mappings: Exercise $8$
- 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts