Injection from Finite Set to Itself is Surjection

Theorem

Let $S$ be a finite set.

Let $f: S \to S$ be an injection.

Then $f$ is also a surjection.

Corollary

Let $S$ be a finite set.

Let $f: S \to S$ be an injection.

Then $f$ is a permutation.

Proof

Let $a \in S$.

We need to show that there exists $b \in S$ such that $a = \map f b$.

Consider what happens when $f$ is applied repeatedly on $S$.

Let $f^2$ denote $f \circ f$ and, generally, $f^n := f \circ f^{n-1}$.

Consider the sequence of elements of $S$:

$a, \map f a, \map {f^2} a, \ldots$

Because $S$ is a finite set, there must be repetitions.

That is, there must exist $r, s \in \N$ such that:

$\map {f^r} a = \map {f^s} a$

where $r \ne s$.

Without loss of generality, assume $r > s$.

$f$ is an injection.

Therefore by Composite of Injections is Injection, $f^s$ is an injection.

By Injection iff Left Cancellable, $f^s$ is left cancellable.

Thus:

That is, $b = \map {f^{r - s - 1} } a$.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $3$. Mappings: Exercise $8$
• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Lemma $3$