Cardinality of Set of Surjections
Theorem
Let $S$ and $T$ be finite sets.
Let $\card S = m, \card T = n$.
Let $C$ be the number of surjections from $S$ to $T$.
Then:
- $C = n! \ds {m \brace n}$
where $\ds {m \brace n}$ denotes a Stirling number of the second kind.
Proof
Let $T$ be the codomain of a surjection $f$ from $S$ to $T$.
By the Quotient Theorem for Surjections, $f$ induces an equivalence $\RR_f$ on $T$:
- $f = r \circ q_{\RR_f}$
where:
- $\RR_f$ is the equivalence induced by $f$ on $T$
- $r: S / \RR_f \to T$ is a bijection from the quotient set $S / \RR_f$ to $T$
- $q_{\RR_f}: S \to S / \RR_f$ is the quotient mapping induced by $\RR_f$.
From the Fundamental Theorem on Equivalence Relations, $\RR_f$ induces a partition on $S$.
From Cardinality of Set of Induced Equivalence Classes of Surjection, $\RR_f$ has $m$ components.
From Number of Set Partitions by Number of Components, there are $\ds {m \brace n}$ different ways of performing such a partitioning.
From Cardinality of Set of Injections, there are $n!$ different bijections from $S / \RR_f \to T$.
The total number of surjections is then the product of these:
- $C = n! \ds {m \brace n}$
$\blacksquare$
Examples
Example: $m = n + 1$
Let $m = n + 1$.
Then:
- $C = \dfrac {n \paren {n + 1}!} 2$
Example: $m = n + 2$
Let $m = n + 2$.
Then:
- $C = \dfrac {n \paren {3 n + 1} \paren {n + 2}!} {24}$