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The number of total surjective functions from $X$ to $Y$ is known to be $T = y!\left\{{x \atop y}\right\}$, with $|X|=x, |Y|=y$.

However, I am interested in the number $P$ of partial surjective functions, that is, elements of $X$ do not necessarily map to an element of $Y$.

An approach I was thinking about was that you can map each partial function $f:X \to Y$ to a total function $f': X \to Y \cup \{\epsilon\}$, where $\forall x \in X \setminus \operatorname{dom}(f), f'(x) = \epsilon$. That is, extending the domain of the partial function such that elements undefined through $f$ map to a special element in $f'$.

Using this, we can say: $$P = (y+1)!\left\{{x \atop y+1}\right\} + T.$$

The added $T$ is to count the partial functions that happen to be total. The first term does not count them since $\left\{{x \atop y+1}\right\}$ counts the number of ways to partition a set of $x$ objects into $y+1$ non-empty subsets. Interrestingly enough, by rearranging a bit, we also have:

$$P = y!\left\{{x+1 \atop y+1}\right\}.$$

Is the reasoning correct? Is there a more elegant approach?

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1 Answers1

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Your reasoning is correct. A more elegant approach to prove that formula might be to apply the original proof to partial functions encoded as total functions between pointed sets:

Adjoin an element $\epsilon$ to both $X$ and $Y$. The partial surjective functions $f: X \to Y$ are in one-to-one correspondence to the total surjective functions $f_\epsilon : X\sqcup\{\epsilon\} \to Y\sqcup\{\epsilon\}$ with $f_\epsilon(\epsilon)=\epsilon$. Their number is given by the number of ways to partition $X\sqcup\{\epsilon\}$ into $|Y\sqcup\{\epsilon\}|$ non-empty subsets times the number of ways to permute $Y\sqcup\{\epsilon\}$ such that $\epsilon$ remains fixed. The first factor is given by $\left\{{x+1 \atop y+1}\right\}$ and the second factor is given by $y!$ (with $|X|=x$ and $|Y|=y$).

Thomas Klimpel
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