"Numerical results" from category theory
I give several examples involving groupoid cardinality in this MO answer. Here's another example which is not in that answer. First I'll quote the brief introduction to groupoid cardinality from over there:
To say it very tersely, if $X$ is a groupoid with at most countably many isomorphism classes of objects, such that each automorphism group $\text{Aut}(x)$ is finite, we can define its groupoid cardinality
$$|X| = \sum_{x \in \pi_0(X)} \frac{1}{|\text{Aut}(x)|}$$
if this sum converges, where $\pi_0(X)$ denotes the set of isomorphism classes of objects. Groupoids such that this sum converges are called tame. Groupoid cardinality has the following properties:
- Normalization: $|\bullet| = 1$, where $\bullet$ denotes the terminal groupoid.
- Additivity: $|X \sqcup Y| = |X| + |Y|$.
- Multiplicativity: $|X \times Y| = |X| |Y|$.
- Covering: If $f : X \to Y$ is an $n$-fold covering map of groupoids then $n |X| = |Y|$.
These properties (in fact just normalization, additivity and covering) uniquely determine $|X|$ for finite groupoids (finitely many objects and morphisms). $|X|$ is even countably additive and this together with normalization and covering determines it for tame groupoids.
Now here is the example. Dobinski's formula for the Bell numbers $B_n$, which count the number of partitions of an $n$-element set, is
$$B_n = \frac{1}{e} \sum_{k \ge 0} \frac{k^n}{k!}.$$
This formula can be given a "bijective proof" using groupoid cardinality as follows; what we will prove is that
$$B_n \sum_{k \ge 0} \frac{1}{k!} = \sum_{k \ge 0} \frac{k^n}{k!}$$
by calculating the groupoid cardinality of a specific groupoid in two different ways. Fix a set $N$ of size $n$. The groupoid we'll consider is the groupoid whose objects are maps $N \to K$ from $N$ to another finite set $K$, and whose morphisms are isomorphisms between these.
On the one hand, if the cardinality $|K| = k$ of $K$ is fixed, the groupoid of maps $N \to K$ and isomorphisms has a $k!$-fold cover given by the groupoid of maps $N \to K$ where $K$ is equipped with a total order and isomorphisms are required to preserve this order. This groupoid is discrete (has no nontrivial automorphisms) and has exactly $k^n$ isomorphism classes of objects, and the groupoid of maps $N \to K$ is the quotient of it by the action of $\text{Aut}(K)$, so the groupoid cardinality is $\frac{k^n}{k!}$. Summing over all $k$ produces the RHS.
On the other hand, a map $f : N \to K$ produces a partition of $N$ given by its fibers, together with a finite set given by the complement of the image of $f$. Our groupoid is in fact equivalent to the groupoid whose objects are pairs consisting of a partition of $N$ and a finite set, and whose morphisms are isomorphisms of the finite set only (the morphisms involving the partition are "canceled out" by the fact that, if the image $K' \subseteq K$ of $f$ is fixed, $\text{Aut}(K')$ acts freely on the set of surjections $N \to K'$, with quotient the set of partitions of $N$ into $|K'|$ nonempty blocks). This groupoid has groupoid cardinality $B_n \sum_{k \ge 0} \frac{1}{k!}$ which is the LHS as desired.
Performing this second identification more carefully gives a bijective proof (in the ordinary, non-groupoid sense) of the identity
$$k^n = \sum_{j=0}^n \left\{ n \atop j \right\} (k)_j$$
where $(k)_j = k(k-1) \dots (k-(j-1))$ is the falling factorial and $\left\{ n \atop j \right\}$ is the Stirling number of the second kind counting the number of partitions of $n$ into $j$ parts; this identity corresponds exactly to counting functions $f : N \to K$ based on the size $j$ of the image. Substituting this expression into $\sum_{k \ge 0} \frac{k^n}{k!}$ gives a "traditional" proof of Dobinski's formula, but to my mind the groupoid cardinality argument is cleaner because it explains the meaning of this proof.