I ~~don't~~ believe the $(2,1)$-category $FinSpan$ has split idempotents.

**Question:** Is there a simple description of the idempotent completion of $FinSpan$?

Foundationally, we may think of $FinSpan$ as an $(\infty,1)$-category, so the idempotent completion exists as an $(\infty,1)$-category. It's in this sense that I mean the idempotent completion. I'm not sure if the idempotent completion is necessarily a $(2,1)$-category.

Here the objects of $FinSpan$ are finite sets, a morphism from $A$ to $B$ is span of finite sets, i.e. consists of maps $A \leftarrow X \to B$ where $X$ is also finite. Composition is given by pullback, and 2-cells are the obvious notion of isomorphism. This $(\infty,1)$-category has been constructed formally as a quasicategory by various authors.

Note that the homotopy category of $FinSpan$ is the category $FinCom_{free}$ of finitely-generated free commutative monoids (an $n$-element set corresponing to $\mathbb N^n$). ~~It's been pointed out to me that an example of an idempotent which doesn't split in $FinCom_{free}$ is given by the matrix $\begin{pmatrix} 0 & 1 \\ 0 & 1\end{pmatrix}$. I haven't actually checked that this lifts to a coherent idempotent in $FinSpan$, but my impression is that $FinComm_{free}$ has ~~*lots* of non-split idempotents, and I suspect that at least some of them must lift to coherent idempotents, which can't split if they don't split in the homotopy category.