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.

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    $\begingroup$ Does not $\left(\begin{smallmatrix}0&1\\0&1\end{smallmatrix}\right)$ split? It is $\binom11\left(0\ 1\right)$, with $\left(0\ 1\right)\binom11=1$ $\endgroup$ – მამუკა ჯიბლაძე Aug 17 '18 at 20:27
  • $\begingroup$ @მამუკაჯიბლაძე Er... of course, you're right. I think it would simplify my life if I'm completely mistaken and $FinSpan$ does have split idempotents after all... $\endgroup$ – Tim Campion Aug 17 '18 at 20:31
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    $\begingroup$ FWIW, I believe the idempotent completion of an $(n,1)$-category is always another $(n,1)$-category: the hom-space into or out of the splitting of an idempotent is a retract of a hom-space in the original category, and $n$-types are closed under retracts. $\endgroup$ – Mike Shulman Aug 18 '18 at 2:15

Sorry, this was much ado about nothing. $FinSpan$ is idempotent-complete.

Proposition: Let $C$ be an $\infty$-category, and let $Ho(C)$ be its homotopy category. If $Ho(C)$ is idempotent-complete, then $C$ is idempotent-complete.

(Of course, the converse is false, with $C=Top$ being a standard counterexample.)

Proof: Let $e$ be a coherent idempotent in $C$. It splits as $e = ir$ in $Ho(C)$, which lifts to a coherent split idempotent $e'$ in $C$, with the same underlying map as $e$. Now the original coherent idempotent $e$ splits in $C$ if and only if the infinite sequence $\xrightarrow e \xrightarrow e \dots$ has a colimit (this is a diagram indexed by the graph $(\mathbb N, +1)$). But this sequence is the same for $e$ as it is for $e'$, so because $e'$ splits in $C$, we have that $e$ splits in $C$.

Lemma: Let $E$ be an idempotent matrix over $\mathbb N$.

  1. Each diagonal entry $E_{ii}$ is either 0 or 1, and the rank of $E$ is the number of diagonal entries which are 1.

  2. For $i \neq j$, if $E_{ij} \neq 0$, then $E_{ji} = 0$.

  3. If $E_{ij} = 0$, then every string $E_{ii_1}E_{i_1 i_2}\cdots E_{i_r j}$ is $0$.

Proof: These are easy to deduce entrywise from the equation $E^{r+1} = E$, using that the entries of $E$ are nonnegative integers.

Proposition: $FinComm_{free}$ has split idempotents.

Proof: Let $E$ be an $n\times n$ idempotent matrix over $\mathbb N$. Assume without loss of generality that the first $k$ diagonal entries are $0$ and all the diagonal entries after that are 1. Then the matrix $E-I$ (where $I$ is the identity matrix) has its first $k$ diagonal entries as $-1$ followed by $0$'s in the last $n-k$ diagonal entries. Use Gaussian elimination to transform $E-I$ into a matrix where the first $k$ rows are lower triangular. Then using the above lemma, one can show that the resulting matrix still has $-1$'s on the diagonal of the first $k$ rows, and thus one has produced a basis over $\mathbb N$ where the last $(n-k)$ basis elements are in the kernel of $E-I$, i.e. are fixed points of $E$. This is equivalent to a splitting of $E$ over $\mathbb N$.

Corollary: $FinSpan$ has split idempotents.

Proof: Since $Ho(FinSpan) = FinComm_{free}$, this follows from the two propositions.


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