# Example of a functor preserving only finite coproducts

What is an example of a functor $$F : \mathsf{Set} \to \mathsf{Set}$$ which preserves finite coproducts, but not infinite coproducts?

The functors preserving infinite coproducts are given by $T \times - : \mathsf{Set} \to \mathsf{Set}$ for some set $T$. If a functor preserves finite coproducts, then the natural map $F(1) \times X \to F(X)$ is an isomorphism if $X$ is finite. Thus, a counterexample will involve infinite sets.

For the category $\mathsf{Ab}$ of abelian groups, a counterexample is simply $X \mapsto X^{\mathbb{N}}$.

Edit.

Jakob Werner has suggested the functor $F(X) = \mathrm{Spec}(A^X)$ for any commutative ring $A$. For fields $A$ this gives the functor of ultrafilters in my answer. I am still curious if there are other, more basic classes of examples.

• Why doesn't the abelian counterexample work in Set? – Asaf Karagila Mar 22 '17 at 14:16
• @AsafKaragila: It doesn't preverve disjoint unions i.e. coproducts. – HeinrichD Mar 25 '17 at 5:18
• Right. Because $1^\Bbb N\coprod 1^\Bbb N\not\cong 2^\Bbb N$? – Asaf Karagila Mar 25 '17 at 6:18
• @AsafKaragila: I assume that you know the answer. ;-) – HeinrichD Mar 26 '17 at 9:58
• I actually don't. But if that is in fact the reason, then I think I have an answer to your question. – Asaf Karagila Mar 26 '17 at 10:05

As noted, the ultrafilter functor is an example; in fact, it is the terminal example, as shown by Reinhard Börger in this paper:

http://www.sciencedirect.com/science/article/pii/0022404987900417

All finite-coproduct-preserving endofunctors of $\mathbf{Set}$ are actually quite closely related to ultrafilters. Another basic example is as follows. If $\mathscr U$ is an ultrafilter on a set $A$, we can consider the ultrapower functor $(-)^{\mathscr U} \colon \mathbf{Set} \to \mathbf{Set}$. This sends a set $X$ to the quotient of $X^A$ by the equivalence relation wherein $\vec{x} \equiv \vec{y}$ just when $\{a \in A : x_a = y_a\} \in \mathscr U$ (i.e., when $\vec x = \vec y$ "$\mathscr U$-almost everywhere").

The ultrapower functors preserve finite coproducts (as well as finite limits). In fact, the ultrapower functors are basic in the sense that any finite-coproduct-preserving functor $\mathbf{Set} \to \mathbf{Set}$ is a colimit in $[\mathbf{Set}, \mathbf{Set}]$ of ultrapower functors.

Here is another example, again involving ultrafilters. Let $\mathbb{T}$ be any first-order theory and let $M$ and $N$ be models for $\mathbb{T}$. We can define a finite-coproduct-preserving functor $F_{M,N} \colon \mathbf{Set} \to \mathbf{Set}$ by taking

$$F_{M,N}(X) = \sum_{\mathscr U \in \beta X} \mathbf{Emb}(M^\mathscr{U}, N)$$

where here $\beta X$ is the set of ultrafilters on $X$; $M^{\mathscr U}$ is the ultrapower model of $\mathbb T$ (which is a model by Łoś's theorem); and $\mathbf{Emb}(M^{\mathscr U}, N)$ denotes the set of elementary embeddings of $M^{\mathscr U}$ in $N$.

The following example seems to work. Consider the composition $$F : \mathsf{Set} \xrightarrow{D} \mathsf{Top} \xrightarrow{\beta} \mathsf{CompHaus} \xrightarrow{U} \mathsf{Set},$$ where $D$ is the discrete topology, $\beta$ is the Stone-Cech-compactification, and $U$ is the forgetful functor. In other words, $F(X)$ is the set of ultrafilters on $X$. Since $D$ and $\beta$ are left adjoint and $U$ preserves finite coproducts, it follows that $F$ preserves finite coproducts. But it does not preserve arbitrary coproducts, since $F(\mathbb{N})$ is not isomorphic to $\coprod_{n \in \mathbb{N}} F(\{n\}) \cong \mathbb{N}$.

• This is more complicated than I expected it to be. Other answers, in particular with more basic examples of functors, are appreciated! – HeinrichD Mar 18 '17 at 15:47
• If $F$ is an example, and $x\in F(X)$ for some set $X$, let $\mathcal{F}$ be the collection of subsets $Y\subseteq X$ such that $x$ is in the image of the map $F(Y)\to F(X)$ induced by inclusion. Then $\mathcal{F}$ is an ultrafilter on $X$. So maybe it's not reasonable to hope for an example much simpler than yours. – Jeremy Rickard Mar 18 '17 at 21:15
• @JeremyRickard: This is interesting. But why is $\mathcal{F}$ closed under intersections? Also, does this imply that the ultrafilter functor is universal in some sense within the category $\hom_{\sqcup}(\mathsf{Set},\mathsf{Set})$ of finite-coproduct-preserving functors? – HeinrichD Mar 19 '17 at 7:53
• It's easier to show that the complement of $\mathcal{F}$ is closed under finite unions. Let $A,B\in\mathcal{F}^c$. To prove $A\cup B\in\mathcal{F}^c$ we can assume $A$ and $B$ are disjoint (otherwise replace $B$ with $B\setminus A$). Then it follows from $F(A\cup B)=F(A)\coprod F(B)$. – Jeremy Rickard Mar 19 '17 at 8:39
• Yes, it's universal. Börger proved the following (Coproducts and ultrafilters, 1987, Theorem 2.1): Each functor reserving finite coproducts has a unique natural transformation to the ultra-filter functor. In other words, the ultra-filter functor is terminal among those functors. – Ben Jun 13 '17 at 17:50

Here is a class of examples, generalizing yours: Look at the composition $$\mathbf{Set}\xrightarrow{\mathfrak{P}}\mathbf{CRing}^{\operatorname{op}}\xrightarrow{\operatorname{Spec}}\mathbf{Set}\,.$$ The first functor takes a set $X$ to the power set ring $\mathfrak{P}(X)\cong\mathbb{F}_2^X$. The interesting thing happens at the second arrow. One can check that the compositions is isomorphic to your functor $F$, as Asaf Karigala points out. But of course you can replace $\mathbb{F}_2$ by any other fixed commutative ring.

• I think that a prime ideal in the power set ring is exactly the co-ideal of an ultrafilter. So you get essentially the same functor as suggested by the OP in his answer. – Asaf Karagila Mar 22 '17 at 13:29
• Yes, that's why I said "for the same reasons". I'll try to make it more clear. – Jakob Werner Mar 22 '17 at 13:32

I think the example that HeinrichD gave himself is probably the canonical one.

But I'll just point out that it's not minimal, in that it has proper subfunctors that also work.

Heinrich's example can be restated as "$F(X)$ is the set of maximal ideals of $\mathbb{F}_2^X$". But you could also take the subset of maximal ideals for which the quotient field is bounded in size by some cardinal.

Or a similar idea: fix a commutative ring $R$ and an integral domain $S$, and let $F(X)$ be the set of ring homomorphisms $R^X\to S$.

• How do you prove that $F(X)=\hom(R^X,S)$ does not preserve coproducts for every integral domain $S$ and $R \neq 0$? – HeinrichD Mar 26 '17 at 18:24
• @HeinrichD Sorry, I didn't mean to imply that it never preserves coproducts (clearly it does for trivial reasons if $R$ and $S$ are fields of different characteristics, and I think it does if $S$ is a finite field), but in general it doesn't (e.g., take $R=\mathbb{F}_p$ and $S$ an algebraically closed field of large transcendence degree of characteristic $p$). – Jeremy Rickard Mar 26 '17 at 18:43