Examples of non-polynomial comonads on Set?

Question

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Question: What are examples of comonads on $\mathbf{Set}$ that are not polynomial?

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Background: polynomial functors and comonads on Set

A functor $F\colon\mathbf{Set}\to\mathbf{Set}$ is called polynomial if it is a sum of representables, i.e. if there is some set $I$ and for each $i:I$ a set $S_i$ and a natural isomorphism $$ F(X)\cong\sum_{i:I}\mathbf{Set}(S_i, X) $$ over $X:\mathbf{Set}$.

For example, $X\mapsto X^2$ is polynomial, because we can take $I:=\{\bullet\}$ and $S_\bullet:=\{1,2\}$ and $$X^2\cong\mathbf{Set}(\{1,2\},X)\cong\sum_{i:\{\bullet\}}\mathbf{Set}(S_i,X).$$ Similarly the constant functor $X\mapsto 2$ is polynomial because we can take $I:=\{1,2\}$ and $S_1=S_2:=\varnothing$.

A comonad on $\mathbf{Set}$ is a tuple $(F,\epsilon,\delta)$ where $F\colon\mathbf{Set}\to\mathbf{Set}$ is a functor, $\epsilon\colon F\to\mathsf{id}_\mathbf{Set}$ and $\delta\colon F\to F\circ F$ are natural transformations, and these satisfy the usual "comonoid" laws, coassociativity and counitality.

Setup: It's easy to find polynomial comonads

There are many polynomial comonads on $\mathbf{Set}$; in fact, a result of Ahman and Uustalu says that they are in bijection with categories: every category can be identified with a polynomial comonad and vice versa.

But what about non-polynomial comonads? Picking up on Mathoverflow: "Big list of comonads", which asks for comonads from all around math, I ask a much more refined question.

Question:

$\quad$ What are examples of comonads on $\mathbf{Set}$ that are not polynomial?

Answer 1

Here's something inspired by Simon's answer that gets a ways beyond surjections out of $\mathbf{Set}.$ This is all very much riding the edge of my comfort zone with topos theory so do let me know if something seems wrong. In short, I aim to show below that a pullback-preserving comonad on $\mathbf{Set}$ is the same thing as a Grothendieck topos with enough points, so that the non-polynomial pullback-preserving comonads correspond to the toposes that aren't presheaf toposes.

Let $X$ be a topological space with underlying set $U(X).$ Then there's a geometric surjection $f:\mathbf{Set}/U(X)\to \mathbf{Sh}(X)$ whose inverse image takes all the stalks and whose direct image takes the product of skyscraper sheaves. This makes $\mathbf{Sh}(X)$ comonadic over $\mathbf{Set}/U(X).$ Now $\mathbf{Set}/U(X)$ is also comonadic over $\mathbf{Set},$ with the forgetful functor $\Sigma$ given by summing over the fibers. These comonads compose: since $\Sigma$ preserves connected limits and $f^*$ preserves finite limits, both preserve coreflexive equalizers so we can just use crude comonadicity.

The composite gives a comonad on $\mathbf{Set}$ sending $A$ to the coproduct of all the stalks of the product of all the $A$-shaped skyscraper sheaves: $\coprod_{x\in X}x^*\prod_{x\in X} x_*A,$ whose coalgebras are sheaves on $X.$ (I'm sure this is well known but I'm just working it out for myself; maybe there's another more familiar description.)

Though the formula I just wrote looks awfully polynomial, I claim this usually is not a polynomial comonad. This is fundamentally because taking the stalk doesn't preserve infinite products, which is equivalent to preserving connected limits for a left exact functor. Specifically, if $S$ is a finite subset of $X$ then the stalk $x^*\prod_{x\in S} x_* A$ is $A,$ if $x\in S,$ and otherwise terminal. That means that the limit, over the (connected) category of finite subsets $S,$ of $x^*\prod_{x\in S} x_*A$ is just $A.$ But the stalks of $\prod_{x\in X}x_*A$ are not necessarily $A$ when $X$ is infinite. (For instance, if $A=\{0,1\}$ and $X=\mathbb R,$ consider the germ at $0$ of the section of $\prod_{x\in\mathbb R}x_*\{0,1\}$ that is $1$ on $x_*\{0,1\}$ on any open containing $x,$ except that it's $0$ on $0_*\{0,1\}.$)

I believe this gives a non-polynomial comonad on $\mathbf{Set}$ for every space $X$ whose sheaf category is not equivalent to a presheaf category. I suspect you could extend this argument even further by letting $X$ be the underlying space of a topological groupoid, which would lead to such a comonad for every Grothendieck topos with enough points not equivalent to a presheaf category. Grothendieck toposes with enough points are, I believe, characterized by inducing a pullback-preserving comonad on $\mathbf{Set},$ so this ought to completely cover that case, so that non-polynomial comonads are exactly the disjoint union of the non-presheaf category Grothendieck toposes and the comonads that don't preserve pullbacks.

Answer 2

In general, given a surjective geometric morphism $f: \mathcal{E} \to \mathcal{T}$ of toposes, then the adjunctions $f^* \vdash f_*$ is comonadic.

In particular, given a topological or localic monoid $M$, we have a canonical surjective geometric morphism $Set \to B M$ where $BM$ is the topos of sets endowed with a continuous (smooth) action of $M$.

So, there is a comonad on sets whose category of coalgebras is the category of (smooth) $BM$-sets.

In the case where $M$ is a localic group $G$, the underlying functor sends a set $X$ to the set of functions $G \to X$ that have an open stabilizer in the translation action of G ( so a kind of uniform continuity condition I guess). I assume a similar formula works for a monoids, but I'm less familiar with it so I don't want to claim it.

I think one recovers a polynomial example only in the case where the monoid is discrete... This suggest maybe there is a class of comonads corresponding to some topological categories...