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

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?

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    $\begingroup$ I edited to make the second iso in the second displayed line an equality, but then realized that depending on how we define unary coproducts it perhaps should be an iso; if the edit was in error I apologize. $\endgroup$ Commented Nov 2, 2023 at 1:27
  • $\begingroup$ Perhaps this should be a community wiki answer instead? $\endgroup$ Commented Nov 2, 2023 at 8:57
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    $\begingroup$ @AlecRhea: I’ve rolled back the edit — whether it’ll actually be an equality or iso here is a slightly fiddly question (since the coproduct here is a special case of an arbitrary set-indexed coproduct, I think under most constructions it won’t be one, but one could reasonably special-casd the unary ones) — but the main point is that it doesn’t matter, the isomorphism is really what we need — so whenever the question is at all subtle, it’s better to just leave the iso rather than wasting time worrying about whether it’s an equality. $\endgroup$ Commented Nov 2, 2023 at 10:29
  • $\begingroup$ Sorry, could not find time to analyze it properly, but it seems there are several comonad structures on $G(X)=X\times2^X$ $\endgroup$ Commented Nov 3, 2023 at 17:21
  • $\begingroup$ I'm assuming that the definition for a comonad to be polynomial is that its functor part is a polynomial. $\endgroup$ Commented Oct 20, 2024 at 19:01

3 Answers 3

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A non-pullback-preserving example:

The category of undirected multigraphs (loops allowed) $\mathbf{Multigraph}$ is comonadic over $\mathbf{Set}$. The forgetful functor $\mathbf{Multigraph} \to \mathbf{Set}$ sends a multigraph to the disjoint union of its vertices and edges, and the right adjoint $\mathbf{Set} \to \mathbf{Multigraph}$ sends a set $X$ to the multigraph with $X$ vertices and $X$ edges between every (unordered) pair of vertices.

The formula for the induced comonad $F\colon\mathbf{Set} \to \mathbf{Set}$ (the composite of these two functors) is $$\small X \times (\text{ways to put $2$ indistinguishable balls in $X$ distinguishable bins}) + X = \frac{X^3 + X^2}{2} +X.$$ This is not a polynomial endofunctor. Morever, it does not preserve pullbacks: for any set $X$, the pullback of $!\colon X \to 1$ along itself is $X^2$, but the pullback of $F(!) \colon F(X) \to F(1)$ along itself is $(\frac{X^3 + X^2}{2})^2 + X^2$, not $F(X^2) = \frac{X^6 + X^4}{2} + X^2$. (That is, squaring every summand is not the same as substituting a squared variable.)

Where does this example come from? As mentioned in the question, polynomial comonads on $\mathbf{Set}$ correspond to categories. So let's aim to understand general comonads on $\mathbf{Set}$ as "generalized categories".

If $C$ is a category, then the category of coalgebras over the corresponding comonad is equivalent to the category of functors $C \to \mathbf{Set}$. The forgetful functor $(C \to \mathbf{Set}) \to \mathbf{Set}$ is defined by $P \mapsto \sum_{c \in C} P(c)$, and the right adjoint $\mathbf{Set} \to (C \to \mathbf{Set})$ is defined by $X \mapsto (c \mapsto X^{|c / C|})$, where $|c / C|$ denotes the set of arrows out of $c$.

For example, the category of quivers $\mathbf{Quiver}$ (a.k.a. directed multigraphs) is comonadic over $\mathbf{Set}$, via a polynomial comonad, as quivers are presheaves over the category $\bullet \rightrightarrows \bullet$. The forgetful functor $\mathbf{Quiver} \to \mathbf{Set}$ sends a quiver to the disjoint union of its vertices and edges, and the right adjoint $\mathbf{Set} \to \mathbf{Quiver}$ sends a set $X$ to the quiver with $X$ vertices and $X$ edges between every (ordered) pair of vertices. The formula for the induced polynomial comonad $F_{\mathrm{directed}}: \mathbf{Set} \to \mathbf{Set}$ is $X^3 + X$.

One way to construct new comonads on $\mathbf{Set}$ is by taking colimits of old ones. (It is a general fact that a colimit in a category of comonoids exists and is computed pointwise if the pointwise colimit exists. Therefore, the category of comonads on $\mathbf{Set}$ has colimits, computed at the level of the underlying endofunctors.) The undirected multigraph comonad $F$ is the coequalizer of the identity morphism $\mathrm{id} \colon F_{\mathrm{directed}} \to F_{\mathrm{directed}}$ and the comonad automorphism $\phi \colon F_{\mathrm{directed}} \to F_{\mathrm{directed}}$ corresponding to the automorphism of $\bullet \rightrightarrows \bullet$ that switches the two parallel arrows. $$F_{\mathrm{directed}} \overset{\mathrm{id}}{\underset{\phi}{\rightrightarrows}} F_{\mathrm{directed}} \twoheadrightarrow F$$

In summary, undirected multigraphs are the "presheaves" over a "generalized category" obtained by forgetting how to distinguish between the two parallel arrows in $\bullet \rightrightarrows \bullet$. (That is, forgetting how to distinguish between the source and the target of an edge.)

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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...

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  • $\begingroup$ How do you show that this is polynomial only when the monoid is discrete? I can see how to show it’s non-polynomial in various specific cases, I think, but I don’t see a general argument. $\endgroup$ Commented Nov 2, 2023 at 22:37
  • $\begingroup$ My reasoning was that for a polynomial comonad the algebra were presheaves categories, and topos of smooth M-Sets aren't. But now that I think about it, it is clear for localic groups, but maybe not so much for localic monoids... I need to think more about it. $\endgroup$ Commented Nov 2, 2023 at 23:01
  • $\begingroup$ (For e.g. topos of smooth G-set are boolean but don't satisfies AC, while presheaf topos are boolean only if they are presheaves on a groupoid, in which case they satisfies internal AC) $\endgroup$ Commented Nov 2, 2023 at 23:03
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    $\begingroup$ For groups, at least, you could also do this without any topos theory: if intersections of open subgroups of $G$ aren't always open then the forgetful functor out of continuous $G$-sets doesn't preserve infinite products, which implies it doesn't preserve connected limits (since it preserves the terminal object), so that neither does the corresponding comonad. $\endgroup$ Commented Nov 3, 2023 at 3:40
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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.

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    $\begingroup$ I was going to extend my answer to include this as well. You get a comonad like this for every Grothendieck topos and every "sufficient" set of points of that topos (So for "Ionads" in the sense of Garner's arxiv.org/abs/0912.1415). Even if the topos is presheaf, this monads will be polynomial only if the chosen points are representble (essentials). $\endgroup$ Commented Nov 3, 2023 at 21:55
  • $\begingroup$ Ahh, thanks for that clarification, I was definitely thinking of the representable points without noticing: $\endgroup$ Commented Nov 3, 2023 at 22:09
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    $\begingroup$ I think this fiber product preserving comonad/"ionads" correspondence is a very neat generalization of Ahman and Uustalu's theorem and should be written out in more details somewhere! and maybe functoriality aspects looked at a little more. Are you Interested in collaborating on this? ( Sorry I couldn't find your current email - I believe mine is on my profile.) $\endgroup$ Commented Nov 3, 2023 at 22:52
  • $\begingroup$ Oops, my profile had gotten quite out of date! Thanks for the suggestion. I'll send you an email. $\endgroup$ Commented Nov 3, 2023 at 22:58
  • $\begingroup$ This is great, thank you! For each such comonad $pts_X: \mathbf{Set}\to\mathbf{Set}$, how does the associated sheaf category $Sh(X)$ compare with algebras of the associated "gluing" monad (which is familial) $\text{glue}_X: Psh(X)\to Psh(X)$ on the presheaf category for the site $Op(X)$? Happy to ask in a new question if that's preferable. And a remedial question if you feel like answering: How does the property of a space's sheaf category being of presheaf type compare with the property that a space is "Alexandrov" over its poset of opens or that of its points? $\endgroup$ Commented Nov 5, 2023 at 18:04

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