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Let $C$ be a cocomplete category, and suppose that it has an object that is colimit dense. Is $C$ automatically monadic over $Set$? And if not, is there an explicit counterexample?

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  • $\begingroup$ I think the category $\mathsf{Cat}$ of small categories is a counterexample. Or Simplicial Sets. More generally, if a category is locally $\lambda$-presentable but its $\lambda$-presentable objects are not generated under $\lambda$-small colimits by a single object, I would not expect it to be monadic over $\mathsf{Set}$. But I'm not that familiar with what can actually be said about categories that are monadic over $\mathsf{Set}$ without rank, so I don't have a proof. $\endgroup$
    – Tim Campion
    Jan 11, 2015 at 23:25
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    $\begingroup$ @arsmath What do you mean by "colimit dense"? Do you just mean dense? $\endgroup$
    – Zhen Lin
    Jan 11, 2015 at 23:34
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    $\begingroup$ @TimCampion One way of seeing that $Cat$ can't be monadic (in any way, shape, or form) over $Set$ is that such categories are regular, and $Cat$ isn't. This result is true without conditions on rank. $\endgroup$
    – Todd Trimble
    Jan 12, 2015 at 0:13
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    $\begingroup$ @Shamisen: are you going to tag every question that asks "is this true or false" with the counterexamples tag? $\endgroup$
    – Yemon Choi
    Jan 12, 2015 at 0:42
  • $\begingroup$ @YemonChoi Sorry... I read the first comment thinking it was already a counterexample; Now to think about it, I guess I have misused this tag in another questions too. Thanks for the advice and sorry for the trouble. $\endgroup$
    – Tadashi
    Jan 12, 2015 at 1:39

2 Answers 2

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I guess I'll go out on a limb and assume arsmath means, when he says an object $c$ of $C$ is colimit-dense, that the full subcategory containing $c$ is a dense subcategory in the usual sense.

There's some categorical lore which is helpful: a necessary condition for a category $C$ to be monadic over $Set$ (that is, for there to exist some monadic functor $U: C \to Set$) is that $C$ be a regular category. I think this result is proved somewhere in Francis Borceux's Handbook of Categorical Algebra, but the quickest reference I can find to hand is here, Proposition 1, part (2) (here $Set$ is a regular category, and any monad $T: Set \to Set$ preserves epis because every epi in $Set$ is already split, by the axiom of choice).

Now I claim the category $Pos$ of posets is not regular, hence cannot be monadic over $Set$. It is not regular because the pullback of a regular epi in $Pos$ need not be a regular epi; see the explicit example described here. On the other hand, it isn't too hard to see that $Pos$ is cocomplete and the two-element chain $\{0 \leq 1\}$ is colimit dense in $Pos$. So this gives an explicit counterexample.

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  • $\begingroup$ The converse direction of the characterization theorem in the linked-to note of Vitale does show that if a cocomplete category with a dense generator is exact, then it is monadic over $\mathsf{Set}$. This surprises me: it implies that every Grothendieck topos is monadic over $\mathsf{Set}$. In particular, contrary to my claim, simplicial sets are monadic over $\mathsf{Set}$. $\endgroup$
    – Tim Campion
    Jan 12, 2015 at 3:42
  • $\begingroup$ By "colimit-dense" I mean that every object is a colimit of some diagram, not necessarily the canonical diagram, but your example works for that as well. I knew Pos wasn't monadic over Set, but I assumed that you couldn't get the 3-element chain as a colimit from the 2-element chain, though now I see that you can. I think I overgeneralized the proof that Pos is not regular (that you link to) to show that the 3-element chain was not a colimit, though it clearly doesn't follow. $\endgroup$
    – arsmath
    Jan 12, 2015 at 7:56
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    $\begingroup$ @TimCampion But it's not true. A Grothendieck topos need not contain a generating object, let alone one that is projective. $\endgroup$
    – Zhen Lin
    Jan 12, 2015 at 8:29
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    $\begingroup$ @TimCampion But I think you're right that simplicial sets are monadic, as the sum $H$ of the representables is (regular) projective and is a dense generator (in this example, every representable is a retract of $H$). I hadn't realized that before. $\endgroup$
    – Todd Trimble
    Jan 12, 2015 at 8:39
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    $\begingroup$ @ZhenLin Ah, I see what I did wrong. I thought the coproduct of the objects of the generator would form a generator. But this won't be true generically because even though an object $X$ admits a map from some object of the generator, it need not admit a map from every object of the generator, so it need not admit a map from the coproduct. It is interesting that this construction does work for simplicial sets. $\endgroup$
    – Tim Campion
    Jan 12, 2015 at 13:30
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I just want to add an argument that $\mathsf{Cat}$ is also a counterexample (partly because I suspect that Todd chose to use $\mathsf{Pos}$ instead on account of it not being obvious that $\mathsf{Cat}$ has a dense generating object :). The same examples that show $\mathsf{Pos}$ is not regular should show that $\mathsf{Cat}$ is not regular. So the question is how to find a dense generating object for $\mathsf{Cat}$.

In fact $\sum_n \Delta^n$, the coproduct of the dense generating set of finite linear orders / simplices is a dense generator in $\mathsf{Cat}$ / $\mathsf{sSet}$. This is not immediate, as Zhen Lin points out above -- the coproduct of the objects of a dense generating set need not even be a generating object in general, as evidenced by sheaves over any nontrivial space (with the representables as the dense generating family).

Nonetheless, $\sum_n\Delta^n$ is a dense generator in $\mathsf{Cat}$ or in $\mathsf{sSet}$ because every representable is a retract of $\sum_n \Delta^n$, and a subcategory is dense iff its closure under retracts is dense (since if $i: C \to \tilde C$ is a full subcategory such that every object of $\tilde C$ is a retract of an object of $\tilde C$, then $\mathsf{Hom}(i,1): [\tilde C^\mathrm{op}, \mathsf{Set}] \to [C^\mathrm{op}, \mathsf{Set}]$ is an equivalence, i.e. $i$ is a Morita equivalence). Actually, in $\mathsf{Cat}$, it's clear that the three-element set $\{[0],[1],[2]\} = \{1,2,3\}$ consisting of the simplices of dimension $\leq 2$ / ordinals $\leq 3$ is dense, but these objects are all retracts of the single simplex $[2]$ / ordinal 3. So this object gives an even simpler dense generator for $\mathsf{Cat}$. The (nerve of) the ordinal $\omega$ would also do for a dense generator in either category.

As Todd observed, by the theorem discussed by Vitale in the notes he linked to,

A category is monadic over $\mathsf{Set}$ if and only if it is exact and contains a regular projective generating object.

(here's the link again), it follows (once we observe that the representable simplicial sets are projective, so their coproduct is too, and that a dense generator is a regular generator) that $\mathrm{Hom}(\sum_n \Delta^n, 1): \mathsf{sSet} \to \mathsf{Set}$ is monadic, surprising as that seems! And categories and simplicial sets are just certain types of $M$-set where $M$ is the endomorphism monoid of $\sum_n \Delta^n$ (via full, reflective inclusions)! I haven't thought about how to characterize the image of these inclusion functors.

In any 2-valued Grothendieck topos (i.e. a topos category such that every non-initial object has a point), the coproduct of a dense generating set similarly contains every member of the generating set as a retract, and so serves as a dense generating object. If this topos is a presheaf category where the base category has a terminal object, then the representables are a projective dense generating family, so their coproduct is a projective dense generating object, so 2-valued presheaf toposes where 1 is projective are monadic over $\mathsf{Set}$.

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  • $\begingroup$ Thanks, that example definitely never would have occurred to me. $\endgroup$
    – arsmath
    Jan 13, 2015 at 10:15
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    $\begingroup$ I just made an edit -- the ordinal 3 is actually already a dense generator of $\mathsf{Cat}$, which certainly seems more natural, given that it represents composition of morphisms. $\endgroup$
    – Tim Campion
    Jan 13, 2015 at 12:25
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    $\begingroup$ No, that's not why I chose $Pos$. I chose $Pos$ partly because it's the first example which came to my mind when I read the question, and partly for the sake of variety after I read your comment, and partly because general coequalizers in $Cat$ are messy, so that the easiest proofs of non-regularity already live in $Pos$. $\endgroup$
    – Todd Trimble
    Jan 13, 2015 at 14:17
  • $\begingroup$ All good reasons. I hope my comment didn't come across as negative. Also it occurs to me that maybe a "minimal" dense generating object of $\mathsf{sSet}$ would be the infinite unit ball: the colimit of a chain of face maps $\Delta^n \to \Delta^{n+1}$. I should also point out that basically everything I've said here is probably best viewed as an unfolding of Todd's comment about $\sum_n \Delta^n$. $\endgroup$
    – Tim Campion
    Jan 13, 2015 at 16:22
  • $\begingroup$ Another obvious thing that just dawned on me: none of the generators considered for $\mathsf{Cat}$ is (regular) projective. Lifting against the coequalizer $1 \overset{\to}{\to} 2 \to \mathbf{B}\mathbb{N}$ fails for all of them, and I suspect that $\mathsf{Cat}$ lacks a projective generating set. The only generator in $\mathsf{sSet}$ which I'm sure is projective is $\sum_n \Delta^n$. $\endgroup$
    – Tim Campion
    Jan 14, 2015 at 0:20

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