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## Free cocommutative commutative Hopf monoids

I have some questions about generalizations of abelian groups, relative to symmetric monoidal categories.

1) Let $C$ be a cocomplete cartesian monoidal category with equalizers. I can show that the forgetful functor $U : \mathsf{Ab}(C) \to C$ has a left adjoint $F$. The idea is to write $U$ as the composition $\mathsf{Ab}(C) \to \mathsf{CMon}(C) \to C$, the second functor has a left adjoint (symmetric algebra), and the first functor has a left adjoint which is a generalization of the Grothendieck construction (should I explain it?). Is there are more direct description $F$? Also I would like to know if the corresponding monad $T : C \to C$ preserves reflexive coequalizers, because then $U$ is monadic, which would be useful.

2) Let $C$ be a cocomplete symmetric monoidal category (perhaps assumed to be presentable). I would like to extend 1) to the category $\mathrm{AbHopf}(C)$ of cocommutative commutative Hopf monoids in $C$, i.e. the category of abelian group objects in the cartesian monoidal category $\mathrm{CoMon}_c(C)$ of cocommutative comonoids in $C$. In particular, I would like to know if $\mathrm{AbHopf}(C) \to C$ is monadic, if the corresponding monad has a nice description and if it preserves reflexive coequalizers. Obviously it would be helpful to understand $\mathrm{CoMon}_c(C) \to C$ first.

There are some papers by Porst, Barr, Takeuchi and others about these sort of questions, but I haven't found an answer there. Actually my goal is to endow $\mathrm{AbHopf}(C)$ with a tensor product such that the left adjoint $C \to \mathrm{AbHopf}(C)$ is symmetric monoidal.

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 Please, can you explain how the first functor has a left adjoint which is a generalization of the Grothendieck construction? It is interesting. – Buschi Sergio Jan 15 at 19:45

As for question (1): assuming cocomplete cartesian monoidal category means that cartesian products distribute over colimits (and I know from past experience that you do mean this, Martin), then it's true that the monad $T$ preserves reflexive coequalizers. The main thing you need is that finitary power functors $c \mapsto c^n$ preserve reflexive coequalizers. This is a corollary of a result that you can find on the first page of chapter 0 of Johnstone's Topos Theory, which is a $3 \times 3$ lemma stating that if the rows and columns are reflexive coequalizer diagrams, then so is the diagonal. It easily follows from this lemma that for example the squaring functor $c \mapsto c \times c$ preserves reflexive coequalizers, and a similar inductive argument allows you to extend this to any finite power $c \mapsto c^n$.

To derive the fact that monads $T: C \to C$ based on a Lawvere algebraic theory $\theta$ preserve reflexive coequalizers, write

$$T(c) = \int^{n \in \mathrm{FinSet}^{op}} \hom_\theta(i(n), i(1)) \cdot c^n$$

where the tensor $S \cdot c$ of a set $S$ with an object $c$ is the coproduct of copies of $c$ indexed over $S$. (Here $i: \mathrm{FinSet}^{op} \to \theta$ denotes the unique (up to isomorphism) map of Lawvere algebraic theories, viewing $\mathrm{FinSet}^{op}$ as the "initial" Lawvere algebraic theory.) Since coend functors and tensor functors $S \cdot -$ preserve reflexive coequalizers, as does $c \mapsto c^{n}$, we see that $T$ does as well.

I can't think of a more direct nice description of the composite left adjoint $F$, nor do I think one is needed because I think the description you gave is plenty nice.

As for (2): the underlying functor is definitely not monadic. It's not even a right adjoint, because for example for $C = \mathrm{Vect}_k$, it fails to preserve the terminal object (which in $\mathrm{AbHopf}(\mathrm{Vect})_k$ is the monoidal unit $k$, as is the case just in $\mathrm{CoMon}(\mathrm{Vect})_k$).

Edit: Since this came up in comments, let me provide an alternative proof of the fact that finite power functors on a cocomplete cartesian monoidal category preserve reflexive coequalizers. Recall that a category $J$ is sifted if the diagonal functor $J \to J \times J$ is final (result due to Gabriel and Ulmer). A prototypical example is where $J$ is the generic parallel pair equipped with a section in common. Then follow Steve Lack's soft proof here, which uses just the assumption that $C$ is cocomplete cartesian monoidal and the finality of the diagonal on $J$, to show the binary product $C^2 \to C$ preserves reflexive coequalizers. Similarly, the $n$-fold product $C^n \to C$ preserves reflexive coequalizers. The $n$-fold power on $C$ is a composite of the diagonal $\Delta: C \to C^n$ (which is a left adjoint, thus colimit-preserving) with the $n$-fold product, so it too preserves reflexive coequalizers.

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 (1) I know this proof, but I thought it only works for the case $C=\mathrm{Set}$. (2) What about a tower of monads/comonads or something like that? – Martin Brandenburg Jan 15 at 19:21 Why do you think it works only over $C = Set$? (Which part do you mean by "it"?) – Todd Trimble Jan 15 at 19:22 As for comment question (2), it's true that cocommutative comonoids are comonadic over $C$ (I'm positive you can find this in work of Porst, for instance), and then $\mathrm{AbHopf}(\mathrm{CoMon}_c(C))$ is monadic over $\mathrm{CoMon}_c(C)$, as we've just been arguing. – Todd Trimble Jan 15 at 20:32 (1) What is $n$ and what is $|n|$? – Martin Brandenburg Jan 21 at 18:23 @Martin: sorry, there's a mistake in what I wrote for the coend. The coend should be over the category $I = \mathrm{FinSet}^{op}$, which is the initial finitary algebraic theory. The "$n$" will denote an object of $I$ which is an $n$-fold product of copies of the 1-element set $1$; you can just write $c^n$ instead of $c^{|n|}$ if you like. The point is that we are taking the coend $\int^{n \in I} F(n) \cdot G(n)$, where $G: I \to C$ is the covariant product-preserving functor taking $1$ to $c$, and $F: I^{op} \to Set$ is $\hom_\theta(i-, i1), where$i: I \to \theta\$ is the unique theory map. – Todd Trimble Jan 21 at 18:52