Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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 '13 at 19:45
add comment

1 Answer

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.

share|improve this answer
Well, I think I can provide details as desired, but let me first suggest you consider it by analogy with monads induced from operads (so that we consider Lawvere algebraic theories as cartesian analogues of operads, or rather or props associated with operads). There is an analogous coend description of the monad $T: C \to C$ induced from a $V$-enriched operad $\theta$ ($C$ a $V$-cocomplete monoidal $V$-category, which looks like $T(c) = \int^{n \in \mathbf{P}} \theta(n) \cdot c^{\otimes n}$, where $\mathbf{P}$ is the category of permutations (made $V$-enriched via appropriate base change ... –  Todd Trimble Jan 22 '13 at 20:37
... $Set \to V$ left adjoint to $\hom(I, -): V \to Set$). So it's exactly analogous, except that instead of working within the doctrine of symmetric monoidal categories (which is the natural environment for permutative operads), we are working within the doctrine of cartesian monoidal categories, which is the natural environment to consider Lawvere algebraic theories. Here $FinSet^{op}$ plays a role exactly analogous to $\mathbf{P}$: it is the free cartesian monoidal category on one generator, just as $\mathbf{P}$ i the free symmetric monoidal category on one generator. –  Todd Trimble Jan 22 '13 at 20:52
If that's not satisfactory, here's another way to view this analogy. If $C$ is a symmetric monoidal $V$-category and $c$ is any object in $C$, recall that there is a tautological $V$-valued operad associated with $c$, whose values are $E_c(n) = \hom(c^{\otimes n}, c)$. Then if $\theta$ is a $V$-operad, a $\theta$-algebra structure on $c$ is tantamount to an operad morphism $\theta \to E_c$. It's the same deal in the doctrine of cartesian monoidal categories: if $C$ is cartesian monoidal and $c$ is an object, there is a tautological Lawvere theory where the objects are natural numbers and ... –  Todd Trimble Jan 22 '13 at 21:03
... where the morphisms $m \to n$ are identified with maps $c^m \to c^n$ in $C$. Call it $E_c$. If $\theta$ is any Lawvere theory, then a $\theta$-algebra structure on $c$ is tantamount to a morphism of Lawvere theories $\theta \to E_c$. Continuing the analogy: on the symmetric monoidal side, the operad morphism $\theta \to E_c$ is required to be a natural transformation between functors $\mathbf{P} \to V$, and the naturality $\theta(n) \to \hom(c^{\otimes n}, c)$ corresponds to a dinatural transformation $\theta(n) \cdot c^{\otimes n} \to c$, or to a map (cont.) –  Todd Trimble Jan 22 '13 at 21:12
$\int^{n \in \mathbf{P}} \theta(n) \cdot c^{\otimes n} \to c$. On the cartesian monoidal side, the morphism of Lawvere theories $\theta \to E_c$ is by definition a product-preserving functor, which can be reformulated in terms of a natural transformation between functors of the form $\mathrm{FinSet}^{op} \to Set$ (taking the POV that a Lawvere theory $\theta$ is determined from its values $\theta(n) := \hom_\theta(n, 1)$). By the same rigmarole, this corresponds to a dinatural transformation and thence to a map out of a coend. Hopefully all this indicates the formal proofs -- is this enough? –  Todd Trimble Jan 22 '13 at 21:18
show 7 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.