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

Let $C$ be a monoidal category, not assumed to be symmetric. Assume that the underlying category of $C$ is nice enough, for example cocomplete, perhaps even presentable. A semigroup object in $C$ is a pair $(X,\mu)$ consisting of an object $X \in C$ and a morphism $\mu : X \otimes X \to X$ satisfying the associativity law $\mu \circ (X \otimes \mu) = \mu \circ (\mu \otimes X)$. Does the forgetful functor from monoid objects in $C$ to semigroup objects in $C$ have a left adjoint? In other words, is there an unitalization internal to $C$?

The cases $C=\mathsf{Set}$ and $C=\mathrm{Mod}(R)$ are well-known. More generally, the answer is yes when $\otimes$ preserves coproducts in each variable. Then the unitalization of $(X,\mu)$ is $(1 \oplus X,\mu',\eta)$ with the obvious morphism $\mu' : (1 \oplus X) \otimes (1 \oplus X) = 1 \oplus X \oplus X \oplus X \otimes X \to 1 \oplus X$ and $\eta : 1 \to 1 \oplus X$.

Actually I am interested in the case that $C=(\mathrm{End}(D),\circ,\mathrm{id})$ for a (nice) category $D$, thus I would like to know if every semi-monad can be made into a monad. Here $\otimes$ preserves colimits in the left variable, but not in the right variable. Actually $D$ is even a presentable symmetric monoidal category and $\mathrm{End}(D)$ refers to enriched endofunctors, i.e. I am interested in strong (semi) monads.

share|improve this question
    
I know how to unitalize operads. This can be done as for algebras. Hence the semimonad associated to an operad can be unitalized. However I don't see a way to extend the construction to arbitrary semimonads. –  Fernando Muro Feb 17 '13 at 9:40
    
In case you have not seen the following mathoverflow question, I would like to direct your attention to mathoverflow.net/questions/19906/are-monads-monadic which is related to this question. In particular, take a look at Tom Leinster's answer and his reference to Kelly's article. –  Ricardo Andrade Feb 17 '13 at 22:51
1  
@Ricardo: Thanks, but Kelly's article doesn't discuss unitalization. –  Martin Brandenburg Feb 18 '13 at 1:59
add comment

1 Answer

Edit: an earlier version of this answer muddled a distinction between lax limit and 2-limit. I've decided to undelete it in case someone sees how to complete the argument at the end.


If $C$ is locally presentable and $S$ is a semi-monad whose underlying functor is accessible, then there exists a unitalization of $S$. Here is a proof modeled after an idea discussed at the nLab at the page free monad.

Define an algebra of a semi-monad $S: C \to C$ in the expected way, as an object $X$ of $C$ equipped with a morphism (an "action") $SX \to X$ satisfying the usual associativity law for an action. Morphisms between algebras are also defined in the expected way, so that there is a full embedding $S$-$\mathrm{Alg}_\mathrm{smd} \hookrightarrow S \downarrow C$ into the comma category. (I use the subscripts "smd" and "mnd" to indicate algebras qua semi-monads and monads.)

The main thing to check is that the forgetful functor $S$-$\mathrm{Alg}_\mathrm{smd} \to C$ is monadic in the "evil" sense, so that there is an isomorphism $F$-$\mathrm{Alg}_\mathrm{mnd} \simeq S$-$\mathrm{Alg}_\mathrm{smd}$ in $Cat/C$ for some monad $F$. The claim is that then $F$ is the free monad on the semi-monad $S$. For in that case, given a monad $M$ on $C$ we have natural bijections between

  • Semi-monad morphisms $S \to M$,

  • $S$-algebra structures $S U_M \to U_M$ where $U_M:$ $M$-$\mathrm{Alg}_\mathrm{mnd} \to C$ is the forgetful functor,

  • Morphisms $M$-$\mathrm{Alg}_\mathrm{mnd} \to S$-$\mathrm{Alg}_\mathrm{smd}$ in $Cat/C$,

  • Morphisms $M$-$\mathrm{Alg}_\mathrm{mnd} \to F$-$\mathrm{Alg}_\mathrm{mnd}$ in $Cat/C$,

  • $F$-algebra structures (qua algebras over a monad) $F U_M \to U_M$,

  • Monad morphisms $F \to M$

so that $F$ is evidently the free monad on the semi-monad $S$.

So now we check monadicity, using the precise monadicity theorem. It is straightforward that the forgetful functor $U: S$-$\mathrm{Alg}_{\mathrm{smd}} \to C$ creates (not just reflects!) $U$-split coequalizers, so we just have to check that $U$ has a left adjoint. However, since the 2-category of locally presentable categories and accessible functors inherits 2-limits from $Cat$, and since $S$-$\mathrm{Alg}_\mathrm{smd}$ is a 2-limit (edit: lax limit) in $Cat$ (for essentially the same reason that Eilenberg-Moore categories for monads are 2-limits lax limits), we see that $U: S$-$\mathrm{Alg}_\mathrm{smd} \to C$ is an accessible functor between locally presentable categories. In this situation, existence of a left adjoint to $U$ is equivalent to preservation of limits by $U$. But limit-preservation is clear. So the conditions of the precise monadicity theorem are satisfied.

Edit: The last paragraph would need to be fixed to make the argument complete, either by somehow showing $U$ lives in the 2-category of accessible categories and accessible functors (note that $S$-$\mathrm{Alg}_{\mathrm{smd}}$ is complete, and so would then be locally presentable), or e.g. by showing that the full inclusion $S$-$\mathrm{Alg}_{\mathrm{smd}} \hookrightarrow S \downarrow C$ is reflective, or by some other means.

share|improve this answer
    
A wonderfully abstract nonsense proof! –  Zhen Lin Feb 18 '13 at 18:41
    
I'm going to delete this answer for the moment, as the last paragraph could be playing a fast one. I may undelete it later, possibly after revisions. –  Todd Trimble Feb 18 '13 at 18:41
    
Todd, thanks a lot! I hope that the argument can be completed. It doesn't restrict to semimonads, right? It also seems to work for semigroup objects in presentable monoidal categories, right? @Zhen: Although it seems to be abstract at the first glance, it is quite simple: It is easy to unitalize semigroups described by generators and relations, one just can add a unit to the presentation. To make this work, one has to construct free semigroups. –  Martin Brandenburg Feb 18 '13 at 20:58
    
Yes, it would apply to semigroup objects as well. I think Mike Shulman wrote most of the pertinent nLab articles; if he sees this question, he might see either how to complete the argument or where it founders. I'll keep thinking about it myself. –  Todd Trimble Feb 18 '13 at 21:17
    
Surely, to show that $U$ is accessible, one simply uses the same kind of argument that shows that the forgetful functor from the Eilenberg–Moore category for an accessible monad creates sufficiently-filtered colimits? –  Zhen Lin Feb 18 '13 at 23:00
add comment

Your Answer

 
discard

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.