# Unitalization internal to monoidal categories

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.

• 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
• @Ricardo: Thanks, but Kelly's article doesn't discuss unitalization. – Martin Brandenburg Feb 18 '13 at 1:59

## 1 Answer

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 lax limits from $$Cat$$, and since $$S$$-$$\mathrm{Alg}_\mathrm{smd}$$ is a lax limit in $$Cat$$ (for essentially the same reason that Eilenberg-Moore categories for monads are 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.

• 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
• It looks good to me! – Mike Shulman Feb 28 at 4:03