# 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.

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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

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.

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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