Unitalization internal to monoidal categories

Question

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.

Answer 1

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.