# Explicit Left Adjoint to Forgetful Functor from Cartesian to Symmetric Monoidal Categories

There is a forgetful functor from the category of (small) Cartesian monoidal categories (a symmetric monoidal category in which the tensor product is given by the categorical product) to the category of symmetric monoidal categories. This functor is known to have a left adjoint and right adjoint. The right adjoint is easy to describe: it takes a symmetric monoidal category to its category of cocommutative coalgebras. I am looking for an explicit description of the left adjoint though. Does anyone know what it is?

There is a MO question about this, but it does not give an explicit description: Cartesian envelope of a symmetric monoidal category

I am also interested in the left adjoint of the forgetful functor with domain semi-Cartesian monoidal categories, which are symmetric monoidal categories in which the tensor unit is terminal.

Another way to look at this is from the point of view of the "commutative algebra of categories," as described (in the quasicategorical case) here: https://arxiv.org/abs/1606.05606. Using that machinery, i.e. knowing that Cartesian monoidal categories are precisely the symmetric monoidal categories which are modules over the solid semi-ring category $$Fin^{op}$$, it would suffice to have an explicit description of the tensor product on (presentable) symmetric monoidal categories (that recovers the structure described in the above reference). So maybe this is known?

• Wouldn’t it be “induction” to the semiring $Fin^{op}$? I.e. tensor with $Fin^{op}$? Oct 24, 2018 at 0:52
• @NoahSnyder maybe so! I guess I don't know what induction would be here. To me, I'd think it'd be something like $C\otimes_{Fin^{iso}}Fin^{op}$? But I'm not sure how this kind of thing works. Oct 24, 2018 at 5:05
• @NoahSnyder Ah I think maybe I misread your comment. Yeah, I agree that it's tensor with $Fin^{op}$, but what is that tensor? I don't think it's just the usual tensor of categories is it? It's some complicated tensor product of $Fin^{iso}$-modules? Oct 24, 2018 at 16:24
• Where did the word "presentable" sneak in? Are you asking about locally presentable categories or about arbitrary ones? It seems to me that the answers might be different. In particular, I believe the tensor product of locally presentable categories does have a somewhat explicit description, but I don't know of one in the general case. Oct 25, 2018 at 20:18
• @MikeShulman I suppose I was just going off of the fact that "presentability" (by which I mean local presentability, since the term has been sadly mucked up) is one of the requirements on the categories addressed in the Berman paper cited in the question. Oct 25, 2018 at 22:15

I don't know the answer, although I've thought about this quite a bit. I think the best that can be said in general is that the free cartesian monoidal category on $$\mathcal{C}^\otimes$$ is a full subcategory of $$\text{Fun}^\otimes(\mathcal{C}^\otimes,\text{Set}^\times)^\text{op}$$, which is the category of symmetric monoidal functors. That is Theorem 3.10 of https://arxiv.org/abs/1606.05606.

There is some hope of giving a more explicit description if we know something about maps into tensor products. Here are three examples. (I would love to understand what they have in common.)

Baby example: If $$\mathcal{C}^\times$$ is already cartesian monoidal, then $$\mathcal{C}_{/Y\times Z}\cong\mathcal{C}_{/X}\times_\mathcal{C}\mathcal{C}_{/Y}$$. That is, a map $$X\rightarrow Y\times Z$$ corresponds to two maps $$X\rightarrow Y$$ and $$X\rightarrow Z$$. In this case, the free cartesian monoidal category on $$\mathcal{C}$$ is $$\mathcal{C}$$ itself.

Example 2: If $$\mathcal{C}^\amalg$$ is cocartesian monoidal, we may ask whether the coproduct behaves like a disjoint union (that is, $$\mathcal{C}$$ is disjunctive) in the following sense: any map $$X\rightarrow Y\amalg Z$$ decomposes canonically as the coproduct of two maps $$X_Y\rightarrow Y$$ and $$X_Z\rightarrow Z$$. That is, $$\mathcal{C}_{/Y\amalg Z}\cong\mathcal{C}_{/Y}\times\mathcal{C}_{/Z}$$. In this case, I believe that the free cartesian monoidal category can be described as a category of spans, something like the following:

Objects are just objects of $$\mathcal{C}$$. Morphisms from $$X$$ to $$Y$$ are spans $$X\leftarrow T\rightarrow Y$$, such that $$T$$ admits a decomposition $$T_1\amalg\cdots\amalg T_n$$ with each $$T_i\rightarrow X$$ an inclusion of a direct summand; that is, $$T_i\amalg T_i^\prime\cong X$$.

For example, when $$\mathcal{C}=\text{Fin}$$, the category of finite sets, the free cartesian monoidal category is spans of finite sets. The $$\mathcal{C}=\text{Fin}$$ case is a main result of the paper cited above (at the level of $$\infty$$-categories). The general case is not written down anywhere, as far as I know.

Example 3: Given an operad $$\mathcal{O}$$, there is a category $$[\mathcal{O}^\otimes]$$, whose objects are finite sets. A map from $$S$$ to $$T$$ is a function $$f:S\rightarrow T$$ and an $$|f^{-1}(t)|$$-ary operation for each $$t\in T$$. Concatenation gives $$[\mathcal{O}^\otimes]$$ a symmetric monoidal structure.

Note that $$[\mathcal{O}^\otimes]$$ is disjunctive (in the sense of Example 2) but not cocartesian monoidal. In this case, the free cartesian monoidal category is the full subcategory of $$\text{Fun}^\otimes([\mathcal{O}^\otimes],\text{Set}^\times)^\text{op}=\text{Alg}_\mathcal{O}(\text{Set})^\text{op}$$ spanned by the free $$\mathcal{O}$$-algebras. Classically, the free $$\mathcal{O}$$-algebra on $$n$$ generators is $$\coprod_{k\geq 0}\mathcal{O}(k)\times_{\Sigma_k}n^k.$$ That effectively computes the free cartesian monoidal category on $$[\mathcal{O}^\otimes]$$ as a type of span construction, similar to Example 2. See the paper above, Remark 5.1 (also Example 3.16). The span construction can be made precise without much difficulty for ordinary categories, but it is an open problem for $$\infty$$-categories.