# Cartesian envelope of a symmetric monoidal category

Is there a left adjoint to the forgetful functor from cartesian monoidal categories to symmetric monoidal categories? And if so: how can it be described explicitely.

In more detail: Given a symmetric monoidal category $\mathbb V$ i am looking for a monoidal functor

$$\chi:\mathbb V \to \mathcal C(\mathbb V)$$

into a cartesian monoidal category $\mathcal C(\mathbb V)$ such that every other monoidal functor

$$\varphi:\mathbb V \to \mathbb C$$

into a cartesian monoidal category $\mathbb C$ factors 'uniquely' over $\chi$.

• what is a cartesian monoidal category? Do you mean a (symmetrical) monoidal closed one? Commented Jun 2, 2013 at 21:31
• "A cartesian monoidal category (usually just called a cartesian category), is a monoidal category whose monoidal structure is given by the category-theoretic product (and so whose unit is a terminal object)." -- ncatlab.org/nlab/show/cartesian+monoidal+category Commented Jun 3, 2013 at 8:15
• You look for a $2$-adjoint, right? See also the comments in Chris' answer. You don't want that $\hom(V,U(C))$ and $\hom(C(V),C)$ are isomorphic, but rather that they are equivalent. Commented Jul 15, 2013 at 20:48

Yes, there is a left adjoint. Both the category $\textbf{Mon}$ of (symmetric) monoidal categories, and the category $\textbf{Cart}$ of Cartesian categories, are algebraic over the category $\textbf{Cat}$ of categories. That is, there are monads on $\textbf{Cat}$ whose categories of algebras are $\textbf{Mon}$ and $\textbf{Cart}$, respectively.
More interestingly, there is also a right adjoint, that sends a monoidal category $\textbf{C}$ to the category of comonoids in $\textbf{C}$. This is a classic result by Thomas Fox.
• As usual with these abstract proofs you can trace them back and find explicit constructions - which are not really useful. Here, when your monoidal category is given by generators and relations, you just replace $\otimes$ by $\times$ in each relation. A basic example is the ($2$-)free symmetric monoidal category on one object, which is the permutation groupoid. Then the left adjoint maps it to the free cartesian category on one object, which should be $\mathsf{FinSet}^{\mathsf{op}}$. Such things get interesting when you really work out specific examples where no presentation is given. Commented Jul 15, 2013 at 21:03