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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 $\chi$.

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  • $\begingroup$ what is a cartesian monoidal category? Do you mean a (symmetrical) monoidal closed one? $\endgroup$
    – Buschi Sergio
    Commented Jun 2, 2013 at 21:31
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    $\begingroup$ "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 $\endgroup$
    – Gerrit Begher
    Commented Jun 3, 2013 at 8:15
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    $\begingroup$ 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. $\endgroup$
    – Martin Brandenburg
    Commented Jul 15, 2013 at 20:48

1 Answer 1

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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.

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    $\begingroup$ Shouldn't Mon and Cart be treated as 2-categories as opposed to 1-categories? Or is there some magic working behind the scenes that ensures that issues with higher coherences do not arise? $\endgroup$
    – Dmitri Pavlov
    Commented Jun 3, 2013 at 0:28
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    $\begingroup$ Dmitri, definitely they are 2-cateogries. I'm pretty sure that 1-categorically those algebraicity statements are false. E.g. associators correspond to associativity conditions in algebraic 2-theory, cartesian structure is encoded in adjunction between diagonal and multiplication, etc. But formally this doesn't seem to change anything, classical adjunction proofs can be repeated 2-categorically. $\endgroup$
    – Anton Fetisov
    Commented Jun 3, 2013 at 11:44
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    $\begingroup$ I should have specified that Fox uses strict monoidal functors as the morphisms. As Dmitri points out and Anton mentions, lax (or even strong) monoidal functors require 2-categories. $\endgroup$
    – Chris Heunen
    Commented Jun 3, 2013 at 14:11
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    $\begingroup$ Thanks for the link; how can said monads - and the left adjoint - be described explicitly? Any reference? $\endgroup$
    – Gerrit Begher
    Commented Jun 4, 2013 at 16:19
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    $\begingroup$ 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. $\endgroup$
    – Martin Brandenburg
    Commented Jul 15, 2013 at 21:03

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