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Suppose $\cal A, B$ are cartesian monoidal categories, that is, categories equipped with a choice of finite products (including the nullary one, $1$). Suppose, moreover, that $F: \cal A \to B$ is a symmetric monoidal functor, that is, it preserves the monoidal structure $\times$ up to natural and coherent isomorphism.

The question is, does this imply $F$ is also a product-preserving functor, meaning, in particular, that it sends projections to projections and diagonals to diagonals?

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    $\begingroup$ If you already knew the answer, then why ask the question? This is not how mathoverflow is normally used. $\endgroup$
    – David White
    Commented Sep 21, 2023 at 13:40
  • $\begingroup$ @DavidWhite the fact there is an explicit UI for doing exactly that makes me think otherwise. I just wanted to share this easy fact I figured out (also for my future self). $\endgroup$
    – seldon
    Commented Sep 21, 2023 at 13:59
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    $\begingroup$ Possibly the explicit UI reflects the fact that it is done more frequently on other SE sites? It certainly does seem to be rare here. $\endgroup$
    – Mike Shulman
    Commented Sep 21, 2023 at 19:51
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    $\begingroup$ It's an interesting problem, 1+. But indeed MO is not a place to share these proofs. Better use a blog or even arxiv for this. $\endgroup$
    – Martin Brandenburg
    Commented Sep 25, 2023 at 16:23
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    $\begingroup$ I’m voting to close this question because it's not actually a question. The OP posted it, then immediately posted the answer. Now it's been bumped to the front page by the software because it has no answer with a positive score. We should close it, rather than waiting around for someone to upvote and incentivize this unusual method of using MathOverflow. $\endgroup$
    – David White
    Commented Nov 22, 2023 at 4:32

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Yes.

Indeed, it is the assumption on $\cal B$ that suffices to promote an apparently weaker condition (preservation of monoidal structure -- a mere algebraic condition) into a stronger one (preservation of cartesian monoidal structure).

To see this, consider the following fact:

A cartesian monoidal category is a monoidal category such that each object is a cocommutative comonoid in a unique way and each map is a comonoid homomorphism.

This is easy to see in one direction: the comonoid structure is given by the diagonal map $\Delta_X:X \to X \times X$ and the terminal map $!_X:X \to 1$, and by universal property these commute with all morphisms in the category. Viceversa, given such a structure, one can obtain a limiting cone $X \overset{\pi_X}\leftarrow X \otimes Y \overset{\pi_Y}\to Y$ by setting $\pi_X = X \otimes !_Y$ and $\pi_Y = !_X \otimes X$ (I'm ignoring the unitors for convenience). To conclude this is a product, one has a (tiny) bit more work to do: see Theorem 6.13 here.

Anyway, this gives a very easy way to check that $F$ preserves products: since it is strong monoidal, it maps the unique comonoid structure on any $A: \cal A$ to a comonoid in $\cal B$. But since $\cal B$ is cartesian monoidal, such a comonoid structure can only be the unique comonoid structure on $FA$. Since projections and diagonals are constructed out of this structure, this shows $F$ preserves products. Indeed, let $A \times A'$ be a product in $\cal A$, then $$ F\pi_A = F(A \times !_{A'}) = FA \times F!_{A'} = FA \times !_{FA'} = \pi_{FA} $$ and likewise for diagonals.

N.B.: it is crucial that $F$ preserves the symmetry, since it is crucial that these comonoids are cocommutative in order to get products.

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