Are there noncartesian monoidal categories with $A \otimes B = A \times B$?

Question

Can we find a non-cartesian monoidal category where for each pair of objects the tensor product $A \otimes B$ is a cartesian product of $A$ and $B$?

Let me explain.

A monoidal category $(\mathsf{C}, \otimes, I, a, r, \ell)$ is a category $\mathsf{C}$ with a tensor product $\otimes$, a unit object $I$, an associator $a$, and left and right unitors $r$ and $\ell$ obeying the pentagon and triangle identities. A cartesian category is one with binary products and a terminal object.

Given a cartesian category $\mathsf{C}$, and choosing a terminal object, and choosing a product cone

$$ A \leftarrow A \times B \rightarrow B$$

for each pair of objects $A,B$, there is a well-known way to make $\mathsf{C}$ into a monoidal category. To create this monoidal category, first we take $A \otimes B$ to be the object $A \times B$ in the chosen product cone for $A$ and $B$. Then we define $\otimes$ on morphisms using the universal property of the chosen product cones. Similarly, we define the associator

$$ a_{A,B,C} \colon (A \otimes B) \otimes C \to A \otimes (B \otimes C)$$

using the universal property of the chosen product cones. Then we take the unit object $I$ to be the chosen terminal object, and define the left and right unitors

$$ \ell_A \colon I \otimes A \to A, \qquad r_A \colon A \otimes I \to A$$

using the universal property of the chosen product cones and the fact that $I$ is terminal. We obtain a monoidal category $(\mathsf{C}, \otimes, I, a, r, \ell)$ called a 'cartesian monoidal category'.

More generally, any monoidal category equivalent as a monoidal category to one of this form is called a cartesian monoidal category.

This leaves open the possibility that there could be a way to make our cartesian category $\mathsf{C}$ into a monoidal category where for each pair $A, B \in \mathsf{C}$ the object $A \otimes B$ equal to the chosen object $A \times B$, but which however is not equivalent, as a monoidal category, to the cartesian monoidal category described above. Is this ever possible?

There are various things we can change: the action of $\otimes$ on morphisms, the associator, the left and right unitors - and prima facie, given the way I've worded the question, even the unit object $I$.

It would be especially fun if we could do this starting from some familiar cartesian category like the category of sets.

[Edit: I have a conjecture here for how we might get a noncartesian monoidal category where $A \otimes B$ is the product of $A$ and $B$ for every pair of objects $A,B$.]

Answer 1

Note: As pointed out by Peter LeFanu Lumsdaine in the comments, the example constructed in this answer does not work in its current form, since naturality of the associator fails. (This only affects the example and not the characterization of cartesian monoidal categories given in the proposition below.)

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As expected, there is a (symmetric) monoidal category in which every tensor $A \otimes B$ is a product of $A$ and $B$, although the monoidal structure is not cartesian. In fact, I'm going to show that this can be done with any nontrivial additive category and a small tweak. This means that my binary products are even biproducts (in most cases). In addition, my associators and unitors will be those of the cartesian monoidal structure.

But before getting to the construction, let me describe a simple way of characterizing cartesian monoidal categories that is manifestly invariant under monoidal equivalence.

Proposition: A monoidal category $\mathsf{C}$ is cartesian monoidal if and only if the monoidal unit $I$ is terminal, and if for any two objects $A$ and $B$, the induced morphisms $$\tag{1}\label{475718_1}\mathrm{id}_A \otimes {!_B} : A \otimes B \longrightarrow A, \qquad !_A \otimes \mathrm{id}_B : A \otimes B \longrightarrow B$$ make $A \otimes B$ into a product of $A$ and $B$.

Here, I've written $!_A : A \to I$ for the unique morphism to $I$ and suppressed the unitors from the notation for simplicity. I think that this proposition reproduces the statement from John's comment on the question in a simpler manner.

Proof sketch: This is easiest to see upon showing first that this property is indeed invariant under monoidal equivalence. Based on this, it's quite straightforward that every cartesian monoidal category satisfies it.

Conversely, if the property holds, then I claim that $\mathsf{C}$ is already of the form described in the question. Indeed $!_A : A \to I$ is natural in $A$, and this implies together with the bifunctoriality of $\otimes$ that the components of any $f \otimes g$ are indeed $f \otimes {!}$ and $! \otimes g$ as required. That the unitors and associators are as required is arguably easiest to see upon assuming without loss of generality that $\mathsf{C}$ is strict, in which case this is quite clear. $\square$

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Now on to the construction of a non-cartesian monoidal category with $A \otimes B = A \times B$. Let $\mathsf{C}$ be any additive category, and assume that it is nontrivial in the sense that there is some nonzero object. For ease of notation, I'll assume that the zero object $0$ is unique. The category that I will actually work with is a tweaked $\mathsf{C}'$, which coincides with $\mathsf{C}$ except in that I declare $\mathsf{C}'(0, A) := \emptyset$ for $A \neq 0$. So I've removed the morphisms from the zero object to any non-zero object. This results in $0$ being a strict terminal object in $\mathsf{C}'$.

Let's call a morphism void if its codomain is zero but its domain is not. Then the strict terminality of $0$ shows that a composite $gf$ is void if and only if $g$ is void or if $g = \mathrm{id}_0$ and $f$ is void.

Note that the binary biproducts from $\mathsf{C}$ are still products in $\mathsf{C}'$ (and in fact still biproducts except when one of the summands is zero). Hence the biproduct functor $\oplus$ from $\mathsf{C}$ also makes $\mathsf{C}'$ into a cartesian monoidal category.

I will now introduce a new monoidal structure $\otimes$ which coincides with $\oplus$ on objects but is not cartesian. To this end, let $\otimes$ coincide with $\oplus$ on objects and on morphisms, except in that $f \oplus g := 0$ if $f$ or $g$ is void. The above property of void morphisms shows that this is bifunctorial.

Now I claim that this $\otimes$ makes $\mathsf{C}'$ into a monoidal category with respect to the same unitors and associators (even symmetric monoidal). Indeed the coherence diagrams are trivial, and the naturality is straightforward to check (for the unitors, use that every void morphism is a zero morphism).

Here comes the punchline: every $!_A : A \to 0$ for nonzero $A$ is void, and therefore with $B = A$, both morphisms in \eqref{475718_1} are zero. In particular, they are not product projections! E.g. because this would require $\mathrm{id}_A$ to factor across the projections. Therefore the monoidal structure is not cartesian.

Answer 2

Take $\newcommand{\bu}{\bullet}\newcommand{\C}{\mathcal{C}}\C$ to be the full subcategory of $\newcommand{\Set}{\mathrm{Set}}\newcommand{\ptSet}{\Set_{\bullet}}\ptSet$ (pointed sets) on the objects $1$ and $\newcommand{\N}{\mathbb{N}}\N$. We can restrict both the cartesian and co-cocartesian monoidal structures of $\ptSet$ to $\C$, since $1$ is the unit for both structures and $\N +_{\bullet} \N \cong \N \cong \N \times_{\bullet} \N$. These two monoidal structures on $\C$ coincide on objects, but the co-cartesian structure $+_{\bullet}$ is not cartesian, as may be seen by the criterion from Tobias’ answer: the maps $\newcommand{\id}{\mathrm{id}}(\id_N +_\bu {!_\N}), (!_\N +_\bu \id_\N) : \N +_\bu \N \to \N$ don’t exhibit $\N +_\bu \N$ as a product, since their induced map to the standard product isn’t an isomorphism (it’s essentially the inclusion of the two axes, $(\N \times \{0\}) \cup (\{0\} \times \N) \hookrightarrow \N \times \N$).