Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: theory: category theory

Topic: Monoidal structure from products

Jean-Baptiste Vienney (Jul 24 2024 at 07:44):

Suppose that $\mathcal{C}$ is a category together with a terminal object $\top$ and a product
$(A \times B,p_{A,B}:A \times B \rightarrow A, q_{A,B}:A \times B \rightarrow B)$
of $A$ and $B$ for every pair of objects $A,B \in \mathcal{C}$.

We obtain a bifunctor $- \times -: \mathcal{C} \times \mathcal{C} \rightarrow \mathcal{C}$ by defining $f \times g=\langle p_{A,B};f, q_{A,B};g\rangle$ for every $f:A \rightarrow C$ and $g:B \rightarrow D$.

We can also check that if we define:
$\lambda_A=q_{\top,A}:\top \times A \rightarrow A$,
$\rho_A=p_{A,\top}:A \times \top \rightarrow A$,
$\alpha_{A,B,C}=\langle p_{A \times B,C};p_{A,B}, \langle p_{A \times B,C};q_{A,B},q_{A \times B,C}\rangle\rangle:(A \times B) \times C \rightarrow A \times (B \times C)$
then, $\lambda,\rho,\alpha$ are natural transformations.

Finally, we can check that $(\mathcal{C},\times,\top,\alpha,\lambda,\rho)$ is a monoidal category.

Let's call such a monoidal category a cartesian monoidal category.

Question: Is a monoidal category $(\mathcal{C},\otimes,I,\alpha,\lambda,\rho)$ such that $I$ is a terminal object and $A \otimes B$ the underlying object of a product of $A$ and $B$ for every $A,B \in \mathcal{C}$ necessarily a cartesian monoidal category?

Jean-Baptiste Vienney (Jul 24 2024 at 08:37):

In fact, I'm more interested into this question for now, where the hypotheses are stronger:

Question 2: Consider a monoidal category $(\mathcal{C},\otimes,I,\alpha,\lambda,\rho)$ such that $I$ is a terminal object, and such that there exists a family of products $(A \times B, p_{A,B}:A \times B \rightarrow B, q_{A,B}:A \times B \rightarrow B)_{A,B \in \mathcal{C}}$ of $A$ and $B$ for every $A,B \in \mathcal{C}$ which verifies:

• $A \otimes B = A \times B$ for every $A,B \in \mathcal{C}$
• $f \otimes g=\langle p_{A,B};f, q_{A,B};g\rangle$ for every $f:A \rightarrow C$ and $g:B \rightarrow D$.

Is $(\mathcal{C},\otimes,I,\alpha,\lambda,\rho)$ necessarily a cartesian monoidal category?

Nathanael Arkor (Jul 24 2024 at 09:11):

In fact, the stronger statement is true that a monoidal category for which each $A \otimes B = A \times B$ is necessarily cartesian. This follows from this characterisation of cartesian monoidal categories.

Jean-Baptiste Vienney (Jul 24 2024 at 10:09):

Thanks. I’m still trying to understand this MO question and its answers.

I trust you that it answers “Yes” to my second question. But does it really answer “Yes” to the first question? I don’t see how the equation $y \otimes z \circ \{f,g\} \circ x = \{y \circ f \circ l(x),z \circ g \circ r(x)\}$ from your MO question is implied by $A \otimes B=A \times B$ without any additional condition on how a tensor product $u \otimes v$ of morphisms is related to the projections and universal properties of the products $A \times B$.

John Baez (Jul 24 2024 at 12:07):

Yeah, I don't see anyone showing that a monoidal category $(C, \otimes, I, a, \ell, r)$ such that $A \otimes B = A \times B$ for each pair of objects is necessarily a cartesian monoidal category! That feels impossible to prove, because there's so little you can do with equations between objects... but I would really like to see a counterexample.

John Baez (Jul 24 2024 at 12:08):

Is it possible to take a cartesian monoidal category and change the associator to get a noncartesian monoidal category? There's a lot known about how you can take a monoidal category and "twist the associator by a cocycle" to get a new, inequivalent monoidal category. But I've never seen this done starting from a cartesian monoidal category.

Nathanael Arkor (Jul 24 2024 at 12:41):

Sorry, that was meant to be in response to question 2. You don't need to require $I$ to be terminal: this is implied.

Jean-Baptiste Vienney (Jul 24 2024 at 12:58):

John Baez said:

Is it possible to take a cartesian monoidal category and change the associator to get a noncartesian monoidal category? There's a lot known about how you can take a monoidal category and "twist the associator by a cocycle" to get a new, inequivalent monoidal category. But I've never seen this done starting from a cartesian monoidal category.

That’s a very good question. I have no idea.

Nathanael Arkor (Jul 24 2024 at 13:02):

A positive answer to question 2 would imply that it is not possible to change the associator (at least not without also changing the tensor products).

Jean-Baptiste Vienney (Jul 24 2024 at 13:34):

Hmm, I see.

Jean-Baptiste Vienney (Jul 24 2024 at 13:36):

(The reason is that the positive answer to question 2 is roughly “binary tensor product induced by binary products” implies “cartesian monoidal category”.)

Nathanael Arkor (Jul 24 2024 at 14:31):

I notice John has asked your second question over on MathOverflow...

Jean-Baptiste Vienney (Jul 24 2024 at 14:53):

@John Baez, if I'm not mistaken, we already know that the answer to your MO question is "No". What we still don't know is whether there are monoidal categories such that $A \otimes B$ is a product of $A$ and $B$ but the monoidal category is not a cartesian monoidal category.

Nathanael Arkor (Jul 24 2024 at 14:58):

I noticed a gap in my reasoning, so I'm no longer sure it is true.

John Baez (Jul 24 2024 at 14:59):

I think I got carried away and specified $\otimes$ not only on objects but on morphisms. Is that what your point was, @Jean-Baptiste Vienney? That if we take a cartesian monoidal category, we can't make it noncartesian by just changing the associator and unitors while keeping $\otimes$ unchanged on objects and morphisms, and keeping the unit object $I$ unchanged?

(Regardless of whether it's true or not, was this your point?)

Jean-Baptiste Vienney (Jul 24 2024 at 15:02):

Yes, it was my point.

I don't know why @Nathanael Arkor is suddenly no longer sure about this. Isn't it a consequence of the answers to your MO question?

Nathanael Arkor (Jul 24 2024 at 15:04):

The MO answer just shows that you do have cartesian monoidal structure, but not necessarily that the cartesian monoidal structure coincides with the original monoidal structure.

Jean-Baptiste Vienney (Jul 24 2024 at 15:06):

I thought it was implied by the end of Tim Campion's answer:
here

Nathanael Arkor (Jul 24 2024 at 15:07):

We would need the identity to be a strict monoidal functor for it to imply the statement you want, no?

Nathanael Arkor (Jul 24 2024 at 15:13):

(This question was asked before on MathOverflow, without any answers.)

Nathanael Arkor (Jul 24 2024 at 15:17):

It's not clear to me why you couldn't have an automorphism on each object, such that $A \times 1 \to A$ is given by postcomposing the projection by the automorphism.

Clémence Chanavat (Jul 24 2024 at 15:21):

it is not a natural isomorphism?

Jean-Baptiste Vienney (Jul 24 2024 at 15:23):

Nathanael Arkor said:

We would need the identity to be a strict monoidal functor for it to imply the statement you want, no?

Yes, that's true. But I'm a bit surprised that we could have two equivalent but nonequal monoidal structures with $A \otimes B$ being the same product of $A$ and $B$ in the two cases. What would be less surprising to me would be to have two equivalent but nonequal monoidal structures with $A \otimes_1 B$ being a product of $A$ and $B$ and $A \otimes_2 B$ being another product of $A$ and $B$.

Jean-Baptiste Vienney (Jul 24 2024 at 15:27):

Nathanael Arkor said:

It's not clear to me why you couldn't have an automorphism on each object, such that $A \times 1 \to A$ is given by postcomposing the projection by the automorphism.

I think you can do this. I was thinking exactly about this a few hours ago. For a concrete example, you can verify that chosing $p:A \times B \rightarrow A$ and $q:B \times A \rightarrow B$ given by $p(a,b)=2a$ and $q(a,b)=2b$ makes $A \times B$ into a product of $A$ and $B$ in $\mathbf{Vec}_{\mathbb{k}}$ for any field $k$.

Jean-Baptiste Vienney (Jul 24 2024 at 15:29):

It will give you a monoidal category with unitors $A \times \top \rightarrow A$ sending $(a,*)$ to $2a$ and $(*,b)$ to $2b$.

Jean-Baptiste Vienney (Jul 24 2024 at 15:31):

It comes from "an object isomorphic to a product is a product". You instanciate the precise version of this proposition to an automorphism.

John Baez (Jul 24 2024 at 15:32):

I have attempted to improve my Math Overflow question, @Jean-Baptiste Vienney. I have gotten rid of the old one, which indeed seems likely to have a negative answer.

Jean-Baptiste Vienney (Jul 24 2024 at 15:32):

Nice!

Jean-Baptiste Vienney (Jul 24 2024 at 15:35):

That's a different story, but now I'm wondering if a monoidal category equivalent as a monoidal category to a cartesian monoidal category is not necessarily a cartesian monoidal category.

Jean-Baptiste Vienney (Jul 24 2024 at 15:36):

Because of the universal property of products which make some morphisms unique.

John Baez (Jul 24 2024 at 15:37):

now I'm wondering if a monoidal category equivalent as a monoidal category to a cartesian monoidal category is not necessarily a cartesian monoidal category.

It depends on how you define 'cartesian monoidal category'. It's interesting that the nLab does not really define a cartesian monoidal category. It says some words, but they are open to several interpretations.

John Baez (Jul 24 2024 at 15:38):

I defined 'cartesian monoidal category in my post in a way that's not elegant, but manifestly invariant under equivalent.

Jean-Baptiste Vienney (Jul 24 2024 at 15:39):

Jean-Baptiste Vienney said:

Suppose that $\mathcal{C}$ is a category together with a terminal object $\top$ and a product
$(A \times B,p_{A,B}:A \times B \rightarrow A, q_{A,B}:A \times B \rightarrow B)$
of $A$ and $B$ for every pair of objects $A,B \in \mathcal{C}$.

We obtain a bifunctor $- \times -: \mathcal{C} \times \mathcal{C} \rightarrow \mathcal{C}$ by defining $f \times g=\langle p_{A,B};f, q_{A,B};g\rangle$ for every $f:A \rightarrow C$ and $g:B \rightarrow D$.

We can also check that if we define:
$\lambda_A=q_{\top,A}:\top \times A \rightarrow A$,
$\rho_A=p_{A,\top}:A \times \top \rightarrow A$,
$\alpha_{A,B,C}=\langle p_{A \times B,C};p_{A,B}, \langle p_{A \times B,C};q_{A,B},q_{A \times B,C}\rangle\rangle:(A \times B) \times C \rightarrow A \times (B \times C)$
then, $\lambda,\rho,\alpha$ are natural transformations.

Finally, we can check that $(\mathcal{C},\times,\top,\alpha,\lambda,\rho)$ is a monoidal category.

Let's call such a monoidal category a cartesian monoidal category.

Like this. I think it is the same definition as in Categories for the Working Mathematician. I'm just being very explicit.

Clémence Chanavat (Jul 24 2024 at 15:39):

Jean-Baptiste Vienney said:

It will give you a monoidal category with unitors $A \times \top \rightarrow A$ sending $(a,*)$ to $2a$ and $(*,b)$ to $2b$.

I believe this does not work since the unit law is not satisfied, as it would mean $(2a, b) = (a, 2b)$?

John Baez (Jul 24 2024 at 15:43):

I believe any expert would tell you that if you have a cartesian monoidal category, any other monoidal category that's equivalent to it as a monoidal category must also be cartesian monoidal. They might, however, not have a specific definition of 'cartesian monoidal category' in mind!

In my MathOverflow question, @Jean-Baptiste Vienney, it seems I started with your definition of 'cartesian monoidal category' - but then I added something to make sure every monoidal category that's monoidally equivalent to a cartesian monoidal category is again cartesian monoidal!

John Baez (Jul 24 2024 at 15:43):

Because I don't want to be using definitions that aren't invariant under equivalence.

John Baez (Jul 24 2024 at 15:44):

Anyway, it's an interesting question whether your definition is already invariant under equivalence.

Jean-Baptiste Vienney (Jul 24 2024 at 15:44):

I see. The two questions are separated as you wrote it, so no worries.

Jean-Baptiste Vienney (Jul 24 2024 at 15:53):

Clémence Chanavat said:

Jean-Baptiste Vienney said:

It will give you a monoidal category with unitors $A \times \top \rightarrow A$ sending $(a,*)$ to $2a$ and $(*,b)$ to $2b$.

I believe this does not work since the unit law is not satisfied, as it would mean $(2a, b) = (a, 2b)$?

We have to be very precise.

• First claim: Let $k$ be a field. In $\mathbf{Vec}_k$, if we define the vector space $A \times B$ as usual, $p_{A,B}:A \times B \rightarrow A$ and $q_{A,B}:A \times B \rightarrow B$ by $p(a,b)=2a$ and $q(a,b)=2b$, then it makes $(A \times B,p_{A,B},q_{A,B})$ into a product of $A$ and $B$.
• Second claim: if we have a product cone for every pair of objects in a category, then it makes this category into a monoidal category by defining the structure exactly like in my first message. I'm sure about this because I've written all the fastidious verifications during the previous days.

We should apply the second claim carefully to the product in the first claim to see what is the structure of monoidal category obtained.

John Baez (Jul 24 2024 at 15:54):

I would like to improve the nLab a little, so it gives an unambiguous definition of cartesian monoidal category.

Note that it does mention a theorem about cartesian monoidal categories:

Moreover, one can show (e.g. Fox 1976 or Heunen-Vicary 2012, p. 79 (p. 85 of the pdf)) that any symmetric monoidal category equipped with suitably well-behaved diagonal and augmentation maps must in fact be cartesian monoidal.

For this theorem to be true, we need a definition of cartesian monoidal category that's invariant under monoidal equivalence... because the 'suitably well-behaved' conditions are invariant.

John Baez (Jul 24 2024 at 15:54):

And presumably these authors stated a definition of cartesian monoidal category, so they could prove the theorem!

Jean-Baptiste Vienney (Jul 24 2024 at 15:54):

Heunen-Vicary define a cartesian monoidal category as one such that $A \otimes B$ is a product of $A$ and $B$ which is rather unsatisfying.

Jean-Baptiste Vienney (Jul 24 2024 at 15:58):

This is page 135 in this book.

John Baez (Jul 24 2024 at 15:58):

Yikes! :face_with_spiral_eyes:

Well, at least that makes it very easy to prove that various kinds of categories are cartesian monoidal.

John Baez (Jul 24 2024 at 15:59):

It just makes it hard to prove that cartesian monoidal categories have other interesting properties, e.g. that their associators are 'what you'd expect'.

John Baez (Jul 24 2024 at 16:02):

Hmm, actually on page 135 they don't seem to officially define 'cartesian monoidal category' - they just have Theorem 4.28 saying that a monoidal category is Cartesian iff something else. But I think you are questioning Theorem 4.28.

Jean-Baptiste Vienney (Jul 24 2024 at 16:03):

Yes but look at this:
Screenshot-2024-07-24-at-12.03.21PM.png

Jean-Baptiste Vienney (Jul 24 2024 at 16:04):

And they don't prove anything more than that.

Jean-Baptiste Vienney (Jul 24 2024 at 16:04):

Which means it is their definition of cartesian monoidal category.

Jean-Baptiste Vienney (Jul 24 2024 at 16:16):

Clémence Chanavat said:

Jean-Baptiste Vienney said:

It will give you a monoidal category with unitors $A \times \top \rightarrow A$ sending $(a,*)$ to $2a$ and $(*,b)$ to $2b$.

I believe this does not work since the unit law is not satisfied, as it would mean $(2a, b) = (a, 2b)$?

I you apply the definitions, you will see that the associator is given by $(x,y,z) \mapsto (2x,y,\frac{1}{2}z)$ and it do makes the triangle diagram of a monoidal category commutes.

Jean-Baptiste Vienney (Jul 24 2024 at 16:18):

For starters, with such projections $\langle f,g \rangle$ associates $(x,y)$ to $(\frac{1}{2}f(x),\frac{1}{2}g(y))$.

Then, the unitors of the monoidal category are given by $(x,*) \mapsto 2x$ and $(*,y) \mapsto 2y$. $f \times g$ is given as usual by $(f \times g)(x,y)=(f(x),g(y))$. And the associator is as written just above.

Jean-Baptiste Vienney (Jul 24 2024 at 16:21):

Note that this is a cartesian monoidal category according to the definition given in the first message.

John Baez (Jul 24 2024 at 16:33):

@Jean-Baptiste Vienney - Now I don't think Heunen and Vicary are proposing a definition of cartesian monoidal category! I think their theorem says "if a monoidal category can be equipped with this extra structure, then it's a cartesian category" - a category with finite products.

Clémence Chanavat (Jul 24 2024 at 16:34):

@Jean-Baptiste Vienney yes my bad, you're right, i just wrongly assumed that the associator would stay the same...

Jean-Baptiste Vienney (Jul 24 2024 at 16:36):

@John Baez , I agree! So I was a bit harsh with them. They don't do anything wrong. But maybe their theorem could be made into a more complete statement.

Jean-Baptiste Vienney (Jul 24 2024 at 16:37):

@Clémence Chanavat No worries, I was not completely confident before checking everything.

Jean-Baptiste Vienney (Jul 24 2024 at 16:56):

I posted an answer to the old MO question.

Amar Hadzihasanovic (Jul 24 2024 at 19:00):

My go-to definition of “cartesian monoidal category” is a monoidal category $(C, \otimes, I)$ such that

• the monoidal unit is a terminal object, so for each object $A$ there is a unique morphism $!_A: A \to I$,
• for each pair of objects $A, B$, the “projections” $\rho_A \circ (A \,\otimes\, !_B): A\otimes B \to A$ and $\lambda_B \circ (!_A \otimes B): A \otimes B \to B$ are product spans.

I know for sure that this definition is one that makes “Fox's theorem” work, that is, a symmetric monoidal category is cartesian if and only if it has a (necessarily unique) supply of natural comonoids.
Of course there may be other more succinct equivalent definitions.

Jean-Baptiste Vienney (Jul 24 2024 at 20:15):

Hmm, this is an interesting definition since it implies that $\rho_A=\pi^1_{A,I}$ and $\lambda_{B}=\pi^2_{I,B}$ (I denote the product spans by $\pi^1_{A,B},\pi^2_{A,B}$).

@Amar Hadzihasanovic, I'm also able to show that it implies $f \otimes g=\langle \pi^1_{A,B};f,\pi^2_{A,B};g\rangle$.

But do you know if it implies that the associator is obtained in the natural way from the projections?

i.e. if it implies that
$\alpha_{A,B,C}=\langle \pi^1_{A \otimes B,C};\pi^1_{A,B}, \langle \pi^1_{A \otimes B,C};\pi^2_{A,B},\pi^2_{A \otimes B,C}\rangle\rangle$?

John Baez (Jul 24 2024 at 21:18):

I think it must, but I don't know the argument.

Jean-Baptiste Vienney (Jul 24 2024 at 21:19):

It would be really cool because then I will finally know a good definition of cartesian monoidal category.

Amar Hadzihasanovic (Jul 25 2024 at 07:09):

Yes, it does, I don't know if there's a good non-explicit argument though. Here's a diagram showing it for the first component

1043bf51-06bd-42a8-a78d-9531706b872d.jpg

Amar Hadzihasanovic (Jul 25 2024 at 07:17):

I also like this definition because its first point on its own (the unit is terminal) is the definition of a semicartesian monoidal category, and one of the good things of semicartesian monoidal categories is that they come equipped with natural “projections” defined in the way that I did.
Some of my favourite monoidal categories are semicartesian.

Amar Hadzihasanovic (Jul 25 2024 at 07:17):

So then being cartesian becomes “semicartesian + the natural spans of projections are universal”.

Amar Hadzihasanovic (Jul 25 2024 at 07:23):

I think there ought to be a coherence result for semicartesian monoidal categories which extends the one for monoidal categories to diagrams also including the $!_A$, but I don't know if it is proven somewhere.

Tobias Fritz (Jul 25 2024 at 09:11):

Amar Hadzihasanovic said:

So then being cartesian becomes “semicartesian + the natural spans of projections are universal”.

I think that this is the condition that's easiest to check in practice to see whether a monoidal category is cartesian.

Tobias Fritz (Jul 25 2024 at 09:13):

I've tried to construct examples of non-cartesian monoidal categories where the monoidal structure on objects is given by products as an MO answer. I'm worried about having missed something though, which happens incredibly easily with this sort of thing. So I'd appreciate some careful checks.

Morgan Rogers (he/him) (Jul 25 2024 at 09:22):

There's a characterization of cartesian monoidal categories involving diagonals, isn't there? It says that if you are semicartesian monoidal (unit object is terminal) and you have a natural transformation with components $\delta_A: A \to A \otimes A$ then you must be cartesian (again with the projections that Amar mentioned)?

John Baez (Jul 25 2024 at 10:38):

Are you alluding to Fox's theorem or something else?

John Baez (Jul 25 2024 at 10:41):

Fox's theorem, translated to sound more like what you just said, says that a category is cartesian iff there's a unique way to give each object $A$ a cocommutative comonoid structure $\delta_A: A \to A \otimes A$, $\eta_A: A \to I$, and these morphisms define natural transformations.

John Baez (Jul 25 2024 at 10:42):

One can imagine ways of trying to strengthen this result by weakening the hypotheses. Can we do it? Like: what if we drop the word "unique"?

Morgan Rogers (he/him) (Jul 25 2024 at 12:32):

John Baez said:

Are you alluding to Fox's theorem or something else?

Something else (but obviously closely related), as discussed here: we don't need to demand uniqueness of the diagonals, only that it is both natural and monoidal (at which point uniqueness must follow, I gather). The difference being that we don't impose coassociativity on $\delta$, for instance.

John Baez (Jul 25 2024 at 12:44):

I should really improve the nLab when it comes to cartesian monoidal categories. If you search for [[cartesian category]] you get sent to [[cartesian monoidal category]], which doesn't really give a precise definition of cartesian monoidal category. And it looks like there are more characterizations that need to be added!

Jean-Baptiste Vienney (Jul 25 2024 at 13:19):

Amar Hadzihasanovic said:

Yes, it does, I don't know if there's a good non-explicit argument though. Here's a diagram showing it for the first component

1043bf51-06bd-42a8-a78d-9531706b872d.jpg

Thanks a lot!

Jean-Baptiste Vienney (Jul 25 2024 at 13:38):

This is curiously (or not curiously) similar to a diagram I wrote to show that the multiplication $\mu_{A,B}:F(A \times B) \rightarrow F(A) \times F(B)$ of an oplax monoidal functor in a cartesian monoidal category is necessarily $\langle F(\pi_1), F(\pi_2)\rangle$.