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: learning: questions

Topic: choosing isomorphisms, vs determined isomorphisms

Alex Kreitzberg (Jun 09 2025 at 17:13):

In the conditions of Mac Lane's coherence theorem, we are required to provide a bifunctor $\otimes$, an identity object $I$, and chosen natural isomorphisms $\alpha : (A \otimes B) \otimes C \rightarrow A \otimes (B \otimes C)$, $\lambda : I \otimes A \rightarrow A$, $\rho : A \otimes I \rightarrow A$, such that the pentagon and triangle identities are satisfied.

In Set, if we chose the cartesian product for our tensor product, and some singleton set for the unit object $I$, then there is only one possible choice for all of the above natural isomorphisms.

What are some nice examples of tensor products and associated natural transformations, where we have more than one choice for $\alpha$, $\lambda$ and $\rho$?

Mike Shulman (Jun 09 2025 at 17:14):

Do you mean, more than one choice satisfying the axioms, or more than one choice of which only one satisfies the axioms?

Alex Kreitzberg (Jun 09 2025 at 17:18):

I'm thinking of $\otimes$ and $I$ as already selected, and I'm assuming it could be the case that there's still a sets worth of possiblities for $\alpha, \lambda$, and $\rho$, which satisfy the pentagon and triangle identity, otherwise why do we need to make a choice? Is that not true?

Mike Shulman (Jun 09 2025 at 17:20):

It's easy to come up with examples in which there is more than one natural isomorphism $\alpha : (A \otimes B) \otimes C \rightarrow A \otimes (B \otimes C)$. For example, in the category of modules over a commutative ring $R$, you can fix any invertible element $r\in R$ and define $\alpha_r((a\otimes b)\otimes c) = r\,a\otimes (b\otimes c)$. But this will not satisfy the pentagon identity unless $r=1$.

Alex Kreitzberg (Jun 09 2025 at 17:22):

Sorry I think I understood your question and reedited my answer to clarify too slowely.

I do mean the pentagon and triangle identities need to be satisfied. Does that then make $\alpha, \lambda$ and $\rho$ uniquely determined?

John Baez (Jun 09 2025 at 17:24):

Of course not! That would mean they're sort of just boring fluff.

Mike Shulman (Jun 09 2025 at 17:24):

There can still be multiple choices that satisfy the axioms, but it's harder to come up with "naturally occurring" examples.

Riley Shahar (Jun 09 2025 at 17:30):

In $G$-graded vector spaces I think possible twists to the associator should be detected by the third cohomology of $G$. The $3$-cocycle condition is the same as the pentagon condition.

Riley Shahar (Jun 09 2025 at 17:31):

(cohomology with coefficients in the units of the field, I guess)

Riley Shahar (Jun 09 2025 at 17:31):

haha, @John Baez I beat you by about two seconds I guess. But having a concrete example is probably good

John Baez (Jun 09 2025 at 17:32):

Yes. Here's a nice example of that 3-cocycle business:

The category of $(\mathbb{Z}/2)^3$-graded vector spaces, with its usual tensor product - I could explain it if asked - has at least two different associators $\alpha$ which together with the 'usual' $\lambda$ and $\rho$ obey the pentagon and triangle identities. So we get two different monoidal categories with the same tensor product. In one of these - the less obvious one! - the octonions are a monoid object!

In short, while the octonions look nonassociative to our ordinary eyes, they are associative in the category of $(\mathbb{Z}/2)^3$-graded vector spaces with its usual tensor product but a funny associator.

Alex Kreitzberg (Jun 09 2025 at 17:33):

John Baez said:

Of course not! That would mean they're sort of just boring fluff.

Right, that was my understanding, Shulman just gave the impression situations where this mattered were "harder" to find which confused me, my question could've been worded better.

Thank you for the example!

John Baez (Jun 09 2025 at 17:33):

I explain the details of that example at the bottom of this page.

John Baez (Jun 09 2025 at 17:34):

Situations like this are indeed "harder to find", since you need to know a bit about group cohomology, the octonions, etc. etc.

Riley Shahar (Jun 09 2025 at 17:36):

John Baez said:

I explain the details of that example at the bottom of this page.

Very cool, thanks. (I think I must have learned about this from your expository paper, which I remember reading at some point, and internalized the fact while forgetting the source. So I should give proper attribution)

John Baez (Jun 09 2025 at 17:41):

There could be two finite groups $G$ and $H$ whose categories of representations are equivalent, and even "equivalent as categories with tensor product functor $\otimes$", but not equivalent as monoidal categories.

I don't know examples, though!

It's vastly easier to find two finite groups $G$ and $H$ whose categories of representations are equivalent, but not equivalent as monoidal categories. You can take $G = \mathbb{Z}/4$ and $H = \mathbb{Z}/2 \times \mathbb{Z}/2$. But their categories of representations are not even isomorphic as "categories with tensor product functor".

Alex Kreitzberg (Jun 09 2025 at 17:44):

I want to add a sentence here as maybe a note to myself.

Even though in Set $I$ and $\times$ determine $\alpha, \lambda$ and $\rho$, you still need Mac Lane's coherence theorem to prove there's exactly one natural isomorphism for any given way to "move parenthesis around".

John Baez (Jun 09 2025 at 17:59):

Do you know a proof that in Set $I$ and $\times$ determine $\alpha, \lambda$ and $\rho$?

John Baez (Jun 09 2025 at 18:00):

I do not. I'd like to know if this is true.

Alex Kreitzberg (Jun 09 2025 at 18:24):

So to be clear, you're serious about there not being a counter example you're aware of? But you are also unaware of a proof?

I must still be misunderstanding something then. Because I would think the proofs would all be very similar to proving there is only one natural transformation between the $F$ and $G$ you just defined here (#learning: questions > AxB vs BxA @ 💬 ).

But interpreting your question as if I'm not confused - to clarify potentially where I am confused - I'm getting this intuition from programming languages. In a programming language with parametric polymorphism, when you provide a certain polymorphic type signature, terms for the programs are often uniquely determined.

And in particular, for each of the following type signatures in haskell, I believe there is only one term that satisfies them:

α :: (a, (b, c)) -> ((a, b), c)
λ :: ((), a) -> a
ρ :: (a, ()) -> a

The way I might "prove" this for $\lambda$, is I'd note I can substitue a one element set into a different than (), and then a's element is what must get returned. Because we have to use the same formula for every type, then projecting on the right pair must be the only formula for this parametric function.

Functions on the singleton element that is associated with the whole language ( in this case () is the syntax for both the singleton type, and its unique element ), necessarily associate the output of the singleton with some object in the category, so can't be the output of our polymorphic function.

And I assumed this style of argument would work for natural isomorphisms.

(Any help getting unconfused is sincerely appreciated!)

Riley Shahar (Jun 09 2025 at 19:40):

Alex Kreitzberg said:

The way I might "prove" this for $\lambda$, is I'd note I can substitue a one element set into a different than (), and then a's element is what must get returned. Because we have to use the same formula for every type, then projecting on the right pair must be the only formula for this parametric function.

What if the formula is something more complicated, that happens to reduce to the identity in the case where a is a one-element set? You cannot necessarily deduce the behavior of the natural transformation just from its value at the singleton. (For instance, in an arbitrary cartesian category $C$, any natural isomorphism $1_C\to 1_C$ will determine a twist $1_C\times (1_C\times 1_C)$ of the canonical associator $\alpha$. There are no such endomorphisms in set, but in general they may exist.)

John Baez (Jun 09 2025 at 19:45):

Alex Kreitzberg said:

So to be clear, you're serious about there not being a counter example you're aware of?

Yes, that's what I said. You claimed something interesting and nontrivial. I asked you if you
knew a proof, I said I don't know one, and I said I'd like to know if this claim is true.

Alex Kreitzberg (Jun 09 2025 at 20:02):

I think then there are more elementary claims here, that I must be confusing with a bigger nontrivial claim, that's incidentally plausible.

I'm going to try and figure out what the elementary claims are, try to prove those properly, and see if I can work up to where the question changes.

For example, to start I'm going to try and make a nice proof for the claim that there is just one natural isomorphism from the functor $F : \text{Set} \times \text{Set} \rightarrow \text{Set}$ defined by $F(A, B) = A \times B$ to the functor $G :\text{Set} \times \text{Set} \rightarrow \text{Set}$ defined by $G(A, B) = B \times A$ (posed here: #learning: questions > AxB vs BxA @ 💬 ).

Riley Shahar (Jun 09 2025 at 20:05):

John Baez said:

Do you know a proof that in Set $I$ and $\times$ determine $\alpha, \lambda$ and $\rho$?

Matteo Capucci (he/him) (Jun 12 2025 at 09:02):

John Baez said:

Do you know a proof that in Set $I$ and $\times$ determine $\alpha, \lambda$ and $\rho$?

I guess being the coherence morphisms of a cartesian monoidal structure they are uniquely determined

John Baez (Jun 12 2025 at 11:46):

Can we change them and get some other possibly noncartesian monoidal structure on $\mathsf{Set}$ - even a nonsymmetric one?

Tobias Fritz (Jun 12 2025 at 12:19):

No, changing the unitors and associators on a cartesian monoidal category does not let you escape cartesianness: you need to at least change the tensor product of morphisms too. This is a consequence of the proposition stated in my answer to your closely related MO question from last summer:

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

Peter Lumsdaine gave a nice example where both the coherence isomorphisms and the action on morphisms change. I tried to find an example where only the action on morphisms changes, but Peter found an error in my attempt and I eventually gave up. It's a really tricky problem!

John Baez (Jun 12 2025 at 12:42):

Thanks for reminding me of all this.

Alex Kreitzberg (Jun 12 2025 at 20:01):

It's a lot harder to state and prove the elementary propositions than I was expecting.

Given my cursory overview of the above, my guess is everything gets even harder when we don't assume the standard universal property of products and initial objects for our tensor product and unit.

So for my elementary argument, I want to take those for granted.

At this point I still needed Riehl to hold my hand. From her text I learned one can define a functor $\lim_J : C^J \rightarrow C$, after assigning an arbitrary member of the limit candidates in $C$ for each diagram in $C^J$.

The proof contains the basic idea, that there's a correspondence between the maps between limits, and the natural transformations between the diagrams they're over.

With that background we can finally state the claim, in Reihl's text a lemma, that:

For any triple of objects $X$, $Y$, $Z$ in a category with binary products, there is a unique natural isomorphism $X\times(Y\times Z) \cong (X\times Y) \times Z$ commuting with the projections to $X$, $Y$, and $Z$.

She leaves this as an exercise for the reader, which I'll try to do here (though the larger context of this discussion has me slightly intimidated)

And Riehl doesn't help me feel less intimidated, immediately following that lemma she writes:

Lemma 3.6.6 asserts that the product is naturally associative. It follows that any iteration of binary products can be used to define n-ary products. However, even in a skeletal category, in which the objects $X \times (Y \times Z)$ and $(X \times Y) \times Z)$ are necessarily equal, the natural isomorphism may not be the identity.

She then provides such a counterexample in the skeleton of Set.

I'm going to work through these more carefully/completely later when I have more time.

But if folks have advice on how to think about this stuff in the mean time, I'd appreciate any help or suggestions.

Kevin Carlson (Jun 12 2025 at 20:50):

My main point would be that all these interesting subtleties about skeletality and whether isos are identities and such aren't really relevant to the actual exercise stated. So you don't need to worry about any of that stuff to do the exercise, which is luckily much more fundamental to understanding the main line of thought in category theory.

John Baez (Jun 12 2025 at 21:00):

I second Kevin's remarks. The exercise is important. Everything about skeletal categories, whether isomorphisms are actually identity morphisms, etc., falls into the realm of 'evil' -that is, the realm of issues that are not invariant under equivalence, which tend to be tricky and fundamentally unimportant, though often entertaining.

Alex Kreitzberg (Jun 18 2025 at 16:24):

Okay, for this lemma, we know $(X \times Y) \times Z$ is universal in the category of cones over $X$, $Y$ and $Z$.

Given any triple of arrows $f : T \rightarrow X$, $g : T \rightarrow Y$ and $h : T \rightarrow Z$, we get a unique arrow $(f, g) : T \rightarrow X\times Y$ which then forms a unique arrow with $h$ to give $((f, g), h) : (X \times Y) \times Z$. We can use the projection arrows of the pair to get any of our original morphisms, which lets us define a triple of projections.

We have a similar story for $X \times (Y \times Z)$ that lets us package the arrows as $(f, (g, h))$ instead.

In particular, because these define a universal property of triple products, then they define an isomorphism between $X \times (Y \times Z)$ and $(X \times Y) \times Z$.

That part I know how to do fairly easily. In set we can even be explicit and show the isomorphism in any particular case just moves parenthesis of elements around $(x, (y, z)) = ((x, y), z)$.

This construction gives us a set of isomorphisms $I_{X, Y, Z}$, what I'm not in the habiting of demonstrating, are that these arrows collectively describe a natural transformation over the triples of objects.

After a lot of thought, the issue seems to be mostly sorted with a particular diagram, let $f : X \rightarrow X'$, $g : Y \rightarrow Y'$ and $h : Z \rightarrow Z'$ be arrows of our category, then I'll explain how the arrows in the following diagram commute:

Functoriality and naturality of associativity of product.jpg

I alluded to the arrows given by functoriality, $f \times (g \times h)$ and $(f \times g) \times h$, in my message from a week ago. The left hand limits are cones over the right hand triple of objects, so there is exactly one arrow from the limits on the left to the limits on the right. This proves the square up top is commutative, because there must be one arrow to get from any one object to any other.

But that square is our naturality square, so we see our uniquely obtained isomorphisms do combine into a natural isomorphism.