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Stream: learning: questions

Topic: Naturality vs “no arbitrary choices”

Ruby Khondaker (she/her) (Dec 21 2025 at 13:05):

I came across an interesting counterexample recently that I wanted to hear people’s opinions of.

Let FinOrd be the category of finite ordered sets, with morphisms as order-preserving maps. Ordinal sum gives us two functors FinOrd^2 -> FinOrd, given by (x, y) -> x + y and (x, y) -> y + x on objects.

Then, there is a unique isomorphism between x + y and y + x (owing to finiteness). However, this is not a natural transformation! One way to see this is by considering constant maps.

So, what gives? We’ve made no arbitrary choices (in fact, no choices at all!) obtaining this isomorphism, yet we still have a failure of naturality.

While I’m here, I may as well give an example of the reverse. The usual proof of “full, faithful and essentially surjective => part of an equivalence” seems to make judicious use of the axiom of choice - namely, for each object X in the codomain, choosing an object A in the domain and an isomorphism F(A) -> X.

The isos can be chosen independently for each object in the codomain, so to me this feels like we have naturality with lots of arbitrary choices!

David Michael Roberts (Dec 22 2025 at 22:27):

Is it a [[core-natural transformation]], though?

David Michael Roberts (Dec 22 2025 at 22:28):

And yes, ff+eso => existence of pseudo-inverse is equivalent to some level of AC

David Michael Roberts (Dec 22 2025 at 22:29):

At, least, in the ordinary 2-cat of categories and functors and nat transformations. Hence [[anafunctors]].

John Baez (Dec 22 2025 at 22:34):

David is hinting that the meaning of "no arbitrary choices" is closer to "core-natural", sometimes called "canonical", than to "natural".

Ruby Khondaker (she/her) (Dec 23 2025 at 04:32):

I think the only automorphism of an ordered set here is the identity, so it is a core-natural transformation?

Since “no arbitrary choices” seems to be neither necessary nor sufficient for naturality, how else should i think about it?

James Deikun (Dec 23 2025 at 06:59):

In the case of an equivalence, you end up making arbitrary choices, but since whatever you choose is connected with whatever you didn't choose by a unique isomorphism, in a real sense the choices don't matter.

That said, "no arbitrary choices" was never anything but a heuristic. The definition of natural transformation is, well, the definition of natural transformation. If the category and the situation are sufficiently nice, the heuristic applies as stated.

That said, there is often an underlying truth in the heuristic even when it doesn't literally apply. For example, in the ordinal sum situation, even though no choices were made, there is still something "arbitrary" about the isomorphism, as it only holds when the underlying sets are (Kuratowski, I think) finite, and when the ordering is total. Specifically, it exists because of the pigeonhole principle, and the fact that total orderings of finite sets are unique. It specifically does not use, nor does it respect, the actual ordering structure on the finite ordered sets, so it is an "arbitrary" facet of the structure of finite sets peeking through into the theory of ordered sets. This is reflected in the lack of naturality.

Patrick Nicodemus (Dec 23 2025 at 08:48):

Something weird: FinOrd is by no means a symmetric monoidal category but if you take Z[FinOrd], which has the same objects and formal linear combinations of morphisms, it should be a symmetric Ab-enriched monoidal category. I have been meaning to write up a description of the symmetry isomorphism here. Since join of simplicial abelian groups is the Day convolution, the join of simplicial Abelian groups is also symmetric, and since Dold-Kan is strong monoidal for appropriately chosen definitions of "Dold" and "Kan" this gives the symmetry isomorphism for the tensor product of chain complexes.

Notification Bot (Jan 01 2026 at 15:02):

Ruby Khondaker (she/her) has marked this topic as resolved.

Alex Kreitzberg (Jan 01 2026 at 22:13):

@Ruby Khondaker (she/her) did you find an intuitive characterization for naturality that you preferred to the commonly given one?

Ruby Khondaker (she/her) (Jan 02 2026 at 06:32):

Alex Kreitzberg said:

Ruby Khondaker (she/her) did you find an intuitive characterization for naturality that you preferred to the commonly given one?

Yes actually, I think I prefer to view naturality as a categorification of “orthogonality/independence” these days, rather than “no arbitrary choices”. I’d be happy to expand on this if you’d like!

Alex Kreitzberg (Jan 02 2026 at 06:34):

Sure! It sounds interesting

Notification Bot (Jan 02 2026 at 06:37):

Ruby Khondaker (she/her) has marked this topic as unresolved.

Ruby Khondaker (she/her) (Jan 02 2026 at 06:38):

So, oftentimes in math you’re dealing with two operations/concepts that are “independent of”/“orthogonal to” each other. You might be aware of breaking up an object’s motion into horizontal and vertical parts, for example, which are literally orthogonal and can be studied separately. And in combinatorics, when you have a bunch of "independent" choices, you multiply them together to get the total number of choices. In a naive geometric sense, the choices in the cartesian product are literally "orthogonal" to each other!

Ruby Khondaker (she/her) (Jan 02 2026 at 06:41):

Another example comes from algebra - you have "internal" operations given by the algebraic structure, and "external" operations given by homomorphisms. These are _defined_ to be orthogonal/independent, in the sense that it doesn't matter in what order you apply the internal/external operations! $\phi(xy) = \phi(x) \phi(y)$ expresses that symbolically.

You can also consider row operations and column operations in linear algebra. These turn out to be orthogonal to/independent of each other, in the sense that it doesn’t matter if you first apply a row operation then a column one, or vice-versa. One way to see this is that row operations correspond to left-multiplication by a fixed matrix, and column operations correspond to right-multiplication by a fixed matrix. So these commute because of associativity! $(PM)Q = P(MQ)$.

Ruby Khondaker (she/her) (Jan 02 2026 at 06:47):

So more generally, you can think of postcomposition and precomposition being orthogonal/independent, because of associativity. $(f \circ g) \circ h = f \circ (g \circ h)$ can be interpreted as requiring postcomposition and precomposition to commute, essentially.

One way I like to intuit this is by viewing postcomposition as "data-changing" and precomposition as "shape-changing". Given a function $f : X \to Y$, you can do some post-processing with a function $g : Y \to Z$ to obtain $g \circ f : X \to Z$. You're acting on the actual output data.

On the other hand, you can precompose with a function $h : W \to X$ to change the shape of your domain, shuffle around values. For each $x \in X$, you "broadcast" the value $f(x)$ to the fiber $h^{-1}(\{x\})$. This gets you $f \circ h : W \to Y$.

So, associativity tells you that "data-changing" operations necessarily commute/are orthogonal to "shape-changing" operations! For example, if you view a list of length 3 as a map $3 \to X$, then mapping over the list is postcomposition with a map $X \to Y$, so "data-changing", while reversing the list is precomposition with a map $3 \to 3$, so "shape-changing".

In general, functorial actions tend to be "data-changing", so the components of natural transformations end up being "shape-changing" in order to be orthogonal/independent of the functorial maps.

Ruby Khondaker (she/her) (Jan 02 2026 at 06:49):

Incidentally, there's a sense in which postcomposition and precomposition are "orthogonal complements". Suppose you have a map that's orthogonal to precomposition, in the sense that $\eta(f \circ g) = \eta(f) \circ g$. Then it must be postcomposition by a fixed map, since $\eta(f) = \eta(\text{id} \circ f) = \eta(\text{id}) \circ f$. Similarly with the other direction - a map that's orthogonal to postcomposition must take the form of a precomposition! If you squint your eyes a little, this is essentially the proof of the yoneda lemma.

Ruby Khondaker (she/her) (Jan 02 2026 at 06:55):

You can make this orthogonality/independence intuition a little more precise by appealing to the bifunctor lemma:

image.png

(Taken from Awodey's book). The idea is that morphisms of C are "orthogonal to/independent of" morphisms of D in the product category C x D, so specifying a bifunctor is equivalent to specifying a separately functorial map whose functorial actions commute.

For example, since precomposition and postcomposition are orthogonal, you immediately get a bifunctor $\text{Hom} : C^\text{op} \times C \to \mathbf{Set}$ for every category. In a monoidal category, you have a tensor product bifunctor $C \times C \to C$ because, in $A \otimes B$, acting with morphisms on the first argument is completely independent of acting with morphisms on the second argument! So you can apply them "in parallel" without issue.

This generalises to the multifunctor lemma, which says that, to define a functor out of $C_1 \times \dots \times C_n$, it suffices to ensure it's functorial in each argument, and that the functorial actions pairwise commute. You can think of each factor $C_i$ as defining a new orthogonal "axis" that the multifunctor must act along!

Ruby Khondaker (she/her) (Jan 02 2026 at 07:06):

Finally, this relates to naturality by observing that natural transformations correspond to functors out of $C \times 2$. If you have $H : C \times 2 \to D$, then $H(-, 0) : C \to D$ and $H(-, 1) : C \to D$ are the functors your natural transformation is between, and $H(c, -) : 2 \to D$ are the components of your natural transformation. Applying the bifunctor lemma then gets you the naturality square!

So I like to think of naturality as a shadow of bifunctoriality/multifunctoriality, where you have many different functorial actions that are "pairwise orthogonal/independent". Thus, naturality categorifies the familiar notion of orthogonality/independence.

I've found this more helpful than "no arbitrary choices":

• In the FinOrd example, the issue is that the canonical iso $x \oplus y \to y \oplus x$ is not "orthogonal" to order-preserving maps acting on $x$ and $y$, since it uses the total order on the finite set in an essential way.
• For finite categories, "no arbitrary choices" doesn't tend to work as well for natural transformations. But, both in a naive geometric way and in the conceptual sense I've laid out, natural maps are necessarily "orthogonal" to whatever functorial action you've chosen! So this extends seamlessly to categories of all sorts of sizes.
• For the equivalence example I laid out at the start of this thread, it's still true that the way you define the inverse functor $G$ is in such a way to be "orthogonal" to the isomorphisms $A \to F(G(A))$ that you've chosen. While you do have freedom to make some arbitrary choices in choosing these isomorphisms, and so arbitrary choices in the functor $G$, these sets of choices are necessarily "orthogonal" to each other!
• Extending the previous example, the reason why naturality often manifests as "no arbitrary choices" is that it has to be orthogonal to your functorial actions - and these are often defined in a canonical way. If you introduce some arbitrary choices in the definitions of your functors, you necessarily obtain some freedom in the choices you can make for your natural transformation. But nevertheless, these sets of choices must remain "orthogonal" to each other, in order to obtain the naturality squares. So naturality is really orthogonal to functorial actions, and it's only because most functors are defined "without arbitrary choices" that the natural maps end up being defined with "no arbitrary choices".

So, what do you think @Alex Kreitzberg ?

Alex Kreitzberg (Jan 02 2026 at 07:20):

Very clever! Thank you for sharing. I have to think a bit more about some of your examples and arguments but my first impression is it seems correct and insightful.

Ruby Khondaker (she/her) (Jan 02 2026 at 08:57):

Perhaps I’m a little geometrically biased, but one thing I enjoy about this framing is that it lets me use existing intuition from linear algebra to think about naturality! I’ve already caught myself making statements like “the maps in a cone are orthogonal to maps within the diagram” as a way to encapsulate the naturality conditions present for limits.

Plus the naturality squares already have the maps being visually “orthogonal” to the functor applications, so it’s easy for me to remember this intuition whenever I see commutative diagrams.

Ruby Khondaker (she/her) (Jan 03 2026 at 16:09):

A silly thing I realised recently is that, in the product category C x D, morphisms of the form “(f, id)” are indeed orthogonal to morphisms of the form “(id, g)” for the pre-existing lifting property sense of morphism orthogonality. So at least in this case, the terminology is consistent!

Alex Kreitzberg (Jan 04 2026 at 03:56):

To try and understand your ideas better, I want to defend the "no arbitrary choice" intuition for naturality slightly.

You can prove naturality is automatically satisfied by formal polymorphic function definitions satisfying certain conditions in some programming languages.

If naturality broadly failed for a given programming language's polymorphic functions I'd interpret that as a weakness of the programming language, not a weakness in the definition of naturality.

So in a sufficiently nice language a given formula being well defined over all objects gives the formula lots of nice features - including naturality whenever that's possible. This is supposed to be unsurprising.

Maybe I'll revisit "Theorems for free" from this point of view you just introduced. Your intuition seems to be apriori about individual commutative squares, indeed commutative structures often can be represented as a product of factors. But then the bifunctor lemma shows, like you said, that categorifying this involves resolving many squares, all the arrows of the "orthogonal" categories must separately commute.

I think your idea is good but I'd like this relationship with logic and "the same formula everywhere" clarified. Which I'll probably do in my own time. Maybe you already addressed this by emphasizing canonically defined functors imply canonically defined natural transformations, but my feeling is there may still be something left worth thinking about.

Thank you for the interesting perspective, I understand commutative squares and naturality better.

Alex Kreitzberg (Jan 04 2026 at 04:02):

Another thought

A natural transformation is supposed to be the natural notion of an arrow between functors. I wonder if your interpretation is working because Cat is cartesian closed.

Ruby Khondaker (she/her) (Jan 04 2026 at 05:42):

Yes that’s fair - for “polymorphism implies naturality”, I like to think of it as follows. A parametrically polymorphic map FX -> GX cannot depend on the “data” of the type X, and so it must necessarily be “shape-changing”. But this is naturally orthogonal to data-changing maps, which are what the functorial actions are! Hence you get naturality for free.

And yes, functors C x 2 -> D naturally correspond to functors 2 -> [C, D], i.e. arrows in the functor category [C, D]. Thus you obtain the familiar “a natural transformation is an arrow between functors”.

You also have this being equivalent to functors C -> [2, D], which you can view as “a natural transformation is an arrow-valued functor”. Sometimes I prefer this perspective because of the way that natural transformations manifest as “a coherent family of arrows” in practice.

Alex Kreitzberg (Jan 04 2026 at 05:51):

Hah! That last point of view is new to me! But I find it immediately evocative to think about, thanks for sharing.

Ryan Wisnesky (Jan 04 2026 at 09:00):

As the recent thread with Mike Shulman showed, that all polymorphic functions between definable functors are natural is not a property of System F (its term model), but only of the parametric models of system F (which the term model is not). So perhaps not 'free', but 'cheap' (since parametric models are well studied)

Ruby Khondaker (she/her) (Jan 04 2026 at 10:38):

Yes that thread was another reason I tried looking for ways to think about naturality other than “no arbitrary choices”. I also used to think of naturality as a way of encoding "polymorphism", but there were too many situations (FinOrd, finite categories) that the intuition didn't quite work for. But so far "orthogonality/independence" has resolved those issues for me!

Alex Kreitzberg (Jan 04 2026 at 16:27):

Ahh okay, you mean the message here (https://categorytheory.zulipchat.com/#narrow/channel/229199-learning.3A-questions/topic/Continuations.2C.20parametricity.2C.20and.20polymorphism/near/525439963)? Thanks for calling that out.

Alex Kreitzberg (Jan 04 2026 at 17:22):

Wow, that's a nasty bit of confusion. I'm now suspicious similar mistakes are made by category theorists in more informal settings. Which I guess is what motivated Ruby to ask the question.

Ruby Khondaker (she/her) (Jan 04 2026 at 17:23):

Yeah that’s accurate, it was precisely because I’d grown dissatisfied with the “no arbitrary choices” and “polymorphism” intuition.

Alex Kreitzberg (Jan 04 2026 at 17:35):

I guess what I liked about (the now treacherous in retrospect) polymorphism intuition, is it seemed "self contained" in the sense that it was an intuition about maps between functors.

Your, as far as I can tell unproblematic, intuition passes via an adjunction from products in Cat to the exponentials.

And I'm just wondering if there's a "direct" intuition for exponential objects in Cat one can then see as satisfying the description you just laid out, because of the adjunction. But maybe I'm being pedantic.

Ruby Khondaker (she/her) (Jan 04 2026 at 17:44):

I think Awodey talks about this in his textbook, where you can imagine determining the structure of the category $[C, D]$ via probing it with “shape” categories (such as the ordinals $1$, $2$ and $3$) to determine objects, morphisms and composition respectively, and then using the universal property of the exponential to reduce this to functors $C \times n \to D$.

Ruby Khondaker (she/her) (Jan 04 2026 at 17:47):

image.png

Yes, this is what I'm referring to. I'm not sure if this is the sort of thing you were looking for? Essentially the argument is that you can already determine the nerve of $[C, D]$ just from knowing product categories - functors $n \to [C, D]$ have to correspond to functors $C \times n \to D$ because of the desired universal property.

Ruby Khondaker (she/her) (Jan 04 2026 at 17:53):

I've used this approach before to give a slick proof of interchange - you have functors $C \times 3 \to D$ and $D \times 3 \to E$, which combine to a functor $C \times 3 \times 3 \to E$. Then since products of preorders are preorders, you get a unique "diagonal" map in $3 \times 3$. This is exactly what expresses the commutativity of horizontal and vertical composition of natural transformations!

To spell that out in a bit more detail - functors $C \times 2 \to D$ correspond to natural transformations, and functors $C \times 3 \to D$ then correspond to vertically composable pairs of natural transformations. On the other hand, the way horizontal composition works is by composing $C \times 2 \to D$ with $D \times 2 \to E$ to get $C \times 2 \times 2 \to E$, and then using the unique "diagonal" map in $2 \times 2$ to obtain a functor $C \times 2 \to E$.

Ruby Khondaker (she/her) (Jan 04 2026 at 17:57):

One thing I like about this approach is that the category $3 \times 3$ automatically visualises all the relationships you have between the various functors and natural transformations involved in the proof of interchange. And the proof really just reduces to "products of preorders are preorders", which I find conceptually satisfying.

Alex Kreitzberg (Jan 04 2026 at 17:58):

Yeah looking at your and Awodey's arguments I guess the case can be made that the usual definitions and arguments involving naturality are worked out in the product of Cat anyway. There's probably no issue.

It's always tough when talking about intuitions, I'm never quite sure I'm "done" or "on to something" until the intuition is captured.

Ruby Khondaker (she/her) (Jan 04 2026 at 18:00):

Oh of course, I think it would be silly of me to view orthogonality/independence as the "One True Perspective" on naturality. Mostly I just try to interrogate and reform my existing intuitions when I start feeling uncomfortable around them.

Ruby Khondaker (she/her) (Jan 04 2026 at 18:07):

It's worth saying that you have to be a little more careful when trying to generalise this to higher categories. To get the correct definitions of natural transformations and their higher-dimensional analogs, you really need to take a "tensor product" of your categories rather than just a cartesian product. The nlab entry on the Gray tensor product is a good starting point. It might also be good to ponder why this issue doesn't show up in 1-category theory.

Ruby Khondaker (she/her) (Jan 04 2026 at 18:44):

I also feel like I should clarify that I'm not an expert on this! I've cobbled together observations I've made through learning category theory on my own, and sometimes I take on more of an authoritative tone than is warranted...

Alex Kreitzberg (Jan 04 2026 at 18:45):

Your tone is fine, it makes the writing easier to understand.

Alex Kreitzberg (Jan 04 2026 at 20:01):

The more I think about this, the more I think my issue may be I don't understand [[closed categories]] as much as I would like. Maybe I'll start a new thread on this.

Alex Kreitzberg (Jan 04 2026 at 20:21):

I'm also reminded of this apparently apocryphal conversation Riehl cited:

Kan:

You have explained how the tensor product can be defined in terms of the hom functor. Can the hom instead be defined in terms of the tensor product?

Eilenberg:

No, of course not. That's absurd.

I bet folks bounce back and force between which of the two sides of this adjunction is "easiest" depending on the category. Regardless I find it funny I seem to find the opposite direction from Eilenberg more confusing.

Ruby Khondaker (she/her) (Jan 04 2026 at 20:23):

Oh yes I’ve heard of this exchange before - so you’re saying you find defining tensor product from Hom the weirder direction? Usually I appeal to vector spaces (or R-modules more generally) to get intuition for that.

Alex Kreitzberg (Jan 04 2026 at 20:30):

Yeah I was thinking "I know what a matrix is independent of the tensor product", I'm curious what the analog would look like, if possible, for natural transformations.