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Stream: learning: questions

Topic: A formal definition of a commuting diagram

Jencel Panic (Jul 26 2023 at 07:43):

I know we have a definition of a diagramin CT and I was wondering, is there a formal definition of a commuting diagram ?

Would it be a diagram in which the index category is thin (poset)?

Seems correct, but there are actually diagrams where there are two morphisms between two objects, such as the diagram for equalizer.

image.png

On the other hand, such diagrams always seemed weird to me, and they look like they don't obey the informal definition of commutative diagram:

"...a commutative diagram is a diagram such that all directed paths in the diagram with the same start and endpoints lead to the same result. " --- Wikipedia

In particular, in the diagram above, it seems that there are two directed paths from $A$ to $B$, consisting of one morphism each ($f$ and $g$) and so they should be equivalent, which wouldn't make sense.

Josselin Poiret (Jul 26 2023 at 08:00):

Jencel Panic said:

I know we have a definition of a diagramin CT and I was wondering, is there a formal definition of a commuting diagram ?

Would it be a diagram in which the index category is thin (poset)?

Seems correct, but there are actually diagrams where there are two morphisms between two objects, such as the diagram for equalizer.

image.png

On the other hand, such diagrams always seemed weird to me, and they look like they don't obey the informal definition of commutative diagram:

"...a commutative diagram is a diagram such that all directed paths in the diagram with the same start and endpoints lead to the same result. " --- Wikipedia

In particular, in the diagram above, it seems that there are two directed paths from $A$ to $B$, consisting of one morphism each ($f$ and $g$) and so they should be equivalent, which wouldn't make sense.

Yes, the diagram above is not a "commuting diagram" in the strict sense! The commuting diagram for the equalizer should probably look more like a pullback with both arms being equal if you want it to fit your definition.

Ralph Sarkis (Jul 26 2023 at 08:57):

When I started writing my notes on CT, I had the ambition to find the perfect formal definition of "commutativity of diagrams". I settled on the following:
image.png

We briefly discussed this here a while ago. I don't think it is extremely useful to have such a definition anymore, but I have yet to agree with myself on how to talk about commutativity in my notes.

Jencel Panic (Jul 26 2023 at 09:12):

@Ralph Sarkis The "greater than one" part I see as problematic, e.g. $f$ equivalent to $id \circ f$.

Ralph Sarkis (Jul 26 2023 at 09:15):

I am not sure what you mean. Can you draw the diagram that you think is problematic?

Jencel Panic (Jul 26 2023 at 09:18):

It's just the idea that every "path of length 1", e.g. $f$, is equivalent to a path of length 2 (e.g. $ID \circ f$, or $f \circ ID$ )

Ralph Sarkis (Jul 26 2023 at 09:19):

Oh I see, but people never draw the identity on diagrams. That should probably be made clear in that definition.

Jencel Panic (Jul 26 2023 at 09:25):

That was just an example of how the definition might be confusing, another thing is that each path that is greater than one is equivalent to a path of length one, as per composition.

Jencel Panic (Jul 26 2023 at 09:25):

More generally, I wonder if this phrase is needed, since, as @Josselin Poiret mentioned, the diagram for equalizer is not a commuting one. The article for equalizers in Wikipedia, also doesn't use the diagram as a definition, and doesn't call it commutative/commuting (as it is done customarily).

Ralph Sarkis (Jul 26 2023 at 09:32):

It is very confusing (and that is the main reason why I don't think it is useful). "Paths" in this definition refer to paths in the diagram, not in the category.

This problem with the equalizer diagram is precisely why I think it'd be great to have an alternative definition to commutativity. I don't think saying the diagram does not commute or drawing a bigger commutative diagram to say the same thing is the right thing to do because when you draw that diagram and say it commutes to a CTist, they know that the parallel arrows are not equal.

Jencel Panic (Jul 26 2023 at 09:43):

OK, thanks, now I read the other topic and got everything about equalizers and other similar diagrams (are there others (besides coequalizer) actually?)

Josselin Poiret (Jul 26 2023 at 09:43):

I don't think you can give a definition that would encompass cases like the equalizer without also adding some more stuff into the diagram

Ralph Sarkis (Jul 26 2023 at 09:47):

Josselin Poiret said:

I don't think you can give a definition that would encompass cases like the equalizer without also adding some more stuff into the diagram

The definition I gave above does not work?

Jencel Panic (Jul 26 2023 at 09:49):

OK, got everything about equalizers and similar diagrams (are there many similar ones, actually?)

Still wondering about my first question:

Because we have:

A diagram in category $C$ is an index category $J$ and a (covariant) functor $D : J → C$.

Does it makes to say that:

A commuting diagram in category $C$ consists of a thin category $J$ and a (covariant) functor $D : J → C$.

If you don't draw parallel arrows that is.

Dylan Braithwaite (Jul 26 2023 at 10:29):

Using a thin category like that you can't express commuting squares for example, since $J$ can only index one side of the square.

I think you could say something like this:

A diagram $D: J \to C$ commutes if for every wide subcategory $J'$ of $J$, the composite $D \circ F$ has the same value for every possible identity-on-objects mono functor $F : J' \to J$

Dylan Braithwaite (Jul 26 2023 at 10:35):

This means you can fix a diagram, then choose a subcategory of $J$ corresponding to 'taking one path through the commuting diagram'. Then the condition says that whatever way you include the path into the larger diagram, the result should be equal.

Ralph Sarkis (Jul 26 2023 at 11:23):

I don't think commutativity is about these formal "diagrams". It is about the things we draw to represent some objects and morphisms of a category, we also call these things "diagrams". The latter are really just directed graphs (if you don't draw parallel arrows), with labels coming from the category in which the diagram is supposed to live.

Going to formal diagrams might go a bit like this: Given a drawing of a graph $G$ with

• the formal diagram represented by this drawing is the functor from the free category on $G$ to $C$ that sends a node/morphism of $G$ to the object/morphism of $C$ labelling it.

Then, I think one could say that $G$ is commutative if the functor above factors through the free thin category generated by $G$.

Dylan Braithwaite (Jul 26 2023 at 12:59):

I think I agree with you in that it probably isn't useful to think about commutativity in terms of functors out of an index category. But if I'm understanding Jencel's original message then I think they were asking how we could define commutativity in terms of this general notion of diagrams as a functor.

Your definition of commutativity is definitely nicer than what I suggested though!

Jencel Panic (Jul 26 2023 at 13:58):

Yeah, I was just wondering if there exists such a definition, as, after all, many concepts in CT are often defined with "such that this diagram commutes."

Jencel Panic (Jul 26 2023 at 14:01):

@Ralph Sarkis Can you elaborate on the last part? What is the free thin category? What does it mean for a functor to factor through a category?

Ralph Sarkis (Jul 26 2023 at 14:24):

The free category on a directed graph $G= (V,E)$ with vertices $V$ and edges $E$ is the category whose objects are $V$ and morphisms $u \to v$ are path in $G$, that is, a finite sequence of edges such that the target of one edge is the source of the next one in the sequence, and the source (resp. target) of the first (resp. last) edge in the path is $u$ (resp. $v$). Identities are empty paths, and composition is done by concatenating paths (sequences of edges). This should look a lot like taking the free monoid on the set of edges but taking care that edges can only be connected when the source and targets match.

Ralph Sarkis (Jul 26 2023 at 14:24):

Taking the free thin category on $G$ would mean doing the exact same process, but quotienting by saying morphisms (paths) are equal when they have the same source and target. More concretely, the objects are still $V$, but now $\mathrm{Hom}(u,v)$ is either empty or a singleton, it contains a morphism if and only if there is a path from $u$ to $v$ in $G$. Identities and composition are completely determined.

Ralph Sarkis (Jul 26 2023 at 14:25):

Taking as an example the graph representing a square (drawn below), the free category will have two diagonal morphisms (two different paths) but the free thing category will only have one.
image.png

Ralph Sarkis (Jul 26 2023 at 14:25):

Now, there is an evident functor $F: \mathbf{FreeCat}(G) \to \mathbf{FreeThinCat}(G)$ that sends a path $u \to v$ in the domain to the unique morphism $u \to v$ in the codomain (that witnesses the existence of at least one path from $u$ to $v$).

Ralph Sarkis (Jul 26 2023 at 14:25):

If you label the nodes and arrows of $G$ with objects and morphisms from a category $\mathbf{C}$, this yields a functor $D: \mathbf{FreeCat}(G) \to \mathbf{C}$. By saying that this factors through the free thin category on $G$, I mean that there is a diagram $D': \mathbf{FreeThinCat}(G) \to \mathbf{C}$ such that $D' \circ F = D$. You can check that this means $D$ sends any two paths with the same source and target to the same morphism in $\mathbf{C}$. Indeed these two paths are equal after applying $F$, so applying $D = D' \circ F$ will also make them equal.

Morgan Rogers (he/him) (Jul 26 2023 at 15:01):

I don't know if this helps or complicates things, but I use commutativity to refer to maximal paths in a diagram. For example, if I draw a diagram consisting of two squares pasted together and tell you it commutes, I am not necessarily asserting that the individual squares commute, but the squares commuting implies the diagram as a whole will commute.

Morgan Rogers (he/him) (Jul 26 2023 at 15:02):

To make that precise, you only need to constrain the indexing directed graph of the diagram to be acyclic. Or you can come up with a convention for cycles, I suppose.

John Baez (Jul 26 2023 at 15:37):

For example, if I draw a diagram consisting of two squares pasted together and tell you it commutes, I am not necessarily asserting that the individual squares commute...

Oh-oh! I'll have to careful when reading your papers. I wouldn't interpret it that way. Why draw the unnecessary edge in the middle if you're just asserting that the outer part commutes:

---->----> | | v v ---->---->

?

Oscar Cunningham (Jul 26 2023 at 15:49):

image.png
In other areas of mathematics we're happy to simply write a string of equalities like $cqp = sbp = sra$. I think this is actually simpler than the diagram. Simple school algebra problems would turn into nightmares if written as commutative diagrams. I think it might be better to use the diagram only to remind the reader of the morphism domains and codomains, and write out the equalities explicitly.

John Baez (Jul 26 2023 at 15:54):

I think it depends on what you're used to. To me the diagram shows clearly what's going on, while writing (say)

ra = bp and sb = cq

looks like a disorganized mass of letters. (Or if I were Morgan: sra = cqp.)

Mike Shulman (Jul 26 2023 at 16:10):

I think it is pretty universally understood that when we say a diagram commutes we mean at least that any two parallel paths anywhere in the diagram are equal if at least one of them has length $>1$, and using it for anything weaker than this is a recipe for confusion.

Nathanael Arkor (Jul 26 2023 at 16:16):

I think that writing that a diagram "commutes" if it has two parallel paths both having length 1, and not intending to mean that both morphisms are equal, is poor practice. (In other words, I think the restriction on path length should be dropped, though in practice I don't think I've seen many commutative diagrams involving parallel arrows.)

Mike Shulman (Jul 26 2023 at 16:45):

I think because of the potential ambiguity, it's best to avoid stating that any diagram containing parallel arrows "commutes" regardless of what you intended to mean by it.

David Egolf (Jul 26 2023 at 16:46):

To give another example of a definition, Riehl says this in "Category Theory in Context":

A diagram in a category $\mathsf{C}$ is a functor $F : \mathsf{J} → \mathsf{C}$ whose domain, the indexing category, is a small category.

and then a few sentences later:

Functoriality requires that any composition relations that hold in $\mathsf{J}$ must hold in the image of the diagram, which is what it means to say that the directed graph defined by the image of the diagram in $\mathsf{C}$ is commutative.

So, for example, if $\mathsf{J}$ has two morphisms $f$ and $g$ with the same source and target but $f \neq g$, then a commutative diagram described by a functor $F: \mathsf{J} \to \mathsf{C}$ would not require $F(f) = F(g)$. (If I understand correctly!)

In general, the paths that are required to commute in the directed graph defined by the image of the diagram I believe depends on the indexing category $\mathsf{J}$. Under this perspective, the phrase "the diagram commutes" is potentially ambiguous unless the indexing category is specified.

Ralph Sarkis (Jul 26 2023 at 18:17):

David Egolf said:

Functoriality requires that any composition relations that hold in $\mathsf{J}$ must hold in the image of the diagram, which is what it means to say that the directed graph defined by the image of the diagram in $\mathsf{C}$ is commutative.

It feels to me like this is only under the hypothesis that $\mathsf{J}$ is a preorder (a thin category).

David Egolf (Jul 26 2023 at 18:40):

Ralph Sarkis said:

David Egolf said:

Functoriality requires that any composition relations that hold in $\mathsf{J}$ must hold in the image of the diagram, which is what it means to say that the directed graph defined by the image of the diagram in $\mathsf{C}$ is commutative.

It feels to me like this is only under the hypothesis that $\mathsf{J}$ is a preorder (a thin category).

If so, as far as I can tell, Riehl doesn't state this hypothesis in "Category Theory in Context". For example, in Lemma 1.6.5, which concerns commutative diagrams, there is no mention of $\mathsf{J}$ being required to be a thin category.

Ralph Sarkis (Jul 26 2023 at 18:52):

The proof is missing some explanations, but I think what is shown is that if $F: J \to C$ is a diagram from a thin category (i.e. a commutative diagram), then $GF : J \to C$ is also a diagram from a thin category, so it is also commutative.

If there is no requirement on $J$, I don't understand what is the definition of "commutative". If $F: J \to C$ is a diagram, when is it commutative according to your interpretation of that excerpt?

Kevin Arlin (Jul 26 2023 at 18:54):

The definition seems to say that $F:J\to C$ is a commutative diagram if and only if it's a functor. This is an interesting approach but I think it's not really convincing, since we specifically want to be able to refer to (for instance) noncommutative squares as functors from the free category generated by the square graph, rather than as, I don't know, non-functors from the non-free square category?

David Egolf (Jul 26 2023 at 19:00):

Ralph Sarkis said:

If there is no requirement on $J$, I don't understand what is the definition of "commutative". If $F: J \to C$ is a diagram, when is it commutative according to your interpretation of that excerpt?

I think Riehl is indicating that any functor $F: \mathsf{J} \to \mathsf{C}$ is a commutative diagram, under the condition that $\mathsf{J}$ is a small category.

That being said, after the initial definition, Riehl goes on to talk about specific examples of commutative diagrams. As far as I can see, all the examples that Riehl calls "commutative" are indexed by a thin category, and the example provided where the indexing category isn't thin leads to what Riehl terms a "not-necessarily commutative square". So, the terminology seems a bit confusing!

David Egolf (Jul 26 2023 at 19:13):

Kevin Arlin said:

The definition seems to say that $F:J\to C$ is a commutative diagram if and only if it's a functor. This is an interesting approach but I think it's not really convincing, since we specifically want to be able to refer to (for instance) noncommutative squares as functors from the free category generated by the square graph, rather than as, I don't know, non-functors from the non-free square category?

Let $\square$ be the category that looks like a square, where there are two paths of composable non-identity morphisms from the top left to the bottom right, and the two paths compose to the same morphism.

Let $\square'$ be the category that looks like a square, where there are two paths of composable non-identity morphisms from the top left to the bottom right, and the two paths do NOT compose to the same morphism.

Then (if I understand you correctly) a "noncommutative square" is a commutative diagram with respect to the indexing category $\square'$. However, it is not a commutative diagram with respect to the indexing category $\square$.

It does seem a bit unfortunate to simultaneously call something a "noncommutative square" and a "commutative diagram"!

Kevin Arlin (Jul 26 2023 at 20:47):

Yeah, we're agreeing on the story here.

Morgan Rogers (he/him) (Jul 27 2023 at 09:06):

I feel that the context in which we actually use commutative diagrams is being buried in this discussion.
When we draw a diagram (commutative or not), it's a directed graph whose vertices are labelled with objects of a category and edges labelled with morphisms. Saying that the diagram is commutative is asserting some equations between those morphisms. If we have a way of asserting which equations externally, like Riehl's functoriality condition, then great, but I fear this is circular since we still need a way of defining the equations in the indexing categories.

There are two reasons we would assert equations that I can think of: as part of a definition (a hypothesis/axiom), or in a diagram-chasing proof when verifying such an equation holds.
In the former case, we would want to make the equation as explicit as possible: if your equations are not identifying maximal paths in the diagram you draw, that's a recipe for confusion (or you should probably say "this diagram and all its subdiagrams commute").
In the second case, the proof may require expanding or decomposing the diagram that you actually want to show is commutative, and in practice one refers to commutativity of subdiagrams of a composite diagram to save space. For example...
John Baez said:

For example, if I draw a diagram consisting of two squares pasted together and tell you it commutes, I am not necessarily asserting that the individual squares commute...

Oh-oh! I'll have to careful when reading your papers. I wouldn't interpret it that way. Why draw the unnecessary edge in the middle if you're just asserting that the outer part commutes:

---->----> | | v v ---->---->

?

Referring to @Oscar Cunningham 's image,
if I were to insist on my definition of commutativity from earlier, I could say "if the diagram commutes and $p$ is an epimorphism then the right-hand square commutes", but ultimately it would be clearer in this case to refer to 'the outer rectangle' and 'the left-hand square' in this argument.

As a final, personal opinion point: I want to be able to refer to the equalizer diagram as a commutative diagram! When i draw parallel arrows in a diagram, I will typically need to say which subdiagrams commute anyway, although it can be deducible from context which are intended (see Proposition 3.5 on this nLab page, for instance; this seems like a reasonable, if informal, use of "commutative diagram", where we can deduce that the squares formed from the left-hand and right-hand parallel arrows are expected to commute but not necessarily the other possible squares).

Ralph Sarkis (Jul 27 2023 at 09:21):

Morgan Rogers (he/him) said:

(see Proposition 3.5 on this nLab page, for instance; this seems like a reasonable, if informal, use of "commutative diagram", where we can deduce that the squares formed from the left-hand and right-hand parallel arrows are expected to commute but not necessarily the other possible squares)

I agree it is reasonable, but I wouldn't mind if the formal definition of commutativity does not encompass this example (maybe I am biased because I don't see these things often). Btw, they are called serially commutative diagrams in Toposes, Triples and Theories.

Morgan Rogers (he/him) (Jul 27 2023 at 09:47):

Agreed, but I think for the sake of formalisation my proposal of maximal paths (plus specifying explicitly which subdiagrams are assumed to commute) covers this.

Graham Manuell (Jul 27 2023 at 13:25):

Morgan Rogers (he/him) said:

In the former case, we would want to make the equation as explicit as possible: if your equations are not identifying maximal paths in the diagram you draw, that's a recipe for confusion (or you should probably say "this diagram and all its subdiagrams commute").

I don't see how this would be confusing? You are the only person I've heard mention anything about maximal paths being important in commuting diagrams. Perhaps if your convention became more common, then it would be confusing.

Graham Manuell (Jul 27 2023 at 13:25):

I think the 'paths of length > 1' definition works fine 97% of the time. For the rest, you can just say what exactly commutes.

Graham Manuell (Jul 27 2023 at 13:26):

I've also actually seen a paper use "the following diagram 'reasonably' commutes" to mean "this diagram doesn't actually commute in the technical sense, but it is clear which paths we want to be equal from context". Perhaps they could have been written this more explicitly, but I did in fact find it obvious what they meant.

John Baez (Jul 27 2023 at 13:55):

I guess everyone here has already noticed, but I think there are at least three things category theorists mean by "diagram":

1) The original meaning, probably, is a diagram that you actually draw on the page - thus, a graph in the category theorist's sense, also called a [[quiver]], with vertices labelled by objects in some category $C$ and edges labelled by morphisms in $C$. I'll use the word "quiver" since graph theorists have another meaning of the word "graph".

2) At least nine times out of ten, when someone actually draws such a diagram, the edges don't cross. Then they're taking a graph and embedding it into the plane - at least up to isotopy: i.e. if they draw it on rubber and you bend the rubber, the diagram shouldn't change meaning. This lets you unambiguously speak of 2-cells bounded by edges, and ask whether each one of these commutes.

3) But sometime around when people started trying to formalize the concept of limit and colimit of a diagram, they decided it was good to define a diagram in a category $C$ to be a functor $F: J \to C$ where $J$ is some category describing the diagram's 'shape'. This concept of diagram reduces to the first definition if we let $J$ be the free category on some quiver.

John Baez (Jul 27 2023 at 13:55):

Meanings 1) and 3) are the most important.

John Baez (Jul 27 2023 at 13:58):

If you want to "ask if a diagram commutes", you can do this. Given a quiver $Q$ there is not only the free category on $Q$, say $F(Q)$, but also the free preorder on $Q$, say $G(Q)$.

John Baez (Jul 27 2023 at 13:59):

There's a natural transformation $F \Rightarrow G$ so we get a functor $F(Q) \to G(Q)$ which forces all parallel pairs of arrows in $F(Q)$ to become equal.

John Baez (Jul 27 2023 at 14:01):

Then a "not necessarily commutative diagram in $C$ shaped like the quiver $Q$" is a functor $\phi: F(Q) \to C$. It "commutes" iff $\phi$ factors through the functor $F(Q) \to G(Q)$.

Mike Shulman (Jul 27 2023 at 14:43):

Graham Manuell said:

Morgan Rogers (he/him) said:

In the former case, we would want to make the equation as explicit as possible: if your equations are not identifying maximal paths in the diagram you draw, that's a recipe for confusion (or you should probably say "this diagram and all its subdiagrams commute").

I don't see how this would be confusing? You are the only person I've heard mention anything about maximal paths being important in commuting diagrams. Perhaps if your convention became more common, then it would be confusing.

Right. And that's one of the reasons why that convention should not be used. (-:O

Morgan Rogers (he/him) (Jul 27 2023 at 17:31):

John Baez said:

Then a "not necessarily commutative diagram in $C$ shaped like the quiver $Q$" is a functor $\phi: F(Q) \to C$. It "commutes" iff $\phi$ factors through the functor $F(Q) \to G(Q)$.

So the equalizer diagram isn't a commutative diagram?

Morgan Rogers (he/him) (Jul 27 2023 at 17:36):

@Mike Shulman I don't understand your last comment, what is the reason?

Mike Shulman (Jul 27 2023 at 17:45):

Because, as I said before, it creates confusion. Everyone else uses "this diagram commutes" to refer both to maximal paths and non-maximal paths.

Kevin Arlin (Jul 27 2023 at 18:09):

I'm curious about the ethnographic question of how often in the literature something like an equalizer diagram is actually referred to as commutative. I'm not sure there's any efficient way to answer such a question though.

Nathanael Arkor (Jul 27 2023 at 18:11):

Morgan Rogers (he/him) said:

John Baez said:

Then a "not necessarily commutative diagram in $C$ shaped like the quiver $Q$" is a functor $\phi: F(Q) \to C$. It "commutes" iff $\phi$ factors through the functor $F(Q) \to G(Q)$.

So the equalizer diagram isn't a commutative diagram?

I don't think an equaliser diagram should be interpreted as a "commutative diagram".

Ralph Sarkis (Jul 27 2023 at 19:00):

But do you agree that when you draw the diagram for an equalizer and you say it commutes, CTists understand what you mean (even after removing the notion of equalizer from their brain somehow)?

I think giving a definition for commutative diagrams is only important for CT learners. As a tool for CTists, commutative diagrams are mostly about condensing information in an intuitive visual representation, and the ambiguity in our diagrams should be handled case by case. However, when teaching category theory, we have to start somewhere for the students to build the same intuition as we have on what is a commutative diagrams and how do people use them. Examples are fine, but I am one to enjoy having a formal definition to fall back to when I am learning things.

Pushing my thoughts to the extreme would be to have a language of diagrams as precise as @Eduardo Ochs's conventions. Unfortunately, it is not representative of the current usage of diagrams in CT, so the benefits are not worth the trouble imo.

Mike Shulman (Jul 27 2023 at 19:05):

For myself, if I saw someone say that the equalizer diagram "commutes" I would feel uncomfortable about the ambiguity and have to look at the context to figure out what they meant.

John Baez (Jul 28 2023 at 08:11):

I've never thought of the equalizer diagram as a commutative diagram so I've never been tempted to define commutativity of a diagram by saying "a commutative diagram is a diagram such that all directed paths in the diagram of length more than 1 with the same start and endpoints lead to the same result. "

Morgan Rogers (he/him) (Jul 28 2023 at 08:37):

Hmm I'm happy to ultimately concede to majority opinion, but what exactly are we trying to formalize here if "commutative" can't be applied to typical diagrams that one encounters in CT texts?

Morgan Rogers (he/him) (Jul 28 2023 at 08:41):

I guess the "puncture mark" notation that appears in Freyd and Scedrov's book works reasonably well; with that, all paths with the same domain and codomain are assumed to have equal composites unless indicated otherwise with a particular symbol; you can draw a parallel pair of non-equal arrows by putting the punture mark between them. However, it is constrained to planar diagrams (and more particularly to those whose inequations can be arranged to bound a region of the diagram).

Nathanael Arkor (Jul 28 2023 at 09:28):

Morgan Rogers (he/him) said:

Hmm I'm happy to ultimately concede to majority opinion, but what exactly are we trying to formalize here if "commutative" can't be applied to typical diagrams that one encounters in CT texts?

Out of curiosity, can you think of specific examples of equaliser diagrams (or similar) in CT texts that are labelled as being commutative?

John Baez (Jul 28 2023 at 10:59):

Morgan Rogers (he/him) said:

Hmm I'm happy to ultimately concede to majority opinion, but what exactly are we trying to formalize here if "commutative" can't be applied to typical diagrams that one encounters in CT texts?

Why should all diagrams one encounters be commutative? When I show a diagram and I'm trying to assert that it commutes, I say it commutes. Otherwise it may not commute.

It's interesting how you reached a really different attitude, but maybe it's because when some authors write a diagram they automatically take that to be an assertion that it commutes. I always thought that was just laziness, not an official convention.

Jencel Panic (Jul 28 2023 at 11:54):

Kevin Arlin said:

I'm curious about the ethnographic question of how often in the literature something like an equalizer diagram is actually referred to as commutative. I'm not sure there's any efficient way to answer such a question though.

Here is one such occurence, in "Category theory for scientists"
image.png

Ralph Sarkis (Jul 28 2023 at 11:58):

Here is one in Toposes, Triples and Theories
image.png

Spencer Breiner (Jul 28 2023 at 12:11):

From Wikipedia:
225366aa-d46f-463d-acf5-8b1bdc680e6f.png

Morgan Rogers (he/him) (Jul 28 2023 at 12:13):

I also pointed to an nLab page earlier with some coequalizer diagrams.

Nathanael Arkor (Jul 28 2023 at 12:15):

Thanks for the examples. It seems this is a somewhat contentious opinion, but I would view all of those instances as being mistakes. On the Wikipedia article, for instance, the diagram is not commutative according to the hyperlinked definition of "commutative diagram" (which I view as being the correct definition).

Nathanael Arkor (Jul 28 2023 at 12:22):

The diagram in Toposes, Triples and Theories is also not commutative according to the definition in the book.

A diagram is said to commute if any two paths between the same nodes compose to give the same morphism.

Nathanael Arkor (Jul 28 2023 at 12:23):

Category theory for scientists doesn't give a precise definition, so it's hard to tell what definition they are taking as fundamental, but they have a more fine-grained notion of commutativity of diagrams than most other references, where commuting subdiagrams are marked by checkmarks, so the equaliser diagram can be said to satisfy this restricted notion of commutativity.

Morgan Rogers (he/him) (Jul 28 2023 at 12:35):

John Baez said:

Morgan Rogers (he/him) said:

Hmm I'm happy to ultimately concede to majority opinion, but what exactly are we trying to formalize here if "commutative" can't be applied to typical diagrams that one encounters in CT texts?

Why should all diagrams one encounters be commutative? When I show a diagram and I'm trying to assert that it commutes, I say it commutes. Otherwise it may not commute.

It's interesting how you reached a really different attitude, but maybe it's because when some authors write a diagram they automatically take that to be an assertion that it commutes. I always thought that was just laziness, not an official convention.

It's not that I assume every diagram commutes, it's that if I expect I should be able to express any (sensibly finite) equation by drawing a diagram and saying it commutes, without that implying a bunch more equations simultaneously. It's possible to express the equation involved in an equalizer by drawing a square instead of parallel arrows, but I struggle to find that a tasteful compromise. I also feel that commutativity of diagrams should be a compositional property wrt pasting, but I see no reason that it should be decompositional, a priori.

I wish I could remember where I got the idea about maximal paths from...

Morgan Rogers (he/him) (Jul 28 2023 at 12:37):

It's probably not interesting for me to argue this any further; besides Wikipedia, nLab also already agrees on the definition settled upon earlier, involving factoring through a poset. However, I'm glad this discussion happened, since it highlights that there is a difference with Riehl's definition (which is a version I had also internalized), and it's something I should beware of when discussing diagrams with people in future!

Mike Shulman (Jul 28 2023 at 15:56):

John Baez said:

I've never thought of the equalizer diagram as a commutative diagram so I've never been tempted to define commutativity of a diagram by saying "a commutative diagram is a diagram such that all directed paths in the diagram of length more than 1 with the same start and endpoints lead to the same result. "

If I were pressed to write a definition, it would probably be something like "a diagram not containing parallel arrows is commutative if all pairs of parallel directed paths are equal," i.e. exclude the equalizer diagram from even having the word applied to it. But if one did want to include it with a condition like "length more than 1", it should be something like "all pairs of parallel directed paths in which at least one of them has length more than 1", to include commutative triangles. (Although if given as a definition, this would have the interesting consequence that commutativity of a triangle with two parallel arrows on one side of it and one arrow on each of the other two composable sides would imply that the two parallel arrows are equal, by transitivity of equality. Which is perhaps counterintuitive.)

Ralph Sarkis (Jul 28 2023 at 17:15):

The latter is my current definition :partying_face:, but now that I think about it, isn't it less confusing to phrase it as "all paths with same source and target are equal, except parallel arrows, which are not necessarily equal" ?

Kevin Arlin (Jul 28 2023 at 17:27):

Thanks for all the examples of people calling an equalizer diagram commutative! It feels like people sometimes have a similar intuition as in the "serial commutativity" terminology that's seemingly not used formally in this case.

Mike Shulman (Jul 28 2023 at 17:57):

Ralph Sarkis said:

now that I think about it, isn't it less confusing to phrase it as "all paths with same source and target are equal, except parallel arrows, which are not necessarily equal" ?

Eh, I find my version more aesthetic as I like positive conditions better than negative ones, but YMMV.

Nathaniel Virgo (Jul 29 2023 at 02:59):

We wouldn't need an exception for parallel paths if we drew the equaliser diagram like this instead:

image.png

I think I actually find this more intuitive than the usual one, because it requires less thought about which paths are supposed to be equal.

Mike Shulman (Jul 29 2023 at 03:08):

I find that version potentially confusing because it can give the impression that an equalizer of $f$ and $g$ is the same as a pullback of $f$ and $g$, which is an easy mistake for a beginner to make.

But also, equalizers are not the only situation involving parallel arrows.

Nathaniel Virgo (Jul 29 2023 at 05:56):

On the other hand you can ask the reader to compare this diagram to the pullback one, which might make it easier to understand what the difference is. (I don't have much teaching experience though, so I can't easily guess how well that would work.) In principle you can do this sort of thing whenever you have parallel arrows.

Mike Shulman (Jul 29 2023 at 06:22):

Maybe in principle, but in practice I think they would get even harder to draw and understand for larger diagrams. For instance, consider this one (from Sketches of an Elephant):
image.png