Talk:Comma category

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So I just expanded this quite a bit - some of the previous phrasing has been incidentally squelched, but everything that was there before is still there now. I'm pretty sure that the new stuff is accurate, but I'm not too happy with the notation: there are far too many different symbols. It could also use a commutative diagram for the most general case. The material on adjoints is currently minimal; I've tried to say just enough to show "look, there's a connection!" without actually going into any details, but perhaps further details are needed. --AlexG 00:44, 19 Jul 2004 (UTC)

Definition of morphisms

I'm new to category theory, so I won't make changes myself here. But I want to suggest the following:

There's no definition of the identity morphism of each object in the comma category.
By the given definition of the morphisms, it is not specified but I suppose it must be obvious that the morphisms between two objects are taken uniquely. For example: if there are objects (a,b,f1) (a',b',f1') (a,b,f2) (a',b',f2') in the comma category, then there could be g:a->a' h:b->b' s.t. (g,h) is a morphism both in hom((a,b,f1), (a',b',f1')) and hom((a,b,f2), (a',b',f2')), but these are two separate morphism with the same notation, isn't it so?

--Itaj Sherman (talk) 23:28, 15 February 2008 (UTC) ’[reply]

The identity morphism for (a, b, f) is (id_a, id_b). You are right that the notation for morphisms is a bit ambiguous. Normally one thinks of the morphism as being given with its domain and codomain, but sometimes writing a map properly — (g, h) : (a, b, f) → (a’, b’, f’) — requires too much notation. I found a reference for the identity morphism in Adámek–Herrlich–Strecker, which does not comment on the ambiguity. It's possible that Mac Lane talks about it, but I don't have my copy handy at the moment. Michael Slone (talk) 01:29, 16 February 2008 (UTC)[reply]

Well, I think it should be at least noted that the meaning of the definition is to be exclusive per domain and codomain, whether the books state it or not, this is what they mean. It would be helpful for people who are new to category theory - I was confused and had to go through all the definitions again to understand what was meant there. --Itaj Sherman (talk) 00:00, 23 February 2008 (UTC)[reply]

There really shouldn't be any confusion here. Two morphisms with different domains (or codomains) are never equal. This isn't special to comma categories. I agree that the language used in the article is a little loose in this regard, but I think to state it formally would add to the confusion rather than subtract from it. -- Fropuff (talk) 05:53, 23 February 2008 (UTC)[reply]

"Common domain"

From the article:

The most general comma category construction involves two functors with the same domain. [...]

Suppose that

{\mathcal {A}}

,

{\mathcal {B}}

, and

{\mathcal {C}}

are categories, and

T

and

S

are functors

File:CommaCategory-04.png

[...]

The diagram defining morphisms is identical to the diagram which defines the components of a natural transformation (assuming the domains of the two functors agree).

It seems to me that it's the codomains that agree, not the domains. Hairy Dude (talk) 18:27, 5 July 2008 (UTC)[reply]

Can't that expression be written using maths mode rather than as an embedded image? Compare:

File:CommaCategory-04.png

{\mathcal {A}}{\xrightarrow {\;\;T\;\;}}{\mathcal {C}}{\xleftarrow {\;\;S\;\;}}{\mathcal {B}}

I don't know if the slight rendering differences are significant, but if not maybe it should be switched to the second version, so that the actual maths is present in the source for the article. --Qef (talk) 11:20, 12 July 2008 (UTC)[reply]

Of course it can be done, and I just did. The difference in rendering is not meaningful. Selinger (talk) 06:39, 6 December 2010 (UTC)[reply]

Lax comma category

When forming a comma category over a 2-category and relaxing commutativity of the squares resp. triangles to the existence of a 2-cell, should we then speak of a lax comma category (resp. oplax comma category, if the direction of the 2-cell is against to that of the 1-cells)? --Tillmo (talk) 15:07, 16 June 2009 (UTC)[reply]

Category of objects under A

The page used to state, in the section "Category of objects under A" and several other places, that if $T$ is a constant functor, then "fixing A makes α irrelevant", meaning that the comma category $(T\downarrow S)$ , where $T$ is a constant functor with value $A$ , is the same as $(A\downarrow S)$ . This is not so. For example, consider the case of

{\mathcal {A}}{\xrightarrow {\;\;T\;\;}}{\textbf {1}}{\xleftarrow {\;\;S\;\;}}{\mathcal {B}}

where ${\textbf {1}}$ is the terminal category (say with object $*$ ), and $T$ and $S$ are the unique (and constant) functors. It is immediately obvious from the definition that in this case the comma category $(T\downarrow S)$ is isomorphic to ${\mathcal {A}}\times {\mathcal {B}}$ . This is not at all the same as $(*\downarrow *)\cong {\textbf {1}}$ .

I fixed this in today's edit. Did I miss something obvious? Selinger (talk) 06:39, 6 December 2010 (UTC)[reply]

Definition

I'm currently learning the basic notions of category theory and it took me some time to understand the definition of comma categories, which I partly blame on the notation used.

Actually, if I'm not totally wrong, the set of objects is just the disjoint union over all objects from the source categories ${\mathcal {A}}$ and ${\mathcal {B}}$ of the morphisms from images of objects from ${\mathcal {A}}$ under $T$ to images of objects from ${\mathcal {B}}$ under $S$ :

$Obj(T\downarrow S)=\biguplus \limits _{(\alpha ,\beta )\in Obj({\mathcal {A\times B}})}{Mor(T(\alpha ),S(\beta ))}$ (writing the pairs of source objects as objects of the product category)

Given that, it's easy to understand that comma categories are about which pairs of morphisms of the target category from images of objects of the first source category to images of objects of the second source category are made commute by the images of which pairs of a morphism of the first source category and a morphism of the second source category. (which pairs of morphisms of ${\mathcal {C}}$ from images of objects of ${\mathcal {A}}$ to images of objects of ${\mathcal {B}}$ are made commute by the images of which pairs of a morphism of ${\mathcal {A}}$ and a morphism of ${\mathcal {B}}$ .)

There is no notation for the "tags" of a disjoint union in Disjoint union (I also haven't seen a otation for this outside of programming languages), so it doesn't seem possible to actually use disjoint union in the definition. I still think it would be worth pointing this out so the definition is easier to understand in the first place. Any other opinions on this?

BTW: Wouldn't it be better to use the letters for the functors in alphabetical order, i.e. $S:{\mathcal {A}}\to {\mathcal {C}},T:{\mathcal {B}}\to {\mathcal {C}}$ instead of $T:{\mathcal {A}}\to {\mathcal {C}},S:{\mathcal {B}}\to {\mathcal {C}}$ ? If the current notations makes sense because of some kind of contravariance or similar, this should be pointed out, too.

-- 132.231.198.153 (talk) 09:05, 26 October 2011 (UTC)[reply]

(Self-reply)

Regarding S and T: S could be associated with "source" and T with "target". Taking into account the definition of slice and coslice category it would make even more sense to switch them, as S is the diagonal functor of the target object of the slice category and T is the diagonal functor of the source category, which seems kind of weird.

I will change that because there have been no objections.

-- 89.121.178.46 (talk) 09:14, 9 January 2012 (UTC)[reply]

Reference

We might include a reference to a physical book, if these are considered more stable than on-line references. In the case of The Joy of Cats, the physical book (in the copy I have) refers the reader to http://katmat.math.uni-bremen.de/acc where the "newest edition of the file of the present book" can be downloaded. Which is the better template to use for cite, "Cite book" or "Cite web"? Model Math (talk) 17:45, 8 April 2017 (UTC)[reply]

Discrete Category

I'm wondering why the Comma Category turns into a Discrete Category when using A=B=1. Won't this be 'only' a Discrete Category of two selected objects (the ones selected by S and T)? The text suggests that it is a general Discrete Category with all objects of C.

Unless we could consider a class of functor pairs (S,T) that selects all objects of C. (Varying S and fix T would also do that job if I understand Category Theory right.) Ron van den Burg (talk) 09:19, 26 December 2023 (UTC)[reply]

Never mind. The objects if the Comma Category are the morphisms from A to B (as stated in the artikel). And there can be many of them. All the morphisms of the Comma Category consist of (id_*, id_*) applied to all the Comma Category Objects, which are their identity morphisms. Ron van den Burg (talk) 11:23, 27 December 2023 (UTC)[reply]