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From Wikipedia:

A concrete category is a category that is equipped with a faithful functor to the category of sets.

This functor makes it possible to think of the objects of the category as sets with additional structure, and of its morphisms as structure-preserving functions.

This definition gives rise to the well-established dichotomy between concrete and abstract categories.

The examples of abstract (= non-concrete) categories I've heard of come in two flavours:

  1. categories with structured sets as objects, and some other sort of structure-preserving maps than standard homomorphisms, e.g. interpretations. (I understand that - basically - interpretations are structure-preserving maps, just involving re-definitions of individuals (as n-tupels) and relations.)

  2. categories with structured set as objects, and equivalence classes of structure-preserving maps as morphisms, e.g. the homotopy category of topological spaces or much simpler: the category $C'$ which collapses all arrows $X \rightarrow Y$ of a (concrete) category $C$ into one (what's its name?)

[Side remark: The trivial equivalence relation on morphisms: $f \sim g :\equiv f = g$ leaves a given category $C$ unchanged.]

Given such a vast variety of possible definitions of structure-preserving maps and equivalence relations between them, I wonder why only classical homomorphisms and the trivial equivalence relation give rise to so-called concrete categories?

The other way around:

What is an example of an abstract category (in the standard sense) with structured sets as objects, such that we cannot think of its morphisms as some equivalence class of some sort of structure-preserving maps?

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Every small category is concrete (that is, falls in your (1)). If all free categories are concrete, then every category is a quotient in the sense of your (2) of a concrete category. –  Mariano Suárez-Alvarez Oct 12 '12 at 23:17
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(I don't think all large free categories are concretizable, because they do not seemto satify Isbell's condition (see Freyd, Peter J. Concreteness. J. Pure Appl. Algebra 3 (1973), 171–191) but I have not thought much about this...) which is necessary and sufficient (according to Isbell and Freyd, respectively) One could sidestep all this and use universes... –  Mariano Suárez-Alvarez Oct 12 '12 at 23:34
    
@Mariano: Could you please be a little more specific about sidestepping by using universes. –  Hans Stricker Oct 14 '12 at 20:29

3 Answers 3

Example: The category whose objects are sets and whose morphisms are relations.

Example: The cobordism category of $n$-manifolds.

Example: The opposite category of your favorite concrete category.

There is certainly no obvious way to think of these as categories of sets with morphisms a slight generalization of standard homomorphisms. There might be some way to construct a set-with-morphism interpretation but it seems unlikely to actually be a useful thing to do.

Certainly, when we do category theory on them, it is useful to think of them as structured sets with homomorphisms. Indeed, that is why they are called "objects" and "morphisms" instead of, say, "vertices" and "edges" - because it is useful to think of nearly any category as consisting of a (perhaps very strange) sort of structured set and a (perhaps very strange) sort of homomorphism. But this is just a process of pretending, that allows us to take intuitions from the common and well-studied sets and morphisms situation and apply them to stranger sorts of categories.

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The relation category is a certain category of lattices and order-preserving maps between them that preserve unions. I think this is a useful thing to do. More generally, for any semiring $R$ we can talk about the category of matrices over $R$, and the (say, finite) relation category is the special case that $R$ is the "truth semiring" (and we consider finite matrices). This shows that the relation category is an example of "matrix mechanics" in the sense of ncatlab.org/nlab/show/matrix+mechanics . –  Qiaochu Yuan Oct 13 '12 at 1:16
    
Alright, that works. Do you have anything for the other two? –  Will Sawin Oct 13 '12 at 2:56
    
Well, any essentially small category is concretizable by the Yoneda lemma, but whether this gives anything useful for e.g. the cobordism category I don't know. I also don't know what my favorite concrete category is, but here are some examples: $\text{Set}^{op}$ is equivalent to a certain category of Boolean algebras (complete and atomic?) and homomorphisms of Boolean algebras (preserving all unions and intersections?). $\text{Ab}^{op}$ is equivalent to the category of compact (Hausdorff) abelian groups by Pontrjagin duality. ... –  Qiaochu Yuan Oct 13 '12 at 4:09
    
$\text{BAlg}^{op}$ is equivalent to the category of Stone spaces by Stone duality. The opposite of the category of commutative unital C*-algebras is equivalent to the category of compact Hausdorff spaces by Gelfand-Naimark. But I agree with your general point that the opposite of a concrete category isn't automatically equipped with a nice concretization, although this doesn't address the question as I interpret it. –  Qiaochu Yuan Oct 13 '12 at 4:14
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The opposite of a concrete category is automatically concrete because the contravariant powerset functor is faithful (even monadic!) –  Zhen Lin Oct 13 '12 at 7:46

The smallest example would be the category whose only object is a 1-element set $A$ (structured just as a set, or, if you don't like that, as a topological space), with two morphisms, the identity $e$ and another morphism $f$, with the composite $ff$ equal to $e$. We can't think of the morphisms as equivalence classes of structure-preserving maps $A\to A$, because there's only one map from $A$ to itself.

I suspect this is not what you were really looking for, but then you should clarify what you had in mind --- perhaps you want to allow the possibility of ignoring the given structured sets and replacing them with others?

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Note that my answer doesn't contradict Mariano's comment on the question. The category I proposed does admit a functor making it concrete. But the actual question refers to the objects being structured sets, and any concretizing functor for my category will have to send $A$ to some other set. –  Andreas Blass Oct 12 '12 at 23:26
    
The way I interpret the question is "does there exist a category which is not a quotient of a concretizable category?" and this doesn't seem to qualify (since it is concretizable). –  Qiaochu Yuan Oct 12 '12 at 23:27
    
@Andreas: I understand your answer. (As you seem to have understood my question.) Do you see a chance to get out of my conceptual "djungle"? Do you see something rightful in my conceptions? Definitely, I thought about replacing "structured sets" by something else. But what by? –  Hans Stricker Oct 12 '12 at 23:34
    
@Qiaouchu: Your comment seems to agree with my comment and the last sentence of my answer, though your interpretation of the question doesn't. The category I proposed is concretizable, but not with the given choice of structured set as the object. And the question did ask for a category with structured sets as objects, so there's more data there than just the category, and the intent of the question seems to be to preserve that additional data. –  Andreas Blass Oct 13 '12 at 0:37
    
@Hans: The only way out that occurs to me at the moment is essentially what I asked about in the last sentence of my answer. If you allow the "abstract category with structured sets as objects" to be modified by changing the structured sets while keeping the abstract category structure, then the question would be as Qiaochu understood it, and Mariano's comments would go a long way toward answering it. –  Andreas Blass Oct 13 '12 at 0:39

A theorem of Kučera asserts that every category arises as a quotient of a concrete category by suitable congruence: see Section 6B of Theory of Mathematical Structures by Adámek.

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Certainly this is true, and it is a comforting fact that one may do this. But their are a few points one should make about this theorem. As stated the theorem is a theorem of GBN rather than ZFC (one might be able to prove it in ZFC however, but I am not sure). The more serious point about this theorem is that to "construct" the concrete category and the appropriate Quotient, one must use an axiom of class choice. For those who are interested, Kucera's original paper is here: sciencedirect.com/science/journal/00224049/1/4 . –  Baby Dragon Sep 6 '13 at 20:59
    
I'm glad you mention that. I'm curious myself about what could be proven for definable large categories in ZFC without access to Global Choice/Well Ordering. One could ask the same about the other concretizability results as well. Do you know if anyone has thought to work this out? –  Adam Epstein Sep 6 '13 at 21:16
    
I do not know too more about this type of problem. I would hazard a guess that some sort of accessibility or presentability properties would allow one to do without choice. This seems like it could be a good question to ask. –  Baby Dragon Sep 7 '13 at 2:12

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