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In Category Theory the $hom(A, B)$ sets are generally formulated to be pairwise disjoint. We know that set theorists do not agree to this day on whether functions are also defined by their codomain or not (e.g. is $f: \mathbb R \to \mathbb R, f(x)=x^2$ different from $g: \mathbb R \to \mathbb R_0^+, f(x)=x^2$?) and on this comment we can read that (as a consequence of the above) many category theorists require functions with different codomains to be distinct objects.

My question is now: is the disjointness such an important assumption? What problems do we face (apart from some proofs that may turn out to be longer) if we relax it? Can Category Theory be formulated without it?

  • Well, how to define domain and codomain of a function/morphism? If homsets are not disjoint then these functions on morphisms are not well defined anymore. That would be quite a loss I think. – drhab Jul 04 '16 at 18:07
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    To a categorically minded person, if I have two sets it doesn't even make sense to ask the question of whether they're disjoint or not, because a question like that isn't invariant under isomorphism of sets. The meaningful question is whether two subsets of a given set are disjoint or not. – Qiaochu Yuan Jul 04 '16 at 19:47
  • If $f\in \mathrm{Hom}(A,C)\cap \mathrm{Hom}(B,C)$ and $g\in\mathrm{Hom}(C,D)$ then $g\circ f$ is ambiguous. – Thomas Andrews Jul 04 '16 at 22:28
  • @ThomasAndrews unless we say it's a single morphism that belongs both to $\mathrm{Hom}(A, D)$ and $\mathrm{Hom}(B, D)$ – Michele De Pascalis Jul 04 '16 at 22:32

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Here I'll assume that we're only talking about small categories; this is a convenience for exposition so I can save some words talking about proper classes. If I start to encroach on a traditionally large category, we'll just say that it lives in a larger universe that lets me think of it as small.

A first observation is that the way a category theorist likes to think about categories, as a collection of arrows with domain and codomain functions to the objects, requires that the sets of morphisms between distinct pairs of objects are disjoint, since $\hom(a,b)$ will just be the members of $\langle \mathsf{dom},\mathsf{cod}\rangle^{-1}(\{\langle a,b\rangle\})$. The usual practice is just to adjust the notion of morphism so that this works.

But we can think instead of starting with some set of potential morphisms $\mathrm{Mor}$, and thinking of a category as being based upon a function $\hom:\mathrm{Ob}(\mathcal{C})\times \mathrm{Ob}(\mathcal{C})\to \mathcal{P}(\mathrm{Mor})$. Then it's possible for two $\hom$-sets to overlap, and it almost looks like you can get into trouble defining the composition operation if we had the following scenario: $\hom(b,c)=\hom(b,c')\neq\varnothing$, but there's an $a$ such that $\hom(a,b)$ is non-empty and $\hom(a,c)\cap\hom(a,c')=\varnothing$. In this case it clearly doesn't work to think of composition as using restrictions of a function $\mathrm{Mor}\times \mathrm{Mor}\to\mathrm{Mor}$. But this doesn't tell us there's a problem with non-disjointness; it tells us that this is the wrong way to think about composition for this way of relating objects and morphisms. If this is how you're doing categories, the sensible way to define composition is by way of a family of functions $\{f_{a,b,c}\}_{a,b,c\in\mathrm{Ob}(\mathcal{C})}$ with $f_{a,b,c}:\hom(b,c)\times\hom(a,b)\to\hom(a,c)$, which works just fine.

It's easy to see that a category described in terms of the first notion of category (morphisms with domain/codomain assignments) can be translated into one using the second notion (mapping pairs of objects to subsets of morphisms). The reverse also works if we take our set of morphisms to be $\{\langle a,b,m\rangle:m\in\hom(a,b)\}$, use the first & second projections as our domain and codomain mappings, and do a little tweaking to turn our family of composition functions into a more standard function on a single domain. It's not a difficult exercise to see that up to isomorphism, these translations are mutually inverse, and establish an equivalence of categories.

  • Though in the case of (small) sets you can actually think of each $f_{a,b,c}$ as being the restriction of a total function $\mathrm{Mor}\times\mathrm{Mor}\to\mathrm{Mor}$ if $\mathrm{Mor}$ is the (large) set of all functional relations. But it's nice to ignore that to get the full generality. – Malice Vidrine Jul 04 '16 at 19:48
  • My question was actually more about the problems with reaching the same conclusions we already reached with our "classic" notion of category... – Michele De Pascalis Jul 04 '16 at 20:24
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    I thought the last paragraph addressed that? There are no problems, they give you the same notion of category differently phrased. – Malice Vidrine Jul 04 '16 at 20:28
  • So there is no property or proof of such that depends on the classic formulation? – Michele De Pascalis Jul 04 '16 at 20:32
  • Not really. While many authors use the first notion above, the authors of Abstract and Concrete Categories, for instance, use essentially the second and the result is an entirely standard category theory. – Malice Vidrine Jul 04 '16 at 20:38
  • @Glaedr - I've tweaked the last paragraph to make a stronger statement relevant to your interest, and removed the side comments. Let me know if there's anything you need clarified. – Malice Vidrine Jul 04 '16 at 20:53
  • I've used that textbook for a while and it uses the second formulation but it does require the Hom sets to be pairwise disjoint and some proofs along the book use that disjunction, hence the main question. – Michele De Pascalis Jul 04 '16 at 22:22
  • They do require it, but remark 3.2 (3), one page later, even states that there's not very much to it. – Malice Vidrine Jul 04 '16 at 22:55
  • I want to point out that the first notion of a small category is an "internal category" to the category of Sets, while the second is a "small Sets-enriched category"; I'd be curious to see a more elementary reference than Mike Shulman's paper on Enriched Indexed Categories at http://www.tac.mta.ca/tac/volumes/28/21/28-21.pdf – Vladimir Sotirov Jul 07 '16 at 14:22