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I understand (based on the relatively few examples of categories I have at my disposal), why distinct pairs of objects should have disjoint hom-sets, but I wanted to know of any structures that are almost categories in the sense that they satisfy all the axioms for a category except that one.

Bachmaninoff
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  • What definition of a category are you using? If I understand you correctly, you are saying that $\hom_\mathcal{C}(a,b)$ and $\hom_\mathcal{C}(a,c)$ should be disjoint when $b\neq c$, but if we take $\mathcal{C}$ to be $\mathbf{Set}$ and $a=\emptyset$, then $\hom_\mathcal{C}(a,b)=\hom_\mathcal{C}(a,c)={\emptyset}$, even though $(a,b)$ and $(a,c)$ are distinct pairs. Whatever the case, I have never seen a definition that requires explicitly hom-sets of distinct pairs of objects to be disjoint. Moreover, there exist definitions that do not mention hom-sets. – Hayden Jul 23 '14 at 15:08
  • @Hayden: I have seen a bunch of texts that define a category without this axiom as well. But I have seen it in my instructor's lecture notes, in Dummit and Foote's Abstract Algebra, and in Paolo Aluffi's Algebra: Chapter 0.

    Other than the axiom in question, my definition of a category is (I think) the standard one: an object class, morphism classes between every pair of objects, a composition, associativity, identities for every object.

     – Bachmaninoff Jul 23 '14 at 15:13
  • I'll look at Dummit and Foote to try and understand the axiom you have in mind. EDIT: Indeed, I see the axiom you have in mind. It should be noted that $(a,b)$ and $(c,d)$ being "distinct" means that both $a\neq c$ and $b\neq d$, so the example I gave above does not apply. – Hayden Jul 23 '14 at 15:15
  • @Hayden: Thanks. In the mean time, I'll look at the example you gave! – Bachmaninoff Jul 23 '14 at 15:17
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    I "seem" recall a question like this, but I cannot remember where. At any rate the answer is that it is convenient to have disjoint hom-sets, but not really matter that much. The reason is that if we have $$hom(a,b)\cap hom(c,d)\neq null, $$ then we can define a new category with hom sets, $$HOM(a,b)={(a,b,f):f\in hom(a,b)}.$$ (notice the capitalization) The two categories are equivalent. – Baby Dragon Jul 23 '14 at 15:19
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    @Bachmaninoff Upon searching I came upon this previous math.se question (http://math.stackexchange.com/questions/651464/is-it-necessary-for-the-homsets-to-be-disjoint-in-a-category), and basically it trys to mirror other constructions where each morphism has a unique domain and unique codomain. – Hayden Jul 23 '14 at 15:20
  • @BabyDragon Yes, it seems to mirror your comments well enough. – Hayden Jul 23 '14 at 15:24
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    If you define a category to be a tuple $(\mathrm{Ob},\mathrm{Mor},s,t,\circ,\mathrm{id})$, where $\mathrm{Ob}$ and $\mathrm{Mor}$ are sets and $s,t \colon \mathrm{Mor} \rightarrow \mathrm{Ob}$, $\mathrm{id} \colon \mathrm{Ob} \rightarrow \mathrm{Mor}$ and $\circ \colon \mathrm{Mor} \times_{\mathrm{Ob}} \mathrm{Mor} \rightarrow \mathrm{Mor}$ are functions (between sets) satisfying certain conditions, then the morphism sets you get satisfy that "morphism disjointness" axiom. – Daniel Gerigk Jul 23 '14 at 15:34
  • So perhaps this axiom is introduced solely in order to make the two definitions of categories "strictly isomorphic". If you don't need that, I see no harm in omitting the "morphism disjointness" axiom. – Daniel Gerigk Jul 23 '14 at 15:34
  • @bachmanimoff Note, that it's not an unusual POV that the notion of intersections/disjointness only makes sense between subsets. So even though some authors don't mention the disjointness axiom, it doesn't necessarily mean they don't require it. They may just "omit" it because they defined $\operatorname{Hom}(a,b)$ and $\operatorname{Hom}(c,d)$ as different (and hence of course disjoint) sets in the first place. – roman Jul 28 '14 at 10:51

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Well, sets and maps. In set theory, a map from $A$ to $B$ is a certain subset of $A \times B$. If the map factors through some subset $B'$ of $B$, we get two maps $A \to B$ and $A \to B'$ which are equal as sets.

For example, for a set theorist, $\emptyset$ is a map from $\emptyset$ to any set. But for a category theorist, a morphism has to have a specified domain and codomain.

For more on this "disjoint-axiom", see SE/651464

Community
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Martin Brandenburg
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