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It appears that there are two different definitions of category. Some authors require the Hom-sets to be pairwise disjoint. Eilenberg and Mac Lane in their original definition require each identity morphism of the category to uniquely determine an object of the category. But some authors (e.g. Kashiwara & Schapira) do not require such conditions in the definition of category. What is the reason for this difference?

Obviously this difference has practical consequences. For instance, take your favorite set (or class) $S$ and take your favorite group $G$. I will attempt to define a category $\mathbf{C}$ as follows: objects of $\mathbf{C}$ are elements of $S$, and for each pair of objects $(X,Y)$ of $\mathbf{C}$ let $\operatorname{Hom}_{\mathbf{C}}(X,Y)=\operatorname{End}(G)$ (endomorphisms of $G$). Is $\mathbf{C}$ as defined above a category? It is not if we require the Hom-sets to be pairwise disjoint (or if we require each identity morphism of the category to uniquely determine an object of the category), but it is a category if we don't have such requirements in our definition of category. So why does this discrepancy exist? Historically, was there a reason for dropping this condition from the original definition of Eilenberg and Mac Lane?

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  • $\begingroup$ Why $\mathrm{End}(G)$ and not just $G$? $\endgroup$
    – Qfwfq
    Commented Apr 12, 2018 at 19:52
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    $\begingroup$ isn't this answered here? math.stackexchange.com/questions/1848926/… $\endgroup$
    – Carlo Beenakker
    Commented Apr 12, 2018 at 19:53
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    $\begingroup$ @CarloBeenakker I was not aware of that post. Thanks for the link. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Apr 12, 2018 at 20:09
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    $\begingroup$ @Qfwfq Yes, $G$ instead of $\operatorname{End}(G)$ would work, too. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Apr 12, 2018 at 20:15
  • $\begingroup$ At least for small categories (I don't want to run into size issues), there is a well-defined set map called "source of a morphism". So $\mathrm{Id}_X=\mathrm{Id}_Y$ implies $X=Y$ and of course $X=Y$ implies $\mathrm{Id}_X=\mathrm{Id}_Y$... $\endgroup$
    – Philippe Gaucher
    Commented Apr 13, 2018 at 8:35

1 Answer 1

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The difference is only cosmetic, not serious. Given a category $\newcommand{\C}{\mathbf{C}}\C$ with not-necessarily-disjoint homsets, we can easily make its homsets disjoint. Precisely, we can define a new isomorphic category $\C'$, with the same objects as $\C$ but with $\hom_{\C'}(x,y) := \hom_{\C}(x,y) \times \{(x,y)\}$. Wrapping this up in a stronger statement: the category of “small disjoint-homsets categories” is equivalent to the category of “small not-necessarily-disjoint-homsets categories”. (The “small” here is just to avoid complications of size issues.)

So no-one really worries about this question: it’s a matter of set-theoretic coding of structures, not a substantive question about the structures themselves.

Why might you want the disjointness condition? So that you can see all hom-sets as subsets of a single set $\mathrm{mor}\,\C$ with well-defined domain and codomain functions $\mathrm{mor}\,\C \to \mathrm{ob}\,\C$. In particular, this approach lets you say “a category consists of two sets, together with certain operations and axioms…” and so fits categories into the long-established setup of algebraic structures.

Why might you want to drop the disjointness condition? Because for many important categories, the most natural presentation gives non-disjoint homsets. E.g. $\mathbf{Set}$, with morphisms as functions, has $\mathrm{hom}(\emptyset,X) = \{\emptyset\}$ for every $X$, due to the set-theoretic representation of functions as sets of ordered pairs.

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    $\begingroup$ Having disjoint homsets is also necessary to make for nicer definitions of functors. For example, if one has a concrete category, say Top, and encodes morphisms non-disjointly, say by identifying a continuous function with its graph, you get well-definedness issues: The same graph may describe morphisms $X\to Y$ and $X\to Z$ where $Y\subseteq Z$, but after applying a functor like (co)homology, $H(Y)$ and $H(Z)$ do not need to have an inclusion between them so that the image of two morphisms is now not compatible any more in the same way. $\endgroup$
    – Johannes Hahn
    Commented Apr 12, 2018 at 21:44
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    $\begingroup$ Concrete example: $X=Y=S^1, Z=\mathbb{R}^2$ and the inclusion/identity map. After applying the functor $H_1$ ($H^1$ wuld work equally well) the identity map becomes the identity $\mathbb{Z}\to\mathbb{Z}$, the inclusion map becomes the zero mapping $\mathbb{Z}\to\{0\}$. $\endgroup$
    – Johannes Hahn
    Commented Apr 12, 2018 at 21:45
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    $\begingroup$ @JohannesHahn: Nice example, but to guard against misinterpretation let me add the caveat that again the difference is just cosmetic. There's no serious obstruction to defining the functor, one just has to make sure one defines functor a in a way that matches the arbitrary-homsets definition of categories, by saying "for all objects x, y, a map hom(x,y) —> hom(Fx,Fy)…” $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Apr 12, 2018 at 22:26
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    $\begingroup$ "Because for many important categories, the most natural presentation gives non-disjoint homsets." (like for the category of sets). No it is wrong at least for me :-). A set map consists of a triple $(X,Y,f)$ where $f$ is a subset of $X\times Y$ such that $(x,y)\in f$ and $(x,y')\in f$ implies $y=y'$. $\endgroup$
    – Philippe Gaucher
    Commented Apr 14, 2018 at 8:39

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