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With given category $\mathcal{C}$ and its objects $A$ and $B$, a hom-set $\hom_\mathcal{C}(A, B)$ is the collection of all morphisms from $A$ to $B$. There is also a related notion of hom-functor from $\mathcal{C}$ to $\mathcal{Set}$, which map objects into related hom-sets.

But why the name "hom"? Who was the first to use this terminology and in what context?

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    $\begingroup$ homomorphism (of rings, groups) I suppose $\endgroup$
    – user99914
    Apr 1, 2015 at 9:54
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    $\begingroup$ It is an apocope of homomorphism, that seems pretty clear… $\endgroup$
    – Bernard
    Apr 1, 2015 at 9:56
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    $\begingroup$ Alternatively, in the fundamental groupoid of a topological space, $\mathsf{hom}(x,y)$ are the "homotopy classes of paths from $x$ to $y$". $\endgroup$
    – JustAskin
    Jan 26, 2016 at 12:04

1 Answer 1

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"Hom" stands for homomorphism, the usual name for structure preserving functions in algebra. I believe the terminology goes back to Eilenberg and Mac Lane's original article on category theory. Of course, in an arbitrary category the objects need not have structure, and the morphisms need not even be functions. Thus the terminology "morphisms" for the arrows in a category. Some texts use $Mor(x,y)$ instead of $Hom(x,y)$. It is also common to be more agnostic and simply write $C(a,b)$ for the morphisms from $a$ to $b$ in the category $C$.

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  • $\begingroup$ What is meant by "Of course, in an arbitrary category the objects need not have structure"? $\endgroup$ Apr 7, 2015 at 13:06
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    $\begingroup$ the objects can be anything at all. For instance, the objects of a category could be the letters $a,b,c$. No structure, just symbols. $\endgroup$ Apr 7, 2015 at 23:10
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    $\begingroup$ I would say that the hom-sets are the structure on the objects. $\endgroup$ Apr 7, 2015 at 23:11
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    $\begingroup$ that may be a bit misleading. The hom sets are the categorical structure on the objects induced by that particular category. Different categories may have the same objects, so the objects themselves do not have structure. $\endgroup$ Apr 7, 2015 at 23:14
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    $\begingroup$ @MartinBrandenburg I think Ittay means that arbitrary categories need not be enriched over other categories. Wouldn't you agree? $\endgroup$
    – AIM_BLB
    Apr 18, 2021 at 16:35

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