I've been told that the category of groups isn't a subcategory of Set. How come?


Let C be a category. A subcategory S of C is given by

a subcollection of objects of C, denoted ob(S), a subcollection of morphisms of C, denoted hom(S), such that:

  • for every $X$ in ob(S), the identity morphism $id_X\in$ hom(S)
  • for every morphism $f : X \to Y$ in hom(S), both the source $X$ and the target $Y$ are in ob(S) for every pair of morphisms $f,g\in$ hom(S) $fg\in$ hom(S) whenever it is defined

Perhaps it's inadequate to consider ob(Grp) as a subcollection of ob(Set)?

In a comment Tobias Kildetoft indicated a condition about non uniqueness: any given set can have more than one group structure, but that condition does not appear to be included in the definition above.

My conclusion from the answer and the comments is that Grp formally is a subcategory of Set but that category theorist don't want to consider it as a subcategory because that is afflicted with a bad intuition.

OK, I realize that Grp isn't a subcategory of Set in the way I thought. The classical picture of a group as a set is not adequate enough for category theory. The equivalent category of set compositions, however, is a subcategory of Set.

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    $\begingroup$ The problem is that any given set can have more than one group structure. $\endgroup$ – Tobias Kildetoft Jul 8 '16 at 5:50
  • $\begingroup$ From the set-theoretical point of view, group is a set, i.e. $\text{Obj}(\mathbf{Grp})\subset\text{Obj}(\mathbf{Set})$ (I have in mind the following definition: group is a pair (two-elemented set) $(G,\bullet)$ etc). But it is not sufficient to say that $\mathbf{Grp}$ is a subcategory of $\mathbf{Set}$. $\endgroup$ – Oskar Jul 8 '16 at 5:58
  • $\begingroup$ @TobiasKildetoft. Yes, but that condition does not appear to be included in the definition? $\endgroup$ – Lehs Jul 8 '16 at 5:59
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    $\begingroup$ I am not sure which condition you mean Lehs. $\endgroup$ – Tobias Kildetoft Jul 8 '16 at 6:07
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    $\begingroup$ To give an example of what I mean, consider the set $\{1,2\}$. There are two ways to make this a group (they just happen to be isomorphic). Which of these would be the one you choose when making groups a subcategory of sets? But then what about the other one? Where did that go? $\endgroup$ – Tobias Kildetoft Jul 8 '16 at 6:24

This definition of subcategory is very poorly behaved; in particular, it's not invariant under equivalence of categories. That is, it's possible to have three categories $S, C, D$ such that

  1. $C$ is equivalent to $D$,
  2. $S$ is equivalent to a subcategory of $C$, but
  3. $S$ is not equivalent to a subcategory of $D$ (exercise).

The well-behaved notion of subcategory is that of full subcategory, for which the above doesn't happen. It's straightforward to prove that $\text{Grp}$ is not equivalent to a full subcategory of $\text{Set}$ (also exercise).

There is a natural forgetful functor $\text{Grp} \to \text{Set}$, but it is not full.

It is possible to exhibit a category equivalent to $\text{Set}$ in which $\text{Grp}$ is a (non-full) subcategory. Namely, take the category whose objects are groups but whose morphisms are arbitrary functions on underlying sets, not respecting the group structure. This category is equivalent to $\text{Set}$ (exercise; you need to know that every set admits at least one group structure). It has a subcategory with the same objects but whose morphisms do respect the group structure, which is of course just $\text{Grp}$. But no category theorist would describe this situation as "$\text{Grp}$ is a subcategory of $\text{Set}$."

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  • $\begingroup$ It shouldn't be a disadvantage being Asperger when reading mathematical definitions, but the real world is far more complicated than the formal. $\endgroup$ – Lehs Jul 8 '16 at 7:12
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    $\begingroup$ If we define a subcategory as a subobject in the 2-category of categories, what we call a “subcategory” will depend on the kind of “monomorphisms” we use to define subobjects. With fully faithful morphisms, we get full subcategories. With faithful morphisms, we get what I would consider as the most basic notion of subcategory, which coincide with concrete categories in the case of subcategories of $\mathsf{Set}$, so that $\mathsf{Grp}$ is a subcategory of $\mathsf{Set}$. (And we can show that any faithful subcategory of $C$ is a “strict” subcategory of a category equivalent to $C$.) $\endgroup$ – Vej Kse Jul 8 '16 at 7:53
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    $\begingroup$ @VejKse The problem is that way you'll also get many things people probably don't want to be subcategories, eg. $ω → 1$, and Grp $→$ Set for that matter. Pseudomonic functors seem like a potentially good compromise. $\endgroup$ – user54748 Jul 8 '16 at 9:24
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    $\begingroup$ @Vej: I'm not aware of any author who thinks of faithful functors as defining a reasonable notion of subobject for categories. There is a general notion of monomorphism in higher category theory, and specialized to the 2-category of categories I believe it reproduces fully faithful functors. $\endgroup$ – Qiaochu Yuan Jul 8 '16 at 9:33
  • $\begingroup$ So you wouldn't consider “sets with injective maps” as a subcategory of “sets with maps”? That the notion of full subcategory is better behaved than the concept of subcategory should not be of a concern here. $\endgroup$ – egreg Jul 8 '16 at 16:07

Even if you take this definition as given, the category of groups is not a subcategory of the category of sets. It's true that if you define "a group" to be an ordered pair consisting of a set and a group structure, and if you define ordered pairs using Kuratowski's trick, then groups "are" sets. However, the set of group homomorphisms between two groups is not a subset of the set of functions between those groups-as-ordered-pairs-as-sets. In particular, Kuratowski ordered pair is always a set with either 1 or 2 elements, so there can be at most 4 functions from one of them to another one; but there are of course many pairs of groups between which there are more than 4 group homomorphisms.

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  • $\begingroup$ OK, I can see the point, but that doesn't prohibit Grp from being a subcategory of Set. You could define a group to be a relation $R$ (the compostion) and a morphism to be certain functions between those relations. What we call a group homomorphism can be revealed from that morphism. mathoverflow.net/questions/231289/a-morphism-revealing-category $\endgroup$ – Lehs Jul 8 '16 at 17:26
  • $\begingroup$ The disagreement seems to originate from a less adequate classical definition of a group and the more consistent viewpoint of the category terrorists. And I like disputes. $\endgroup$ – Lehs Jul 8 '16 at 17:34
  • $\begingroup$ If you like disputes, maybe you would be better off reading philosophy. Mathematicians generally like to resolve disputes. (-: Defining a group to be a relation is a clever trick, but that isn't the standard definition of "group". So you haven't shown that Grp "is" a subcategory of Set, but rather that you can cook up a subcategory of Set that is equivalent (perhaps even isomorphic) to Grp, analogously to Qiaochu's example of how you can cook up a category equivalent to Set that contains Grp as a subcategory. $\endgroup$ – Mike Shulman Jul 8 '16 at 18:17
  • $\begingroup$ Yeah! But also I like to resolve the disputes. Thanks for helping me understand! $\endgroup$ – Lehs Jul 8 '16 at 18:21

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