28

The category $\mathbf{Set}$ contains as its objects all small sets and arrows all functions between them. A set is "small" if it belongs to a larger set $U$, the universe.

Let $\mathbf{Grp}$ be the category of small groups and morphisms between them, and $\mathbf{Abs}$ be the category of small abelian groups and its morphisms.

I don't see what it means to say there is no functor $f: \mathbf{Grp} \to \mathbf{Abs}$ that sends each group to its center, when $U$ isn't even specified. Can anybody explain?

Arturo Magidin
  • 398,050
Corey545
  • 289
  • 7
    The claim you're asking about should be true for every universe $U$ and should have nothing to do with universes; there should be a counterexample using a few finite groups. – Qiaochu Yuan Jun 14 '12 at 22:19
  • 14
    General advice: don't worry about universes. They're only there to pacify logicians and set theorists.. – Zhen Lin Jun 14 '12 at 23:04
  • 16
    Well, there is a time and a place to worry about universes, but this isn't it. – Qiaochu Yuan Jun 14 '12 at 23:07
  • This is not a category problem, you should look for why this does not make sense by group homomorphisms. – Bombyx mori Jun 15 '12 at 03:23
  • 2
    There also is this nice post http://qchu.wordpress.com/2012/02/06/centers-2-categories-and-the-eckmann-hilton-argument/ by Qiaochu Yuan above – Bogdan Jun 15 '12 at 15:06
  • If we restrict our category to groups with monomorphisms, and $f:G\rightarrow H$ is a homomorphism of groups, then it does seem to be the case that $f^{-1}(Z(H)) \subseteq Z(G)$. Viewing this containment as a monomorphism, this does give a contravariant functor from Groups with monos to Ab with monos. – tghyde Jan 01 '15 at 16:38
  • There's an answer here: https://math.stackexchange.com/questions/1862171/what-are-the-subcategories-of-ordered-sets-groups/ – user557 Mar 19 '21 at 01:29
  • Does this answer your question? What are the subcategories of ordered sets / groups? – user557 Mar 19 '21 at 01:30
  • @user634426 This question has more upvotes and is older, so the question you linked should be marked as a duplicate of this one. – Toby Mak Mar 21 '21 at 06:52
  • @TobyMak I thought about this too, but the linked question contains this question together with some other questions. The answer in this question doesn't answer all questions raised in the linked question. – user557 Mar 21 '21 at 14:58

3 Answers3

50

The problem with such a functor is group theoretical, not categorical. The problem arises because morphisms between groups need not map centers to centers. It doesn't have anything to do with universes, smallness, or foundational issues.

Consider for example $G=C_2$, $H=S_3$, $K=C_2$, and the maps $f\colon G\to H$ sending the nontrivial element of $G$ to $(1,2)$, and $g\colon H\to K$ by viewing $S_3/A_3$ as the cyclic group of order $2$.

Since $Z(G) = Z(K) = C_2$, and $Z(H) = \{1\}$, such a putative functor $\mathcal{F}$ would give that $\mathcal{F}(f)\colon C_2\to\{1\}$ is the zero map $\mathbf{z}$, and $\mathcal{F}(g)\colon \{1\}\to C_2$ is the inclusion of the trivial group into $C_2$. But $g\circ f=\mathrm{id}_{C_2}$, so $$\mathrm{id}_{C_2} = \mathcal{F}(\mathrm{id}_{C_2}) = \mathcal{F}(gf) = \mathcal{F}(g)\mathcal{F}(f) = \mathbf{z}$$ where $\mathbf{z}\colon C_2\to C_2$ is the zero map.

Thus, no such functor $\mathcal{F}$ can exist.

Arturo Magidin
  • 398,050
  • Can you explain why giving this nice example proves that no such functor exists? Does $\mathbf{Grps}$ contain as its objects every single group? – Corey545 Jun 15 '12 at 18:05
  • 2
    @Corey545: $\mathsf{Grps}$ will contain, at the very least, an isomorphic copy of every countable group. But, if you are concerned about foundational and universe issues, note that a universe is, by definition, a model for ZFC, and therefore must contain $\omega$ (an inductive set), and hence contain an isomorphic copy of any countable group (by constructing it with underlying set contained in $\omega$). In fact, you can guarantee much more: you can guarantee groups of arbitrarily large cardinality, since for any ordinal $X$ in the universe there must exist $Y$ with $X\lt Y$. – Arturo Magidin Jun 15 '12 at 23:50
  • 2
    @Corey545: So while it need not contain every finite group (because the collection of, say, all sets with two elements will not be an element of the universe), it will definitely contain at least one isomorphic copy of every finite group. The example above shows that there is no way to define a functor, because a functor must map identity functions to identity functions, and the functor above then yields that the identity is not the identity. – Arturo Magidin Jun 15 '12 at 23:52
  • 6
    @Corey545: Why do you ask if Grp contains all groups? This example only uses $1$, $C_2$, and $S_3$ and surely you wouldn't call something the category of groups if it didn't at the very very least have a subcategory isomorphic to the category formed by those 3 groups and all homomorphisms between them. – Omar Antolín-Camarena Jun 19 '12 at 13:54
8

This is very similar to Arturo Magidin's answer, but offers another point of view.

Consider the dihedral group $D_n=\mathbb Z_n\rtimes \mathbb Z_2$ with $2\nmid n$ (so the $Z(D_n)=1$). From the splitting lemma we get a short exact sequence $$1\to\mathbb Z_n\rightarrow D_n\xrightarrow{\pi} \mathbb Z_2\to 1$$ and an arrow $\iota\colon \mathbb Z_2\to D_n$ such that $\pi\circ \iota=1_{\mathbb Z_2}$.

Hence the composite morphism $$\mathbb Z_2\xrightarrow{\iota} D_n\xrightarrow{\pi}\mathbb Z_2$$ is an iso and would be mapped by the centre to an iso $$\mathbb Z_2\to 1\to \mathbb Z_2$$ what is impossible. (One can also recognize a split mono and split epi above and analyze how they behave under an arbitrary functor).

Therefore the centre can't be functorial.

Paweł Czyż
  • 3,238
5

I like this example: consider the free groups on $1$ and $2$ generators, $F_1=\langle x\rangle$ and $F_2=\langle y,z\rangle$. Then obviously their centers are $Z(F_1)=F_1$ and $Z(F_2)=1$. Now consider the homomorphisms $F_1\to F_2$, $x\mapsto y$ and $F_2\to F_1$, $y,z\mapsto x$. Then their composition $F_1\to F_2\to F_1$ is the identity of $F_1$, but at the level of centers it cannot be the identity since it factors through the trivial group, $F_1\to 1\to F_1$. Of course, it's similar to the previous ones; just a matter of taste.

Dry Bones
  • 697