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Some of my friends and I were trying to discover a universal mapping property that characterizes the integers $\mathbb{Z}$ in the category of groups without referring to the underlying sets (So it is a no no to say it is the free group on a one element set). One of the big uses of $\mathbb{Z}$ is that it is a separator, i.e. for any two distinct pair of parallel arrows $f,g:A \rightarrow B$, there is at least one morphism $x:\mathbb{Z} \rightarrow A$ such that $f\circ x \neq g\circ x$. Unfortunately, any free group satisfies this property. I have two questions:

What are the separators in the category of groups (I think they will be just the free groups, but I have not proved this yet). Given this I think I can write down a universal property for Z which stays inside the category of groups.

Whether or not the above claim is correct, does anyone have a UMP that does the job?

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  • $\begingroup$ Is UMP synonymous with "universal property"? $\endgroup$
    – Qiaochu Yuan
    Commented Nov 2, 2009 at 22:50
  • $\begingroup$ Yes, sorry for the nonstandard terminology, UMP = universal mapping property $\endgroup$
    – Steven Gubkin
    Commented Nov 2, 2009 at 23:07

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According to your definition, the separators will be exactly those groups $G$ with a surjection to $\mathbb{Z}$. One direction: if $f(x)\neq g(x)$, then take the composition $G \twoheadrightarrow \mathbb{Z} \rightarrow A$ where the latter map sends the generator of $\mathbb{Z}$ to $x$. For the other direction, to distinguish the maps $f,g: \mathbb{Z} \rightarrow \mathbb{Z}$ defined by $f(x)=x$ and $g(x)=2x$, your separator must surject to $\mathbb{Z}$.

This is a huge class of groups which has no particularly nice description beyond the definition, as far as I know.

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  • $\begingroup$ Thanks! I was not thinking. Do you have any insight on the second question? $\endgroup$
    – Steven Gubkin
    Commented Nov 2, 2009 at 23:09
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    $\begingroup$ "Every separator has a map to it that is an epimorphism" ia a property but is not universal. It seems the most promising way to proceed is consider the set of groups and assignments of morphisms $x$ to pairs $f,g$, satisfying some kind of naturality/commutativity/coherence condition, such that these correspond to the category of "groups with a surjection onto $\mathbb Z$", of which $\mathbb Z$ is of course a final element. But I can't think of a way to do this right now. $\endgroup$
    – Will Sawin
    Commented Feb 14, 2012 at 1:54
  • $\begingroup$ "This is a huge class of groups which has no particularly nice description beyond the definition, as far as I know." If you don't need the description to be nice, just use the yoneda lemma. $\endgroup$
    – Christopher King
    Commented Jan 16, 2015 at 22:53
  • $\begingroup$ Wait, if one takes $G = \mathbb{Z}$ and then you have $s : G \rightarrow \mathbb{Z} = x \mapsto 3*x$ Is a separator for you $f$ and $g$ but no an epimorphism I think. I am probably wrong. If so, how? $\endgroup$
    – Christopher King
    Commented Jan 16, 2015 at 23:08
  • $\begingroup$ @PyRulez: a group that maps nontrivially to $\mathbb{Z}$ surjects to $\mathbb{Z}$ (take the image of a nontrivial map). $\endgroup$
    – Tom Church
    Commented Jan 21, 2015 at 8:00
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Let $H$ be any group and let $G$ the direct product of $\mathbb{Z}$ and $H$. By sending all elements of $H$ to the identity, you can see that $G$ is also a separator. It is easy to construct lots more examples along these lines (any group with $\mathbb{Z}$ as a quotient will do, for starters).

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A group surjects onto $\mathbb{Z}$ if and only if its abelianization tensored with the rationals is non-trivial.

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    $\begingroup$ The group of rationals doesn't surject to Z, yet its abelianization (namely itself) tensored with the rationals (still itself) is non-trivial. $\endgroup$
    – Andreas Blass
    Commented Feb 3, 2012 at 18:29
  • $\begingroup$ You're right. I was thinking of the finitely generated case. $\endgroup$
    – HJRW
    Commented Feb 3, 2012 at 19:33
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It is the minimal separator in the sense that it corepresents the forgetful functor Grp $\rightarrow$ Set, but this uses sets so probably isn't what you are after. In fact it is the same as the statement that $\mathbb{Z}$ is the free group on one generator.

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In category theory there is the notion of a set $\mathcal S$ of generators for a category $\mathcal C$. The defining property is that for any morphisms $f,g:A \to B$ in $\mathcal C$ $f=g$ if and only $fa=ga$ for all morphisms $a:S \to A$ for all $S \in \mathcal S$.

Thus for a set of generators morphisms with the same domain and codomain in $\mathcal C$ can be distinguished by morphisms from elements of $\mathcal S$.

A more subtle property which is especially relevant to algebraic categories is that of density. A small subcategory $D$ of $\mathcal C$ is dense in $\mathcal C$ of for any objects $A,B$ of $\mathcal C$ the natural function

$$\mathcal C(A,B) \to Nat_D(\mathcal C(-,A), \mathcal C(-,B))$$

which assigns to a morphism $f$ the natural transformation of functors $D\to \mathcal C$ which sends $\mathcal C(-,A) \to \mathcal C(-,B)$ by composition. Intuitively, this says that morphisms $A \to B$ can be recovered from morphisms from objects of $D$.

A dense subcategory of the category of groups is that generated by the free groups on a 1 and 2 elements.

For more discussion, see an article by Vaughan Pratt http://boole.stanford.edu/pub/yon.pdf.

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