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Let $C$ be a category, say with finite products. What can be said about the category $Ab(C)$ of abelian group objects of $C$? Is it always an abelian category? If not, what assumptions on $C$ have to be made? What happens when $C$ is the category of smooth proper geometrically integral schemes over some locally noetherian scheme $S$?

For example if $C=Set$, we get of course the abelian category of abelian groups. If $C=Ring/R$ for some ring $R$, then we get the abelian category $Mod(R)$ (cf. nlab). In general, I have already trouble to show that $Hom(A,B) \times Hom(B,C) \to Hom(A,C)$ is linear in the left coordinate if $A,B,C$ are abelian group objects.

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Wow, thanks for all these answers :) – Martin Brandenburg Nov 9 '10 at 18:26
up vote 9 down vote accepted

Is $\mathscr{C}$ is regular/(exact in Barr sense) then for any algeraic theory $T$ the category $T$-$Alg(\mathscr{C})$ of internal $T$-algebras is regular/(exact), in particular for $\mathscr{C}$ exact and $Ab$= "commutative groups theory" we have $Ab(\mathscr{C})$ is exat, this is also addittive (i.e. abelian, a category is abelian iff is additive and exact ) :

give $f, g: A\to B$ in $Ab(\mathscr{C})$ get $f+g$ in any of these following two way:

1) appling the Yoneda valutation $h^X,\ X\in \mathscr{C}$ to $f, g: A\to B$ (and considers Yoneda Lemma).

2) $f+g: G \xrightarrow{(f, g)} G\times G \xrightarrow{+}G$

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Do you know a reference for the statement that if $C$ is regular, then $T$-$Alg(C)$ is regular for any algebraic theory $T$? I found related material in work of Barr, and I looked at work of Adámek, Rosický, Vitale, and Pedicchio, but couldn't find the statement. For the record, I wrote down a proof here: <>; (Proposition 3.23), at least for one-sorted algebraic theories. – Martin Frankland Sep 3 '14 at 15:02

No, it fails badly in general. A simple example might be where $C = Top$: topological abelian groups do not form an abelian category. For example, this category is not balanced.

In general one will need some exactness assumptions on $C$; I'll have to check into this carefully later (have to run now), but I think Barr-exactness of $C$ (regular, and equivalence relations are kernel pairs of their coequalizers) may be close to an ideal assumption.

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The abelian group objects in any elementary topos form an abelian category, see P.T. Johnstone's "Topos Theory", Theorem 8.11 (page 259).

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I'll point out that toposes are Barr-exact, so all three answers that have appeared so far are in confluence. – Todd Trimble Nov 9 '10 at 15:24
I know I'm commenting on a very old question here - I hope you see it! Question: Is the converse true? That is, is every abelian category equivalent to the category of abelian group objects in some topos? – Alex Kruckman Nov 15 '14 at 21:37
I very much doubt that this is true. The category of abelian groups in a topos has lots of very nice properties and general abelian categories can probably get much wilder than that. Take for example the category of finitely generated R-modules for a noncommutative left-noetherian ring R. It is abelian and has no obvious tensor structure, while abelian group objects in a topos have the obvious closed monoidal structure with all its good properties. I can't think of a proof that this is a counterexample, though... – Peter Arndt Nov 16 '14 at 17:08

If $C$ is any additive category then every object has a unique structure as an abelian group object so $Ab(C)=C$; but typically $C$ is not abelian. For example, this applies to the category of free abelian groups. One can also think about triangulated categories, which are usually not abelian, although a nice theorem of Freyd gives a canonical embedding in an abelian category. One example that has been studied extensively is the category of spectra in the sense of stable homotopy theory. Similarly, one can consider abelian group objects in the homotopy category of spaces, otherwise known as commutative H-spaces.

The question also says:

In general, I have already trouble to show that $Hom(A,B)\times Hom(B,C)\to Hom(A,C)$ is linear in the left coordinate

Surely this is formal? I have drawn the diagram but sadly I cannot get MathJax to display it.


Just to be clear about notation, I'll write $\mathcal{C}(X,Y)$ for morphism sets in $\mathcal{C}$, and $Hom(A,B)$ for morphism sets in $Ab(\mathcal{C})$. An object $A\in Ab(\mathcal{C})$ has a natural abelian group structure on $\mathcal{C}(T,A)$ for all $T\in\mathcal{C}$. Naturality means that $q\circ p+r\circ p=(q+r)\circ p$ for all $p:S\to T$ and $q,r:T\to A$. Now let $B$ be another object of $Ab(\mathcal{C})$. A morphism in $Ab(\mathcal{C})$ from $A$ to $B$ is just a morphism $f:A\to B$ in $\mathcal{C}$ with the property that $f\circ(p+q)=f\circ p+f\circ q$ for all $T$ and all $p,q\in\mathcal{C}(T,A)$. Now suppose we have such morphisms $f,g:A\to B$ and $h,k\:B\to C$. We then have

$ (f+g)\circ(p+q) = f\circ(p+q) + g\circ(p+q) = f\circ p + g\circ p + f\circ q + g\circ q = $ $ (f+g)\circ p + (f+g)\circ q $

(using the naturality of addition, the homomorphism property of $f$ and $g$, and then naturality again). This shows that $f+g$ is again a homomorphism. A similar argument shows that $h\circ f$, $h\circ g$ and $h\circ(f+g)$ are homomorphisms. We have $h\circ(f+g)=h\circ f+h\circ g$ by the homomorphism property of $h$. We also have $(h+k)\circ f=h\circ f+k\circ f$ by the naturality of addition.

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Neil, if you ever have trouble getting mathjax to display a diagram or something, just leave it as far as you can get it, and someone here will come along and fix it. – Harry Gindi Nov 9 '10 at 16:22
Concerning the edit: Ah of course, I used the hom sets of C instead of the sets of group hom. ... Thanks. – Martin Brandenburg Nov 9 '10 at 22:41

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