Tensored Over Abelian Groups?

Suppose I have an category additive category C (i.e. the hom sets are enriched in abelian groups and there are finite direct sums). Suppose further that C has cokernels. Then I can make C tensored over finitely presentable Abelian groups by the following ad hoc construction:

First define $\mathbb{Z}^n \otimes X := \oplus_n X$. Now given a finitely presentable abelian group A, choose a presentation, i.e. realize $A$ as the cokernel of $f:\mathbb{Z}^r \to \mathbb{Z}^g$. Define $A \otimes X$ as the cokernel of the induced map:

$\mathbb{Z}^r \otimes X \to \mathbb{Z}^g \otimes X$

My questions: Is there a way to do the same thing which feels more canonical and less ad hoc? Under what conditions will C be tensored over all abelian groups?

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Given an object $X$ in an additive category $C$ and an abelian group $A$, define the object $A\otimes X$ in $C$ by the rule $Hom_C(A\otimes X,\:Y) = Hom_{Ab}(A,Hom_C(X,Y))$, where $Ab$ denotes the category of abelian groups. If arbitrary direct sums and cokernels (arbitrary colimits, in other words) exist in an additive category $C$, the tensor product $A\otimes X$ exists in $C$ for any abelian group $A$ and any $X\in C$. It can be constructed just as you describe in your question, except that finite direct sums should be replaced with infinite direct sums.

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I thought of this as I was writing the question, but wanted to see what other people would say, anyway. When you say "infinite direct sum" you mean coproduct, right? When I hear "direct sum" I usually think something which is simultaneously a product and a coproduct. Even for the category of abelian groups infinite direct sums in that sense don't exists. – Chris Schommer-Pries Dec 9 '09 at 15:00
"Direct sum" and "coproduct" (in an additive category) are the same thing for me. I thought it was the standard terminology. As to the product and the coproduct of an infinite number of copies of a certain object, these two never coincide in an abelian category, unless the object is zero. – Leonid Positselski Dec 9 '09 at 15:47
There's already a term for something which is simultaneously a product and a coproduct: biproduct. – Qiaochu Yuan Dec 9 '09 at 17:07
In certain contexts "biproduct" can have a different meaning (e.g. in the context of bicategories). But arguing over terminology is a little ridiculous. The main point is that this is a beautiful answer. – Chris Schommer-Pries Dec 9 '09 at 17:40
I like this definition for the tensor product! It's uniquely determined by Yoneda's Lemma, correct? (I just want to check that I'm not missing something important.) – Manny Reyes Dec 9 '09 at 19:21

This question feels a little meta-mathematical; I'm not sure whether you mean a construction that is less ad-hoc, or some description of the properties of said object that makes it clear that it's not an ad-hoc object. Leonid gave a description in the latter terms above.

One construction is that you can take as an index category I the category of finitely generated free abelian groups F equipped with a basis and a map F → A, with morphisms being commuting triangles that ignore the basis. Then A⊗X is the colimit of F⊗X as F ranges over I. (You need choices of basis in order to define a functor, and this assumes that you actually have a direct sum functor.)

A related description is that if C is your category, D is the category of finitely generated abelian groups, and E is the category of finitely generated free abelian groups with a basis (and maps ignoring the basis), then I'd like to say that you have a diagram of functors

C×D ← C×E → C

given on the left by forgetting and on the right by tensoring, and your desired "tensor product" functor is a left Kan extension C×D → C.

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sometimes I wish I could accept two answers... – Chris Schommer-Pries Dec 9 '09 at 15:07

Here is a general context for the same answer. Let V be a bicomplete symmetric monoidal closed category such that the underlying-set functor $\hom_V(I,-)$ is conservative, where I is the unit object, and which is "extremally well-copowered" in the sense that the isomorphism classes of extremal epimorphisms out of any object form a set. If C is a V-enriched category whose underlying ordinary category is cocomplete, then C is tensored over V. This is Prop. 3.46 in Kelly's book "Basic concepts of enriched category theory", http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html

In particular, V = abelian groups satisfies these conditions, as does V = R-modules. The "extremally well-copowered" condition is quite mild, but conservativity of the underlying-set functor is quite strong and fails in many other cases.

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Is "extremely well-copowered" implied by "locally presentable"? – Reid Barton Dec 9 '09 at 17:06
What is "conservative"? – Chris Schommer-Pries Dec 9 '09 at 17:41
"Conservative" means "reflects isomorphisms", i.e., if $U = \mathrm{hom}_V(I, -)$ sends $f$ to an isomorphism, then $f$ was already an isomorphism. A non-example is the category of sheaves of abelian groups on something with the objectwise tensor product, where $U$ sends a sheaf to the underlying set of its global sections. – Reid Barton Dec 9 '09 at 18:47
Yes, locally presentable implies well-copowered wrt all epimorphisms, hence a fortiori also wrt extremal epimorphisms; this is 1.58 of Adamek-Rosicky. – Mike Shulman Dec 9 '09 at 20:55
Topological spaces are another good non-example, where U sends a space to its underlying set. A continuous bijection need not be a homeomorphism. Chain complexes are another non-example, where U sends a chain complex to the set of 0-cycles. – Mike Shulman Dec 9 '09 at 20:56

Incidentally, this is part of a more general story about algebraic theories and relates to Tall-Wraith monoids (surprise, surprise).

Take an algebraic theory, say $V$, (which we identify with its category of models in Set) and a category $D$ with "sufficient structure". Then we can consider co-$V$-objects in $D$. These represent covariant functors $D \to V$. Let $H$ be such. Now if we take a co-$V$-object in $V$, say $B$, then by composition we get a covariant functor $D \to V \to V$. Under the "sufficient structure" assumption on $D$, representability is equivalent to having a left adjoint. As both $D \to V$ and $V \to V$ are representable, both have left adjoints. Thus their composition has a left adjoint and so is representable. Hence there is a co-$V$-algebra object representing $B_* H_*$ which we may as well write as $B \otimes H$. Lots of obvious naturality then implies that there is a corresponding bifunctor $VV^c \times DV^c \to DV^c$. In the particular case that $D = V$ we see that $VV^c$ is monoidal - which is the starting point of the construction of Tall-Wraith $V$-monoids - and more naturality then implies that the bifunctor $VV^c \times DV^c \to DV^c$ is an action of $VV^c$ on $DV^c$.

This generalises even further to give a - slightly odd-looking - action of $VV^c$ on the category of $V$-objects in $D$.

In the specific case in question, $V$ is the category of abelian groups and as $D$ is an abelian category, every object in $D$ is automatically a co-$V$-object in $D$.

(Bits of this story are in the Hunting of the Hopf Ring, other bits will be in a forthcoming paper with Sarah Whitehouse on Tall-Wraith monoids.)

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