Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question

4 Answers 4

up vote 14 down vote accepted

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.

share|improve this answer
    
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
1  
"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
2  
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.

share|improve this answer
    
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.

share|improve this answer
    
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
1  
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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.