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

Let $k$ be a field. Let $\mathcal{C},\mathcal{D}$ be finitely cocomplete $k$-linear categories, which are essentially small. Then Kelly's tensor product $\mathcal{C} \boxtimes \mathcal{D}$ is a finitely cocomplete $k$-linear category together with an universal functor from $\mathcal{C} \times \mathcal{D}$ which is right exact and $k$-linear in each variable (also denoted by $(A,B) \mapsto A \boxtimes B$). For an overview of this construction, see section 2.3 in Schäppi's paper on ind-abelian categories.

Roughly, it is constructed as a full subcategory of the category $L$ of $k$-linear functors $F : (\mathcal{C} \otimes_k \mathcal{D})^{op} \to \mathsf{Vect}_k$ with the property that for every exact sequence $A'' \to A' \to A \to 0$ in $\mathcal{C}$ and all $B \in \mathcal{D}$ the sequence $0 \to F(A,B) \to F(A',B) \to F(A'',B)$ is exact, and similarly for the other variable. Representable functors lie in $L$. The crucial step is to observe that $L$ is an orthogonal class in the category of all $k$-linear functors, hence reflective. In particular, it is cocomplete. Now $\mathcal{C} \boxtimes \mathcal{D}$ is the closure of the representable functors under finite colimits taken in $L$.

In my research I need a more explicit description of the objects in $\mathcal{C} \boxtimes D$. Unfortunately, the reflector is dreadful and isn't useful at all.

First of all, is it true that every object $M$ in $\mathcal{C} \boxtimes \mathcal{D}$ can be written as a cokernel of a map of the form $\oplus_j (A'_j \boxtimes B'_j) \to \oplus_i (A_i \boxtimes B_i)$? The answer is yes when $C,D$ are ind-abelian (Lemma 6.5 in Schäppi's paper). Secondly, and more important for me: Given that we know the objects, what are the morphisms? If $A \in \mathcal{C}, B \in \mathcal{D}$, how can we describe $\hom(A \boxtimes B,M)$ explicitly in terms of such a presentation of $M$?

share|improve this question
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.