What does it mean for a category to be “tensored over” another category?

What does it mean for a category to be "tensored over" another category?

I was reading "Stable model categories are categories of modules" by Schwede and Shipley (http://www.math.uni-bonn.de/people/schwede/stablemodelcats.pdf), and found the term "a category $C$ tensored over another category", I googled the term "tensored over" and found this reference: http://math.univ-lille1.fr/~fresse/OperadHomotopyBook/SimplicialModelCategories.pdf, on page 49 they say that a category $C$ is said to be tensored over simplicial sets if there's a tensor product operation that defines a bifunctor from the category of pairs ($A.K$) $\in C \times$ sSet towards $C$, given by ($A.K$) $\mapsto A \otimes K$, satisfying certain conditions, they give examples: the category of topological spaces is tensored over simplicial sets with $A \otimes K = A \times | K |$ , for $A \in$ Top and $K \in$ sSet.

What's $A \otimes K$ supposed to be for arbitrary categories? Is there a canonical way in which this tensor product is defined? Or is it supposed to be defined explicitly by the person referring to it?

The grammar is "a category is tensored over a monoidal category"; this is a generalization of a set being equipped with an action of a monoid, or an abelian group being equipped with an action of a ring. In full generality you should provide the tensoring, but sometimes if you require enough it exists uniquely.

The general pattern of the uniqueness results I know is as follows: let $V$ be a closed symmetric monoidal category and let $C$ be a $V$-enriched category, so that we have a hom functor $[-, -]$ taking values in $V$. $C$ is tensored or copowered over $V$ if the functor $[x, -] : C \to V$ always has an enriched left adjoint; that is, if there is an "action" functor $(-) \otimes x : V \times C \to C$ fitting into an enriched tensor-hom adjunction

$$[v \otimes x, y]_C \cong [v, [x, y]]_V.$$

This strong notion of being tensored over $V$ implies that the tensoring is unique given the $V$-enriched structure, since it is determined by the $V$-enriched structure by the above adjunction. Examples:

1. Every cocomplete category $C$ is tensored over $\text{Set}$ (with the cartesian monoidal structure) in this sense. If $v$ is a set and $x \in c$, then $v \otimes x$ is the coproduct of $v$ copies of $x$. This tensoring is uniquely determined by the condition that $1 \otimes x$ is the identity functor and that $(-) \otimes x$ is cocontinuous.

2. Every cocomplete $\text{Ab}$-enriched category $C$ is tensored over $\text{Ab}$ (with monoidal structure given by the usual tensor product $\otimes$) in this sense. For example, $\mathbb{Z}^n \otimes x$ is the direct sum of $n$ copies of $x$, and $\mathbb{Z}/n\mathbb{Z} \otimes x$ is the cokernel of the multiplication-by-$n$ map $x \xrightarrow{n} x$. This tensoring is again uniquely determined by the condition that $\mathbb{Z} \otimes x$ is the identity functor and that $(-) \otimes x$ is cocontinuous. If $C = \text{Mod}(R)$ is the category of modules over a ring, then this notion of tensoring with an abelian group agrees with the obvious notion.

Hence one way to exhibit a tensoring over simplicial sets is to exhibit an enrichment over simplicial sets and then ask for its left adjoint. There are various ways of doing this, including but not limited to Hammock localization with respect to some collection $W$ of weak equivalences.

In a convenient (in particular cartesian closed) category of topological spaces, we can get an enrichment over simplicial sets by taking the singular simplicial set of the mapping space. Then the tensoring is given by taking the product (in the convenient category) with the geometric realization: I believe this follows from the fact that geometric realization is left adjoint to taking the singular simplicial set.

• Thanks for your answer Qiaochu, very detailed, I'll read that and get back to you. – Samuel M Apr 12 '15 at 22:10
• Actually, there's a subtlety in the enriched setting: the adjunction itself must be enriched. Otherwise the action is not "associative". – Zhen Lin Apr 12 '15 at 22:29
• @Zhen: really? Are you sure it doesn't suffice to ask for $V$ to be closed? – Qiaochu Yuan Apr 13 '15 at 4:04
• If $\mathcal{V}$ is symmetric monoidal closed then $\mathcal{V}$ is $\mathcal{V}$-tensored. But I don't see how you're going to get "associativity" for free like that for a general $\mathcal{V}$-category. So there's a tradeoff: either you impose "associativity" by hand in the definition of "tensor", or you require an enriched adjunction. – Zhen Lin Apr 13 '15 at 7:22
• @Zhen: oh, I see where I went wrong. You're right, I want to require an enriched adjunction. – Qiaochu Yuan Apr 13 '15 at 7:30