Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is standard to define the tensor product $M\otimes_R N$ of $R$-modules as a universal object of bilinear maps from $M\times N$. Now, suppose that $\mathscr{F}$, $\mathscr{G}$ are sheaves of $\mathscr{O}$-modules on a topological space $X$. I'm trying to give a categorical definition of $\mathscr{F}\otimes_{\mathscr{O}}\mathscr{G}$ as an object in the category of sheaves on $X$, satisfying some universal property. Presumably it should begin with some kind of "bilinear morphism" from $\mathscr{F}\times\mathscr{G}$, but how should I define "bilinear" (or "balanced") morphism in category theory language? (i.e., without reference to elements of sets, or open subsets of $X$.)

share|cite|improve this question
Dear ashpool, Why are you describing $\mathcal F \times \mathcal G$ as a category, rather than as a sheaf? Regards, – Matt E Sep 11 '12 at 2:47
Oops, sorry. That is a typo. I'll edit it. – ashpool Sep 11 '12 at 2:48
I suspect the way you really want to think about it is not as defining the tensor product intrinsically to $\mathscr{O}$-Mod, but instead as an extension of the usual notion tensor product of modules. The usual setting probably yields a "geometric theory of a ring, two modules, and their tensor product" which can then be applied in the topos of sheaves (of sets) on $X$, thus giving the tensor product on $\mathscr{O}$-mod. Of course, after unwinding everything, this surely just boils down to an abstract way of saying "do what John did". – Hurkyl Sep 11 '12 at 4:00
up vote 1 down vote accepted

I'm pretty tired, as I normally would have been asleep for over an hour by now, but I'll take a stab at this. We need to basically use $\mathscr{O}$-bilinear morphisms(I haven't defined them and won't! Not publically anyway...), and we unfortunately use sets in this. I don't really think there's a way around this, as there is no convenient categorical definition of the tensor product for modules. The tensor product certainly does have one for algebras, but this is not the case we are interested in here.

So this should have some sort of property(starting with the presheaf tensor product) that any $\mathscr{O}$-bilinear natural transformation from the presheaf of sets $\mathscr{F}\times \mathscr{G}$ to a presheaf $\mathscr{H}$ should factor through(via the 'natural' projection) the presheaf tensor product $\mathscr{F}\otimes_\mathscr{O} \mathscr{G}$. When $\mathscr{H}$ is a sheaf, it factors through the sheafification of the presheaf tensor product(by property of sheafification!) so we only need to show this universal property for presheaves. We see first that $\mathscr{F}\otimes \mathscr{G}: U \mapsto \mathscr{F}(U)\otimes_{\mathscr{O}(U)} \mathscr{G}(U)$ at least gives an abelian group structure on it, and the restriction morphism sending $\rho_\mathscr{F} \otimes \rho_\mathscr{G}: f\otimes g \mapsto (\rho_\mathscr{F})\otimes(\rho_\mathscr{G})$ gives a compatible $\mathscr{O}$-module structure.

Now let's say we're given a $\mathscr{O}$-bilinear natural transformation of presheaves from $\mathscr{F} \times \mathscr{G}$ to $\mathscr{H}$. In particular for each $U$ we get a bilinear map from $\mathscr{F}(U) \times \mathscr{G}(U)$ to $\mathscr{H}(U)$, so we get a morphism $\mathscr{F}(U)\otimes\mathscr{G}(U) \rightarrow \mathscr{H}(U)$. The commutativity of the bilinear natural transformation diagram guarantees that this is a natural transformation of functors, i.e. a morphism of presheaves.

Again, if we are working in the category of sheaves then the sheafification of this construction will have the universal property.

share|cite|improve this answer
There's a perfectly convenient categorical definition of the tensor product of modules! See . – Qiaochu Yuan Sep 11 '12 at 3:56
@Qiaochu: Of course, you need to define Hom first.... – Hurkyl Sep 11 '12 at 4:02
@Hurkyl: sure, but given that you're already willing to define morphisms, it's not a big step from there to defining Hom. – Qiaochu Yuan Sep 11 '12 at 4:03
So there is! I'm still adding adjointness to my diet, the thought doesn't come naturally to me. Looks like this follows from a 'standard' theorem on the existence of adjoints, once we show $\mathrm{Hom}(\mathscr{G},-)$ is continuous. – John Stalfos Sep 11 '12 at 11:11
You mean $\mathscr{H}om(\mathscr{G}, -)$. $\textrm{Hom}(\mathscr{G}, -)$ is continuous almost by definition. But even if you show $\mathscr{H}om(\mathscr{G}, -)$ is continuous you still have to verify the conditions of the adjoint functor theorem, and this seems unpleasant because the usual trick doesn't work. (It can probably be made to work once you figure out the appropriate factorisation system to use...) – Zhen Lin Sep 12 '12 at 16:55

Your Answer


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.