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

I'm familiar with the tensor product of modules, but I've also come across functor tensor product (in emily riehls paper on homotopy limits), what are they, and how are they (if they are) related to traditional tensor products? (Emily shows that they can be defined as a particular coend, but that doesn't really provide any intuition for me).

share|improve this question
4  
The ordinary tensor product can be written as a coend, so this may be the source of the terminology. Uses go back (at least) to work by tom Dieck on geometric realisation (which is manifestly a coend) as a tensor product of functors. –  David Roberts Jul 26 '12 at 1:55
    
Have a look a MacLane's "Categories for the working mathematician" –  Fernando Muro Jul 27 '12 at 0:56

3 Answers 3

up vote 10 down vote accepted

It is easy to be explicit. Not in full generality, given a (small) closed symmetric monoidal category $\mathcal C$ with coequalizers, a covariant functor $M\colon \mathcal C\to \mathcal C$ and a contravariant functor $N\colon \mathcal C \to \mathcal C$, the tensor product $N\otimes_{\mathcal C} M$ is the coequalizer of the diagram

$$\coprod_{(c,d)} N(d) \otimes \mathcal C(c,d) \otimes M(c) \implies \coprod_{e} N(e)\otimes M(e).$$

Here $c,d,e$ range over the objects of $\mathcal C$ and $\implies$ indicates a pair of arrows; one is given by the evaluation maps $N(d) \otimes \mathcal C(c,d)\longrightarrow N(c)$ of $N$ and the other by the evaluation maps $\mathcal C(c,d)\otimes M(c) \longrightarrow M(d)$ of $M$. The similarity to Mike's special case should be clear. This is of course an example of a coend, but I prefer to use the tensor product notation in this special case to make the intuition clear.

share|improve this answer
    
Thanks for an explicit answer indeed! –  Yannic Jul 27 '12 at 3:07

The enriched version of the functor tensor product does literally generalize the tensor product of modules. A ring is the same as an $\mathbf{Ab}$-enriched category with one object, a (covariant or contravariant) $\mathbf{Ab}$-enriched functor from a ring to $\mathbf{Ab}$ is a (left or right) module, and the $\mathbf{Ab}$-enriched tensor product of such functors is exactly the classical tensor product of a left and a right module.

share|improve this answer

A very comprehensive account on the tensor product of functors can be found in Section 2.4 of the paper "The fundamental pro-groupoid of an affine 2-scheme" by Alex Chirvasitu and Theo Johnson-Freyd (arXiv).

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.