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Stream: learning: questions

Topic: Tensor products as coends

Keith Elliott Peterson (Jan 05 2026 at 22:50):

Is the tensor product we all learn about in linear algebra really a coend?

Ruby Khondaker (she/her) (Jan 05 2026 at 22:56):

Silly answer - by the right kan extension formula for left adjoints, it should be an end :)

fosco (Jan 06 2026 at 09:48):

Yes, it is a coend. A high-level explanation is that a module ${}_R M_S$ is a profunctor $M : S^{op}\otimes R \to Ab$ and tensor product with ${}_S N_T$ coincides with composition in the appropriate bi-/double-category

fosco (Jan 06 2026 at 09:51):

If you don't like this explanation, however, there's a more analytic one. The tensor product is a quotient of $M\otimes_{\mathbb Z} N$ (the tensor product as abelian groups, which is a way bigger module) by the relation that balances scalars in $S$ imposing $(m.s)\otimes n = m \otimes (s.n)$.

fosco (Jan 06 2026 at 09:52):

As such, it's an instance of the functor tensor product https://ncatlab.org/nlab/show/tensor+product+of+functors of this nLab page when $\cal C$ has a single object.

Keith Elliott Peterson (Jan 08 2026 at 22:30):

So bilinearity is actually the "push-pull" quotient one gets from a coend. Neat.

Keith Elliott Peterson (Jan 08 2026 at 22:45):

Does it hold that every bimodule/profunctor is given as a colimit of tensor product of functors? That is to say, is there a density theorem for profunctors, just as there are for (co)presheaves?

Ruby Khondaker (she/her) (Jan 08 2026 at 23:12):

I guess you have $P(a, b) \cong \int^c \int^d \text{Hom}(a, c) \times \text{Hom}(d, b) \times P(c, d)$?

Ruby Khondaker (she/her) (Jan 08 2026 at 23:12):

So $P(-, +) \cong \int^c \int^d \text{Hom}(-, c) \times \text{Hom}(d, +) \times P(c, d)$