MathOverflow is a question and answer site for professional mathematicians. 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

What kinds of Yoneda-like situations induce an embedding that preserves the tensor product for some arbitrary monoidal category?

The cases where the monoidal product is given by a limit or colimit give this immediately for the usual Yoneda embedding, but this breaks down for "real" monoidal categories like $(Vect, \otimes)$.

Are there $V$-enriched cases where the generalised embedding

$$ Y : C \to V^{C^{op}} $$

does preserve the tensor product for interesting monoidal categories $C$?

share|cite|improve this question
up vote 4 down vote accepted

Day showed that, for suitable V, any monoidal structure on a (V-)functor category $[C^{\mathrm{op}}, V]$ is essentially determined by its restriction to the representables as $$ F \otimes G = \int^{A,B} F A \otimes G B \otimes P(A,B,-) $$ where $P(A,B,-) = C(-, A) \otimes C(-, B)$ is a profunctor $C \otimes C \otimes C^{\mathrm{op}} \to V$. P (together with a unit and the usual structural isos) is said to endow C with a promonoidal structure.

If C is already a monoidal V-category, then there is a canonical promonoidal structure on it given by $$ C(-, A) \otimes C(-, B) = C(-, A \otimes B) $$ In that case, the Yoneda embedding is strong monoidal by definition. In fact it is the unit for the monoidal cocompletion of C.

share|cite|improve this answer
Could you give a reference to the paper of Day? – Harry Gindi Aug 20 '10 at 17:42
Yes, sorry, it's in his 1970 thesis, to be found at – Finn Lawler Aug 20 '10 at 18:10
I think you mean any closed monoidal structure on a functor category? – Mike Shulman Aug 21 '10 at 2:47
Yes, I should also have said that this assumes that the tensor on both V and [C^op, V] is cocontinuous in both variables. – Finn Lawler Aug 21 '10 at 19:58

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.