# Profunctors and multicategories

I've been told that there is a way to link profunctors and multicategories, probably obtaining a multicategory from $\bf Prof$; I feel I didn't understand the meaning of this claim.

Can you provide me with an explanation?

• There is some information at ncatlab.org/nlab/show/generalized+multicategory which indicates that multicategories can be defined as suitable monads in the $2$-category $\mathbf{Prof}$. I don't know if this is what you mean. Oct 20 '16 at 17:57
• Perhaps the link is the fact that each promonoidal category (i.e. pseudomonoid in Prof) gives a multicategory. Oct 21 '16 at 5:40

Hyland's Elements of a theory of algebraic theories describes a precise connection between multicategories and $$\mathbf{Prof}$$ in Section 4.3. I shall briefly describe the intuition; a complete picture may be found ibid.
First, note that $$\mathbf{Prof}$$ is the Kleisli bicategory $$\mathrm{Kl}(\mathrm{Psh})$$ of the free cocompletion relative pseudomonad $$\mathrm{Psh} : \mathbf{Cat} \to \mathbf{CAT}$$. Now let $$S$$ be an appropriate 2-monad on $$\mathbf{CAT}$$ (intuitively such that there is a relative pseudodistributive law $$\mathrm{Psh} \circ S \Rightarrow S \circ \mathrm{Psh}$$). Then $$S$$-multicategories are precisely monads $$M : \mathbb A \not\to S(\mathbb A)$$ in the bicategory $$\mathrm{Kl}(\mathrm{Psh} \circ S)$$, such that either:
• $$\mathbb A$$ is discrete.
• $$\mathbb A$$ is the underlying category of the multicategory $$M$$.
When $$S$$ is the free (symmetric) monoidal category 2-monad, then $$S$$-multicategories correspond exactly to (symmetric) multicategories. Hence multicategories are (certain) monads in $$\mathbf{Prof}$$.