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?
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1$\begingroup$ 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. $\endgroup$– HeinrichDCommented Oct 20, 2016 at 17:57
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$\begingroup$ Perhaps the link is the fact that each promonoidal category (i.e. pseudomonoid in Prof) gives a multicategory. $\endgroup$– Alexander CampbellCommented Oct 21, 2016 at 5:40
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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 $\mathbf{Prof}$.
