biprofunctor

If $K$ and $L$ are bicategories, then a **biprofunctor** $H \colon K ⇸ L$ is a pseudofunctor $\bar H \colon L^{op} \times K \to Cat$.

To define the tricategory $BiProf$ of biprofunctors, we need to know that $P K = [K^{op}, Cat]$ is the free 2-cocompletion of $K$. Then $BiProf$ can be defined as having objects bicategories $K,L,\ldots$ and hom-bicategories $\hom(K,L)$ the strict 2-categories of cocontinuous pseudofunctors $P K \to P L$.

Given $H \colon P K \to P L$ and $G \colon P L \to P M$, their composite $G H$ corresponds to the pseudofunctor $\bar{G H}(m,k) = (G H y k) m = \bar H(-, k) \star \bar G(m,-)$, the colimit of $\bar G(m,-)$ weighted by $\bar H(-,k)$. Using the bicategorical co-Yoneda lemma and a couple of other tricks from Kelly section 3.3, we can write this as

$\bar{G H}(m,k) = \hom_L \star (\bar H(-,k) \times \bar G(m,-))$

showing that the composite $\bar{G H}$ of profunctors is indeed a ‘coend’ $\int^\ell H(\ell,-) \times G(-,\ell)$.

The co-Yoneda lemma then shows that if $F \colon K \to L$ and $G \colon J \to M$ are functors, and $H \colon L ⇸ M$ is a profunctor, then

$M(G,1) \circ H \circ L(1,F) \simeq H(G,F)$

If $L(1,F)$ is taken as a functor $K \to P L$, then the corresponding cocontinuous functor is its left Kan extension along the Yoneda embedding, which by the usual nerve and realization business has a right adjoint given by the pullback-along-$F$ functor $F^* \colon P L \to P K \colon V, k \mapsto V F k$. By the co-Yoneda lemma this latter is $V \star L(F k, -)$, so that the right adjoint of $\bar{L(1,F)}$ is equivalently $\bar{L(F,1)}$. Hence in $BiProf$

$L(1,F) \dashv L(F,1)$

as in $Prof$.

Suppose $T \colon K ⇸ K$ is a pseudomonad in $BiProf$. Its Kleisli object is the bicategory $K_T$ with objects those of $K$ and hom-categories $K_T(k,\ell) = T(k,\ell)$, with composition defined using the multiplication of $T$. The unit of $T$ supplies a functor $F_T \colon K \to K_T$ that is the identity on objects, and clearly $T \simeq K_T(F_T, F_T)$.

The statement that $K_T$ is the Kleisli object of $T$ is the statement that $K_T(1,F_T)$ is the universal right $T$-module (or right $K_T(F_T,F_T)$-module).

We have as before the adjunction $K_T(1,F_T) \dashv K_T(F_T,1)$, and precomposition gives an adjunction between $BiProf(K, L)$ and $BiProf(K_T, L)$, with left adjoint $H \mapsto H \circ K_T(F_T,1)$. Then $RMod(T,L)$ is equivalent to the category of algebras for the monad induced by this adjunction, so there is a comparison functor $BiProf(K_T,L) \to RMod(T,L)$, given by composition with the right $T$-module $(K_T(1,F_T), \epsilon \circ K_T(1,F_T))$. This comparison functor is an equivalence if the right adjoint $U \colon G \mapsto G \circ K_T(1,F_T)$ is monadic in the sense of LMV, that is if it preserves $U$-split codescent objects and reflects adjoint equivalences. $BiProf$ has local colimits, which because composition is given by coends are stable under composition on both sides, so $BiProf(K_T,L)$ has, and $U$ preserves, the required codescent objects. Suppose $\alpha \colon G \Rightarrow H$ is a transformation; then because $F_T$ is the identity on objects, the components of $\alpha$ are exactly the components of $\alpha \circ K_T(1,F_T)$, and so if the latter are all equivalences then so are the former. Hence $U = - \circ K_T(1,F_T)$ is monadic, and $BiProf(K_T,L) \sim RMod(T,L)$.

Revised on March 1, 2012 07:19:56
by Finn Lawler?
(86.41.17.185)