Finn Lawler
biprofunctor

Definition

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

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

Given $H:PK\to PL$ and $G:PL\to PM$, their composite $GH$ corresponds to the pseudofunctor $\overline{GH}(m,k)=(GHyk)m=\overline{H}(-,k)\star \overline{G}(m,-)$, the colimit of $\overline{G}(m,-)$ weighted by $\overline{H}(-,k)$. Using the bicategorical co-Yoneda lemma and a couple of other tricks from Kelly section 3.3, we can write this as

$$\overline{GH}(m,k)={\mathrm{hom}}_L\star (\overline{H}(-,k)\times \overline{G}(m,-))$$\bar{G H}(m,k) = \hom_L \star (\bar H(-,k) \times \bar G(m,-))

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

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

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

If $L(1,F)$ is taken as a functor $K\to PL$, 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 $F^{*}:PL\to PK:V,k\mapsto VFk$. By the co-Yoneda lemma this latter is $V\star L(Fk,-)$, so that the right adjoint of $\overline{L(1,F)}$ is equivalently $\overline{L(F,1)}$. Hence in $\mathrm{BiProf}$

$$L(1,F)⊣L(F,1)$$L(1,F) \dashv L(F,1)

as in $\mathrm{Prof}$.

Kleisli objects

Suppose $T:K⇸K$ is a pseudomonad in $\mathrm{BiProf}$. 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 $F_T:K\to K_T$ 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 $K_T(1,F_T)⊣K_T(F_T,1)$, and precomposition gives an adjunction between $\mathrm{BiProf}(K,L)$ and $\mathrm{BiProf}(K_T,L)$, with left adjoint $H\mapsto H\circ K_T(F_T,1)$. Then $\mathrm{RMod}(T,L)$ is equivalent to the category of algebras for the monad induced by this adjunction, so there is a comparison functor $\mathrm{BiProf}(K_T,L)\to \mathrm{RMod}(T,L)$, given by composition with the right $T$-module $(K_T(1,F_T),ϵ\circ K_T(1,F_T))$. This comparison functor is an equivalence if the right adjoint $U:G\mapsto G\circ K_T(1,F_T)$ is monadic in the sense of LMV, that is if it preserves $U$-split codescent objects and reflects adjoint equivalences. $\mathrm{BiProf}$ has local colimits, which because composition is given by coends are stable under composition on both sides, so $\mathrm{BiProf}(K_T,L)$ has, and $U$ preserves, the required codescent objects. Suppose $\alpha :G\Rightarrow H$ 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 $\mathrm{BiProf}(K_T,L)\sim \mathrm{RMod}(T,L)$.