## Adjunction up to distributor

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Suppose we have functors $F : C \to D_1$ and $G : D_2 \to C$, together with a distributor (profunctor) $D : D_1^{\rm op} \times D_2 \to {\rm Set}$. We could define "$G$ is right adjoint to $F$ up to $D$" as the existence of a natural isomorphism $C(c, Gd) \cong D(Fc, d)$. We could also consider generalizing this by replacing $C$ by two categories $C_1$ and $C_2$ related by a distributor. Question: has this sort of "adjunction up to distributor" been studied somewhere, and/or is there a better way of formulating it?

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Noam, you were right about my answer; I wrote too soon. I am retracting it. Regarding my remark about being a special case of representable profunctor goes: the condition that "$G$ is right adjoint to $F$ up to $D$" is that the profunctor $D \circ F^{op}$ is representable by the functor $G$; see the nLab page on profunctor.  – Todd Trimble Nov 3 2010 at 19:22

## 1 Answer

I will slightly modify my earlier answer which I retracted. There is the notion of collage of a profunctor $R: C^{op} \times D \to Set$, a category whose collection of objects is $Ob(C) \sqcup Ob(D)$, and where $\hom(x,y) = \hom(x,y)$ if $x$ and $y$ are both objects of $C$ or both objects of $D$, where $\hom(x,y) = R(x,y)$ if $x \in Ob(C)$ and $y \in Ob(D)$, and $\hom(x,y)$ is empty if $x \in Ob(D)$ and $y \in Ob(C)$. Composition is just as you'd expect.

Now, in Noam's notation, consider taking the collage of the profunctor $R = D \circ F^{op}: C^{op} \times D_2 \to Set$ (the composition here is profunctor composition). There is an obvious inclusion functor $i: C \to Coll(R)$ (acting as the identity on objects and morphisms). Then Noam's "right adjoint $G$ of $F$ up to $D$" is essentially equivalent to an ordinary right adjoint $G'$ to the inclusion $i$. For such a $G': Coll(D \circ F^{op}) \to C$, there are natural isomorphisms

$$Coll(D \circ F^{op})(ic, c') \cong C(c, G'c')$$

$$Coll(D \circ F^{op})(ic, d') \cong C(c, G'd')$$

($c' \in Ob(C)$, $d' \in Ob(D_2)$), and following the definition of collage, we calculate that $G'c'$ is $c'$ up to isomorphism, and $C(c, G'd') \cong (D \circ F^{op})(ic, d') = D(Fc, d') \cong C(c, Gd')$. So $G'$ is canonically isomorphic to the evident functor

$$(1_C, G): Coll(D \circ F^{op}) \to C$$

where $G$ is a right adjoint to $F$ up to $D$.

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thanks for this explanation, and for your comment above about representability. – Noam Zeilberger Nov 6 2010 at 9:44
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