MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 '10 at 19:22

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 $D$.

share|cite|improve this answer
thanks for this explanation, and for your comment above about representability. – Noam Zeilberger Nov 6 '10 at 9:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.