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Stream: learning: questions

Topic: Two functor pairs out of a profunctor

Peva Blanchard (Mar 19 2024 at 08:18):

Let $R : Y^{op} \times X \rightarrow V$ be a profunctor. I will write $\hat{A} = [A^{op}, V]$ the category of presheaves on $A$, and $\check{A} = [A, V]^{op}$ the category of op-co-presheaves.

Following this post, from $R$, I obtain an adjoint pair $R^{\star} : \hat{Y} \rightarrow \check{X}$ and $R_{\star} : \check{X} \rightarrow \hat{Y}$ given by the ends, for all $\phi : \hat{Y}$ and $\psi : \check{X}$:

$$( R^{\star} \phi ) x = \int_y [\phi y, R(y,x)] \\ ( R_{\star} \psi ) y = \int_x [\psi x, R(y, x)]$$

But, by twiddling around, I found another pair of functors out of $R$. Namely, $\hat{R} : \hat{X} \rightarrow \hat{Y}$, and $\check{R} : \check{Y} \rightarrow \check{X}$, defined by coends, for all $\alpha : \check{Y}$ and $\beta : \hat{X}$:

$$(\check{R} \alpha) x = \int^y \alpha y \otimes R(y, x) \\ (\hat{R} \beta) y = \int^x R(y, x) \otimes \beta x$$

They are just the right and left profunctor compositions of $R$ with $\alpha$ and $\beta$ (with presheaves and op-co-presheaves considered as profunctors).

My blunt question is: what kind of structure is at play here? Are there special relations between these four functors?

fosco (Mar 19 2024 at 08:43):

Your observation that they are composition (the second two) and right extensions/lifts (the right adjoints to compositions) when co/presheaves are profunctors to/from the point is correct. What other additional (deeper?) characterization would you like?

Peva Blanchard (Mar 19 2024 at 08:47):

I think you may have answered the question. It is the "right extension/lift" that I don't know about, I think.

Could you expand a bit more on that part? or just a reference to the definition?

fosco (Mar 19 2024 at 08:51):

Simplifying a bit: in a bicategory $\cal K$, for a 1-cell $f : X \to Y$,

• for every 0-cell $A$ postcomposing with $f$ is a functor ${\cal K}(A,X)\to{\cal K}(A,Y)$; if this functor has a right adjoint, you call it the right lifting along $f$
• for every 0-cell $A$ precomposing with $f$ is a functor ${\cal K}(Y,A)\to{\cal K}(X,A)$; if this functor has a right adjoint, you call it the right extension along $f$.

Peva Blanchard (Mar 19 2024 at 12:57):

Thank you!