1
$\begingroup$

$\def\Prof{\mathsf{Prof}}\def\Set{\mathsf{Set}}\def\tobar{\mathrel{\mkern3mu \vcenter{\hbox{$\scriptscriptstyle+$}}\mkern-12mu{\to}}}$Let $A$ and $B$ be categories. Define a profunctor $A\tobar B$ to be a functor $B^{op}\times A \rightarrow \Set$. We can form the category of profunctors $\Prof(A,B)$ from $A$ to $B$ with natural transformations as morphisms. For $F:A\tobar B$, we can take the right Kan extension of $id_A:A⇸A$ along $F$ to get a profunctor $\mathfrak{L}(F):B \tobar A$. Explicitly, $$ \mathfrak{L}(F)(a,b) = \int_{a'} \Set(F(b,a'),A(a,a')). $$

See these (1, 2) questions for some more discussion of this map.

Note that $\mathfrak{L}(F)$ is contravariant in $F$, so it's a functor $\Prof(A,B) \rightarrow \Prof(B,A)^{op}$. Note also that $\mathfrak{L}(F)$ is sometimes adjoint to $F$, but not always. A sufficient condition is that $F$ is a representable profunctor.

Question 1: Is $\mathfrak{L}$ an equivalence of categories?

Question 2: If the answer to Question 1 is no, is $\Prof(A,B)$ equivalent to $\Prof(B,A)^{op}$ or $\Prof(B,A)$ at all?

Right now, I'm primarily looking at the ordinary $\Set$-enriched case. If there's a reference that looks at these questions more generally, I'd be glad to read it.

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ Take both $A$ and $B$, or only $A$, or only $B$ trivial. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented May 28, 2017 at 20:19
  • $\begingroup$ @მამუკაჯიბლაძე Thanks. I should have thought of that. $\endgroup$
    – SCappella
    Commented May 28, 2017 at 21:31
  • 1
    $\begingroup$ The question about in what sense are $\mathrm{Set}^A$ and $\mathrm{Set}^{A^{op}}$ dual to each other is very interesting. Lawvere has several deep considerations on that, built around the analogy with the duality between functions and distributions. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented May 29, 2017 at 5:40
  • $\begingroup$ It's probably worth commenting that $\mathbf{Prof}(A, B) \simeq \mathbf{Prof}(B°, A°)$, so a similar formula is true, but you need to introduce opposites. $\endgroup$
    – varkor
    Commented Jul 11, 2022 at 12:57

0

Reset to default

You must log in to answer this question.