# How can I think of an invertible profunctor (distributor)?

Bénabou introduced the notion of profunctor quite some time ago advocating that they would be some sort of categorical flavoured relations (or at least that is what I understand of the story). An ivertible morphism in $\cal Rel$ being just a bijection between sets, I was wondering if such analogy still holds in the case of profunctors. Indeed, if $F : {\cal A} \times {\cal B}^{op} \to {\cal Set}$ is an invertible profunctor with inverse $G : {\cal B} \times {\cal A}^{op} \to {\cal Set}$ then $$\forall a_1, a_2 \in {\cal A}, (G \circ F)(a_1, a_2) \cong {\cal A}(a_1, a_2)$$ and $$\forall b_1, b_2 \in {\cal B}, (F \circ G)(b_1, b_2) \cong {\cal B}(b_1, b_2)$$ As composition of profunctors can be seen as an appropriate quotient of a coproduct, we can extract that for any $a : a_1 \to a_2$ there exists some $b \in {\cal B}, x \in F(a_2, b)$ and $y \in G(b, a_1)$ such that the equivalence class of $(x, y)$ correspond to $a$. Of course we also have the same thing exchanging the positions of the categories $\cal A$ and $\cal B$.

edit : The end of my post did not make any sense so I have corrected it.

In the case where both profunctors $F$ and $G$ happens to be representable, let's say $F = {\cal B}(-, F^0 -)$ and $G = {\cal A}(-, G^0 -)$ for $F^0 : {\cal A \to B}, G^0 : \cal B \to A$, then we can show that $$\forall a_1, a_2 \in {\cal A}, {\cal A}(a_1, G^0F^0(a_2)) \cong (G \circ F)(a_2, a_1) \cong {\cal A}(a_1, a_2)$$ and also $$\forall b_1, b_2 \in {\cal B}, {\cal B}(b_1, F^0G^0(b_2)) \cong (F \circ G)(b_2, b_1) \cong {\cal B}(b_1, b_2)$$

Then, since $id_{\cal A} \dashv G^0F^0$ and $id_{\cal B} \dashv F^0G^0$, the pair $(F^0, G^0)$ is an equivalence of category between $\cal A$ and $\cal B$.

So this gives me a partial answer for representables profunctor, but I am wondering if anything similar can be said about non-representable ones.

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If memory serve correctly, $F$ being invertible is equivalent to the Cauchy completions of $A$ and $B$ being equivalent. At the very least, you have the categories of presheaves $Pre(A)$ and $Pre(B)$ being equivalent. –  David Roberts Jul 25 '13 at 8:02
YOu find adequate explanation (in the line of what David ROberts said) in "Handbook of Categorical Algebra" Volume 1, Borceux Francis –  Buschi Sergio Jul 25 '13 at 9:55
Regarding David's comment: it might help to think of (small) categories and profunctors as the Kleisli bicategory of free cocompletion, so that $Prof$ is equivalent to the 2-category of presheaf categories and cocontinuous functors between them. Then invertibility of a profunctor $F: A\to B$ trivially translates into the corresponding cocontinuous $Pre(A)\to Pre(B)$ being an equivalence. This induces an equivalence between Cauchy completions $\bar{A}\to\bar{B}$, which is immediate if $\bar{A}$ is defined as the category of left adjoints $Set\to Pre(A)$ of cocontinuous functors $Pre(A)\to Set$. –  Todd Trimble Jul 25 '13 at 11:27