BĂ©nabou 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.