Comonoid homomorphisms in the bicategory of profunctors

Question

Cartesian bicategories axiomatize the intuitively evident but mathematically elusive "cartesian" product on bicategories such as Rel, Span, and Prof. An important concept for cartesian bicategories are maps, or 1-cells with right adjoints. In the ur-examples Rel and Span, maps are isomorphic to functions, as one would hope. An apparently less nice example is Prof since I gather from the nLab that the maps in Prof are not functors but "semifunctors"!

In a cartesian bicategory, every object has the structure of a commutative comonoid (with laws holding up to iso) and so in addition to asking whether a 1-cell is map, one can ask whether it is a comonoid homorphism (again up to iso). My understanding is that maps are always comonoid homomorphisms but that the converse need not hold unless the "Frobenius condition" is satisfied. Moreover, my impression, again from the nLab, is that Prof does not satisfy this condition unless the objects are restricted to be groupoids. If all this is true, then it stands to reason that Prof could have comonoid homomorphisms that are not maps. Does it? If so, is there a nice characterization of the comonoid homorphisms in Prof?

Answer 1

It's helpful to consider the case of posets and $\mathbf{2}$-enriched profunctors between them, where $\mathbf{2}$ is the cartesian closed poset $(0 \leq 1)$. (This obviates the complications of coherence conditions when considering what "cocommutative comonoid" should mean.) Maps, i.e., left adjoints in this cartesian bicategory, are simply poset maps in the usual sense -- again, without any attendant complications of having to consider objects up to "Morita equivalence" (cf. semifunctors), since every poset is automatically Cauchy complete in this $\mathbf{2}$-enriched context.

You're quite right that posets maps $f: X \to Y$ are automatically comonoid homomorphisms, but it's worth spelling out some things in detail. First, $f$ considered as a $\mathbf{2}$-profunctor from $X$ to $Y$ is the relation $R: Y \times X \to \mathbf{2}$ where $R(y, x) = 1$ iff $y \leq f(x)$. In general, a relation $R$ is a $\mathbf{2}$-profunctor if

$$R(y, x') \wedge (x' \leq x) \vdash R(y, x), \qquad (y' \leq y) \wedge R(y, x) \vdash R(y', x).$$

The condition that a $\mathbf{2}$-profunctor $R: X \to Y$ is a comonoid homomorphism boils down to the two conditions

$$(\exists y \in Y)\ R(y, x) = 1,$$

$$R(y', x) \wedge R(y'', x) \vdash (\exists y \in Y)\ (y' \leq y) \wedge (y'' \leq y) \wedge R(y, x).$$

Clearly the first condition amounts to a "totality" condition, but the second condition does not imply "well-definedness". For example, if $X$ is a single point $\ast$ and $Y$ is any poset with binary joins, then the maximal relation $R$ where $R(y, \ast) = 1$ for all $y \in Y$ satisfies both conditions.

I don't know any particularly nice characterizations of comonoid maps, even in the simplified setting where we consider just posets. But I will remark that if the "Frobenius conditions" hold in a cartesian bicategory, i.e., if the canonical 2-cells

$$\delta_X \delta_X^\dagger \to (1_X \times \delta_X^\dagger)(\delta_X \times 1_X), \qquad \delta_X \delta_X^\dagger \to (\delta_X^\dagger \times 1_X)(1 \times \delta_X)$$

that are mated to the coassociativity isomorphisms $(1 \times \delta_X)\delta_X \cong (\delta_X \times 1_X)\delta_X$ and its inverse (respectively), are themselves isomorphisms, then it may be shown that every object $X$ is monoidally dual to itself, and that the "opposite" $R^{op}: Y \to X$ of a comonoid homomorphism $R: X \to Y$ is its right adjoint (making $R$ a map) -- much as in the case of Cartesian Bicategories I by Carboni and Walters. This happens for example in the case of groupoids and profunctors between them.