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In Bénabou's Les distributeurs, in which the bicategory of profunctors is introduced, Bénabou remarks (page 17, quoted below) that $\mathbf{Prof}$ may be viewed as the construction of a bicategory from $\mathbf{Cat}$ by freely adjoining adjoints to functors. (Indeed, for every functor $F : A \to B$ between small categories, we have profunctors $F_* : A \nrightarrow B$ and $F^* : B \nrightarrow A$ given by $F_*(b, a) = B(b, Fa)$ and $F^*(a, b) = B(Fa, b)$, such that $F_* \dashv F^*$.) Is $\mathbf{Prof}$ the free bicategory with this property?

More precisely, does the inclusion $\mathbf{Cat} \to \mathbf{Prof}$ exhibit $\mathbf{Prof}$ as the free bicategory for which every 1-cell in $\mathbf{Cat}$ has a right adjoint in $\mathbf{Prof}$?

If not, how close is this statement to being true?

Le but visé en construisant Dist était d'ajouter des adjoints pour tous les foncteurs (les flèches de Cat). Ce but est atteint de la façon la plus économique possible : tout distributeur peut être représenté par un couple formé d'un foncteur et du distributeur adjoint d'un autre foncteur.

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  • $\begingroup$ I don't know the answer, but that's definitely what the cited passage suggests ! $\endgroup$ – Maxime Ramzi Mar 3 at 20:54
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    $\begingroup$ I can believe something like "any (pseudo?) functor F:Cat -> K, with every Ff having a right adjoint in K, factors uniquely (up to isomorphism?) through Cat -> Prof". To construct the factorisation F': Prof -> K you use that any profunctor J: A -|-> B factors as $P_*$: A -|-> <J> followed by $Q^*$: <J> -|-> B, where <J> is the graph of J and P and Q are the projections (See Benabou's proposition 2.3.2.) $\endgroup$ – Roald Koudenburg Mar 4 at 13:15
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    $\begingroup$ @RoaldKoudenburg: I can also believe this. I had been hoping that, if it was true, it was surely proven explicitly somewhere, because it's such an elegant characterisation – but perhaps I simply ought to take the time to work it out myself. $\endgroup$ – varkor Mar 4 at 13:51
  • $\begingroup$ @RoaldKoudenburg: is there also a factorisation property for 2-cells? This also seems to be necessary, but I'm not sure what form it should take. $\endgroup$ – varkor Apr 9 at 21:14
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    $\begingroup$ @varkor: yes you're right, defining F' on cells requires some work... What might work: any profuctor J: A -|-> B has, besides its graph <J>, also a cograph [J], with inclusions S:A -> [J] and T: B -> [J], such that $J \simeq T^* S_*$ (this is also in Benabou's 2.3.2). In the double category of functors and profunctors between categories it is clear that composing any transformation $\phi\colon J \Rightarrow K$ of profunctors with the universal cells defining <J> and [K] gives a transformation of functors $\psi\colon S_KP_J \Rightarrow T_KQ_J$. $\endgroup$ – Roald Koudenburg Apr 10 at 12:47
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The idea that passing from $Cat$ to $Prof$ is a great way to give functors adjoints features prominently in Richard Wood's theory of proarrow equipments. Rosebrugh and Wood showed in Proarrows and cofibrations that there is a general 2-categorical construction which takes a 2-category like $V-Cat$ for an enriching category $V$ and outputs $V-Prof$. (Incidently, they also showed that the same construction applied to the 2-category of topoi and geometric morphisms yields the 2-category of topoi and left exact functors). There's another construction which does something similar for internal categories. But neither of these constructions claim to have precisely the universal property of adjoining adjoints.

In Adjoining adjoints, Dawson, Pare, and Pronk study the 2-categorical construction which freely adjoins adjoints to 1-cells. If I remember correctly, they find that this construction does not quite turn $Cat$ into $Prof$, nor $Set$ into $Span$. But I think they formulate a base-change condition to add to the universal property which comes closer. More recently Yanovski and Horev have studied a similar construction in $\infty$-categories.

I apologize for being too lazy to look through the papers I'm linking to and fish out the actual answer!

I should also mention that the same authors showed that freely adjoining adjoints is in general undecidable in undecidability of free adjoints. I'm not a computability person, so I don't know have a good handle on the "computability of the input" assumptions going into this. In general this construction does seem to be ill-behaved (beyond cases like the free adjunction on a 1-morphism). I thought I'd read about getting better behavior by imposing some additional compatibility on the adjunction, but I can't seem to find it now. It might be related to Free extensions of double categories by the same authors.

EDIT: Aha! In The span construction, the same authors give a universal property of spans in terms of companions and conjoints rather than pure adjointness. I suspect similar considerations might be relevant to $Cat$ and $Prof$.

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    $\begingroup$ In truth, the motivation behind my question was proarrow equipments, but I couldn't even find a reference for Prof. As I understand it, Rosebrugh–Wood establish that every PAE satisfying Axiom C is uniquely determined by its domain. Certainly freely adding right adjoints to a bicategory will produce a proarrow equipment (namely, the free PAE on the domain). So Cat → RAdj(Cat) is a PAE, as is Cat → Prof, the latter of which satisfies Axiom C. Perhaps the right question to ask is when RAdj(K) satisfies Axiom C in general. $\endgroup$ – varkor Mar 3 at 23:39
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    $\begingroup$ Thanks for these references. Unfortunately, Dawson–Paré–Pronk don't give any examples as far as I can see (they also work at the level of 1-categories, though it's not clear to me how much of a difference this would make in the case of Cat). $\endgroup$ – varkor Mar 3 at 23:42

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