"Closed bicategories"

Question

I am interested in the following property that a bicategory may or may not have.

Let $\mathbf{B}$ be a bicategory. Every one-morphism $f\colon x\rightarrow y$ defines a functor $\mathbf{B}(y,z)\rightarrow \mathbf{B}(x,z)$ for an arbitrary object $z$ via precomposition $g\mapsto g\ast f$. Similarly, there is a postcomposition functor.

We claim that

($\ast$) The pre- and postcompositions with any $1$-morphism $f$ (with respect to an arbitrary object $z$) both have right adjoints $f_{!}$ and $f^{!}$.

Examples: 1. A bicategory with one object aka monoidal category has property ($\ast$) if and only if it is closed (sometimes called biclosed). Thus the title of the question.

2. The bicategory with rings as objects, $(R,S)$-bimodules as $1$-morphisms (composition: tensor product) and $(R,S)$-linear maps as $2$-morphisms also has this property. If $M$ is an $(R,S)$-bimodule and $N$ is an $(R,T)$-bimodule, then $M_{!}(N)$ is given by $\mathrm{Hom}_R(M,N)$ which is an $(S,T)$-bimodule.

3. The bicategory of profunctors $\mathrm{Prof}(\mathcal{V})$ where $\mathcal{V}$ is a cosmos (similar construction to 2).

Question: Has this notion been studied?

Answer 1

Section 4 of Street's 1974 paper Elementary cosmoi is about "extension systems", which are like bicategories but have only one of the "adjoints to composition" (not composition itself), just like a closed category has "only" the internal-hom of a closed monoidal category and not the monoidal structure itself. He remarks as a special case that a bicategory in which all right extensions exist (a "right (Kan) extension" is another name for one of the adjoints to composition, the other one being a "right (Kan) lifting") yields an extension system, and that a bicategory of this sort is called a "closed bicategory" by "some authors". So the notion is at least that old (Wood's paper mentioned in Fosco's answer is from 1982), although unfortunately this doesn't help find the oldest usage.

Such a use of "closed bicategory" for a bicategory with only one sided adjoints to composition is in line with the use of "biclosed bicategory" when both adjoints exist, although I personally don't really like using the prefix "bi-" with two different meanings in the same phrase, and the plain "closed bicategory" is ambiguous as to which adjoint is meant. I prefer the terminology used e.g. in May-Sigurdsson, section 16.3 where a "closed bicategory" has both adjoints, and when only one is present we speak of "left closed" or "right closed" bicategories (although there's probably no canonical way to decide which is "left" and which is "right").

Answer 2

$\def\rightproarrow{\mathrel{\ooalign{$\hfil\mapstochar\hfil$\cr$\longrightarrow$}}}$Every bicategory $\mathcal B$ such that each $f * -$ and $-*f$ have right adjoints is called biclosed.

The oldest reference I can think of is Wood's Abstract Proarrows where being biclosed is a request on a bicategory $\mathcal M$ that equips a 2-category $\mathcal K$ with proarrows.

It is also possible to export this notion to the more modern setting of virtual double categories, where a proarrow equipment is a virtual equipment (i.e. a virtual double category having identities and enough cartesian cells to close all "pre-cells" like $$ \begin{array}{ccc} A & & B\\ f\downarrow && \downarrow g \\ C &\underset{p}\rightsquigarrow & D \end{array} $$ into cells $$ \begin{array}{ccc} A & \overset{p(f,g)}\rightsquigarrow & B\\ f\downarrow && \downarrow g \\ C &\underset{p}\rightsquigarrow & D \end{array} $$ ) with all compositions (i.e. every cell $$ \begin{array}{cccccc} X_0 &\overset{p_1}\rightsquigarrow&\overset{p_2}\rightsquigarrow&\dots&\overset{p_n}\rightsquigarrow&X_n\\ || &&&&& || \\ X_0 &&&& &X_n \end{array} $$ has an "opcartesian composite" $p_n\odot p_{n-1}\odot\dots\odot p_1 : X_0 \rightsquigarrow X_n$.

You might be interested in noting that in many situation an equipment $p : \mathcal K \to \mathcal M$ has the following form: $\mathcal M$ is the Kleisli bicategory of a relative pseudomonad $P : \mathcal K\to \mathcal C$ so that equipping $\mathcal K$ with proarrows coincides with taking free $P$-algebras.