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First, there are two possible generalization of the notion of closed category, vertical and horizontal.

I'm interested in the vertical one, something saying, I guess, that a monoidal bicategory $\mathcal B$ is closed if tensoring with any object $-\otimes X$ has a right pseudoadjoint $[X,-]$.

I attempted to prove that such hypothesis suffices in order to have an enrichment over the bicategory itself. This being a structure consisting of hom objects $$\mathcal B(X,Y)\colon= [X,Y];$$ unit $$u_X\colon\mathbb 1\rightarrow [X,X]$$ defined to be the transpose of the monoidal unitor $\ell_X\colon\mathbb 1\otimes X\to X$ and composition $$c_{X,Y,Z}\colon[Y,Z]\otimes[X,Y]\rightarrow[X,Z]$$ defined to be the transpose of the composed counits $[Y,Z]\otimes[X,Y]\otimes X\to[Y,Z]\otimes Y\to Z$; higher associator and higher unitors, everything satisfying identity and associativity coherence axioms.

My problem arise in defining the higher unitors, let's say for example the left one $\lambda$, which should be a 2-morphism in $\mathcal B$ filling the diagram

$$\require{AMScd} \require{cancel} \def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1} \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}} \begin{CD} && [Y,Y]\otimes[X,Y]\\ & \diaguparrow{u\otimes\mathrm{id}} @VVcV \\ \mathbb 1\otimes[X,Y] @>>\ell> [X,Y] \end{CD}$$

So, how to I define this? I wrote down the definitions of $c$ and $u$ as transpose enter image description here but I can't see any natural construction leading to an isomorphism of this shape. Of course, the structure to use should be the one provided by the pseudoadjunction,namely by the triangulators (the 2-isomorphisms fitting the triangular "identities"), but I can't find a way to use them! Could anyone help me by pointing me out this construction, or even better providing me a reference for a proof of such a bicategory being enriched over itself?

Thank you so much in advance

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  • $\begingroup$ I am not sure that I understand what you need, but here is how I resolved similar problem some time ago. $\endgroup$ Commented Feb 22 at 15:38
  • $\begingroup$ Ok I did it, actually it suffices to find the stucture not directly from the diagram I draw above but to transpose it first, that means tensoring right with $X$ and then whiskering with the counit $[X,Y]\otimes X\to Y$. $\endgroup$
    – Nikio
    Commented Mar 11 at 13:23

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