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What are all the different levels of 'isomorphism/equivalence/adjunction' we can have between bicategories? Do any of them 'collapse' to one-another?

When working with $1$-categories, we have four basic levels of 'sameness' to consider -- in decreasing order of strictness: equality, isomorphism, equivalence and adjunction.

Moving up to bicategories the landscape becomes signifigantly more diverse (for this question I will focus on bicategories specifically, not strict $2$-categories or lax ones). In particular, at each level below equality we can now ask for multiple versions of each notion depending on wether we use strict, pseudo- or lax $2$-functors between our bicategories. Even further, at the level of equivalence or adjunction we can use strict, pseudo- or lax natural transformations between these functors in our data.

Naively this leads me to believe that there are $1+3+2\cdot3^2=22$ possible notions of 'sameness' for two bicategories, but I expect there to be some 'degeneracy' between these notions. Even if they're all distinct, I wonder what the implication strength hierarchy is -- if we have a 'fully strict' $2$-adjunction would this still be weaker than a 'fully lax' $2$-equivalence, in the sense that a fully lax $2$-equivalence 'is' a fully strict $2$-adjunction? Would a fully strict $2$-adjunction maybe surprise us and give rise to a fully lax $2$-equivalence? Are they not comparable? (this seems most likely)

What would be really nice is a partial (maybe total?) order on these differing notions, with the ordering given by "one of these 'is' one of those" -- in the $1$-categorical setting we have that an equality 'is' an isomorphism (the identity), an isomorphism of categories 'is' an equivalence (with identity natural transformations), and an equivalence 'is' an adjunction (with invertible unit and counit) -- there is something to be said here about adjoint equivalences, but let's ignore that for now.

The $1$-categorical ordering would thusly be total and look like

$$ {\sf Equality>Isomorphism>Equivalence>Adjunction}. $$

What does this poset look like for bicategories?

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    $\begingroup$ I would not call an adjunction a "notion of sameness". $\endgroup$
    – Mike Shulman
    Commented Apr 22, 2021 at 14:46
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    $\begingroup$ I've been thinking a lot about this lately, though I haven't quite found an answer yet. I think the weakest possible such notion is that of an "op/lax biadjunction", which I think can be equivalently formulated in three ways (and maybe four): [...] $\endgroup$
    – Emily
    Commented Apr 22, 2021 at 19:48
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    $\begingroup$ "Functors whose nerve is a weak homotopy equivalence" are certainly a notion of sameness for categories. But I wouldn't call adjunctions a notion of sameness, even though they lie "in between" equivalences and weak homotopy equivalences, because for instance they're not symmetric: if there is an adjunction $C\rightleftarrows D$ there need not be an adjunction $D\rightleftarrows C$. $\endgroup$
    – Mike Shulman
    Commented Apr 23, 2021 at 2:37
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    $\begingroup$ Hah, actually I should probably have said that "having weakly homotopy equivalent nerves" is a notion of sameness. Functors that are weak homotopy equivalences aren't necessarily symmetric either; you have to pass to zigzags. (-: So "being connected by a zigzag of adjunctions" could be a notion of sameness, but it doesn't seem likely to be a very interesting one. (However, this is rather a digression!) $\endgroup$
    – Mike Shulman
    Commented Apr 23, 2021 at 2:50
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    $\begingroup$ I also just noticed a mistake: it's not that the above different definitions of "op/lax biadjunctions" need not agree if $F$ and $G$ are merely lax, it's that they can only work if $F$ and $G$ are pseudo: for the first we need whiskerings of functors with pseudonatural transformations, and these can't be defined for lax functors, while for the second (and similarly for the third) the assignment $A,B\mapsto\mathsf{Hom}_{\mathcal{D}}(F(A),B))$ won't define a pseudofunctor $\mathcal{C}^\mathsf{op}\times\mathcal{C}\to\mathsf{Cats}$ if $F$ is lax! $\endgroup$
    – Emily
    Commented Apr 23, 2021 at 22:10

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