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Kobe Wullaert (Jan 12 2023 at 13:28):Hi everyone. I am looking for a reference on adjunctions between tri-categories.
More precisely, I am looking for a characterization of a tri-functor having a left adjoint in terms of bi-equivalences of hom-bicategories/universal arrows (instead of using unit/counit transformations and coherences).
In the case of (1/bi)categories, one knows that a (pseudo)functor R:C->D has a left (bi)adjoint if for any d : D, there is an object L(d) : C together with (for any object c : C), an isomorphism/equivalence (of sets, resp. categories) between C(L d, c) and D(d, R c).
Following this, one could expect that a tri-functor R:C->D has a left tri-adjoint if for any d : D, there is an object L(d) : C together with (for any object c : C), a biequivalence between the bicategories C(L d, c) and D(d, R c).
However, if one looks at the definition of an adjunction between (infty,1)-categories (https://ncatlab.org/nlab/show/adjoint+%28infinity%2C1%29-functor#CharacterizationInTermsOfHomEquivalences), one does already require that L is a functor and that the unit is already a transformation. However, in the case of 1 and bi-categories, this automatically follows from the equivalence hom-sets/categories.
As a summary, I would like to know whether it is sufficient to only give the the assignment ob(D) -> ob(C) (on objects) and the bi-equivalence of hom-bicategories, or that more data/properties have to be given (as in the (infty,1) case).
Taichi Uemura (Jan 13 2023 at 08:47):It is the case that a functor between -categories has a left adjoint iff for every there are an object and an equivalence (natural in ). See e.g. Proposition 6.1.11 of Cisinski's book. This only depends on the Yoneda Lemma, so it is likely to be true for tri-adjunctions, but I don't know a reference.
Fernando Yamauti (Jan 13 2023 at 20:36):Taichi Uemura said:
It is the case that a functor $R : C \to D$ between $(\infty, 1)$-categories has a left adjoint iff for every $d \in D$ there are an object $L d \in C$ and an equivalence $C(L d, c) \simeq D(d, R c)$ (natural in $c$). See e.g. Proposition 6.1.11 of Cisinski's book. This only depends on the Yoneda Lemma, so it is likely to be true for tri-adjunctions, but I don't know a reference.
Maybe I'm missing something, but I think that the point of question is exactly that you cannot strictify 3-categories. So the analogous problem in the higher case would be for -categories since you can't strictify it into some sort of simplicially enriched strict 2-category. The -case is easier precisely because it's completely determined by its 2-truncation (a strictly simplicially enriched category). Feel free to correct me if I'm committing any mistake.
Nathanael Arkor (Jan 13 2023 at 20:54):I think Taichi's point is that Kobe said that:
if one looks at the definition of an adjunction between (infty,1)-categories [...], one does already require that L is a functor and that the unit is already a transformation
but in fact this is not required.
Fernando Yamauti (Jan 13 2023 at 21:06):Nathanael Arkor said:
I think Taichi's point is that Kobe said that:
if one looks at the definition of an adjunction between (infty,1)-categories [...], one does already require that L is a functor and that the unit is already a transformation
but in fact this is not required.
My bad. I didn't pay attention to that. So the reply was only adressing that part.
Taichi Uemura (Jan 13 2023 at 22:10):Nathanael Arkor said:
I think Taichi's point is that Kobe said that:
if one looks at the definition of an adjunction between (infty,1)-categories [...], one does already require that L is a functor and that the unit is already a transformation
but in fact this is not required.
Yes, indeed. And functors, adjunctions, etc are all in the (∞,1)-sense, so no strictification is involved.
Fernando Yamauti (Jan 14 2023 at 01:16):Taichi Uemura said:
Nathanael Arkor said:
I think Taichi's point is that Kobe said that:
if one looks at the definition of an adjunction between (infty,1)-categories [...], one does already require that L is a functor and that the unit is already a transf
but in fact this is not required.
Yes, indeed. And functors, adjunctions, etc are all in the (∞,1)-sense, so no strictification is involved.
Sorry for the incovenience. You were indeed answering the question. I commented impulsively after reading only the title and summary, so that I wrongly assumed the meaning of "additional data/property" mentioned at the end. I need to change my habit of not reading the whole text whenever I'm on the cellphone.