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At the $1$-categorical level we can 'demote' a category to a set by letting it be discrete, and every category has a canonical discrete subcategory that we can view as its 'demotion' to a set given by all the objects and only identity arrows.

Is there a similar notion of 'demotion' for bicategories to $1$-categories?

We can define a 'discrete' bicategory to be a bicategory with only identity $2$-cells, but this immediately forces it to be strict since certain $2$-cells are part of the defining data of a bicategory; if the associators and unitors are identities then horizontal composition commutes on the nose and $1$-cell identities vanish under composition.

This means that $2$-categories (strict bicategories) have canonical demotions to $1$-categories given by the above definition, but bicategories don't.

Specifically, this prevents us from having a canonical discrete sub-$2$-category $\mathcal{C}$ (as defined above) for an arbitrary bicategory $\mathfrak{C}$ since the unitors and associators in $\mathfrak{C}$ may not be identities, so we'll have to choose new composition/identity functors in $\mathcal{C}$. I believe that the new functors will be a postcomposition of the previous composition/identity functors with the isomorphism quotient mapping $q:1-cell_\mathfrak{C}\to[1-cell_\mathfrak{C}]$, but this means that we have no canonical embedding back since (for example) $f\circ(g\circ h)$ and $(f\circ g)\circ h$ will be equal in $\mathcal{C}$ but not (in general) equal in $\mathfrak{C}$.

Is there some other, more well-behaved definition of a 'discrete' bicategory such that all bicategories have a canonical discrete sub-bicategory we can view as it's demotion to a $1$-category? If not, is this because the 'weakness' involved in the definition of a bicategory is not 'visible' at the $1$-categorical level?

We might be able to circumvent these issues at the $2$-categorical level by appealing to a coherence theorem and obtaining an equivalent strict $2$-category, then taking the discrete sub-$2$-category of the strictification, but this answer is less interesting to me because full strictification stops working at the $2$-categorical level.

Is there some notion of a 'discrete' weak $n$-category in the literature such that all weak $n$-categories have canonical discrete sub-$n$-categories we can view as their 'demotion' to an $n-1$-category? If not, is this somehow because the new 'weakness' that appears at each level is invisible from the previous level?

It seems like the step from $2$- to $3$-categories actually introduces some fundamental new weakness that stepping from $1$- to $2$-categories doesn't, so an answer for $3$-categories (perhaps involving Gray sub-$3$-categories?) may suffice to answer both questions.

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    $\begingroup$ Going up a dimension there are sesquicategories, and: "A Gray-category does not have an underlying strict 2-category, but it does have an underlying strict sesquicategory." This seems to answer your last paragraph to some extent. $\endgroup$ – theHigherGeometer Nov 28 '20 at 13:11
  • $\begingroup$ @DavidRoberts Interesting, thank you for the reference, I'll have a look. $\endgroup$ – Alec Rhea Nov 28 '20 at 15:48
  • $\begingroup$ Note that the adjective "discrete" is more commonly used for categories that have only identity $k$-cells for all $k>0$. Your "discrete bicategories" would more commonly be called something like "1-truncated". $\endgroup$ – Mike Shulman Nov 28 '20 at 17:16
  • $\begingroup$ @MikeShulman Thank you for the correction. $\endgroup$ – Alec Rhea Nov 28 '20 at 19:21
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every category has a canonical discrete subcategory that we can view as its 'demotion' to a set given by all the objects and only identity arrows.

This construction is already fishy; note that it's not invariant under equivalence of categories.

Working invariantly, the notion of a discrete category is also not invariant under equivalence of categories. An invariant notion (something that reproduces exactly the categories equivalent to discrete categories) would be the $0$-truncated groupoids / setoids / equivalence relations. The inclusion of setoids into categories has a left adjoint, which one could call $\pi_0$, sending a small category to the equivalence relation on its objects generated by the relation of being connected by a morphism. It does not, as far as I can tell, have a right adjoint, so there is no canonical setoid mapping into a category.

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    $\begingroup$ Thank you Qiaochu. To clarify, it isn't invariant because equivalence of categories cares about isomorphism classes and taking a discrete subcategory puts each object in its own isomorphism class alone, even if it wasn't originally? $\endgroup$ – Alec Rhea Nov 28 '20 at 15:51
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    $\begingroup$ And perhaps it goes without saying, but this picture is equally true in higher dimensions: the inclusion of 0-truncated n-groupoids into n-categories has a left adjoint $\pi_0$. $\endgroup$ – Mike Shulman Nov 28 '20 at 17:05
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    $\begingroup$ And, more directly addressing the question asked, the inclusion of $(n-1)$-truncated $n$-categories into $n$-categories also has a left adjoint $\tau_{n-1}$. $\endgroup$ – Mike Shulman Nov 28 '20 at 17:15
  • $\begingroup$ @Alec: yes, that’s right. $\endgroup$ – Qiaochu Yuan Nov 28 '20 at 19:10
  • $\begingroup$ Much appreciated. $\endgroup$ – Alec Rhea Nov 29 '20 at 1:41

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