# Discrete bicategories/$n$-categories

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.

• 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. – theHigherGeometer Nov 28 '20 at 13:11
• @DavidRoberts Interesting, thank you for the reference, I'll have a look. – Alec Rhea Nov 28 '20 at 15:48
• 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". – Mike Shulman Nov 28 '20 at 17:16
• @MikeShulman Thank you for the correction. – Alec Rhea Nov 28 '20 at 19:21

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.
• 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 $\pi_0$. – Mike Shulman Nov 28 '20 at 17:05
• And, more directly addressing the question asked, the inclusion of $(n-1)$-truncated $n$-categories into $n$-categories also has a left adjoint $\tau_{n-1}$. – Mike Shulman Nov 28 '20 at 17:15