Given a set $X$, we can promote it to a discrete category $\mathcal{C}_X$ by considering a set containing identities on all the objects of $X$ and trivial composition/identity selecting functions.

What is a natural way to view this process of adding identities?

Assuming we're working in $ZFC$ (or some other background set theory without atoms) we can take the identiy arrows to be literal identity functions $1_X$ on the sets $x\in X$ and composition to be literal composition of functions, but this seems ugly from a categorical perspective.

It seems we could also add some primitive objects $1_x$ to our theory satisfying idempotence under composition, but this feels like we're doing a lot to accomplish something simple.

The motivation for this question is convenience in defining a notion of functor between different dimensions of category. It is common practice in descent theory to consider pseudofunctors from a category $\mathcal{C}$ to a bicategory $\mathfrak{C}$, and these can be defined by promoting ${\bf Hom}_\mathcal{C}$ to a category as above to promote $\mathcal{C}$ to a $1$-truncated strict bicategory $\mathfrak{C}_\mathcal{C}$, then considering a pseudofunctor $P:\mathfrak{C}_\mathcal{C}\to\mathfrak{C}$.

Although I haven't seen it in the literature anywhere (maybe because it isn't interesting), we could also use promotion to define a pseudofunctor from a bicategory $\mathfrak{C}$ to a regular category $\mathcal{C}$ by considering a pseudofunctor $Q:\mathfrak{C}\to\mathfrak{C}_\mathcal{C}$.

This extends easily to higher dimensions and is all based on this process of 'adjoining identity arrows to a set', so I was hoping for a clear explanation of this process.

It was pointed out in the answer a recent question of mine that trying to 'demote' a category to a set doesn't respect equivalence, but this process of promotion pretty trivially does so hopefully it can be understood easily from a categorical perspective

Any assistance is appreciated.

  • 4
    $\begingroup$ Isn't it exactly right adjoint to the truncation functor ? Which would mean that your pseudofunctor $Q: \mathfrak{C\to C}_\mathcal C$ is exactly a functor $h\mathfrak C\to \mathcal C$, where $h\mathfrak C$ is the category where we mod out by the relation on $1$-arrows generated by "$f\sim g$ if there is an arrow $f\implies g$" (of course this truncation is more reasonable if you're dealing with a $(2,1)$-category, or more generally an $(n+1,n)$-category if you want to add identities from $n$ to $n+1$) $\endgroup$ Dec 5, 2020 at 18:21
  • $\begingroup$ @MaximeRamzi That sounds like a very nice interpretation, thank you; could you post it as an answer to close the question? $\endgroup$
    – Alec Rhea
    Dec 5, 2020 at 19:14

1 Answer 1


Let me preface this answer by noting that I don't know whether this is true all known models for $n$-categories; certainly it is true in the quasicategorical model for $(n,1)$-categories, and it seems to me like it should be true in all reasonable models and for $(n,n)$-categories as well.

The construction you describe from $n$-categories to $(n+1)$-categories should be right adjoint to the truncation functor from $(n+1)$-categories to $n$-categories - this truncation functor is maybe not so sensible for $(n+1)$-categories that aren't $(n+1,n)$-categories, but it should still be defined - or at the very least, that could give an adjunction between $(n+1,n)$-categories and $n$-categories (and then we could just follow the right adjoint with the inclusion of $(n+1,n)$ into $(n+1)$)

The reason this should be true is of course that if all $n+1$-morphisms in $\mathfrak C_\mathcal C$ are identities, then any functor $\mathfrak{C\to C}_\mathcal C$ must factor through the truncation, and uniquely so.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.