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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.

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    $\begingroup$ 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

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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.

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