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Nov 29, 2020 at 1:41 comment added Alec Rhea Much appreciated.
Nov 28, 2020 at 19:10 comment added Qiaochu Yuan @Alec: yes, that’s right.
Nov 28, 2020 at 17:15 comment added Mike Shulman And, more directly addressing the question asked, the inclusion of $(n-1)$-truncated $n$-categories into $n$-categories also has a left adjoint $\tau_{n-1}$.
Nov 28, 2020 at 17:05 comment added Mike Shulman 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 $\pi_0$.
Nov 28, 2020 at 15:51 comment added Alec Rhea 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?
Nov 28, 2020 at 15:43 vote accept Alec Rhea
Nov 28, 2020 at 8:51 history answered Qiaochu Yuan CC BY-SA 4.0