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Oct 27, 2020 at 11:57 comment added varkor My understanding is that it is bicomplete (which follows from the definition of CFS). However, I should note that there is an issue with terminology, in that you'll find it stated (e.g. in this answer) that the Isbell envelope may not have small limits or colimits: but this instead refers to the construction arising from the Isbell adjunction (the distinction is discussed here). I don't know whether the functor you mention has adjoints, either.
Oct 27, 2020 at 11:14 comment added Simon Henry I'd like to know ! When $\mathbb{C}$ is small the Isbell completion is complete and cocomplete correct ? So we always have a limite & colimit preserving functor $\mathcal{B}(\mathbb{C}) \to \mathcal{I}(\mathbb{C})$. Do you know if this functor has right and left adjoints ? I suspect not, but that's the only way I can think to send the Isbell completion to the Bi-completion.
Oct 26, 2020 at 20:34 history edited varkor CC BY-SA 4.0
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Oct 26, 2020 at 20:14 history asked varkor CC BY-SA 4.0