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Stream: learning: questions

Topic: existence of set-valued Kan extensions

Joshua Meyers (Jan 05 2022 at 04:15):

Let $I:C\to{\sf Set}$ and $F:C\to D$ be functors. I know that ${\sf Lan}_F(I):D\to {\sf Set}$ exists when $C$ is small, but are there any weaker sufficient conditions for its existence?

Patrick Nicodemus (Jan 05 2022 at 04:33):

If F has an adjoint G (left or right? off the top of my head i can't quite remember) then the left Kan extension of I along F is given by GI.

Patrick Nicodemus (Jan 05 2022 at 04:37):

The "pointwise" Kan extension, by definition, requires certain colimits to exist and these are the most important ones from both a theoretical and practical pov (theoretically they have certain richer properties than an arbitrary Kan extension) so you definitely want these colimits to exist, either you need C to be small or you need a way of reducing large limits to small limits by imposing some kind of well-behavedness on the category, a kind of solution set condition like in the case of the adjoint functor theorem or perhaps a "filtering" condition like when we assume that a class of morphisms in a category admits a calculus of fractions (this amounts to saying certain kinds of diagrams that we want to take colimits of are filtered)

Mike Shulman (Jan 05 2022 at 05:56):

The Kan extensions arising from adjoints are also pointwise, and even absolute.

Dylan McDermott (Jan 05 2022 at 14:05):

Instead of requiring $C$ to be small, it is enough to require each $D(F-, d) : C^{\mathrm{op}} \to \mathsf{Set}$ to be a small functor (the left Kan extension of its restriction to some small full subcategory $H_d : C_d \hookrightarrow C$, see https://web.science.mq.edu.au/~slack/papers/small.pdf). This suffices for existence of pointwise left Kan extensions along $F$, because

$$\mathsf{Lan}_F (I) (d) \cong \mathsf{colim}_{(c, \alpha) \in F \downarrow d} (I c) \cong \mathsf{colim}_{(c, \alpha) \in (F \circ H^{\mathrm{op}}_d) \downarrow d} (I (H_d c))$$

which is a small colimit.

This includes the case where $F$ has a right adjoint $G$, because $D(F-, d) \cong C(-,Gd)$, which is the left Kan extension of its restriction to $\{Gd\} \hookrightarrow C$.

Joshua Meyers (Jan 05 2022 at 17:33):

Follow-up question: Is it true that whenever a set-valued Kan extension exists it is then pointwise?

Mike Shulman (Jan 05 2022 at 17:38):

It's hard to come up with examples of non-pointwise Kan extensions, but I certainly don't see any reason why that would be true.

Joshua Meyers (Jan 05 2022 at 17:41):

Because ${\sf Set}$ is nice I guess...I don't have a clear intuition