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Stream: theory: category theory

Topic: Extending copresheaves along functors

Kevin Arlin (May 16 2023 at 18:52):

I've recently gotten into a situation where I want to extend a functor $F: C\to \mathbf{Set}$ along a functor $u:C\to D.$ (Clarification: I mean a functor $G:C\to\mathbf{Set}$ such that $G\circ u = F.$) Basically the only case when I really know that this is possible is when $u$ is fully faithful (when both Kan extensions are really extensions of $F$), but that's far too strong for my problem. Beyond that case, I don't really expect to be able to extend every $F.$

Does anybody know anything else about this problem? More abstractly, it's about finding the essential image of the inverse image functor of a geometric morphism. More concretely, a fun case seems to be this: given a functor $\mathrm{ob}(C)\to \mathbf{Set},$ can you extend it to $C$? That is, can you make $C$ act on a given family $(S_c)_{c\in C}$ of sets indexed by its objects? The only constraints I can see for this come from the facts that there are no functions from a nonempty set into an empty set and functors preserve split monos and epis. Could it be that there's always such an extension if, (1) for every $f:c_1\to c_2,$ if $S_{c_1}$ is nonempty then so is $S_{c_2}$ and, (2), if $f$ is a split mono then there exists a mono $S_{c_1}\to S_{c_2}$ and vice versa for split epis?

Nathanael Arkor (May 16 2023 at 23:02):

If $C$ is small, then the left extension $\mathrm{Lan}_u F \colon D \to \mathrm{Set}$ always exists, because $\mathrm{Set}$ has small colimits.

Kevin Arlin (May 16 2023 at 23:10):

Yeah I know, but I meant an extension that actually restricts back to $F.$ Probably phrased that badly.

fosco (May 21 2023 at 07:59):

Nathanael Arkor said:

If $C$ is small, then the left extension $\mathrm{Lan}_F u \colon D \to \mathrm{Set}$ always exists, because $\mathrm{Set}$ has small colimits.

Maybe you meant $\mathrm{Lan}_u F$? I also think that in most cases (for example when all left Kan extensions exist) the fact that $\mathrm{Lan}_u F \circ u \cong F$ is equivalent to $u$ being fully faithful, because $\mathrm{Lan}_u F \circ u = u^*(\mathrm{Lan}_u F)\Leftarrow F$ is the unit of the adjunction $\mathrm{Lan}_u\dashv u^*$...

fosco (May 21 2023 at 08:00):

mmmh, no, now I'm not sure that this implies u is f.f. from the fact that Lan_u is ...