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Jun 2, 2020 at 18:17 comment added fosco arxiv.org/abs/1501.02503 see 3.1.1 here; not because there are no other reference, just because that's the most convenient source to quote for me :) in case $A'$ is not small, see here: ncatlab.org/nlab/show/small+presheaf
Jun 2, 2020 at 17:32 comment added gregodom @Fosco I see. Do you have a reference for the detailed treatment of what you said? Anyways what I am really interested in is that $A'$ is the fppf site over a fixed base scheme which I think is not small in any sense and $A$ the category of sheaves $A'$? can we still say something similar? thanks btw!
Jun 1, 2020 at 23:48 history became hot network question
Jun 1, 2020 at 19:39 answer added Jiří Rosický timeline score: 7
Jun 1, 2020 at 19:26 comment added fosco I actually meant $[(A')^{op},Set]$, sorry for the typo! The category $[(A')^{op},Set]$ has objects the contravariant functors from $A'$ to the category of sets. The fact that every functor $F : X \to B$ with cocomplete domain has an extension to a cocontinuous functor $\bar F : [X^{op},Set]\to B$ "is the Yoneda lemma" in a suitable sense.
Jun 1, 2020 at 17:55 comment added gregodom @Fosco What's your definition of $[A', Set]$ here? Btw do you have a reference for such related results?
Jun 1, 2020 at 17:37 comment added gregodom Hi @Fosco . One example in my mind: let $C$ be a category (say category of schemes over a fixed base), $A$ be its category of presheaves of sets and $A'$ the subcategory of representable presheaves. Does $F'$ extend to colimit preserving $F$ in this case?
Jun 1, 2020 at 17:29 comment added fosco If $A'$ is full in $A$ and skeletally small, then yes: even more is true, there is also an adjunction $[A',Set]\leftrightarrows B$; replace $A'$ with its small skeleton, and use Yoneda lemma. Otherwise, subtle set theory comes in, and I guess you might want to say a bit more on the context of the question :-)
Jun 1, 2020 at 17:24 history edited gregodom CC BY-SA 4.0
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Jun 1, 2020 at 17:24 comment added gregodom @JeremyRickard Uhhh, sorry for the typos.. right, I meant cocomplete and $A'$ a dense subcategory of $A$. Maybe I need to say $F'$ preserves colimits.
Jun 1, 2020 at 16:45 comment added Jeremy Rickard Do you want some condition on $F'$? Otherwise you could just take $A'=A$, with $F'$ some functor that doesn't preserve colimits.
Jun 1, 2020 at 16:41 comment added Jeremy Rickard I think you mean "cocomplete", rather than "complete"? And $A'$ is a dense subcategory of $A$, rather than $B$?
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Jun 1, 2020 at 16:21
Jun 1, 2020 at 15:47 history asked gregodom CC BY-SA 4.0