Timeline for Characterizing freely adjoining K-filtered colimits as K-continuous presheaves
Current License: CC BY-SA 3.0
10 events
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| Dec 20, 2017 at 18:57 | comment | added | Yonatan Harpaz | For the example with the doctrine of countable products, indeed, it seems to also give a counter-example in the $\infty$-categorical setting: the terminal presheaf ${\cal C}^{\rm op} \to {\rm Spaces}$ is also not a homotopy colimit of representables, because then (by the same argument) ${\cal C}$ would be ${\cal K}$-filtered in the $\infty$-categorical sense, and hence also ${\cal K}$-filtered 1-categorically. | |
| Dec 20, 2017 at 18:51 | comment | added | Yonatan Harpaz | The colimit of $f:{\cal J} \to {\cal C} \to P({\cal C})$ being the terminal presheaf is sometimes taken as the definition of cofinality. This can also be extended to $\infty$-categories, by realizing the colimit in presheaves of spaces, instead of sets. If one defines cofinality by the comma categories ${\cal J}_{x/}$ being connected (or contractible in the $\infty$-case) then there is a simple passage between the two definitions, as ${\rm colim}_{j \in {\cal J}}R_{f(j)}(x) = {\rm colim}_{j \in {\cal J}}{\rm Hom}(x,f(j)) =\pi_0(|{\cal J}_{x/}|)$ (or without the $\pi_0$ in the $\infty$-case). | |
| Dec 19, 2017 at 21:39 | vote | accept | KotelKanim | ||
| Dec 19, 2017 at 21:31 | comment | added | KotelKanim | It indeed seems to be a counterexample to my question in the 1-categorical setting (as the terminal presheaf preserves countable products). I didn't check the details too carefully, but it also seems that everything here goes through to $\infty$-categories. Am I right? | |
| Dec 19, 2017 at 21:25 | comment | added | KotelKanim | I see now... it took me a few minutes to convince myself that the diagram $\mathcal{J} \to \mathcal{C}$ is indeed cofinal. This uses the fact that after composing with the Yoneda embedding, the colimit is the terminal presheaf, right? I didn't know this criterion for cofinality. Is it obvious? The argument I found for this is not difficult, but it uses something (e.g. that the colimit of a co-representable functor is a point etc.). | |
| Dec 19, 2017 at 15:48 | comment | added | Yonatan Harpaz | @KotelKanim, in Example 2.3(vi), ${\cal C}$ is not two parallel arrows, it is the category generated from two parallel arrows by freely adding countable coproducts. It is then claimed that ${\cal C}$ is not ${\cal K}$-filtered, where ${\cal K}$ is the doctrine of countable products. This implies that the terminal presheaf is not a ${\cal K}$-filtered colimit of representables. Indeed, if it were then this diagram of representables would be encoded by a cofinal map ${\cal J} \to {\cal C}$ from a ${\cal K}$-filtered category ${\cal J}$, which would imply that ${\cal C}$ is ${\cal K}$-filtered. | |
| Dec 19, 2017 at 12:05 | comment | added | KotelKanim | I meant "all countable coproducts". | |
| Dec 19, 2017 at 6:41 | comment | added | KotelKanim | Hi Yonatan! It might be the motivation, though I couldn't find any counterexample to my question in this paper. As for Example 2.3(vi), the category $\mathcal{C}$ (two parallel arrows) does not have all countable products, so it doesn't strictly fit into the framework, right? Anyway, thanks for the reference, but I am still looking for an explicit and convincing $\infty$-categorical counterexample. | |
| Dec 18, 2017 at 21:10 | history | edited | Yonatan Harpaz | CC BY-SA 3.0 |
added 240 characters in body
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| Dec 18, 2017 at 21:01 | history | answered | Yonatan Harpaz | CC BY-SA 3.0 |