Timeline for Given fiber functors on all the subcategories of the form $\langle M\rangle$, can we obtain a fiber functor on the whole category?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Feb 15, 2022 at 10:43 | vote | accept | Gabriel | ||
| Feb 15, 2022 at 1:15 | answer | added | user477095 | timeline score: 2 | |
| Feb 14, 2022 at 19:53 | comment | added | Gabriel | One example I have in mind is differential Galois theory. If $\mathsf{T}$ is the category of differential modules, then fiber functors $\omega_M$ are determined by Picard-Vessiot extensions. Then we can take the direct limit of all such extensions to obtain a fiber functor of the whole category. But I'm not sure if this works in other cases. | |
| Feb 14, 2022 at 18:02 | comment | added | Aidan | I don't quite know if there is a canonical way, but the obvious thing would be that if, for every pair of objects $M,N$,$\omega_M(X)\cong\omega_N(X)$ for every object in both $\langle M\rangle$ and $\langle N\rangle$. Which seems something like a "global section of the sheaf of functors to vector spaces", in a non rigourous sense. Then in order to be canonical, you would want the isomorphism $\omega_M(X)\cong\omega_N(X)$ to be canonical in some sense, which would probably be classified by some sort of cohomology-like object | |
| Feb 14, 2022 at 17:36 | history | asked | Gabriel | CC BY-SA 4.0 |