Timeline for Criteria for representability of a functor from schemes to sets
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 27, 2019 at 21:55 | vote | accept | Joakim Færgeman | ||
| Sep 25, 2019 at 18:46 | comment | added | Joakim Færgeman | I agree that if the functor is a Zariski Sheaf, the answer would be yes. I wonder if it is even true in general. | |
| Sep 25, 2019 at 15:51 | comment | added | Fred Rohrer | I think (without checking all the details) that it follows from the result in EGA I mentioned above that the answer to your question is yes if your functor is a Zariski sheaf on the category of all schemes. I have no idea about the general case. | |
| Sep 25, 2019 at 14:44 | comment | added | Joakim Færgeman | I mean, when it is seen as a contravariant functor on the affine schemes, there exists an affine scheme X such that X represents the restricted functor | |
| Sep 25, 2019 at 13:25 | comment | added | Fred Rohrer | @Joakim: What precisely do you mean by "it is representable when restricted to affine schemes"? | |
| Sep 25, 2019 at 11:24 | comment | added | Joakim Færgeman | Maybe I was unclear in my question, or maybe you have already answered it, and I dont understand. In any case, I am dealing with the following situation: I have a contravariant functor from Schemes to Sets. I know that it is representable when restricted to affine schemes. Can I conclude that it is also representable on Schemes? | |
| Sep 25, 2019 at 6:14 | comment | added | Denis Nardin | @JoakimFærgeman Maybe this helps: the restriction functor induces an equivalences of categories between Zariski sheaves on $\mathrm{Sch}$ and Zariski sheaves on $\mathrm{AffSch}$, so to give one is the same thing as to give the other. | |
| Sep 24, 2019 at 23:40 | comment | added | Arun Debray | We don't know that a priori. But if $F$ is represented by some scheme $X$, then $F$ is determined by its restriction to affines. Let $Y$ be another scheme and $\mathfrak U$ be an affine open cover of $Y$; then the set $F(Y) = \mathrm{Hom}(Y, X)$ is the subset of $\prod_{U\in\mathfrak U}\mathrm{Hom}(U, X)$ consisting of tuples of maps which agree on all pairwise intersections. | |
| Sep 24, 2019 at 23:08 | comment | added | Joakim Færgeman | Thanks for the answer. I am still a bit confounded. How is it that it suffices to define our functor $F:Sch^{op}\rightarrow Set$ on affine schemes? How do we know it has an extension to all schemes, then? | |
| Sep 24, 2019 at 22:54 | history | answered | Arun Debray | CC BY-SA 4.0 |