Timeline for Recover a morphism from its pullback
Current License: CC BY-SA 2.5
22 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Sep 6, 2010 at 15:58 | vote | accept | Martin Brandenburg | ||
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| Sep 1, 2010 at 8:09 | comment | added | Martin Brandenburg | Yes you're right. | |
| Aug 31, 2010 at 16:30 | comment | added | t3suji | Dear Martin, isn't your compatibility condition vacuous? After all, colimit is defined by a universal property; so a colimit diagram for $colim f^*M_i$ would get mapped to the colimit diagram for $colim g^*M_i$. | |
| Aug 31, 2010 at 11:32 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| Aug 31, 2010 at 11:03 | vote | accept | Martin Brandenburg | ||
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| Aug 31, 2010 at 10:06 | answer | added | user8907 | timeline score: 0 | |
| Aug 31, 2010 at 9:08 | answer | added | Laurent Moret-Bailly | timeline score: 5 | |
| Aug 31, 2010 at 9:05 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| Aug 31, 2010 at 4:39 | comment | added | BCnrd | Martin, sure, assume whatever you want if it makes a proof work. My main point was that in view of Lurie's result and its proof, probably you should keep track of tensor compatibility and nothing more (at least under some finiteness hypotheses). He uses quasi-compactness to get a smooth cover by an affine scheme (in your case even an etale cover: disjoint union of constituents of a finite open affine covering) and then needs affine diagonal for some effective descent argument (if I remember correctly). By the way, the LaTeX command \rightrightarrows is very useful. | |
| Aug 30, 2010 at 23:28 | comment | added | Martin Brandenburg | Hm, isn't also reasonable to assume that the isomorphism is compatible with colimits, i.e. $colim_i f^* M_i = f^* (colim_i M_i) \cong g^* (colim_i M_i) = colim_i g^* M_i$ is induced by $f^* M_i \cong g^* M_i$? | |
| Aug 30, 2010 at 22:30 | comment | added | BCnrd | Martin, you almost certainly want to assume the tensor compatibility. All it means is the obvious necessary (and extremely reasonable) condition that the isomorphism of functors respects the formation of tensor products on both sides (in a manner respecting the associativity, symmetry, and identity objects for the tensor products on both sides). It sounds like you throw away too much structure if you ignore that aspect in general. Again, look at Lurie's arguments (specialized to your situation with schemes, so some of his complications will go away). | |
| Aug 30, 2010 at 19:01 | comment | added | Martin Brandenburg | Thanks Brian, I will read your literature when I know the basics. Concerning your questions: 1) I'm interested in rather general schemes, but some finiteness conditions (and counterexamples to the general case) would be also great. I only consider the Zariski topology. 2) I just want the functors to be isomorphic in the usual sense. What is meant by "isomorphic as functors between tensor product"? And is this a reasonable assumption? | |
| Aug 30, 2010 at 18:36 | comment | added | BCnrd | Dear Martin: one more small technical point is that you need to clarify in what sense you're saying that the pullback functors are isomorphic. That is, as functors between mere abelian categories, or as tensor functors, etc. It seems plausible that to make a proof you should assume the isomorphism is as functors between tensor categories. (I'm not saying I see a counterexample if you weaken to just an isomorphism of functors between abelian categories, but that kind of hypergenerality does seem a bit silly, and perhaps false.) | |
| Aug 30, 2010 at 18:20 | answer | added | Sasha | timeline score: 3 | |
| Aug 30, 2010 at 17:34 | comment | added | BCnrd | Martin, small addendum: in section 6 Lurie gives the argument for faithfulness (it is the much easier part), which is all you need. But his notion of "geometric stack" covers only those schemes which are quasi-compact and have affine diagonal. So I should make some caveats on it's relevance to your situation. But if you look at his arguments then perhaps you'll get some idea for what to do (and surely his arguments become even simpler in the scheme setting, especially with the Zariski topology). Good luck. | |
| Aug 30, 2010 at 17:24 | comment | added | BCnrd | Probably. Definitely if use q-coh sheaves on etale sites, not Zar. sites. Functor assigning to any scheme its Zar. (or etale) loc. ringed topos is fully faithful (proved in SGA4 VIII, & explained in Prop. 3.1.1 and Thm. 3.1.3 in draft of "Univ. property of non-arch. analytification" on my webpage), so suffices that functor from loc. ringed topos to tensor category of q-coh sheaves is faithful. By main result (Thm. 5.11) of Lurie's "Tannaka duality & geom. stacks", fully faithful in etale case. Surely pf works in Zar. case for schemes. It's been a while since I read it. Check for yourself. | |
| Aug 30, 2010 at 16:18 | comment | added | Martin Brandenburg | For example, if you take an automorphism $\sigma$ of a ring $R$, which maps an ideal $I$ to a different ideal $J$, then the pullback of $R/I$ with $\sigma$ is $R/J$, and these $R$-modules are not isomorphic. | |
| Aug 30, 2010 at 15:51 | comment | added | Martin Brandenburg | @Harry: I also fell for this trap ;). The module structure changes when you pull it back with an isomorphism. | |
| Aug 30, 2010 at 15:38 | comment | added | Harry Gindi | @Martin: I haven't checked, but have you tried to see what happens on the pullbacks when you compose with a nontrivial automorphism? I haven't checked, but I suspect that the pullbacks will remain isomorphic. | |
| Aug 30, 2010 at 15:32 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| Aug 30, 2010 at 15:26 | history | asked | Martin Brandenburg | CC BY-SA 2.5 |