Timeline for Can Inequivalent Topologies Have Same Sheaves/Cohomology?
Current License: CC BY-SA 3.0
12 events
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| May 9, 2013 at 22:57 | history | edited | Igor Minevich | CC BY-SA 3.0 |
Removed the last edit; the question has pretty much completely been answered now.
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| Apr 25, 2013 at 9:48 | history | edited | Igor Minevich | CC BY-SA 3.0 |
Edited since the first question was answered
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| Apr 25, 2013 at 9:47 | history | rollback | Igor Minevich |
Rollback to Revision 1
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| Apr 25, 2013 at 2:31 | vote | accept | Igor Minevich | ||
| Apr 25, 2013 at 2:14 | history | edited | Igor Minevich | CC BY-SA 3.0 |
$T_1$ should be not just subordinate to $T_2$ if $g$ is to be continuous.
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| Apr 24, 2013 at 15:13 | vote | accept | Igor Minevich | ||
| Apr 25, 2013 at 2:31 | |||||
| Apr 24, 2013 at 7:14 | comment | added | Zhen Lin | The short answer is, Grothendieck topologies are uniquely determined by the subcategory of sheaves. This is not so for pretopologies. | |
| Apr 24, 2013 at 6:58 | answer | added | Andrej Bauer | timeline score: 7 | |
| Apr 24, 2013 at 6:52 | answer | added | Angelo | timeline score: 4 | |
| Apr 24, 2013 at 2:36 | comment | added | Igor Minevich | I mean, literally the same sheaves. For example, say I have a topology $T$ that is subcanonical and I can prove that there is a universal regular epimorphism (so a covering in the canonical topology) which does not have a refinement in $T$. Is there a specific sheaf in $T$ that is not representable, or at least a sheaf that is not a sheaf for the canonical topology? | |
| Apr 24, 2013 at 2:08 | comment | added | Qiaochu Yuan | What do you mean by "the same" here? (There are two things this could mean, namely just isomorphic or literally the same sheaves.) | |
| Apr 24, 2013 at 1:37 | history | asked | Igor Minevich | CC BY-SA 3.0 |