Timeline for On the ordered set of real numbers, does sheaf+cosheaf imply constant?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 17, 2016 at 4:36 | vote | accept | David Treumann | ||
| Mar 15, 2016 at 20:12 | comment | added | Dmitri Pavlov | @DavidTreumann: I added a new paragraph that explains how to eliminate homotopies from the discussion. This should make reading the rest of the argument easier. | |
| Mar 15, 2016 at 17:14 | history | edited | Dmitri Pavlov | CC BY-SA 3.0 |
added 888 characters in body
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| Mar 15, 2016 at 16:32 | comment | added | Dmitri Pavlov | As for r=t, it is somewhat poorly formulated there: we start with the existing nullhomotopy for r=t, use the argument in the third paragraph to construct some lift for r'>r=t, then construct a compatible system of lifts for all q such that r'>q≥t. | |
| Mar 15, 2016 at 16:30 | comment | added | Dmitri Pavlov | The criterion for weak equivalence of spectra is a variation of a classical criterion for weak equivalences of (fibrant) simplicial sets, see Proposition 4.1 in Dugger and Isaksen “Weak equivalences of simplicial presheaves”. | |
| Mar 15, 2016 at 15:49 | comment | added | David Treumann | Thanks Dmitri. I think I haven't understood your criterion for isomorphism of spectra yet, in the second paragraph. But before I dig in let me check something: in the third paragraph you take a supremum over a set of numbers $r$ that obey $r > q \geq t$, which implies $r > t$. Then in parentheses you argue that this set of numbers is nonempty because it contains $r = t$. Is that a problem? | |
| Mar 15, 2016 at 14:20 | history | answered | Dmitri Pavlov | CC BY-SA 3.0 |