Timeline for Is the subobject functor really a presheaf?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Jul 28, 2010 at 12:16 | comment | added | Finn Lawler | Incidentally, a category whose every object has a small set of subobjects is called well-powered. (See e.g. ncatlab.org/nlab/show/well-powered+category). | |
| Jul 28, 2010 at 11:26 | comment | added | Torsten Ekedahl | "defining the notion of presheaf so that it doesn't matter" That was just a thought I had. The idea is to use the Grothendieck construction to turn a presheaf into a fibred category and then use that the fibred category works even if the purported presheaf doesn't make sense. Hence one should look at the category whose objects are pairs $(A,c)$ where $A$ is an object of the category and $c$ an element of what morally should be $F(A)$. | |
| Jul 28, 2010 at 11:05 | vote | accept | fosco | ||
| Jul 28, 2010 at 10:56 | comment | added | fosco | (P.S.: How can I indent the code as if it were a quotation? And how can I link stuff from wiki or google or something? The tips in editing help page don't work... :( ) | |
| Jul 28, 2010 at 10:55 | answer | added | Peter Arndt | timeline score: 11 | |
| Jul 28, 2010 at 10:55 | comment | added | fosco | > defining the notion of presheaf so that it doesn't matter How can I do that? "My" definition af a presheaf on C is "a contavariant functor between C and Sets". I obviously thought about throwing everything in a suitable universe, but MacLane never mention the Grothendieck's universes, so I believe there is another way. | |
| Jul 28, 2010 at 10:54 | answer | added | Martin Brandenburg | timeline score: 4 | |
| Jul 28, 2010 at 10:24 | comment | added | Torsten Ekedahl | By adding the condition that it doesn't happen (people do that), by throwing in a universe or two or by defining the notion of presheaf so that it doesn't matter (I haven't checked details but I think that that is possible). Another, more common, problem (usually more easily solvable) is that the members of $\mathrm{Sub}_{\mathbb C}(D)$ are proper classes so that $\mathrm{Sub}_{\mathbb C}(D)$ doesn't even exist. | |
| Jul 28, 2010 at 9:53 | history | asked | fosco | CC BY-SA 2.5 |