Timeline for Monomorphisms of sheaves
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Sep 12, 2010 at 19:46 | comment | added | Nikita | Thanks Dustin (you wrote that while I composed the edit above)! | |
| Sep 12, 2010 at 19:44 | history | edited | Nikita | CC BY-SA 2.5 |
added 627 characters in body
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| Sep 12, 2010 at 19:37 | comment | added | Dustin Clausen | An open U in X can be viewed as a sheaf on X: sections over V are inclusions of V in U (so either empty or one-element). We then have Hom(U,F) = F(U) for any sheaf F. Thus if F --> G is a monomorphism, it is injective on sections over any open, hence injective on stalks. | |
| Sep 12, 2010 at 19:02 | comment | added | Nikita | Let me be a little more precise. In Hartschorne, a morphism of sheaves is called injective if its kernel is 0 (by the way, of course that only makes sense for sheaves of abelian groups, not sets as I have in my question). Thus when it is proven that the inverse image functor is exact, it is proven that if a map of sheaves has trivial kernel, so does the image of the map under the inverse image. But for this to give what I want, you would have to first prove that if a morphism of sheaves is a categorical monomorphism, then it has trivial kernel, which is exactly what I asked about! | |
| Sep 12, 2010 at 18:56 | comment | added | Nikita | That seems a little unfair, Dan. The questions on math.stackexchange.com seem mostly undergraduate and high school level. This one is at least is at the graduate level. Also, at least as I understand the foundations, your answer is a little circular. In all the sources I'm familiar with, the exactness of the inverse image functor is proven by studying what it does on stalks (eg this is how it is done in Hartschorne, which AFAIK never talks about the categorical version of a monomorphism). | |
| Sep 12, 2010 at 18:45 | comment | added | Dan Petersen | This is indeed standard and should probably be asked at the sister site math.stackexchange.com Anyway it follows because the stalk of F at x is just the inverse image sheaf under the inclusion ${x}\to X$, and the inverse image is an exact functor. | |
| Sep 12, 2010 at 18:33 | history | asked | Nikita | CC BY-SA 2.5 |