Timeline for Eilenberg-Steenrod axioms of sheaf cohomology
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 19, 2012 at 16:53 | comment | added | Liviu Nicolaescu | If you are interested in the cohomology of sheaves you do not need pairs. For example if $A\subset X$ is an open subset with complement $C$, a closed subset then $H^*\bullet(X,A, \mathscr{S})$ is the local cohomology of $\matscr{S}$ along $C$, which is the the cohomology of $X$ with coefficients in a complex of sheaves naturally associated to $C$ and $\matscr{S}$. If $A$ is closed and $i:A\to X$ is the natural inclusion, then $H^\bullet(X,A,\mathscr{S})=H^\bullet(X,i_*\mathscr{S})$. | |
| Nov 19, 2012 at 16:44 | comment | added | Mark Grant | It should be noted that the classical Eilenberg-Steenrod Uniqueness theorem requires some additional assumptions on the spaces involved (eg compact polyhedra, or polyhedra if we include Milnor's additivity axiom). Another comment: there is a sort of uniqueness theorem for sheaf cohomology when we fix the space $X$ and look at the category of sheaves on $X$. It is in Bredon's "Sheaf Theory", sections II.6-7. | |
| Nov 19, 2012 at 15:52 | comment | added | Dylan Wilson | Some axioms that might be included (for the ordinary case) would be that the case of a point should reduce to a projection onto the global sections, for example. | |
| Nov 19, 2012 at 15:50 | comment | added | Dylan Wilson | No need to apologize- I just didn't understand. Anyway, probably Dan means something like this: Are there axioms we can put on functors out of the category of pairs equipped with sheaves of abelian groups to the category of abelian groups such that the functors satisfying these axioms are precisely sheaf cohomology of abelian sheaves? Are there axioms so that the functors we get have a right to be called "generalized sheaf cohomology"? | |
| Nov 19, 2012 at 15:41 | comment | added | Fernando Muro | I'm really sorry for my bad explanations. Let me clarify that I don't claim to have offered any answer. I'm just remarking that I don't know how to make precise sense of the question, and trying to explain why. If any of you guys can at least make a guess, that would be great. | |
| Nov 19, 2012 at 15:17 | comment | added | Dylan Wilson | @Dan: I think there was some mention of cohomology theories with sheaves of coefficients in "A Geometric Approach to homology theory" but I'm not sure how well-behaved their set-up is. | |
| Nov 19, 2012 at 15:09 | comment | added | Dylan Wilson | (I realize that you could ask for your cohomology to spit out sheaves, but it doesn't seem like this is what Dan is asking.) | |
| Nov 19, 2012 at 15:06 | comment | added | Dylan Wilson | I'm confused, Fernando. Dan is talking abelian group valued sheaves on a space. Global sections thus spit out an abelian group, and the derived functors of global sections also spit out an abelian group. So we will still always land inside abelian groups. Notice, in particular, that we are not considering the case where the space has a structure sheaf and so don't have competing module structures; perhaps this is the case where your objection applies? | |
| Nov 19, 2012 at 14:05 | comment | added | Fernando Muro | The Eilenberg-Steenrod axioms can be restated as follows: there is a unique homological functor from the Spanier-Whitehead category (triangulated) to abelian groups (abelian) taking $S^0$ to a given abelian group $\mathbb{A}$. If you try to rephrase your problem in these terms you can't because there's no candidate for target abelian category. The problem is not created by putting the axioms in this form. | |
| Nov 19, 2012 at 13:27 | comment | added | Dan Petersen | A colleague just pointed out this older MO question which raises the same question: mathoverflow.net/questions/32689 But the question still seems to be open. Fernando, I don't really understand your objection. | |
| Nov 19, 2012 at 11:51 | comment | added | Fernando Muro | How would you start to write the axioms? The abelian category of coefficients varies with the space whose sheaf cohomology you want to compute. In classical cohomology, coefficients always live in the same abelian category (abelian groups) and this is used in the statement of the Eilenberg-Steenrod axioms. | |
| Nov 19, 2012 at 10:52 | history | asked | Dan Petersen | CC BY-SA 3.0 |