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Cohomology of a space is often defined axiomatically: a cohomology theory is a functor from pairs of spaces to abelian groups satisfying the Eilenberg-Steenrod axioms. Is there a similar characterization of sheaf cohomology, where the domain of the functor is now a category of pairs $(A,X,\mathcal F)$ with $A \subset X$ a pair and $\mathcal F$ an abelian sheaf on $X$ (with the obvious morphisms)?

Are there extraordinary sheaf cohomology theories?!

flag	asked Nov 19 at 10:52 Dan Petersen 10.1k●1●21●46

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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. – Fernando Muro Nov 19 at 11:51

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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. – Fernando Muro Nov 19 at 14:05

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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? – Dylan Wilson Nov 19 at 15:06

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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"? – Dylan Wilson Nov 19 at 15:50

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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. – Mark Grant Nov 19 at 16:44

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Eilenberg-Steenrod axioms of sheaf cohomology

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