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Mar 30, 2010 at 6:32 answer added Dev Sinha timeline score: 6
Mar 30, 2010 at 2:18 answer added Sean Tilson timeline score: 2
Jan 12, 2010 at 15:03 comment added Martin Brandenburg yes, this seems to be my question..
Jan 12, 2010 at 12:47 comment added Andrew Stacey Cohomology isn't necessarily a sequence of functors. Cohomology theories can be ungraded (or graded by some other set than the integers). So that leads me to the "abelian group" part and makes me think that your question is: "Are there any interesting cohomology-like theories that are Set-valued rather than abelian group-valued.". Is that the correct interpretation?
Jan 12, 2010 at 12:28 comment added Martin Brandenburg again I don't understand what's unclear in my question. the theorem is: every functor satisfying these two properties is representable. cohomology consists of a sequence of functors, which also are valuend in abelian groups.
Jan 12, 2010 at 12:11 comment added Andrew Stacey I have difficulty understanding the question. It seems to be saying "I want applications of the theorem 'Every cohomology theory is representable' that don't use the words 'cohomology theory'.". It feels a bit like "Applications of the Fundamental Theorem of Algebra without mentioning polynomials or complex numbers.".
Jan 12, 2010 at 1:24 history edited Martin Brandenburg CC BY-SA 2.5
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Jan 11, 2010 at 22:54 comment added Martin Brandenburg again, my question does not seem to be clear :-(. I repeat it: "these examples should really differ from the ones mentioned above" (cohomology/bundles)!
Jan 11, 2010 at 21:28 answer added Ryan Budney timeline score: 11
Jan 11, 2010 at 21:07 answer added Scott Carter timeline score: 3
Jan 11, 2010 at 19:30 history asked Martin Brandenburg CC BY-SA 2.5