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Nov 17, 2011 at 2:05 comment added algori Dear David -- you are right. A spectacular example of dyslexia on my part.
Nov 16, 2011 at 22:15 comment added David Carchedi Dear Algori. Thanks, but in fact, this is why I said "this amounts to computing the cohomology of the interior of any arbitrary closed subset of $mathbb{R}^n$." I am well aware of this result, however, it does not mean necessarily that the properties of the function $f$ cannot be used to describe the cohomology. I'm guessing it still can't, but, I wanted to see what was known.
Nov 16, 2011 at 18:58 history answered algori CC BY-SA 3.0