Is the (n-1)th de Rham Cohomology of U\{x} Non-Zero?

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SUMMARY

The (n-1)th de Rham cohomology of the open subset U \ {x} in R^n, where n ≥ 2 and x is not in U, is indeed non-zero. This conclusion is supported by applying the excision theorem from singular homology. The discussion emphasizes the importance of considering the restriction of forms from a small sphere S centered at x to the space U \ {x}. An example of a closed (n-1)-form on the (n-1)-sphere that is not exact is also relevant for this proof.

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seydunas
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Hi,

Let U be an open subset of R^n and n=>2 and x /in U. I want to show that (n-1)th de rham cohomology of U\ {x} is non zero. I suppose i can solve this question by using excision theorem from singular homology. But i have a hint for this problem: Consider the restrictions S--->U\ {x}----> R^n \{x} where S is a small sphere centered at x. I have to use hint. Can you help me?
 
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Presumably you know an example of a closed (n-1)-form on the (n-1)-sphere that isn't exact?
 

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