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May 24 at 12:30 vote accept Jaehwan Kim
May 23 at 5:11 comment added Jaehwan Kim @ChrisGerig I think my version is outdated
May 23 at 5:10 comment added Jaehwan Kim @ChrisSchommer-Pries The sheaf associated with the given presheaf is locally constant, but I do not understand the significance of that sheaf.
May 22 at 5:59 comment added Chris Gerig As GregFriedman highlights, different editions of the book may be more/less accurate -- I hope your version is older? (My version is "revised third printing" which agrees with Friedman's post.)
May 22 at 5:29 answer added Greg Friedman timeline score: 9
May 22 at 5:22 history edited Greg Friedman CC BY-SA 4.0
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May 21 at 14:40 comment added Chris Schommer-Pries This is a good question. As your example shows, the map $\rho_V^U$ need not be an isomorphisms. That seems to be a mistake. I think it is still true that $\mathcal{F}$ is a locally constant presheaf, but this might depend on the definition that Bott and Tu are using. Is it enough for the sheafification to be locally constant?
May 21 at 10:29 comment added Francesco Polizzi Actually, using the notation in the book for the trivial line bundle on $\mathbb{R}^2$, if $U=D^2$ and $V=D^2-\{(0, \, 0)\}$, I do not see how the pullback ("restriction") map $$H^1( U \times \mathbb{R}) \to H^1(V \times \mathbb{R})$$ can be the identity, since $$H^1(U \times \mathbb{R})= 0, \; H^1(V \times \mathbb{R})= \mathbb{R}$$
May 21 at 9:01 comment added Francesco Polizzi If I understand the question correctly, it seems that the OP's doubt is that a contractible open set $U$ can contain a non-contractible open set $V$. For instance, a disk in $\mathbb{R}^2$ contains an open annulus, whose cohomology is non-trivial.
May 21 at 7:48 review Close votes
May 27 at 3:03
May 21 at 7:40 history edited gmvh
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S May 21 at 7:25 review First questions
May 21 at 7:40
S May 21 at 7:25 history asked Jaehwan Kim CC BY-SA 4.0