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Let $X$ be a topological space and $G$ an abelian group. Denote by $\mathcal{S}$ the skyscraper sheaf with group $G$ at the point $x\in X$. How I can prove that $\mathcal{S}$ has not cohomology, i.e $H^i(X,\mathcal{S})=0, \: \forall i>0$ ?

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    $\begingroup$ Do you assume $G$ is an abelian group? $\endgroup$ – Andrew Jan 11 '17 at 5:39
  • $\begingroup$ Yes. Thanks fo the remark :) $\endgroup$ – Vincenzo Zaccaro Jan 11 '17 at 18:06
  • $\begingroup$ Can I have a moment to understand the answers? $\endgroup$ – Vincenzo Zaccaro Jan 11 '17 at 19:13
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    $\begingroup$ Yes! -:) ${}{}{}$ $\endgroup$ – Georges Elencwajg Jan 11 '17 at 20:23
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    $\begingroup$ There is also a cheaper way for showing this : you can build a discrete partition of unity adapted to your sheaf. This is done in the book of Miranda, Algebraic curves and Riemann surfaces. $\endgroup$ – user171326 Jan 16 '17 at 1:22
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A skyscraper sheaf is flasque, hence has no cohomology: Hartshorne Chap. III, Prop.2.5, page 208.

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You can also see this using Čech cohomology:

Consider an open cover $\mathfrak{U}=(U_i)_{i\in I}$ of your space $X$. You can always refine this cover so that only one of the sets $U_i$ contains the point $x$: Pick a set $U_0$ containing $x$ and consider the cover $\mathfrak{U}'$ consisting of $U_0$ and $U_i\setminus\{x\}$ for all $i\in I$. Then $\mathfrak{U}'$ is a refinement of $\mathfrak{U}$ and $x$ is only contained in the set $U_0$.

In particular, $x$ is not contained in any intersection of two or more distinct sets in $\mathfrak{U}'$, so the scyscraper sheaf has no sections over these. Hence all higher Čech-cocylces are $0$ and all higher cohomology groups vanish.

(I believe this approach can also be found in Forster's book on Riemann Surfaces.)

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  • $\begingroup$ I got a regarding the equivalence of Cech cohomology and the normal cohomology group H^i. In Hartshorne III 4.5, it said the two cohomology are equivalent only if we assume the open covering used for the Cech consideration is affine. How can you guarantee your specially chosen open covering to be affine? (if I am considering from the algebraic geometry perspective) $\endgroup$ – IvanSo Nov 26 '19 at 6:45
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    $\begingroup$ Dear @IvanSo, the question as asked was in the setting of topological spaces, where there is no notion of affine. If you are in the situation of, say, schemes, then you may mimic this approach by starting with an affine cover (which exists by definition of a scheme) and refine this as described, noting that intersections and open subsets of affines remain affine. $\endgroup$ – Jonathan Nov 26 '19 at 12:42
  • $\begingroup$ But I suppose affine intersect affine is affine is only true for X being seperated (hausdorff)? $\endgroup$ – IvanSo Nov 26 '19 at 13:38
  • $\begingroup$ @IvanSo You are right, there are subtleties involved. One should probably require Noetherian schemes, as Hartshorne does (if I am not mistaken). Maybe qcqs is enough, though... $\endgroup$ – Jonathan Nov 26 '19 at 14:07
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These notes probably have enough detail to give you what you want. I'll assume you mean $G$ to be an abelian group. Basically the argument goes like this: if $G$ is an abelian group, take an injective resolution $G \to I^\bullet$ of abelian groups. Then $\pi_*G \to \pi_*I^\bullet$ is an injective resolution, where $\pi:{\ast} \to X$ is the inclusion of a point. Note that $\pi_*G$ is your skyscraper sheaf. To compute $H^i(X,G)$ take global sections of $\pi_*G \to \pi_*I^\bullet$, which just gives back the resolution $G\to I^\bullet$, which shows that the higher cohomology of $\pi_*G$ vanishes.

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  • $\begingroup$ The link is not working. Do you have a new link for the notes? $\endgroup$ – Fawzy Hegab Dec 26 '20 at 10:34

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