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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, 2017 at 5:39
  • $\begingroup$ Yes. Thanks fo the remark :) $\endgroup$ Jan 11, 2017 at 18:06
  • $\begingroup$ Can I have a moment to understand the answers? $\endgroup$ Jan 11, 2017 at 19:13
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    $\begingroup$ Yes! -:) ${}{}{}$ $\endgroup$ Jan 11, 2017 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, 2017 at 1:22

3 Answers 3

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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) $\endgroup$
    – Ivan So
    Nov 26, 2019 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, 2019 at 12:42
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    $\begingroup$ But I suppose affine intersect affine is affine is only true for X being seperated (hausdorff)? $\endgroup$
    – Ivan So
    Nov 26, 2019 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, 2019 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? $\endgroup$
    – FNH
    Dec 26, 2020 at 10:34

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