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As we know,a fine sheaf is also soft.So,I need some examples of the sheaves that are soft but not fine.Can the holomorphic sheaf $\mathcal O(X)$ be one?Any help and comments are accepted.Thanks a lot!

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No, the sheaf of holomorphic maps can be not soft!

Example. Let $\displaystyle X=\left\{z\in\mathbb{C}:|z|\leq\frac{1}{2}\right\}$ be the "half unitary disk"; the holomorphic maps \begin{equation*} f(z)=\sum_{n=0}^{+\infty}z^n,\,g(z)=\sum_{n=1}^{+\infty}z^{n!} \end{equation*} can not be extendend to the whole of $\mathbb{C}$, so $\mathcal{O}_{\mathbb{C}}$ is neither soft nor fine. $\triangle$

Considering the constant sheaf $\mathcal{F}$ to $\mathbb{Z}$ on $\mathbb{A}^1_{\mathbb{C}}$ with Zariski topology; $\mathcal{F}$ is flabby but not fine (of course), so it is soft not fine.

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    $\begingroup$ Dear Armando,thanks a lot! $\endgroup$
    – Steve
    Aug 13, 2019 at 1:33
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    $\begingroup$ I don't think the constant sheaf on $\mathbb{A}^1$ is soft : a section over the closed subset $\{0,1\}$ cannot necessarily be extended to the whole space. $\endgroup$
    – Roland
    Aug 15, 2019 at 13:04
  • $\begingroup$ @Roland is absolutely right. $\endgroup$ Aug 16, 2019 at 19:33
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    $\begingroup$ @Armandoj18eos Sorry for my late reply. I checked several sources and it seems that there is two different notions of soft sheaves, or more accurately, there is two different notions of $\Gamma(Z,\mathcal{F})$ when $Z$ is a closed subset. One is $\varinjlim_{U\supset Z}\Gamma(U,\mathcal{F})$, the other is $\Gamma(Z,\mathcal{F}|_Z)$ (this is the one used for example in Kashiwara-Shapira, Godement...). These two notions agree if $Z$ has a fundamental system of paracompact (hence Hausdorff) neighborhood. The problem is : the first notion is ill-behaved for other spaces. $\endgroup$
    – Roland
    Aug 21, 2019 at 19:28
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    $\begingroup$ If we use the first definition, this is true that a flabby sheaf is soft, but with the other one this is false. The only proofs I know that soft sheaves are acyclic either use the second definition or restrict themself to a space where the two definitions agree. $\endgroup$
    – Roland
    Aug 21, 2019 at 19:32

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