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Grothendieck gives in his Tohoku paper in example 3.8.3 an example for that $\check{\mathrm{H}}^{2}(X,\mathcal{F}) \neq \mathrm{H}^{2}(X,\mathcal{F})$. In the beginning he states that there exisits an exact sequence

$0 \longrightarrow \check{\mathrm{H}}^{2}(X,\mathcal{F}) \longrightarrow \mathrm{H}^{2}(X,\mathcal{F}) \longrightarrow \check{\mathrm{H}}^{1}(X, \mathcal{H}^{1}(\mathcal{F})) \longrightarrow 0$

which results from the Cech-to-derived functor spectral sequence $\check{\mathrm{H}}^{p}(X, \mathcal{H}^{q}(\mathcal{F})) \Rightarrow \mathrm{H}^{p+q}(X,\mathcal{F})$.

My question ist: Does this short exact sequence exist for an arbitrary topological space $X$ and a sheaf of abelian groups on $X$ or just in his concrete example? In the example there is $X = \mathbb{A}_{\mathbb{C}}^{2}$, $Y = Y_{1} \cup Y_{2}$ the union of two irreducible closed subsets $Y_{1}, Y_{2}$ and $\mathcal{F} = \mathbb{Z}_{X \backslash Y} = j_{*}(\mathbb{Z}_{X}|_{X \backslash Y})$ where $j: X \backslash Y \hookrightarrow X$ denotes the open inclusion.

I have shown that in the general setting there exists an exact sequence $0 \longrightarrow \check{\mathrm{H}}^{2}(X,\mathcal{F}) \longrightarrow \mathrm{H}^{2}(X,\mathcal{F}) \longrightarrow \check{\mathrm{H}}^{1}(X, \mathcal{H}^{1}(\mathcal{F})) \longrightarrow \check{\mathrm{H}}^{3}(X,\mathcal{F})$, and would therefore assume that Grothendieck is already in the special case of the example.

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  • $\begingroup$ Haven't you answered your own question --- in general, there is no 0 at the right. $\endgroup$
    – anon
    Commented Jul 18, 2019 at 23:41
  • $\begingroup$ Yes, but why is there a $0$ at the right in this concrete example? $\endgroup$
    – mathcourse
    Commented Jul 19, 2019 at 13:52

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