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Let $X$ be a scheme with an finite affine open covering $\mathcal{U} = (U_i)_{i=1}^k$ such that all intersections of $U_i$'s are affine too. Why does a short exact sequence of quasicoherent $\mathcal{O}_X$-modules $$ 0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0$$ induces a short exact sequence $$ 0 \to C^q(\mathcal{U},\mathcal{F}) \to C^q(\mathcal{U},\mathcal{G}) \to C^q(\mathcal{U},\mathcal{H}) \to 0$$ where $C^q(\mathcal{U},\mathcal{F}) = \prod_{i_0 < \ldots < i_q} \mathcal{F}(U_{i_0} \cap \ldots \cap U_{i_q})$ is the Cech-Complex?

This is an intermediate step to show that a short exact sequence of quasicoherent sheafs on a separated quasicompact sheaf induces a long exact cech-cohomology sequence.

What I've tried: I know that $H^q(U_{i_0} \cap \ldots \cap U_{i_q},\mathcal{F}) = 0$ since the intersection is affine and $\mathcal{F}$ is quasicoherent, so Leray's theorem tells me that $H^q(X,\mathcal{F}) = H^q(\mathcal{U},\mathcal{F})$. However this does not help with the $C^q$, does it? Or is this something much simpler?

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You essentially already found the reason yourself: For any $i_0<\ldots<i_q$ you have an exact sequence $$0\to{\mathscr F}(U_{i_0}\cap\dots\cap U_{i_q})\to {\mathscr G}(U_{i_0}\cap\dots\cap U_{i_q})\to{\mathscr H}(U_{i_0}\cap\dots\cap U_{i_q})\to\text{H}^1(U_{i_0}\cap\dots\cap U_{i_q};{\mathscr F}),$$ and the last term vanishes since $U_{i_0}\cap\dots\cap U_{i_q}$ is affine. Hence, you get a short exact sequence $$0\to{\mathscr F}(U_{i_0}\cap\dots\cap U_{i_q})\to {\mathscr G}(U_{i_0}\cap\dots\cap U_{i_q})\to{\mathscr H}(U_{i_0}\cap\dots\cap U_{i_q})\to 0,$$ and taking the product over all choices of $i_0<\ldots<i_q$ gives the exact sequence you are after.

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  • $\begingroup$ Cool, so it works because taking finite products is exact. $\endgroup$ – legacytron Mar 24 '15 at 9:30
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    $\begingroup$ @legacytron: Yes, and since you are taking products of abelian groups, infinite products would be fine as well. $\endgroup$ – Hanno Mar 24 '15 at 9:32
  • $\begingroup$ That sounds interesting, but I don't quite understand. Would I have to check by hand that taking products is exact or is there some nice argument? $\endgroup$ – legacytron Mar 24 '15 at 9:34
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    $\begingroup$ @legacytron: It's a distinguished feature of abelian categories with a projective generator, like module categories - but you can of course also check it by hand. Note that without the assumption of enough projectives, even the nicest (locally Noetherian Groethendieck) categories such as $\text{QCoh}({\mathbb P}^1_{\mathbb k})$ fail to have exact products. $\endgroup$ – Hanno Mar 24 '15 at 9:40

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