Lemma 20.11.3. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be a covering. Denote $j_{i_0\ldots i_ p} : U_{i_0 \ldots i_ p} \to X$ the open immersion. Consider the chain complex $K(\mathcal{U})_\bullet$ of presheaves of $\mathcal{O}_ X$-modules

$\ldots \to \bigoplus _{i_0i_1i_2} (j_{i_0i_1i_2})_{p!}\mathcal{O}_{U_{i_0i_1i_2}} \to \bigoplus _{i_0i_1} (j_{i_0i_1})_{p!}\mathcal{O}_{U_{i_0i_1}} \to \bigoplus _{i_0} (j_{i_0})_{p!}\mathcal{O}_{U_{i_0}} \to 0 \to \ldots$

where the last nonzero term is placed in degree $0$ and where the map

$(j_{i_0\ldots i_{p + 1}})_{p!}\mathcal{O}_{U_{i_0\ldots i_{p + 1}}} \longrightarrow (j_{i_0\ldots \hat i_ j \ldots i_{p + 1}})_{p!} \mathcal{O}_{U_{i_0\ldots \hat i_ j \ldots i_{p + 1}}}$

is given by $(-1)^ j$ times the canonical map. Then there is an isomorphism

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(K(\mathcal{U})_\bullet , \mathcal{F}) = \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$

functorial in $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\textit{PMod}(\mathcal{O}_ X))$.

Proof. We saw in the discussion just above the lemma that

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}( (j_{i_0\ldots i_ p})_{p!}\mathcal{O}_{U_{i_0\ldots i_ p}}, \mathcal{F}) = \mathcal{F}(U_{i_0\ldots i_ p}).$

Hence we see that it is indeed the case that the direct sum

$\bigoplus \nolimits _{i_0 \ldots i_ p} (j_{i_0 \ldots i_ p})_{p!}\mathcal{O}_{U_{i_0 \ldots i_ p}}$

represents the functor

$\mathcal{F} \longmapsto \prod \nolimits _{i_0\ldots i_ p} \mathcal{F}(U_{i_0\ldots i_ p}).$

Hence by Categories, Yoneda Lemma 4.3.5 we see that there is a complex $K(\mathcal{U})_\bullet$ with terms as given. It is a simple matter to see that the maps are as given in the lemma. $\square$

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