Lemma 20.24.1. Let $X$ be a ringed space. Let $\mathcal{U} : X = \bigcup _{i \in I} U_ i$ be an open covering of $X$. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Denote $\mathcal{F}_{i_0 \ldots i_ p}$ the restriction of $\mathcal{F}$ to $U_{i_0 \ldots i_ p}$. There exists a complex ${\mathfrak C}^\bullet (\mathcal{U}, \mathcal{F})$ of $\mathcal{O}_ X$-modules with

${\mathfrak C}^ p(\mathcal{U}, \mathcal{F}) = \prod \nolimits _{i_0 \ldots i_ p} (j_{i_0 \ldots i_ p})_* \mathcal{F}_{i_0 \ldots i_ p}$

and differential $d : {\mathfrak C}^ p(\mathcal{U}, \mathcal{F}) \to {\mathfrak C}^{p + 1}(\mathcal{U}, \mathcal{F})$ as in Equation (20.9.0.1). Moreover, there exists a canonical map

$\mathcal{F} \to {\mathfrak C}^\bullet (\mathcal{U}, \mathcal{F})$

which is a quasi-isomorphism, i.e., ${\mathfrak C}^\bullet (\mathcal{U}, \mathcal{F})$ is a resolution of $\mathcal{F}$.

Proof. We check

$0 \to \mathcal{F} \to \mathfrak {C}^0(\mathcal{U}, \mathcal{F}) \to \mathfrak {C}^1(\mathcal{U}, \mathcal{F}) \to \ldots$

is exact on stalks. Let $x \in X$ and choose $i_{\text{fix}} \in I$ such that $x \in U_{i_{\text{fix}}}$. Then define

$h : \mathfrak {C}^ p(\mathcal{U}, \mathcal{F})_ x \to \mathfrak {C}^{p - 1}(\mathcal{U}, \mathcal{F})_ x$

as follows: If $s \in \mathfrak {C}^ p(\mathcal{U}, \mathcal{F})_ x$, take a representative

$\widetilde{s} \in \mathfrak {C}^ p(\mathcal{U}, \mathcal{F})(V) = \prod \nolimits _{i_0 \ldots i_ p} \mathcal{F}(V \cap U_{i_0} \cap \ldots \cap U_{i_ p})$

defined on some neighborhood $V$ of $x$, and set

$h(s)_{i_0 \ldots i_{p - 1}} = \widetilde{s}_{i_{\text{fix}} i_0 \ldots i_{p - 1}, x}.$

By the same formula (for $p = 0$) we get a map $\mathfrak {C}^{0}(\mathcal{U},\mathcal{F})_ x \to \mathcal{F}_ x$. We compute formally as follows:

\begin{align*} (dh + hd)(s)_{i_0 \ldots i_ p} & = \sum \nolimits _{j = 0}^ p (-1)^ j h(s)_{i_0 \ldots \hat i_ j \ldots i_ p} + d(s)_{i_{\text{fix}} i_0 \ldots i_ p}\\ & = \sum \nolimits _{j = 0}^ p (-1)^ j s_{i_{\text{fix}} i_0 \ldots \hat i_ j \ldots i_ p} + s_{i_0 \ldots i_ p} + \sum \nolimits _{j = 0}^ p (-1)^{j + 1} s_{i_{\text{fix}} i_0 \ldots \hat i_ j \ldots i_ p} \\ & = s_{i_0 \ldots i_ p} \end{align*}

This shows $h$ is a homotopy from the identity map of the extended complex

$0 \to \mathcal{F}_ x \to \mathfrak {C}^0(\mathcal{U}, \mathcal{F})_ x \to \mathfrak {C}^1(\mathcal{U}, \mathcal{F})_ x \to \ldots$

to zero and we conclude. $\square$

Comment #937 by correction_bot on

Replace direct sum with product in the definition of $\mathfrak{C}^{\bullet}(\mathcal{U},\mathcal{F})$; then it looks like following works without any local finiteness assumption on the cover. We check is exact on stalks. Let $x \in X$ and fix $i$ such that $x \in U_i$. Then define as follows: If $s \in \mathfrak{C}^{p}(\mathcal{U},\mathcal{F})_x$, take a representative defined on some neighborhood $V$ of $x$, and set (By the same formula we get a map $\mathfrak{C}^{0}(\mathcal{U},\mathcal{F})_x \to \mathcal{F}_x$.) A computation shows $h$ is a homotopy from the identity map of the complex in question to zero. (This proof is the same as Tag 01X9.)

Comment #958 by on

Thanks very much. The original idea for this lemma was to write the stalk as a product of stalks, which only works in the locally finite case (because taking stalks does not commute with products). But of course what you say is much better and now it works in general. Thanks! The changes can be found here.

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