Lemma 20.10.4. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be a covering. Let $\mathcal{O}_\mathcal {U} \subset \mathcal{O}_ X$ be the image presheaf of the map $\bigoplus j_{p!}\mathcal{O}_{U_ i} \to \mathcal{O}_ X$. The chain complex $K(\mathcal{U})_\bullet $ of presheaves of Lemma 20.10.3 above has homology presheaves
\[ H_ i(K(\mathcal{U})_\bullet ) = \left\{ \begin{matrix} 0
& \text{if}
& i \not= 0
\\ \mathcal{O}_\mathcal {U}
& \text{if}
& i = 0
\end{matrix} \right. \]
Proof.
Consider the extended complex $K^{ext}_\bullet $ one gets by putting $\mathcal{O}_\mathcal {U}$ in degree $-1$ with the obvious map $K(\mathcal{U})_0 = \bigoplus _{i_0} (j_{i_0})_{p!}\mathcal{O}_{U_{i_0}} \to \mathcal{O}_\mathcal {U}$. It suffices to show that taking sections of this extended complex over any open $W \subset X$ leads to an acyclic complex. In fact, we claim that for every $W \subset X$ the complex $K^{ext}_\bullet (W)$ is homotopy equivalent to the zero complex. Write $I = I_1 \amalg I_2$ where $W \subset U_ i$ if and only if $i \in I_1$.
If $I_1 = \emptyset $, then the complex $K^{ext}_\bullet (W) = 0$ so there is nothing to prove.
If $I_1 \not= \emptyset $, then $\mathcal{O}_\mathcal {U}(W) = \mathcal{O}_ X(W)$ and
\[ K^{ext}_ p(W) = \bigoplus \nolimits _{i_0 \ldots i_ p \in I_1} \mathcal{O}_ X(W). \]
This is true because of the simple description of the presheaves $(j_{i_0 \ldots i_ p})_{p!}\mathcal{O}_{U_{i_0 \ldots i_ p}}$. Moreover, the differential of the complex $K^{ext}_\bullet (W)$ is given by
\[ d(s)_{i_0 \ldots i_ p} = \sum \nolimits _{j = 0, \ldots , p + 1} \sum \nolimits _{i \in I_1} (-1)^ j s_{i_0 \ldots i_{j - 1} i i_ j \ldots i_ p}. \]
The sum is finite as the element $s$ has finite support. Fix an element $i_{\text{fix}} \in I_1$. Define a map
\[ h : K^{ext}_ p(W) \longrightarrow K^{ext}_{p + 1}(W) \]
by the rule
\[ h(s)_{i_0 \ldots i_{p + 1}} = \left\{ \begin{matrix} 0
& \text{if}
& i_0 \not= i_{\text{fix}}
\\ s_{i_1 \ldots i_{p + 1}}
& \text{if}
& i_0 = i_{\text{fix}}
\end{matrix} \right. \]
We will use the shorthand $h(s)_{i_0 \ldots i_{p + 1}} = (i_0 = i_{\text{fix}}) s_{i_1 \ldots i_ p}$ for this. Then we compute
\begin{eqnarray*} & & (dh + hd)(s)_{i_0 \ldots i_ p} \\ & = & \sum _ j \sum _{i \in I_1} (-1)^ j h(s)_{i_0 \ldots i_{j - 1} i i_ j \ldots i_ p} + (i = i_0) d(s)_{i_1 \ldots i_ p} \\ & = & s_{i_0 \ldots i_ p} + \sum _{j \geq 1}\sum _{i \in I_1} (-1)^ j (i_0 = i_{\text{fix}}) s_{i_1 \ldots i_{j - 1} i i_ j \ldots i_ p} + (i_0 = i_{\text{fix}}) d(s)_{i_1 \ldots i_ p} \end{eqnarray*}
which is equal to $s_{i_0 \ldots i_ p}$ as desired.
$\square$
Comments (2)
Comment #5859 by Alex Scheffelin on
Comment #6071 by Johan on