The Stacks project

Lemma 20.9.3. Let $X$ be a topological space. Let $\mathcal{F}$ be an abelian presheaf on $X$. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. If $U_ i = U$ for some $i \in I$, then the extended Čech complex

\[ \mathcal{F}(U) \to \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \]

obtained by putting $\mathcal{F}(U)$ in degree $-1$ with differential given by the canonical map of $\mathcal{F}(U)$ into $\check{\mathcal{C}}^0(\mathcal{U}, \mathcal{F})$ is homotopy equivalent to $0$.

Proof. Fix an element $i \in I$ with $U = U_ i$. Observe that $U_{i_0 \ldots i_ p} = U_{i_0 \ldots \hat i_ j \ldots i_ p}$ if $i_ j = i$. Let us define a homotopy

\[ h : \prod \nolimits _{i_0 \ldots i_{p + 1}} \mathcal{F}(U_{i_0 \ldots i_{p + 1}}) \longrightarrow \prod \nolimits _{i_0 \ldots i_ p} \mathcal{F}(U_{i_0 \ldots i_ p}) \]

by the rule

\[ h(s)_{i_0 \ldots i_ p} = s_{i i_0 \ldots i_ p} \]

In other words, $h : \prod _{i_0} \mathcal{F}(U_{i_0}) \to \mathcal{F}(U)$ is projection onto the factor $\mathcal{F}(U_ i) = \mathcal{F}(U)$ and in general the map $h$ equals the projection onto the factors $\mathcal{F}(U_{i i_1 \ldots i_{p + 1}}) = \mathcal{F}(U_{i_1 \ldots i_{p + 1}})$. We compute

\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 i_0 \ldots i_ p}\\ & = \sum \nolimits _{j = 0}^ p (-1)^ j s_{i 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 i_0 \ldots \hat i_ j \ldots i_ p} \\ & = s_{i_0 \ldots i_ p} \end{align*}

This proves the identity map is homotopic to zero as desired. $\square$

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