The Stacks project

Lemma 20.23.7. 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 alternating Čech complex

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

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 $0$. Similarly, for any total ordering on $I$ the extended ordered Čech complex

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

is homotopy equivalent to $0$.

Second proof. Since the alternating and ordered Čech complexes are isomorphic it suffices to prove this for the ordered one. We will use standard notation: a cochain $s$ of degree $p$ in the extended ordered Čech complex has the form $s = (s_{i_0 \ldots i_ p})$ where $s_{i_0 \ldots i_ p}$ is in $\mathcal{F}(U_{i_0 \ldots i_ p})$ and $i_0 < \ldots < i_ p$. With this notation we have

\[ d(x)_{i_0 \ldots i_{p + 1}} = \sum \nolimits _ j (-1)^ j x_{i_0 \ldots \hat i_ j \ldots i_ p} \]

Fix an index $i \in I$ with $U = U_ i$. As homotopy we use the maps

\[ h : \text{cochains of degree }p + 1 \to \text{cochains of degree }p \]

given by the rule

\[ h(s)_{i_0 \ldots i_ p} = 0 \text{ if } i \in \{ i_0, \ldots , i_ p\} \text{ and } h(s)_{i_0 \ldots i_ p} = (-1)^ j s_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p} \text{ if not} \]

Here $j$ is the unique index such that $i_ j < i < i_{j + 1}$ in the second case; also, since $U = U_ i$ we have the equality

\[ \mathcal{F}(U_{i_0 \ldots i_ p}) = \mathcal{F}(U_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p}) \]

which we can use to make sense of thinking of $(-1)^ j s_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p}$ as an element of $\mathcal{F}(U_{i_0 \ldots i_ p})$. We will show by a computation that $d h + h d$ equals the negative of the identity map which finishes the proof. To do this fix $s$ a cochain of degree $p$ and let $i_0 < \ldots < i_ p$ be elements of $I$.

Case I: $i \in \{ i_0, \ldots , i_ p\} $. Say $i = i_ t$. Then we have $h(d(s))_{i_0 \ldots i_ p} = 0$. On the other hand we have

\[ d(h(s))_{i_0 \ldots i_ p} = \sum (-1)^ j h(s)_{i_0 \ldots \hat i_ j \ldots i_ p} = (-1)^ t h(s)_{i_0 \ldots \hat i \ldots i_ p} = (-1)^ t (-1)^{t - 1} s_{i_0 \ldots i_ p} \]

Thus $(dh + hd)(s)_{i_0 \ldots i_ p} = -s_{i_0 \ldots i_ p}$ as desired.

Case II: $i \not\in \{ i_0, \ldots , i_ p\} $. Let $j$ be such that $i_ j < i < i_{j + 1}$. Then we see that

\begin{align*} h(d(s))_{i_0 \ldots i_ p} & = (-1)^ j d(s)_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p} \\ & = \sum \nolimits _{j' \leq j} (-1)^{j + j'} s_{i_0 \ldots \hat i_{j'} \ldots i_ j i i_{j + 1} \ldots i_ p} - s_{i_0 \ldots i_ p} \\ & + \sum \nolimits _{j' > j} (-1)^{j + j' + 1} s_{i_0 \ldots i_ j i i_{j + 1} \ldots \hat i_{j'} \ldots i_ p} \end{align*}

On the other hand we have

\begin{align*} d(h(s))_{i_0 \ldots i_ p} & = \sum \nolimits _{j'} (-1)^{j'} h(s)_{i_0 \ldots \hat i_{j'} \ldots i_ p} \\ & = \sum \nolimits _{j' \leq j} (-1)^{j' + j - 1} s_{i_0 \ldots \hat i_{j'} \ldots i_ j i i_{j + 1} \ldots i_ p} \\ & + \sum \nolimits _{j' > j} (-1)^{j' + j} s_{i_0 \ldots i_ j i i_{j + 1} \ldots \hat i_{j'} \ldots i_ p} \end{align*}

Adding these up we obtain $(dh + hd)(s)_{i_0 \ldots i_ p} = - s_{i_0 \ldots i_ p}$ as desired. $\square$

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