Lemma 15.29.4. Let $R$ be a ring. Let $f_1, \ldots , f_ r \in R$. Let $M$ be an $R$-module. If there exists an $i \in \{ 1, \ldots , r\} $ such that $f_ i$ is a unit, then the extended alternating Čech complex of $M$ is homotopy equivalent to $0$.
Proof. We will use the following notation: a cochain $x$ of degree $p + 1$ in the extended alternating Čech complex of $M$ is $x = (x_{i_0 \ldots i_ p})$ where $x_{i_0 \ldots i_ p}$ is in $M_{f_{i_0} \ldots f_{i_ p}}$. With this notation we have
As homotopy we use the maps
given by the rule
Here $j$ is the unique index such that $i_ j < i < i_{j + 1}$ in the second case; also, since $f_ i$ is a unit we have the equality
which we can use to make sense of thinking of $(-1)^ j x_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p}$ as an element of $M_{f_{i_0} \ldots f_{i_ p}}$. 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 $x$ a cochain of degree $p + 1$ and let $1 \leq i_0 < \ldots < i_ p \leq r$.
Case I: $i \in \{ i_0, \ldots , i_ p\} $. Say $i = i_ t$. Then we have $h(d(x))_{i_0 \ldots i_ p} = 0$. On the other hand we have
Thus $(dh + hd)(x)_{i_0 \ldots i_ p} = -x_{i_0 \ldots i_ p}$ 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
On the other hand we have
Adding these up we obtain $(dh + hd)(x)_{i_0 \ldots i_ p} = - x_{i_0 \ldots i_ p}$ as desired. $\square$
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