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
d(x)_{i_0 \ldots i_{p + 1}} = \sum \nolimits _ j (-1)^ j x_{i_0 \ldots \hat i_ j \ldots i_{p + 1}}
As homotopy we use the maps
h : \text{cochains of degree }p + 2 \to \text{cochains of degree }p + 1
given by the rule
h(x)_{i_0 \ldots i_ p} = 0 \text{ if } i \in \{ i_0, \ldots , i_ p\} \text{ and } h(x)_{i_0 \ldots i_ p} = (-1)^ j x_{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 f_ i is a unit we have the equality
M_{f_{i_0} \ldots f_{i_ p}} = M_{f_{i_0} \ldots f_{i_ j} f_ i f_{i_{j + 1}} \ldots f_{i_ p}}
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
d(h(x))_{i_0 \ldots i_ p} = \sum (-1)^ j h(x)_{i_0 \ldots \hat i_ j \ldots i_ p} = (-1)^ t h(x)_{i_0 \ldots \hat i \ldots i_ p} = (-1)^ t (-1)^{t - 1} x_{i_0 \ldots i_ p}
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
\begin{align*} h(d(x))_{i_0 \ldots i_ p} & = (-1)^ j d(x)_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p} \\ & = \sum \nolimits _{j' \leq j} (-1)^{j + j'} x_{i_0 \ldots \hat i_{j'} \ldots i_ j i i_{j + 1} \ldots i_ p} - x_{i_0 \ldots i_ p} \\ & + \sum \nolimits _{j' > j} (-1)^{j + j' + 1} x_{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(x))_{i_0 \ldots i_ p} & = \sum \nolimits _{j'} (-1)^{j'} h(x)_{i_0 \ldots \hat i_{j'} \ldots i_ p} \\ & = \sum \nolimits _{j' \leq j} (-1)^{j' + j - 1} x_{i_0 \ldots \hat i_{j'} \ldots i_ j i i_{j + 1} \ldots i_ p} \\ & + \sum \nolimits _{j' > j} (-1)^{j' + j} x_{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)(x)_{i_0 \ldots i_ p} = - x_{i_0 \ldots i_ p} as desired. \square
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