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. $\square$

Comment #6238 by Owen on

same thing as comment #6205: rather than

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