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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