Lemma 20.13.6. Let $X$ be a topological space. Let $\mathcal{F}$ be an abelian sheaf on $X$. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. Assume the restriction mappings $\mathcal{F}(U) \to \mathcal{F}(U')$ are surjective for $U'$ an arbitrary union of opens of the form $U_{i_0 \ldots i_ p}$. Then $\check{H}^ p(\mathcal{U}, \mathcal{F})$ vanishes for $p > 0$.

Proof. Let $Y$ be the set of nonempty subsets of $I$. We will use the letters $A, B, C, \ldots$ to denote elements of $Y$, i.e., nonempty subsets of $I$. For a finite nonempty subset $J \subset I$ let

$V_ J = \{ A \in Y \mid J \subset A\}$

This means that $V_{\{ i\} } = \{ A \in Y \mid i \in A\}$ and $V_ J = \bigcap _{j \in J} V_{\{ j\} }$. Then $V_ J \subset V_ K$ if and only if $J \supset K$. There is a unique topology on $Y$ such that the collection of subsets $V_ J$ is a basis for the topology on $Y$. Any open is of the form

$V = \bigcup \nolimits _{t \in T} V_{J_ t}$

for some family of finite subsets $J_ t$. If $J_ t \subset J_{t'}$ then we may remove $J_{t'}$ from the family without changing $V$. Thus we may assume there are no inclusions among the $J_ t$. In this case the minimal elements of $V$ are the sets $A = J_ t$. Hence we can read off the family $(J_ t)_{t \in T}$ from the open $V$.

We can completely understand open coverings in $Y$. First, because the elements $A \in Y$ are nonempty subsets of $I$ we have

$Y = \bigcup \nolimits _{i \in I} V_{\{ i\} }$

To understand other coverings, let $V$ be as above and let $V_ s \subset Y$ be an open corresponding to the family $(J_{s, t})_{t \in T_ s}$. Then

$V = \bigcup \nolimits _{s \in S} V_ s$

if and only if for each $t \in T$ there exists an $s \in S$ and $t_ s \in T_ s$ such that $J_ t = J_{s, t_ s}$. Namely, as the family $(J_ t)_{t \in T}$ is minimal, the minimal element $A = J_ t$ has to be in $V_ s$ for some $s$, hence $A \in V_{J_{t_ s}}$ for some $t_ s \in T_ s$. But since $A$ is also minimal in $V_ s$ we conclude that $J_{t_ s} = J_ t$.

Next we map the set of opens of $Y$ to opens of $X$. Namely, we send $Y$ to $U$, we use the rule

$V_ J \mapsto U_ J = \bigcap \nolimits _{i \in J} U_ i$

on the opens $V_ J$, and we extend it to arbitrary opens $V$ by the rule

$V = \bigcup \nolimits _{t \in T} V_{J_ t} \mapsto \bigcup \nolimits _{t \in T} U_{J_ t}$

The classification of open coverings of $Y$ given above shows that this rule transforms open coverings into open coverings. Thus we obtain an abelian sheaf $\mathcal{G}$ on $Y$ by setting $\mathcal{G}(Y) = \mathcal{F}(U)$ and for $V = \bigcup \nolimits _{t \in T} V_{J_ t}$ setting

$\mathcal{G}(V) = \mathcal{F}\left(\bigcup \nolimits _{t \in T} U_{J_ t}\right)$

and using the restriction maps of $\mathcal{F}$.

With these preliminaries out of the way we can prove our lemma as follows. We have an open covering $\mathcal{V} : Y = \bigcup _{i \in I} V_{\{ i\} }$ of $Y$. By construction we have an equality

$\check{C}^\bullet (\mathcal{V}, \mathcal{G}) = \check{C}^\bullet (\mathcal{U}, \mathcal{F})$

of Čech complexes. Since the sheaf $\mathcal{G}$ is flasque on $Y$ (by our assumption on $\mathcal{F}$ in the statement of the lemma) the vanishing follows from Lemma 20.13.4. $\square$

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