The Stacks project

Proof. This is a reformulation of Lemma 20.8.1. $\square$

Proof. We will prove this using Derived Categories, Lemma 13.15.6. Since every injective module is flasque we see that we can embed every $\mathcal{O}_ X$-module into a flasque module, see Injectives, Lemma 19.4.1. Thus it suffices to show that given a short exact sequence

with $\mathcal{F}$, $\mathcal{G}$ flasque, then $\mathcal{H}$ is flasque and the sequence remains short exact after taking sections on any open of $X$. In fact, the second statement implies the first. Thus, let $U \subset X$ be an open subspace. Let $s \in \mathcal{H}(U)$. We will show that we can lift $s$ to a section of $\mathcal{G}$ over $U$. To do this consider the set $T$ of pairs $(V, t)$ where $V \subset U$ is open and $t \in \mathcal{G}(V)$ is a section mapping to $s|_ V$ in $\mathcal{H}$. We put a partial ordering on $T$ by setting $(V, t) \leq (V', t')$ if and only if $V \subset V'$ and $t'|_ V = t$. If $(V_\alpha , t_\alpha )$, $\alpha \in A$ is a totally ordered subset of $T$, then $V = \bigcup V_\alpha $ is open and there is a unique section $t \in \mathcal{G}(V)$ restricting to $t_\alpha $ over $V_\alpha $ by the sheaf condition on $\mathcal{G}$. Thus by Zorn's lemma there exists a maximal element $(V, t)$ in $T$. We will show that $V = U$ thereby finishing the proof. Namely, pick any $x \in U$. We can find a small open neighbourhood $W \subset U$ of $x$ and $t' \in \mathcal{G}(W)$ mapping to $s|_ W$ in $\mathcal{H}$. Then $t'|_{W \cap V} - t|_{W \cap V}$ maps to zero in $\mathcal{H}$, hence comes from some section $r' \in \mathcal{F}(W \cap V)$. Using that $\mathcal{F}$ is flasque we find a section $r \in \mathcal{F}(W)$ restricting to $r'$ over $W \cap V$. Modifying $t'$ by the image of $r$ we may assume that $t$ and $t'$ restrict to the same section over $W \cap V$. By the sheaf condition of $\mathcal{G}$ we can find a section $\tilde t$ of $\mathcal{G}$ over $W \cup V$ restricting to $t$ and $t'$. By maximality of $(V, t)$ we see that $V \cup W = V$. Thus $x \in V$ and we are done. $\square$

Proof. The presheaves $\underline{H}^ q(\mathcal{F})$ used in the statement of Lemma 20.11.5 are zero by Lemma 20.12.3. Hence $\check{H}^ p(U, \mathcal{F}) = H^ p(U, \mathcal{F}) = 0$ by Lemma 20.12.3 again. $\square$

Proof. Immediate from Lemma 20.7.3 and Lemma 20.12.3. $\square$

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

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

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

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

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

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

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

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

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 20.12.4. $\square$