The Stacks project

Lemma 20.12.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Any flasque $\mathcal{O}_ X$-module is acyclic for $R\Gamma (X, -)$ as well as $R\Gamma (U, -)$ for any open $U$ of $X$.

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

\[ 0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0 \]

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. $\square$


Comments (1)

Comment #772 by Anfang Zhou on

Typos.

  1. The first word in the 8th line. It should be "section of over ".

  2. The last word in the 7th line from the bottom. It should be "We can find a small open neighbourhood of and ..."

  3. At the beginning of the 4th line from the bottom, it should be .

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