
## 20.13 Flasque sheaves

Here is the definition.

Definition 20.13.1. Let $X$ be a topological space. We say a presheaf of sets $\mathcal{F}$ is flasque or flabby if for every $U \subset V$ open in $X$ the restriction map $\mathcal{F}(V) \to \mathcal{F}(U)$ is surjective.

We will use this terminology also for abelian sheaves and sheaves of modules if $X$ is a ringed space. Clearly it suffices to assume the restriction maps $\mathcal{F}(X) \to \mathcal{F}(U)$ is surjective for every open $U \subset X$.

Lemma 20.13.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Then any injective $\mathcal{O}_ X$-module is flasque.

Proof. This is a reformulation of Lemma 20.9.1. $\square$

Lemma 20.13.3. 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 13.16.6. 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 $V \cap W = V$. Thus $x \in V$ and we are done. $\square$

The following lemma does not hold for flasque presheaves.

Lemma 20.13.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. Let $\mathcal{U} : U = \bigcup U_ i$ be an open covering. If $\mathcal{F}$ is flasque, then $\check{H}^ p(\mathcal{U}, \mathcal{F}) = 0$ for $p > 0$.

Proof. The presheaves $\underline{H}^ q(\mathcal{F})$ used in the statement of Lemma 20.12.5 are zero by Lemma 20.13.3. Hence $\check{H}^ p(U, \mathcal{F}) = H^ p(U, \mathcal{F}) = 0$ by Lemma 20.13.3 again. $\square$

Lemma 20.13.5. Let $(X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. If $\mathcal{F}$ is flasque, then $R^ pf_*\mathcal{F} = 0$ for $p > 0$.

Proof. Immediate from Lemma 20.8.3 and Lemma 20.13.3. $\square$

The following lemma can be proved by an elementary induction argument for finite coverings, compare with the discussion of Čech cohomology in [FOAG].

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. $\square$

## Comments (4)

Comment #2058 by Reimundo on

In the proof of 20.13.3 there are some small typos: "We can find a small open neighborhood $W \subset U$ of $x$ and $t'\in H(W)$..." should be $t' \in G(W)$

similarly below, should be restricting to $r' \in W \cap V$ and not $W' \cap V$

Comment #2995 by Sandor Kovacs on

Another typo: in the proof of 20.13.3 "We will show that we can lift s to a sequence of G over U." should be "We will show that we can lift s to a section of G over U."

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09SV. Beware of the difference between the letter 'O' and the digit '0'.