# The Stacks Project

## Tag 09SV

### 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. $\square$

The code snippet corresponding to this tag is a part of the file cohomology.tex and is located in lines 1825–2032 (see updates for more information).

\section{Flasque sheaves}
\label{section-flasque}

\noindent
Here is the definition.

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

\noindent
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 $U \subset X$.

\begin{lemma}
\label{lemma-injective-flasque}
Let $(X, \mathcal{O}_X)$ be a ringed space.
Then any injective $\mathcal{O}_X$-module is flasque.
\end{lemma}

\begin{proof}
This is a reformulation of Lemma \ref{lemma-injective-restriction-surjective}.
\end{proof}

\begin{lemma}
\label{lemma-flasque-acyclic}
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 $X$.
\end{lemma}

\begin{proof}
We will prove this using
Derived Categories, Lemma \ref{derived-lemma-subcategory-right-acyclics}.
Since every injective module is flasque we see that we can embed
every $\mathcal{O}_X$-module into a flasque module, see
Injectives, Lemma \ref{injectives-lemma-abelian-sheaves-space}.
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.
\end{proof}

\noindent
The following lemma does not hold for flasque presheaves.

\begin{lemma}
\label{lemma-flasque-acyclic-cech}
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 $p > 0$.
\end{lemma}

\begin{proof}
The presheaves $\underline{H}^q(\mathcal{F})$ used in the statement
of Lemma \ref{lemma-cech-spectral-sequence} are zero by
Lemma \ref{lemma-flasque-acyclic}.
Hence $\check{H}^p(U, \mathcal{F}) = H^p(U, \mathcal{F}) = 0$
by Lemma \ref{lemma-flasque-acyclic} again.
\end{proof}

\begin{lemma}
\label{lemma-flasque-acyclic-pushforward}
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 $p > 0$.
\end{lemma}

\begin{proof}
Immediate from
Lemma \ref{lemma-describe-higher-direct-images} and
Lemma \ref{lemma-flasque-acyclic}.
\end{proof}

\noindent
The following lemma can be proved by an elementary induction
argument for finite coverings, compare with the discussion
of {\v C}ech cohomology in \cite{FOAG}.

\begin{lemma}
\label{lemma-vanishing-ravi}
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 $p > 0$.
\end{lemma}

\begin{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 $V$.

\medskip\noindent
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 $J_{t_s} = J_t$.

\medskip\noindent
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 $\mathcal{F}$.

\medskip\noindent
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 {\v C}ech 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 \ref{lemma-flasque-acyclic-cech}.
\end{proof}

Comment #2058 by Reimundo on June 9, 2016 a 7:04 pm UTC

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 #2088 by Johan (site) on June 16, 2016 a 1:48 pm UTC

Thanks, fixed here.

Comment #2995 by Sandor Kovacs on November 12, 2017 a 6:56 pm UTC

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."

Comment #3118 by Johan (site) on February 1, 2018 a 12:58 am UTC

@#2995 Thanks, fixed here.

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