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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 sequence 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 sequence 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}

    Comments (3)

    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."

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