I think the following is true. (Sheaves are of abelian groups) Let $\mathscr{F}$ be a sheaf and $\mathscr{F'}$ is a subsheaf of $\mathscr{F}$ such that $\mathscr{F'}$ is flasque. (The restriction maps of $\mathscr{F'}$ are surjective. ) Show that the presheaf $U \mapsto \mathscr{F}(U)/\mathscr{F'}(U)$ is a sheaf.

Please prove it directly without using the exactness properties.(I want to use this to prove one of them) I am having problems proving sheaf property (II).

What I have tried is this (for showing sheaf property II):

Let $V_i$, $i \in I$, be an open covering for an open subset $U$ of $X$. We have to show that if there are $x_i \in \mathscr{F}(V_i)$ for every $i \in I$ such that $x_i \mid_{V_i \cap V_j} - x_j\mid_{V_i \cap V_j} \in \mathscr{F'}({V_i \cap V_j})$ for every $i,j \in I$. Then we have to show that there exists an $s \in \mathscr{F}(U)$ such that $s\mid _{V_i} - x_i \in \mathscr{F}'(V_i)$ for every $i \in I$.

I will be done if I can prove that there exists $t_i \in \mathscr{F}'(V_i)$ such that $x_i \mid_{V_i \cap V_j} - x_j\mid_{V_i \cap V_j} = t_i \mid_{V_i \cap V_j} - t_j \mid_{V_i \cap V_j} $ for all $i,j$. I'm unable to prove this.


Let us write $\mathcal{G} = \mathcal{F}/\mathcal{F}^{'}$ an remember that it is a presheaf so far.

We want to show that it is a sheaf. Take two intersecting open sets $U_{1}, U_{2}$ and take sections $s_{i} \in \mathcal{G}(U_{i}), i =1,2$ which agree on $\mathcal{G}(U_{1} \cap U_{2})$. Now we can lift these sections to $\widetilde{s_{i}} \in \mathcal{F}(U_{i})$, but they will in general not agree on $U_{1}\cap U_{2}$. Their difference restricted to $U_{1}\cap U_{2}$, $\widetilde{s}_{12} =\widetilde{s_{1}} - \widetilde{s_{2}} |_{$U_{1}\cap U_{2}} \in \mathcal{F}^{'}(U_{1} \cap U_{2})$. Using flasque property you can find a section $t \in \mathcal{F}^{'}(U_{1} \cup U_{2})$ that restricts to $\widetilde{s}_{12}$. Now the sections $\widetilde{s_{i}} -t \in \mathcal{F}(U_{i})$ patch nicely to a section of $\mathcal{F}(U_{1} \cup U_{2})$ and its image in $\mathcal{G}(U_{1} \cup U_{2})$ is the section extending $s_{i} \in \mathcal{G}(U_{i})$. The uniqueness follows from the fact that everything is unique mod $\mathcal{F}^{'}(U_{1} \cup U_{2})$.

  • $\begingroup$ How do you show that it is compatible with any open covering of $U$ ? $\endgroup$ – Sami Jul 9 '13 at 19:07
  • $\begingroup$ You will need to use some commutative diagrams for that.If you consider an open cover $\{U_{i} \}_{i \in I}$ of $U$, then you want to show that $$0 \rightarrow\mathcal{G}(U) \rightarrow \prod_{i} \mathcal{G}(U_{i}) \rightarrow \prod_{ij}\mathcal{G}_{U_{ij}}$$ is exact. But the arrows are defined component wise. So to show exactness you argue at each component $U_{i}, U_{j}, U_{i}\cap U_{j}$ and that is the argument above (non intersecting sets cause no problem). If you have seen Mayer Vietoris in topology this is exactly the same idea. $\endgroup$ – DBS Jul 9 '13 at 19:29
  • $\begingroup$ Thanks for your reply. I will try to look into your comment in detail. I've edited my post adding my work. Could you please tell me how to construct the $t_i$s in my post? $\endgroup$ – Sami Jul 9 '13 at 19:44
  • 1
    $\begingroup$ Because in your notation $(x_{i} -x_{j}) |_{V_{i} \cap V_{j}} \in \mathcal{F}^{'}$ and $\mathcal{F}^{'}$ is flasque so we can lift (x_{i} -x_{j}) |_{V_{i} \cap V_{j}} it to $t \in\mathcal{F}^{'}{V_{i} \cup V_{j}}$. Then $t_{i}$ = restriction of $t$. $\endgroup$ – DBS Jul 9 '13 at 19:52
  • $\begingroup$ I realized my formatting is sub-par above. I hope it is not too confusing. $\endgroup$ – DBS Jul 9 '13 at 20:14

Let $\mathscr{F}$ be a sheaf and $\mathscr{F'}$ is a subsheaf of $\mathscr{F}$ such that $\mathscr{F'}$ is flasque. Show that the presheaf $U \mapsto \mathscr{F}(U)/\mathscr{F'}(U)$ is a sheaf.

Pf. Let $\{U_i\}_{i\in I}$ cover $U$ and let $s_i\in\mathscr F(U_i)/\mathscr F(U_i)$ such that $\left.s_i\right|_{U_i\cap U_j}=\left.s_j\right|_{U_i\cap U_j}$. This is equivalent to the existence of $t_i\in\mathscr F(U_i)$ and $q_{ij}\in\mathscr F'(U_i\cap U_j)$ such that $\left.t_i\right|_{U_i\cap U_j}-\left.t_j\right|_{U_i\cap U_j}=q_{ij}$ for all $i,j\in I$.

Since $\mathscr F'$ is flasque, there exists a non-canonical distinguished section $\gamma\in\mathscr F(Q)$ over $Q:=\bigcup_{i,j\in I}U_i\cap U_j$ such that $\left.\gamma\right|_{U_i\cap U_j}/\mathscr F'(U_i\cap U_j) = \left.s_i\right|_{U_i\cap U_j}\forall i$, and sections $\ell_{ij}\in\mathscr F'(U_i\cap U_j)$ such that $\left.t_i\right|_{U_i\cap U_j}+\ell_{ij}=\left.\gamma\right|_{U_i\cap U_j}$.

Notice that, for each fixed $i$, the sections $\{\ell_{ij}\}_{j\in I}$ collate to a section $\ell_i\in\mathscr F'(Q\cap U_i)$. Replace $t_i$ by $t_i+\tilde\ell_i$ where $\tilde\ell_i$ is a section in $\mathscr F'(U_i)$ restricting to $\ell_i$ on $Q\cap U_i$, whose existence is guaranteed as $\mathscr F'$ is flasque. Then $\left\{t_i+\tilde \ell_i\right\}_{i\in I}$ collate to an element $t\in\mathscr F(U)$ such that $\overline t\in\mathscr F(U)/\mathscr F'(U)$ restricts to $s_i$ on $U_i.\quad\square$

A note about the distinguished section $\gamma$. Why do we need $\mathscr F'$ to be flasque to assure the existence of $\gamma$? To see this, fix some $i\in I$ and suppose the restriction of $\gamma$ to $U_i\cap U_j$ is specified. It may be that $U_i\cap U_j\cap U_k\ne\emptyset$, in which case the restriction of $\gamma$ to $U_i\cap U_k$ is already specified on an open subset. But now, starting with $t_i$ (or $t_k$), there is an element $\eta\in\mathscr F'(U_i\cap U_j\cap U_k)$ such that $\left.t_i\right|_{U_i\cap U_j\cap U_k}+\eta=\left.\gamma\right|_{U_i\cap U_j\cap U_k}$. We require that $\mathscr F'$ be flasque so that we are assured of an element $\tilde\eta\in\mathscr F'(U_i\cap U_k)$ restricting to $\eta$; we then define $\left.\gamma\right|_{U_i\cap U_k}:=\left.t_i\right|_{U_i\cap U_k}+\tilde\eta$ if $j$ is the sole index such that $U_i\cap U_k\cap U_j\ne\emptyset$, or, more generally, we can ensure that $\gamma$ is defined on $U_i\cap U_k$ in a manner compatible with any previous specified restrictions on an open cover of $U_i\cap U_k$ by sets of the form $U_i\cap U_k\cap U_j$, again since $\mathscr F'$ is flasque (in such a case, $\tilde\eta$ will be already specified on a possibly-partial open cover of $U_i\cap U_k$). (Note that we won't need to look over intersections finer than the intersection of three different $U_i$, as if $U_i\cap U_j$ is covered by sets of the form $U_i\cap U_j\cap_{s\in S}U_s$ with $|S|<|\mathbf N|$, it is covered by sets of the form $\{U_i\cap U_j\cap U_s\}_{s\in S}$.) This is enough to establish the existence of the distinguished section $\gamma$ in the case $\mathscr F'$ is flasque.


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