Local Property of Flasque Sheaves

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Let $X$ be a topological space, and let $\mathscr{F}$ be a sheaf on $X$. Show that if $\mathscr{U}$ is an open cover of $X$ such that the restriction $\mathscr{F}|_U$ is flasque for every $U\in \mathscr{U}$, then $\mathscr{F}$ is flasque.

Note: A sheaf $\mathscr{G}$ on $X$ is flasque if for all open subsets $U\subset X$, the restriction map $\mathscr{G}(X) \to \mathscr{G}(U)$ is surjective.

 

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For this problem you will need to use Zorn's lemma.

 

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Spoiler

Fix an open subset $U_0\subset X$ and an $s\in \mathscr{F}(U_0)$. Let $\Sigma$ be the collection of all pairs $(W,t)$, where $W$ is open in $X$ with $W\supset U_0$ and $t\in \mathscr{F}(W)$ such that $t|_{U_0} = s$. Partially order $\Sigma$ by declaring $(W,t) \le (W',t')$ if $U_0 \subset W \subset W'$ and $t'|_{W} = t$. Then $\Sigma$ is a nonempty inductive set, and by Zorn's lemma there is a maximal element $(V,r)$ of $\Sigma$. Suppose $V \neq X$. There is an $x\in X\setminus V$; let $U\in \mathscr{U}$ be an open neighborhood of $x$. Since $\mathscr{F}|_U$ is flasque, there is an $\alpha \in \mathscr{F}(U)$ such that $\alpha|_{U\cap V} = r|_{U\cap V}$. The sheaf property produces a $\beta\in \mathscr{F}(U\cap V)$ such that $\beta|_V = r$. The pair $(U\cup V, \beta) > (V,r)$ in $\Sigma$, contradicting maximality of $(V,r)$. Hence, $V = X$ and $\mathscr{F}$ is flasque.