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It is easy to show that if $X$ is an irreducible topological space, then the constant sheaf $\mathbb{Z}$ is flasque. Is the converse true?

user127542
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1 Answers1

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Flasqueness is a local condition, but irreducibility is not. Indeed, let $X = X_1 = X_2$ be an irreducible topological space and let $Y = X_1 \amalg X_2$. For any sheaf $\mathscr{F}$ on $Y$ and any open $V \subseteq Y$, $$\Gamma (V, \mathscr{F}) \cong \Gamma (V \cap X_1, \mathscr{F}) \times \Gamma (V \cap X_2, \mathscr{F})$$ so if the restrictions $\mathscr{F} |_{X_1}$ and $\mathscr{F} |_{X_2}$ are flasque, then so is $\mathscr{F}$. In particular, $\mathbb{Z}$ is flasque on $Y$, but $Y$ is not irreducible (because it is not connected!).

Zhen Lin
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  • Thanks for your answer! If you add the additional condition that X is connected, is it true? – user127542 Feb 10 '14 at 13:45
  • If you know that $X$ is connected and locally connected, then you can deduce that every open subset of $X$ is connected (or empty). I don't see why this implies $X$ is irreducible, however. – Zhen Lin Feb 10 '14 at 16:14
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    Is this correct? If every open subset of the connected space $X$ is connected and $\mathbb{Z}$ is flasque, but $X$ is not irreducible, then there exist disjoint nonempty open subsets $U,V \subseteq X$. Then the restriction map $\mathbb{Z}(X)=\mathbb{Z} \to \mathbb{Z}(U \cup V)=\mathbb{Z} \oplus \mathbb{Z}$ cannot be surjective. So every two nonempty open subsets intersect after all. – user4601931 Feb 12 '14 at 00:50
  • @ZhenLin Is it true that, if for an open cover ${U_i}$ we have that $\mathscr{F}\mid_{U_i}$ is flasque for each $i$, then $\mathscr{F}$ is Flasque? – Grobber Feb 11 '18 at 10:16