Flasque Constant Sheaf

Question

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?

Answer 1

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!).