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Quick question, is the sheaf of locally constant functions flasque?

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    No, since otherwise the whole subject of ordinary cohomology theory would be trivial: The sheaf of locally constant functions is also called "the constant sheaf $\underline{\mathbb{R}}$". If it were flasque, then sheaf cohomology with values in $\underline{\mathbb{R}}$ would be trivial in degress > 0, since flasque sheaves are acyclic. But for nice spaces, for instance manifolds, sheaf cohomology with values in $\underline{\mathbb{R}}$ is the same as ordinary $\mathbb{R}$-valued cohomology (singular or de Rham).2014-05-31

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No, take two disjoint open sets $U$ and $V$ lying in the same connected component $X_0$ of the entire space $X$. Then define a section on $U \cup V$ by the function being $0$ on $U$ and 1 on $V$. Then this section cannot extend to $X$.

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    In the Zariski topology on an irreducible space, I meant.2012-04-18
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This is true if your space is irreducible. Kedlaya gives the example of $U = \mathbf{R} - \{0\}$ sitting inside of $\mathbf{R}$ with the Euclidean topology, showing that a constant sheaf need not be flasque in general.