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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). – Ingo Blechschmidt May 31 '14 at 17:06
up vote 6 down vote accepted

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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That's so weird you know. Take a topological space $X$, let $S$ be the Borel sigma-algebra generated by the topology, a measurable function can always be extended to a measurable function outside its original domain, so if we have a locally constant function defined on a subset $A$, we can extend it to a function with value 0 outside of $A$, under this transformation, the locally constant function would stay locally constant no? Am I missing something here? – Mario Carrasco Oct 16 '11 at 0:17
I'm not sure I know exactly what you're saying but I think the extension will be measurable but not locally constant. In the counterexample I gave you can extend that function to all of $X$ in many ways and it will be measurable but no longer locally constant (indeed it will no longer be continuous). – Eric O. Korman Oct 16 '11 at 0:46
Thanks Eric, now check this out: Let $X$ and $Y$ be finite-dimensional topological vector spaces. Let $F$ be the sheaf of $Y$-valued locally constant functions. Let $S_1$ and $S_2$ be their Borel sigma-algebras respectivelly. Now since every measurable function $f$ defined on a measurable subset $A$ of a measurable space can be extended to a measurable function $f'$ on $X\A$ by defining it to be 0 (hence linear) on $X\A$, this means $f'$ is continuous as linear maps between finite-dimensional TVSs are continuous, so the sheaf of locally constant functions on $X$ is flasque? in this case no? – Mario Carrasco Oct 21 '11 at 17:10
Also worth noting: in the Zariski topology, generally open sets are dense, you can't find disjoint opens, and the sheaf of constant functions is flasque. – ziggurism Apr 18 '12 at 13:19
In the Zariski topology on an irreducible space, I meant. – ziggurism Apr 18 '12 at 15:56

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.

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Thank you Dylan – Mario Carrasco Oct 21 '11 at 17:05

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