Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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 at 17:06

2 Answers 2

up vote 5 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 $X$.

share|improve this answer
    
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. –  Joe Hannon Apr 18 '12 at 13:19
    
In the Zariski topology on an irreducible space, I meant. –  Joe Hannon 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.

share|improve this answer
    
Thank you Dylan –  Mario Carrasco Oct 21 '11 at 17:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.