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Let $S$ be a sheaf over $X$ and $r$ an element in $S_x$ for some $x$ in $X$. Must there exist a section $s$ in $S(X)$ such that such that $s$ equals $r$ when mapped to $S_x$ by the canonical map?

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@erwin: I seem to vaguely remember an example of a sheaf with $S(X)$ trivial but with nontrivial stalks; perhaps it was a sheaf of functions over the complex plane, where there is no nonzero function with the desired properties over the entire plane, but there are plenty of locally defined functions that have the right property, so the stalks are nontrivial? I could be misremembering, though, and my old Alg. Geom. homeworks are at home. –  Arturo Magidin Nov 12 '10 at 21:22

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No: take the sheaf of holomorphic functions $S=\mathcal O$ on $X=\mathbb C$. Then the stalk at $x=1$ of $1/z$ does not come from $\mathcal O(\mathbb C)$.

@Arturo: no you are not misremembering. Here is an example. Take $X=\mathbb P_n (\mathbb C)$ and $S=\mathcal O (-1)$= tautological bundle. Then every stalk $S_x$ is isomorphic with $\mathcal O_x$ so is not zero but $S(\mathbb P_n (\mathbb C))=0$. (By the GAGA principle of Serre you can choose to think in holomorphic way or algebraic way in this example)

@Robin Chapman: Yes, there are sheaves $S$ with all morphisms $S(X) \to S_x$ surjective, that are not flabby. For example, take $X=\mathbb R$ and $S=C$ = sheaf of continuous functions. Then for $x\in \mathbb R$ and $f_x \in C_x$ take a bump function $\phi$ equal to $1$ in vicinity of $x$ and support smaller than the domain of definition of $f$. Then $\phi f$ (extended by zero) is a global function in $C(\mathbb R)$ and its germ at $x$ is equal to $f_x$. But of course $C$ is not flabby.

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@evganiamerkulova: Thanks! –  Arturo Magidin Nov 13 '10 at 0:39
    
Nice examples, especially the one of non-flabby sheaf. Thanks. –  user1119 Nov 14 '10 at 10:00

This is only true in pathological cases. Neither in complex analysis (see above) nor in algebraic geometry: If we take the structure sheaf $\mathcal{O}_{Spec(A)}$ of an affine scheme, then the question is whether the localzation maps $A \to A_{\mathfrak{p}}$ are surjective, which is almost never true.

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But it is surjective... so in a way the question is the wrong question :) –  Mariano Suárez-Alvarez Nov 13 '10 at 15:17
    
It is an epimorphism, but not surjective. –  Martin Brandenburg Nov 14 '10 at 8:06

This raises another question. If the restriction maps from $S(X)$ to $S(U)$ are surjective for all open sets $U$ (a sheaf with this property is called flasque or flabby) then certainly the maps from $S(X)$ to each stalk $S_x$ are surjective. So are there sheaves $S$ with the property that each map $S(X)\to S_x$ is surjective but which are not flabby? I expect there are, but cannot see an example immediately.

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Yes, you are correct: exists not flabby such sheaves. I have added example for you in my answer. –  evgeniamerkulova Nov 13 '10 at 14:05

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