A non-zero sheaf that vanishes on an open covering
$begingroup$
Let $X$ be a topological space, $mathcal F$ a sheaf of abelian groups on $X$ and $X=bigcup_iU_i$ an open covering such that $mathcal F(U_i)=0$ for all $i$. Does this imply $mathcal F=0$?
I think it doesn't, but I haven't managed to come up with a counterexample.
abstract-algebra algebraic-geometry sheaf-theory
asked Dec 24 '18 at 13:39
Layer CakeLayer Cake
263111
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space, $mathcal F$ a sheaf of abelian groups on $X$ and $X=bigcup_iU_i$ an open covering such that $mathcal F(U_i)=0$ for all $i$. Does this imply $mathcal F=0$?
I think it doesn't, but I haven't managed to come up with a counterexample.
abstract-algebra algebraic-geometry sheaf-theory
asked Dec 24 '18 at 13:39
Layer CakeLayer Cake
263111
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space, $mathcal F$ a sheaf of abelian groups on $X$ and $X=bigcup_iU_i$ an open covering such that $mathcal F(U_i)=0$ for all $i$. Does this imply $mathcal F=0$?
I think it doesn't, but I haven't managed to come up with a counterexample.
abstract-algebra algebraic-geometry sheaf-theory
asked Dec 24 '18 at 13:39
Layer CakeLayer Cake
263111
$endgroup$
Let $X$ be a topological space, $mathcal F$ a sheaf of abelian groups on $X$ and $X=bigcup_iU_i$ an open covering such that $mathcal F(U_i)=0$ for all $i$. Does this imply $mathcal F=0$?
I think it doesn't, but I haven't managed to come up with a counterexample.
abstract-algebra algebraic-geometry sheaf-theory
abstract-algebra algebraic-geometry sheaf-theory
asked Dec 24 '18 at 13:39
Layer CakeLayer Cake
263111
asked Dec 24 '18 at 13:39
Layer CakeLayer Cake
263111
asked Dec 24 '18 at 13:39
Layer CakeLayer Cake
263111
asked Dec 24 '18 at 13:39
Layer CakeLayer Cake
263111
asked Dec 24 '18 at 13:39
Layer CakeLayer Cake
263111
263111
add a comment |
add a comment |
1 Answer
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$begingroup$
No. Suppose $X = Bbb P^1,$ and let $mathcal{F} = mathcal{O}(-1).$ $mathcal{F}$ is nontrivial, as $mathcal{F}(Bbb A^1)neq 0$ for any $Bbb A^1subseteqBbb P^1,$ but $mathcal{F}(X) = {0}.$ $X$ is of course a one-element open cover of itself.
More generally, take any nontrivial sheaf which has no nonzero global sections.
answered Dec 24 '18 at 13:45
StahlStahl
16.7k43455
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1 Answer
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1 Answer
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$begingroup$
No. Suppose $X = Bbb P^1,$ and let $mathcal{F} = mathcal{O}(-1).$ $mathcal{F}$ is nontrivial, as $mathcal{F}(Bbb A^1)neq 0$ for any $Bbb A^1subseteqBbb P^1,$ but $mathcal{F}(X) = {0}.$ $X$ is of course a one-element open cover of itself.
More generally, take any nontrivial sheaf which has no nonzero global sections.
answered Dec 24 '18 at 13:45
StahlStahl
16.7k43455
$endgroup$
add a comment |
$begingroup$
No. Suppose $X = Bbb P^1,$ and let $mathcal{F} = mathcal{O}(-1).$ $mathcal{F}$ is nontrivial, as $mathcal{F}(Bbb A^1)neq 0$ for any $Bbb A^1subseteqBbb P^1,$ but $mathcal{F}(X) = {0}.$ $X$ is of course a one-element open cover of itself.
More generally, take any nontrivial sheaf which has no nonzero global sections.
answered Dec 24 '18 at 13:45
StahlStahl
16.7k43455
$endgroup$
add a comment |
$begingroup$
No. Suppose $X = Bbb P^1,$ and let $mathcal{F} = mathcal{O}(-1).$ $mathcal{F}$ is nontrivial, as $mathcal{F}(Bbb A^1)neq 0$ for any $Bbb A^1subseteqBbb P^1,$ but $mathcal{F}(X) = {0}.$ $X$ is of course a one-element open cover of itself.
More generally, take any nontrivial sheaf which has no nonzero global sections.
answered Dec 24 '18 at 13:45
StahlStahl
16.7k43455
$endgroup$
No. Suppose $X = Bbb P^1,$ and let $mathcal{F} = mathcal{O}(-1).$ $mathcal{F}$ is nontrivial, as $mathcal{F}(Bbb A^1)neq 0$ for any $Bbb A^1subseteqBbb P^1,$ but $mathcal{F}(X) = {0}.$ $X$ is of course a one-element open cover of itself.
More generally, take any nontrivial sheaf which has no nonzero global sections.
answered Dec 24 '18 at 13:45
StahlStahl
16.7k43455
answered Dec 24 '18 at 13:45
StahlStahl
16.7k43455
answered Dec 24 '18 at 13:45
StahlStahl
16.7k43455
answered Dec 24 '18 at 13:45
StahlStahl
16.7k43455
16.7k43455
add a comment |
add a comment |
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