Exact sequence of sheaves from the exactness of sections
$begingroup$
Let $X$ be a topological space, and $mathcal{F}_1$, $mathcal{F}_2$ and $mathcal{F}_3$ be sheaves on $X$. Suppose for all $U$ open in $X$ we have,
$0longrightarrow mathcal{F}_1(U)longrightarrowmathcal{F}_2(U)longrightarrowmathcal{F}_3(U)longrightarrow 0$ is exact, then I think that it is not necessarily true that the sheaf sequence $0longrightarrowmathcal{F}_1longrightarrowmathcal{F}_2longrightarrowmathcal{F}_3longrightarrow 0$ is exact.
But if we take direct limit over all open sets $U$ in the first sequence, since direct limit preserves exactness, we get
$0longrightarrowmathcal{F}_{1_x}longrightarrowmathcal{F}_{2_x}longrightarrowmathcal{F}_{3_x}longrightarrow 0$ is exact for all $x$.
That is we have stalk level exactness, therefore the sheave sequence is exact too. Is this right?
sheaf-theory exact-sequence
asked Nov 5 '14 at 19:27
gradstudentgradstudent
1,303720
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space, and $mathcal{F}_1$, $mathcal{F}_2$ and $mathcal{F}_3$ be sheaves on $X$. Suppose for all $U$ open in $X$ we have,
$0longrightarrow mathcal{F}_1(U)longrightarrowmathcal{F}_2(U)longrightarrowmathcal{F}_3(U)longrightarrow 0$ is exact, then I think that it is not necessarily true that the sheaf sequence $0longrightarrowmathcal{F}_1longrightarrowmathcal{F}_2longrightarrowmathcal{F}_3longrightarrow 0$ is exact.
But if we take direct limit over all open sets $U$ in the first sequence, since direct limit preserves exactness, we get
$0longrightarrowmathcal{F}_{1_x}longrightarrowmathcal{F}_{2_x}longrightarrowmathcal{F}_{3_x}longrightarrow 0$ is exact for all $x$.
That is we have stalk level exactness, therefore the sheave sequence is exact too. Is this right?
sheaf-theory exact-sequence
asked Nov 5 '14 at 19:27
gradstudentgradstudent
1,303720
$endgroup$
$begingroup$
Exactness of the sequences of sections immediately results in exactness of the sequence of sheaves. Compute each $ker$ and $operatorname{im}$ and verify this.
$endgroup$
– Ayman Hourieh
Nov 5 '14 at 19:34
$begingroup$
Ah thanks, it was quite obvious! Sorry for asking a trivial question.
$endgroup$
– gradstudent
Nov 5 '14 at 20:03
add a comment |
$begingroup$
Let $X$ be a topological space, and $mathcal{F}_1$, $mathcal{F}_2$ and $mathcal{F}_3$ be sheaves on $X$. Suppose for all $U$ open in $X$ we have,
$0longrightarrow mathcal{F}_1(U)longrightarrowmathcal{F}_2(U)longrightarrowmathcal{F}_3(U)longrightarrow 0$ is exact, then I think that it is not necessarily true that the sheaf sequence $0longrightarrowmathcal{F}_1longrightarrowmathcal{F}_2longrightarrowmathcal{F}_3longrightarrow 0$ is exact.
But if we take direct limit over all open sets $U$ in the first sequence, since direct limit preserves exactness, we get
$0longrightarrowmathcal{F}_{1_x}longrightarrowmathcal{F}_{2_x}longrightarrowmathcal{F}_{3_x}longrightarrow 0$ is exact for all $x$.
That is we have stalk level exactness, therefore the sheave sequence is exact too. Is this right?
sheaf-theory exact-sequence
asked Nov 5 '14 at 19:27
gradstudentgradstudent
1,303720
$endgroup$
Let $X$ be a topological space, and $mathcal{F}_1$, $mathcal{F}_2$ and $mathcal{F}_3$ be sheaves on $X$. Suppose for all $U$ open in $X$ we have,
$0longrightarrow mathcal{F}_1(U)longrightarrowmathcal{F}_2(U)longrightarrowmathcal{F}_3(U)longrightarrow 0$ is exact, then I think that it is not necessarily true that the sheaf sequence $0longrightarrowmathcal{F}_1longrightarrowmathcal{F}_2longrightarrowmathcal{F}_3longrightarrow 0$ is exact.
But if we take direct limit over all open sets $U$ in the first sequence, since direct limit preserves exactness, we get
$0longrightarrowmathcal{F}_{1_x}longrightarrowmathcal{F}_{2_x}longrightarrowmathcal{F}_{3_x}longrightarrow 0$ is exact for all $x$.
That is we have stalk level exactness, therefore the sheave sequence is exact too. Is this right?
sheaf-theory exact-sequence
sheaf-theory exact-sequence
asked Nov 5 '14 at 19:27
gradstudentgradstudent
1,303720
asked Nov 5 '14 at 19:27
gradstudentgradstudent
1,303720
asked Nov 5 '14 at 19:27
gradstudentgradstudent
1,303720
asked Nov 5 '14 at 19:27
gradstudentgradstudent
1,303720
asked Nov 5 '14 at 19:27
gradstudentgradstudent
1,303720
1,303720
$begingroup$
Exactness of the sequences of sections immediately results in exactness of the sequence of sheaves. Compute each $ker$ and $operatorname{im}$ and verify this.
$endgroup$
– Ayman Hourieh
Nov 5 '14 at 19:34
$begingroup$
Ah thanks, it was quite obvious! Sorry for asking a trivial question.
$endgroup$
– gradstudent
Nov 5 '14 at 20:03
add a comment |
$begingroup$
Exactness of the sequences of sections immediately results in exactness of the sequence of sheaves. Compute each $ker$ and $operatorname{im}$ and verify this.
$endgroup$
– Ayman Hourieh
Nov 5 '14 at 19:34
$begingroup$
Ah thanks, it was quite obvious! Sorry for asking a trivial question.
$endgroup$
– gradstudent
Nov 5 '14 at 20:03
$begingroup$
Exactness of the sequences of sections immediately results in exactness of the sequence of sheaves. Compute each $ker$ and $operatorname{im}$ and verify this.
$endgroup$
– Ayman Hourieh
Nov 5 '14 at 19:34
$begingroup$
Exactness of the sequences of sections immediately results in exactness of the sequence of sheaves. Compute each $ker$ and $operatorname{im}$ and verify this.
$endgroup$
– Ayman Hourieh
Nov 5 '14 at 19:34
$begingroup$
Ah thanks, it was quite obvious! Sorry for asking a trivial question.
$endgroup$
– gradstudent
Nov 5 '14 at 20:03
$begingroup$
Ah thanks, it was quite obvious! Sorry for asking a trivial question.
$endgroup$
– gradstudent
Nov 5 '14 at 20:03
add a comment |
1 Answer
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$begingroup$
If the sequence $0to mathcal{F}_1(U)tomathcal{F}_2(U)tomathcal{F}_3(U)to 0$ is exact for every open $U$ then the sequence of sheaves $0to mathcal{F}_1tomathcal{F}_2tomathcal{F}_3to 0$ is certainly exact. What is not true is the opposite. That is if the sequence of sheaves $0to mathcal{F}_1tomathcal{F}_2tomathcal{F}_3to 0$ is exact then it doesn't follows that for every open $U$ the sequence $0to mathcal{F}_1(U)tomathcal{F}_2(U)tomathcal{F}_3(U)to 0$ is exact. For example $X=mathbb{S}^1$ (a circle), let $C^infty$ be the sheaf of (real valued) infinitely differentiable functions on $X$ and let $underline{mathbb{R}}$ be the subsheaf of $C^infty$ of constant functions (i.e. constant on every connected component), the sequence $0to underline{mathbb{R}}to C^infty xrightarrow{frac{d}{dx}} C^infty to 0$ is exact, however the sequence of global section (for $U=X$) is not. The remedy is sheaf cohomology.
answered Jan 24 at 10:29
ЖекаЖека
835
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add a comment |
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1 Answer
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$begingroup$
If the sequence $0to mathcal{F}_1(U)tomathcal{F}_2(U)tomathcal{F}_3(U)to 0$ is exact for every open $U$ then the sequence of sheaves $0to mathcal{F}_1tomathcal{F}_2tomathcal{F}_3to 0$ is certainly exact. What is not true is the opposite. That is if the sequence of sheaves $0to mathcal{F}_1tomathcal{F}_2tomathcal{F}_3to 0$ is exact then it doesn't follows that for every open $U$ the sequence $0to mathcal{F}_1(U)tomathcal{F}_2(U)tomathcal{F}_3(U)to 0$ is exact. For example $X=mathbb{S}^1$ (a circle), let $C^infty$ be the sheaf of (real valued) infinitely differentiable functions on $X$ and let $underline{mathbb{R}}$ be the subsheaf of $C^infty$ of constant functions (i.e. constant on every connected component), the sequence $0to underline{mathbb{R}}to C^infty xrightarrow{frac{d}{dx}} C^infty to 0$ is exact, however the sequence of global section (for $U=X$) is not. The remedy is sheaf cohomology.
answered Jan 24 at 10:29
ЖекаЖека
835
$endgroup$
add a comment |
$begingroup$
If the sequence $0to mathcal{F}_1(U)tomathcal{F}_2(U)tomathcal{F}_3(U)to 0$ is exact for every open $U$ then the sequence of sheaves $0to mathcal{F}_1tomathcal{F}_2tomathcal{F}_3to 0$ is certainly exact. What is not true is the opposite. That is if the sequence of sheaves $0to mathcal{F}_1tomathcal{F}_2tomathcal{F}_3to 0$ is exact then it doesn't follows that for every open $U$ the sequence $0to mathcal{F}_1(U)tomathcal{F}_2(U)tomathcal{F}_3(U)to 0$ is exact. For example $X=mathbb{S}^1$ (a circle), let $C^infty$ be the sheaf of (real valued) infinitely differentiable functions on $X$ and let $underline{mathbb{R}}$ be the subsheaf of $C^infty$ of constant functions (i.e. constant on every connected component), the sequence $0to underline{mathbb{R}}to C^infty xrightarrow{frac{d}{dx}} C^infty to 0$ is exact, however the sequence of global section (for $U=X$) is not. The remedy is sheaf cohomology.
answered Jan 24 at 10:29
ЖекаЖека
835
$endgroup$
add a comment |
$begingroup$
If the sequence $0to mathcal{F}_1(U)tomathcal{F}_2(U)tomathcal{F}_3(U)to 0$ is exact for every open $U$ then the sequence of sheaves $0to mathcal{F}_1tomathcal{F}_2tomathcal{F}_3to 0$ is certainly exact. What is not true is the opposite. That is if the sequence of sheaves $0to mathcal{F}_1tomathcal{F}_2tomathcal{F}_3to 0$ is exact then it doesn't follows that for every open $U$ the sequence $0to mathcal{F}_1(U)tomathcal{F}_2(U)tomathcal{F}_3(U)to 0$ is exact. For example $X=mathbb{S}^1$ (a circle), let $C^infty$ be the sheaf of (real valued) infinitely differentiable functions on $X$ and let $underline{mathbb{R}}$ be the subsheaf of $C^infty$ of constant functions (i.e. constant on every connected component), the sequence $0to underline{mathbb{R}}to C^infty xrightarrow{frac{d}{dx}} C^infty to 0$ is exact, however the sequence of global section (for $U=X$) is not. The remedy is sheaf cohomology.
answered Jan 24 at 10:29
ЖекаЖека
835
$endgroup$
If the sequence $0to mathcal{F}_1(U)tomathcal{F}_2(U)tomathcal{F}_3(U)to 0$ is exact for every open $U$ then the sequence of sheaves $0to mathcal{F}_1tomathcal{F}_2tomathcal{F}_3to 0$ is certainly exact. What is not true is the opposite. That is if the sequence of sheaves $0to mathcal{F}_1tomathcal{F}_2tomathcal{F}_3to 0$ is exact then it doesn't follows that for every open $U$ the sequence $0to mathcal{F}_1(U)tomathcal{F}_2(U)tomathcal{F}_3(U)to 0$ is exact. For example $X=mathbb{S}^1$ (a circle), let $C^infty$ be the sheaf of (real valued) infinitely differentiable functions on $X$ and let $underline{mathbb{R}}$ be the subsheaf of $C^infty$ of constant functions (i.e. constant on every connected component), the sequence $0to underline{mathbb{R}}to C^infty xrightarrow{frac{d}{dx}} C^infty to 0$ is exact, however the sequence of global section (for $U=X$) is not. The remedy is sheaf cohomology.
answered Jan 24 at 10:29
ЖекаЖека
835
answered Jan 24 at 10:29
ЖекаЖека
835
answered Jan 24 at 10:29
ЖекаЖека
835
answered Jan 24 at 10:29
ЖекаЖека
835
835
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$begingroup$
Exactness of the sequences of sections immediately results in exactness of the sequence of sheaves. Compute each $ker$ and $operatorname{im}$ and verify this.
$endgroup$
– Ayman Hourieh
Nov 5 '14 at 19:34
$begingroup$
Ah thanks, it was quite obvious! Sorry for asking a trivial question.
$endgroup$
– gradstudent
Nov 5 '14 at 20:03