Hartshorne Ex. II 1.16 b) Flasque sheaves and exact sequences
$begingroup$
the exercise states that when we have an exact sequence
$0tomathcal{F}'tomathcal{F}tomathcal{F}''to 0$
of sheaves (say of Abelian groups) over a topological space $X$, and when $mathcal{F}'$ is flasque, then for any open set $Usubset X$, the sequence
$0tomathcal{F}'(U)tomathcal{F}(U)tomathcal{F}''(U)to 0$
is again exact.
By a previous exercise it is enough to show surjectivity.
I could also figure out that when we have $sinmathcal{F}''(U)$ and open subsets $V_1,V_2subset U$, such that $s$ on both there is a lift of $s$ (i.e. there is $t_iinmathcal{F}(V_i)$, s.t. the image of $t_i$ in $mathcal{F}''(V_i)$ is equal to the restriction $s|_{V_i}$), then $s$ can be lifted on their union.
It is also clear to me that there is an open cover of $U$ consisting of sets on which $s$ can be lifted. From here on I do not know how to proceed.
I have looked at other solutions and they want to apply Zorns Lemma, but it is not clear to me how this works here. They seem to use that given a chain (w.r.t. inclusion) of open subsets ($U_iota$) on which $s$ can be lifted, then there is a lift of $s$ on $bigcup U_iota$ because $mathcal{F}$ is a sheaf. However, I think this does not work, because we have no reason to assume that all the different lifts are compatible.
Can anybody help out here?
algebraic-geometry sheaf-theory
asked Nov 17 '14 at 12:59
Nikolas KuhnNikolas Kuhn
301110
$endgroup$
add a comment |
$begingroup$
the exercise states that when we have an exact sequence
$0tomathcal{F}'tomathcal{F}tomathcal{F}''to 0$
of sheaves (say of Abelian groups) over a topological space $X$, and when $mathcal{F}'$ is flasque, then for any open set $Usubset X$, the sequence
$0tomathcal{F}'(U)tomathcal{F}(U)tomathcal{F}''(U)to 0$
is again exact.
By a previous exercise it is enough to show surjectivity.
I could also figure out that when we have $sinmathcal{F}''(U)$ and open subsets $V_1,V_2subset U$, such that $s$ on both there is a lift of $s$ (i.e. there is $t_iinmathcal{F}(V_i)$, s.t. the image of $t_i$ in $mathcal{F}''(V_i)$ is equal to the restriction $s|_{V_i}$), then $s$ can be lifted on their union.
It is also clear to me that there is an open cover of $U$ consisting of sets on which $s$ can be lifted. From here on I do not know how to proceed.
I have looked at other solutions and they want to apply Zorns Lemma, but it is not clear to me how this works here. They seem to use that given a chain (w.r.t. inclusion) of open subsets ($U_iota$) on which $s$ can be lifted, then there is a lift of $s$ on $bigcup U_iota$ because $mathcal{F}$ is a sheaf. However, I think this does not work, because we have no reason to assume that all the different lifts are compatible.
Can anybody help out here?
algebraic-geometry sheaf-theory
asked Nov 17 '14 at 12:59
Nikolas KuhnNikolas Kuhn
301110
$endgroup$
4
$begingroup$
Yes it works because in the chain $(U_i,s_i)$ the $s_i$'s are mutually compatible by definition of the partial order relation on the pairs $(U_i,s_i)$.
$endgroup$
– Georges Elencwajg
Nov 17 '14 at 13:08
add a comment |
$begingroup$
the exercise states that when we have an exact sequence
$0tomathcal{F}'tomathcal{F}tomathcal{F}''to 0$
of sheaves (say of Abelian groups) over a topological space $X$, and when $mathcal{F}'$ is flasque, then for any open set $Usubset X$, the sequence
$0tomathcal{F}'(U)tomathcal{F}(U)tomathcal{F}''(U)to 0$
is again exact.
By a previous exercise it is enough to show surjectivity.
I could also figure out that when we have $sinmathcal{F}''(U)$ and open subsets $V_1,V_2subset U$, such that $s$ on both there is a lift of $s$ (i.e. there is $t_iinmathcal{F}(V_i)$, s.t. the image of $t_i$ in $mathcal{F}''(V_i)$ is equal to the restriction $s|_{V_i}$), then $s$ can be lifted on their union.
It is also clear to me that there is an open cover of $U$ consisting of sets on which $s$ can be lifted. From here on I do not know how to proceed.
I have looked at other solutions and they want to apply Zorns Lemma, but it is not clear to me how this works here. They seem to use that given a chain (w.r.t. inclusion) of open subsets ($U_iota$) on which $s$ can be lifted, then there is a lift of $s$ on $bigcup U_iota$ because $mathcal{F}$ is a sheaf. However, I think this does not work, because we have no reason to assume that all the different lifts are compatible.
Can anybody help out here?
algebraic-geometry sheaf-theory
asked Nov 17 '14 at 12:59
Nikolas KuhnNikolas Kuhn
301110
$endgroup$
the exercise states that when we have an exact sequence
$0tomathcal{F}'tomathcal{F}tomathcal{F}''to 0$
of sheaves (say of Abelian groups) over a topological space $X$, and when $mathcal{F}'$ is flasque, then for any open set $Usubset X$, the sequence
$0tomathcal{F}'(U)tomathcal{F}(U)tomathcal{F}''(U)to 0$
is again exact.
By a previous exercise it is enough to show surjectivity.
I could also figure out that when we have $sinmathcal{F}''(U)$ and open subsets $V_1,V_2subset U$, such that $s$ on both there is a lift of $s$ (i.e. there is $t_iinmathcal{F}(V_i)$, s.t. the image of $t_i$ in $mathcal{F}''(V_i)$ is equal to the restriction $s|_{V_i}$), then $s$ can be lifted on their union.
It is also clear to me that there is an open cover of $U$ consisting of sets on which $s$ can be lifted. From here on I do not know how to proceed.
I have looked at other solutions and they want to apply Zorns Lemma, but it is not clear to me how this works here. They seem to use that given a chain (w.r.t. inclusion) of open subsets ($U_iota$) on which $s$ can be lifted, then there is a lift of $s$ on $bigcup U_iota$ because $mathcal{F}$ is a sheaf. However, I think this does not work, because we have no reason to assume that all the different lifts are compatible.
Can anybody help out here?
algebraic-geometry sheaf-theory
algebraic-geometry sheaf-theory
asked Nov 17 '14 at 12:59
Nikolas KuhnNikolas Kuhn
301110
asked Nov 17 '14 at 12:59
Nikolas KuhnNikolas Kuhn
301110
asked Nov 17 '14 at 12:59
Nikolas KuhnNikolas Kuhn
301110
asked Nov 17 '14 at 12:59
Nikolas KuhnNikolas Kuhn
301110
asked Nov 17 '14 at 12:59
Nikolas KuhnNikolas Kuhn
301110
301110
4
$begingroup$
Yes it works because in the chain $(U_i,s_i)$ the $s_i$'s are mutually compatible by definition of the partial order relation on the pairs $(U_i,s_i)$.
$endgroup$
– Georges Elencwajg
Nov 17 '14 at 13:08
add a comment |
4
$begingroup$
Yes it works because in the chain $(U_i,s_i)$ the $s_i$'s are mutually compatible by definition of the partial order relation on the pairs $(U_i,s_i)$.
$endgroup$
– Georges Elencwajg
Nov 17 '14 at 13:08
4
4
$begingroup$
Yes it works because in the chain $(U_i,s_i)$ the $s_i$'s are mutually compatible by definition of the partial order relation on the pairs $(U_i,s_i)$.
$endgroup$
– Georges Elencwajg
Nov 17 '14 at 13:08
$begingroup$
Yes it works because in the chain $(U_i,s_i)$ the $s_i$'s are mutually compatible by definition of the partial order relation on the pairs $(U_i,s_i)$.
$endgroup$
– Georges Elencwajg
Nov 17 '14 at 13:08
add a comment |
1 Answer
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yeah, it works. you can always substitute the original lift of a chain to get a compatible lift, then you can define the maximal element of the chain.
answered Mar 9 '15 at 10:19
zhang wuzhang wu
11
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$begingroup$
Would you provide a bit more detail, please? :)
$endgroup$
– Shaun
Mar 9 '15 at 10:56
add a comment |
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$begingroup$
yeah, it works. you can always substitute the original lift of a chain to get a compatible lift, then you can define the maximal element of the chain.
answered Mar 9 '15 at 10:19
zhang wuzhang wu
11
$endgroup$
$begingroup$
Would you provide a bit more detail, please? :)
$endgroup$
– Shaun
Mar 9 '15 at 10:56
add a comment |
$begingroup$
yeah, it works. you can always substitute the original lift of a chain to get a compatible lift, then you can define the maximal element of the chain.
answered Mar 9 '15 at 10:19
zhang wuzhang wu
11
$endgroup$
$begingroup$
Would you provide a bit more detail, please? :)
$endgroup$
– Shaun
Mar 9 '15 at 10:56
add a comment |
$begingroup$
yeah, it works. you can always substitute the original lift of a chain to get a compatible lift, then you can define the maximal element of the chain.
answered Mar 9 '15 at 10:19
zhang wuzhang wu
11
$endgroup$
yeah, it works. you can always substitute the original lift of a chain to get a compatible lift, then you can define the maximal element of the chain.
answered Mar 9 '15 at 10:19
zhang wuzhang wu
11
answered Mar 9 '15 at 10:19
zhang wuzhang wu
11
answered Mar 9 '15 at 10:19
zhang wuzhang wu
11
answered Mar 9 '15 at 10:19
zhang wuzhang wu
11
11
$begingroup$
Would you provide a bit more detail, please? :)
$endgroup$
– Shaun
Mar 9 '15 at 10:56
add a comment |
$begingroup$
Would you provide a bit more detail, please? :)
$endgroup$
– Shaun
Mar 9 '15 at 10:56
$begingroup$
Would you provide a bit more detail, please? :)
$endgroup$
– Shaun
Mar 9 '15 at 10:56
$begingroup$
Would you provide a bit more detail, please? :)
$endgroup$
– Shaun
Mar 9 '15 at 10:56
add a comment |
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Yes it works because in the chain $(U_i,s_i)$ the $s_i$'s are mutually compatible by definition of the partial order relation on the pairs $(U_i,s_i)$.
$endgroup$
– Georges Elencwajg
Nov 17 '14 at 13:08