Exercise on monomorphism of sheaves
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The following is exercise 6, chapter 0 from Johnstone's Topos Theory:
Suppose $Frightarrowtail G$ a monomorphism of sheaves in $Shv(X)$ [$X$ being a topological space], and $sigmain G(U)$ for some open $Usubseteq X$. Prove that there is a unique largest open set $Vsubseteq U$ such that $sigma_V$ is in the image of $F(V)$.
Unicity is easy, because given a family of such subsets $V_i_iin I$ then the union $bigcup_I V_i$ would share the same property. However, I can't find a reason for the existence to hold. I tried with the following counterexample:
$X=*$ the singleton space
$Fin Shv(X)$ the constant sheaf sending all opens to the set $a$
$Gin Shv(X)$ the constant sheaf sending all opens to $a,sigma$
Then $Frightarrowtail G$, but a $V$ as above does not exist for any choice of $Usubseteq X$ and taking $sigmain G(U)$.
sheaf-theory
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The following is exercise 6, chapter 0 from Johnstone's Topos Theory:
Suppose $Frightarrowtail G$ a monomorphism of sheaves in $Shv(X)$ [$X$ being a topological space], and $sigmain G(U)$ for some open $Usubseteq X$. Prove that there is a unique largest open set $Vsubseteq U$ such that $sigma_V$ is in the image of $F(V)$.
Unicity is easy, because given a family of such subsets $V_i_iin I$ then the union $bigcup_I V_i$ would share the same property. However, I can't find a reason for the existence to hold. I tried with the following counterexample:
$X=*$ the singleton space
$Fin Shv(X)$ the constant sheaf sending all opens to the set $a$
$Gin Shv(X)$ the constant sheaf sending all opens to $a,sigma$
Then $Frightarrowtail G$, but a $V$ as above does not exist for any choice of $Usubseteq X$ and taking $sigmain G(U)$.
sheaf-theory
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
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The following is exercise 6, chapter 0 from Johnstone's Topos Theory:
Suppose $Frightarrowtail G$ a monomorphism of sheaves in $Shv(X)$ [$X$ being a topological space], and $sigmain G(U)$ for some open $Usubseteq X$. Prove that there is a unique largest open set $Vsubseteq U$ such that $sigma_V$ is in the image of $F(V)$.
Unicity is easy, because given a family of such subsets $V_i_iin I$ then the union $bigcup_I V_i$ would share the same property. However, I can't find a reason for the existence to hold. I tried with the following counterexample:
$X=*$ the singleton space
$Fin Shv(X)$ the constant sheaf sending all opens to the set $a$
$Gin Shv(X)$ the constant sheaf sending all opens to $a,sigma$
Then $Frightarrowtail G$, but a $V$ as above does not exist for any choice of $Usubseteq X$ and taking $sigmain G(U)$.
sheaf-theory
The following is exercise 6, chapter 0 from Johnstone's Topos Theory:
Suppose $Frightarrowtail G$ a monomorphism of sheaves in $Shv(X)$ [$X$ being a topological space], and $sigmain G(U)$ for some open $Usubseteq X$. Prove that there is a unique largest open set $Vsubseteq U$ such that $sigma_V$ is in the image of $F(V)$.
Unicity is easy, because given a family of such subsets $V_i_iin I$ then the union $bigcup_I V_i$ would share the same property. However, I can't find a reason for the existence to hold. I tried with the following counterexample:
$X=*$ the singleton space
$Fin Shv(X)$ the constant sheaf sending all opens to the set $a$
$Gin Shv(X)$ the constant sheaf sending all opens to $a,sigma$
Then $Frightarrowtail G$, but a $V$ as above does not exist for any choice of $Usubseteq X$ and taking $sigmain G(U)$.
sheaf-theory
edited Aug 16 at 10:06
asked Aug 16 at 9:38
TheMadcapLaughs
341211
341211
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1 Answer
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If $A$ is a sheaf, then $A(varnothing)$ is a one-point set.
So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = a, sigma $ and $G(varnothing) = * $.
Going back to the generic problem, $sigma|_varnothing$ is the unique element of $G(varnothing)$, which is the image of the unique element of $F(varnothing)$, so there's your existence proof.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $A$ is a sheaf, then $A(varnothing)$ is a one-point set.
So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = a, sigma $ and $G(varnothing) = * $.
Going back to the generic problem, $sigma|_varnothing$ is the unique element of $G(varnothing)$, which is the image of the unique element of $F(varnothing)$, so there's your existence proof.
add a comment |Â
up vote
1
down vote
accepted
If $A$ is a sheaf, then $A(varnothing)$ is a one-point set.
So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = a, sigma $ and $G(varnothing) = * $.
Going back to the generic problem, $sigma|_varnothing$ is the unique element of $G(varnothing)$, which is the image of the unique element of $F(varnothing)$, so there's your existence proof.
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $A$ is a sheaf, then $A(varnothing)$ is a one-point set.
So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = a, sigma $ and $G(varnothing) = * $.
Going back to the generic problem, $sigma|_varnothing$ is the unique element of $G(varnothing)$, which is the image of the unique element of $F(varnothing)$, so there's your existence proof.
If $A$ is a sheaf, then $A(varnothing)$ is a one-point set.
So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = a, sigma $ and $G(varnothing) = * $.
Going back to the generic problem, $sigma|_varnothing$ is the unique element of $G(varnothing)$, which is the image of the unique element of $F(varnothing)$, so there's your existence proof.
answered Aug 16 at 9:46
Hurkyl
108k9113254
108k9113254
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