Constant Presheaf not necessarily a sheaf. Proof?
$begingroup$
Let $mathcal{X}$ be a topological space and $mathcal{F}$ the constant presheaf, that assigns to each open set $mathcal{U}$ of $mathcal{X}$ the set $A$. The restriction map is the identity $A rightarrow A$. I want to show that this presheaf is not necessarily a sheaf and I am trying to understand the argument in this wikipedia article: http://en.wikipedia.org/wiki/Constant_sheaf.
In particular, let $emptyset = cup_i mathcal{U}_i, , mathcal{U}_i=emptyset$, be a covering of the empty set. Take two sections over $emptyset$, i.e. take $s,s' in mathcal{F}(emptyset)=A$. The wikipedia article says that the restrictions of $s,s'$ to any $mathcal{U}_i$ must be equal. I don't understand this. For example, since $mathcal{U}_i subset emptyset$, then we have a restriction map $mathcal{F}(emptyset) rightarrow mathcal{F}(mathcal{U}_i)$ which sends $s$ to $s$ and $s'$ to $s'$. So why $s=s'$?
Something else, possibly related to what I am missing: since the elements of $A$ are abstract, we can not really say anything about $s|_{mathcal{U}_i}$, right?
Thanks.
algebraic-geometry
edited Dec 11 '18 at 23:05
dantopa
6,46942243
asked Sep 13 '12 at 19:14
ManosManos
13.9k33084
$endgroup$
add a comment |
$begingroup$
Let $mathcal{X}$ be a topological space and $mathcal{F}$ the constant presheaf, that assigns to each open set $mathcal{U}$ of $mathcal{X}$ the set $A$. The restriction map is the identity $A rightarrow A$. I want to show that this presheaf is not necessarily a sheaf and I am trying to understand the argument in this wikipedia article: http://en.wikipedia.org/wiki/Constant_sheaf.
In particular, let $emptyset = cup_i mathcal{U}_i, , mathcal{U}_i=emptyset$, be a covering of the empty set. Take two sections over $emptyset$, i.e. take $s,s' in mathcal{F}(emptyset)=A$. The wikipedia article says that the restrictions of $s,s'$ to any $mathcal{U}_i$ must be equal. I don't understand this. For example, since $mathcal{U}_i subset emptyset$, then we have a restriction map $mathcal{F}(emptyset) rightarrow mathcal{F}(mathcal{U}_i)$ which sends $s$ to $s$ and $s'$ to $s'$. So why $s=s'$?
Something else, possibly related to what I am missing: since the elements of $A$ are abstract, we can not really say anything about $s|_{mathcal{U}_i}$, right?
Thanks.
algebraic-geometry
edited Dec 11 '18 at 23:05
dantopa
6,46942243
asked Sep 13 '12 at 19:14
ManosManos
13.9k33084
$endgroup$
add a comment |
$begingroup$
Let $mathcal{X}$ be a topological space and $mathcal{F}$ the constant presheaf, that assigns to each open set $mathcal{U}$ of $mathcal{X}$ the set $A$. The restriction map is the identity $A rightarrow A$. I want to show that this presheaf is not necessarily a sheaf and I am trying to understand the argument in this wikipedia article: http://en.wikipedia.org/wiki/Constant_sheaf.
In particular, let $emptyset = cup_i mathcal{U}_i, , mathcal{U}_i=emptyset$, be a covering of the empty set. Take two sections over $emptyset$, i.e. take $s,s' in mathcal{F}(emptyset)=A$. The wikipedia article says that the restrictions of $s,s'$ to any $mathcal{U}_i$ must be equal. I don't understand this. For example, since $mathcal{U}_i subset emptyset$, then we have a restriction map $mathcal{F}(emptyset) rightarrow mathcal{F}(mathcal{U}_i)$ which sends $s$ to $s$ and $s'$ to $s'$. So why $s=s'$?
Something else, possibly related to what I am missing: since the elements of $A$ are abstract, we can not really say anything about $s|_{mathcal{U}_i}$, right?
Thanks.
algebraic-geometry
edited Dec 11 '18 at 23:05
dantopa
6,46942243
asked Sep 13 '12 at 19:14
ManosManos
13.9k33084
$endgroup$
Let $mathcal{X}$ be a topological space and $mathcal{F}$ the constant presheaf, that assigns to each open set $mathcal{U}$ of $mathcal{X}$ the set $A$. The restriction map is the identity $A rightarrow A$. I want to show that this presheaf is not necessarily a sheaf and I am trying to understand the argument in this wikipedia article: http://en.wikipedia.org/wiki/Constant_sheaf.
In particular, let $emptyset = cup_i mathcal{U}_i, , mathcal{U}_i=emptyset$, be a covering of the empty set. Take two sections over $emptyset$, i.e. take $s,s' in mathcal{F}(emptyset)=A$. The wikipedia article says that the restrictions of $s,s'$ to any $mathcal{U}_i$ must be equal. I don't understand this. For example, since $mathcal{U}_i subset emptyset$, then we have a restriction map $mathcal{F}(emptyset) rightarrow mathcal{F}(mathcal{U}_i)$ which sends $s$ to $s$ and $s'$ to $s'$. So why $s=s'$?
Something else, possibly related to what I am missing: since the elements of $A$ are abstract, we can not really say anything about $s|_{mathcal{U}_i}$, right?
Thanks.
algebraic-geometry
algebraic-geometry
edited Dec 11 '18 at 23:05
dantopa
6,46942243
asked Sep 13 '12 at 19:14
ManosManos
13.9k33084
edited Dec 11 '18 at 23:05
dantopa
6,46942243
asked Sep 13 '12 at 19:14
ManosManos
13.9k33084
edited Dec 11 '18 at 23:05
dantopa
6,46942243
edited Dec 11 '18 at 23:05
dantopa
6,46942243
edited Dec 11 '18 at 23:05
dantopa
6,46942243
6,46942243
asked Sep 13 '12 at 19:14
ManosManos
13.9k33084
asked Sep 13 '12 at 19:14
ManosManos
13.9k33084
asked Sep 13 '12 at 19:14
ManosManos
13.9k33084
13.9k33084
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The argument in the Wikipedia article states that you should take the empty covering of the empty set, i.e. your indexing set (where $i$ comes from) is the empty set. The empty union is the empty set, so it is a covering.
Then you pick any $s neq s^prime in A$ and consider them to be sections over the empty set ($A$ needs to have at least two elements). The condition that they agree after restricting is an empty one (because there is no set to restrict to), so by the local identity axiom, you would need $s=s^prime$, a contradiction.
answered Sep 13 '12 at 19:34
Alexander KörschgenAlexander Körschgen
47626
$endgroup$
$begingroup$
What do you mean that "the condition is an empty one"? Why an empty condition is true?
$endgroup$
– Manos
Sep 13 '12 at 20:11
$begingroup$
Ok, call the indexing set $I = emptyset$. Formally, the condition of the local identity axiom says: for all $i in I$, if $s$ and $s^prime$ get mapped to the same element in $mathcal{F}(U_i)$, then they should agree. But since $I$ is empty, there is nothing to check.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:14
$begingroup$
I agree that there is nothing to check. But then why does that mean that $s,s'$ get mapped to the same element in $mathcal{F}(U_i)$?
$endgroup$
– Manos
Sep 13 '12 at 20:23
1
$begingroup$
Let's denote the statement "$s$ and $s^prime$ agree when restricted to $i$ by $phi(i)$. Then the local identity axiom (for fixed $s$ and $s^prime$ says: $(forall i in I: phi(i)) => (s=s^prime)$. But the condition $(forall i in I: phi(i))$ is a true statement, regardless of what $s$ and $s^prime$ are. It is a general fact that a statement that is quantified over the empty set is always true.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:54
1
$begingroup$
To make this even more precise, let's think about what the statement $(forall i in I: phi(i))$ means. In first-order logic, one would have to write $(forall i) ((i in I) => phi(i))$, and $i in I$ will always evaluate to false, because $I=emptyset$, so $((i in I) => phi(i))$ will always be true (one may deduce anything from a wrong statement ;)). Hope that helps to clarify things.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:55
|
show 3 more comments
$begingroup$
I think you are taken the wrong definition of the constant presheaf. Since the presheaf $mathcal{F}$ is a functor on the category of open sets of $X$, the constant presheaf means that $mathcal{F}$ is a constant function, i.e., $mathcal{F}(U)= mathcal{F}(V)$ for all $U,V$ open in $X$.
answered Sep 18 '12 at 21:09
CC_AzusaCC_Azusa
1
$endgroup$
$begingroup$
The definition of constant presheaf in the question is correct. Specifically, the OP is considering the constant presheaf where $mathcal(U)=A$ for some fixed set $A$ for all $U$ opin in $mathcal X$.
$endgroup$
– Alex Becker
Sep 19 '12 at 18:44
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The argument in the Wikipedia article states that you should take the empty covering of the empty set, i.e. your indexing set (where $i$ comes from) is the empty set. The empty union is the empty set, so it is a covering.
Then you pick any $s neq s^prime in A$ and consider them to be sections over the empty set ($A$ needs to have at least two elements). The condition that they agree after restricting is an empty one (because there is no set to restrict to), so by the local identity axiom, you would need $s=s^prime$, a contradiction.
answered Sep 13 '12 at 19:34
Alexander KörschgenAlexander Körschgen
47626
$endgroup$
$begingroup$
What do you mean that "the condition is an empty one"? Why an empty condition is true?
$endgroup$
– Manos
Sep 13 '12 at 20:11
$begingroup$
Ok, call the indexing set $I = emptyset$. Formally, the condition of the local identity axiom says: for all $i in I$, if $s$ and $s^prime$ get mapped to the same element in $mathcal{F}(U_i)$, then they should agree. But since $I$ is empty, there is nothing to check.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:14
$begingroup$
I agree that there is nothing to check. But then why does that mean that $s,s'$ get mapped to the same element in $mathcal{F}(U_i)$?
$endgroup$
– Manos
Sep 13 '12 at 20:23
1
$begingroup$
Let's denote the statement "$s$ and $s^prime$ agree when restricted to $i$ by $phi(i)$. Then the local identity axiom (for fixed $s$ and $s^prime$ says: $(forall i in I: phi(i)) => (s=s^prime)$. But the condition $(forall i in I: phi(i))$ is a true statement, regardless of what $s$ and $s^prime$ are. It is a general fact that a statement that is quantified over the empty set is always true.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:54
1
$begingroup$
To make this even more precise, let's think about what the statement $(forall i in I: phi(i))$ means. In first-order logic, one would have to write $(forall i) ((i in I) => phi(i))$, and $i in I$ will always evaluate to false, because $I=emptyset$, so $((i in I) => phi(i))$ will always be true (one may deduce anything from a wrong statement ;)). Hope that helps to clarify things.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:55
|
show 3 more comments
$begingroup$
The argument in the Wikipedia article states that you should take the empty covering of the empty set, i.e. your indexing set (where $i$ comes from) is the empty set. The empty union is the empty set, so it is a covering.
Then you pick any $s neq s^prime in A$ and consider them to be sections over the empty set ($A$ needs to have at least two elements). The condition that they agree after restricting is an empty one (because there is no set to restrict to), so by the local identity axiom, you would need $s=s^prime$, a contradiction.
answered Sep 13 '12 at 19:34
Alexander KörschgenAlexander Körschgen
47626
$endgroup$
$begingroup$
What do you mean that "the condition is an empty one"? Why an empty condition is true?
$endgroup$
– Manos
Sep 13 '12 at 20:11
$begingroup$
Ok, call the indexing set $I = emptyset$. Formally, the condition of the local identity axiom says: for all $i in I$, if $s$ and $s^prime$ get mapped to the same element in $mathcal{F}(U_i)$, then they should agree. But since $I$ is empty, there is nothing to check.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:14
$begingroup$
I agree that there is nothing to check. But then why does that mean that $s,s'$ get mapped to the same element in $mathcal{F}(U_i)$?
$endgroup$
– Manos
Sep 13 '12 at 20:23
1
$begingroup$
Let's denote the statement "$s$ and $s^prime$ agree when restricted to $i$ by $phi(i)$. Then the local identity axiom (for fixed $s$ and $s^prime$ says: $(forall i in I: phi(i)) => (s=s^prime)$. But the condition $(forall i in I: phi(i))$ is a true statement, regardless of what $s$ and $s^prime$ are. It is a general fact that a statement that is quantified over the empty set is always true.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:54
1
$begingroup$
To make this even more precise, let's think about what the statement $(forall i in I: phi(i))$ means. In first-order logic, one would have to write $(forall i) ((i in I) => phi(i))$, and $i in I$ will always evaluate to false, because $I=emptyset$, so $((i in I) => phi(i))$ will always be true (one may deduce anything from a wrong statement ;)). Hope that helps to clarify things.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:55
|
show 3 more comments
$begingroup$
The argument in the Wikipedia article states that you should take the empty covering of the empty set, i.e. your indexing set (where $i$ comes from) is the empty set. The empty union is the empty set, so it is a covering.
Then you pick any $s neq s^prime in A$ and consider them to be sections over the empty set ($A$ needs to have at least two elements). The condition that they agree after restricting is an empty one (because there is no set to restrict to), so by the local identity axiom, you would need $s=s^prime$, a contradiction.
answered Sep 13 '12 at 19:34
Alexander KörschgenAlexander Körschgen
47626
$endgroup$
The argument in the Wikipedia article states that you should take the empty covering of the empty set, i.e. your indexing set (where $i$ comes from) is the empty set. The empty union is the empty set, so it is a covering.
Then you pick any $s neq s^prime in A$ and consider them to be sections over the empty set ($A$ needs to have at least two elements). The condition that they agree after restricting is an empty one (because there is no set to restrict to), so by the local identity axiom, you would need $s=s^prime$, a contradiction.
answered Sep 13 '12 at 19:34
Alexander KörschgenAlexander Körschgen
47626
answered Sep 13 '12 at 19:34
Alexander KörschgenAlexander Körschgen
47626
answered Sep 13 '12 at 19:34
Alexander KörschgenAlexander Körschgen
47626
answered Sep 13 '12 at 19:34
Alexander KörschgenAlexander Körschgen
47626
47626
$begingroup$
What do you mean that "the condition is an empty one"? Why an empty condition is true?
$endgroup$
– Manos
Sep 13 '12 at 20:11
$begingroup$
Ok, call the indexing set $I = emptyset$. Formally, the condition of the local identity axiom says: for all $i in I$, if $s$ and $s^prime$ get mapped to the same element in $mathcal{F}(U_i)$, then they should agree. But since $I$ is empty, there is nothing to check.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:14
$begingroup$
I agree that there is nothing to check. But then why does that mean that $s,s'$ get mapped to the same element in $mathcal{F}(U_i)$?
$endgroup$
– Manos
Sep 13 '12 at 20:23
1
$begingroup$
Let's denote the statement "$s$ and $s^prime$ agree when restricted to $i$ by $phi(i)$. Then the local identity axiom (for fixed $s$ and $s^prime$ says: $(forall i in I: phi(i)) => (s=s^prime)$. But the condition $(forall i in I: phi(i))$ is a true statement, regardless of what $s$ and $s^prime$ are. It is a general fact that a statement that is quantified over the empty set is always true.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:54
1
$begingroup$
To make this even more precise, let's think about what the statement $(forall i in I: phi(i))$ means. In first-order logic, one would have to write $(forall i) ((i in I) => phi(i))$, and $i in I$ will always evaluate to false, because $I=emptyset$, so $((i in I) => phi(i))$ will always be true (one may deduce anything from a wrong statement ;)). Hope that helps to clarify things.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:55
|
show 3 more comments
$begingroup$
What do you mean that "the condition is an empty one"? Why an empty condition is true?
$endgroup$
– Manos
Sep 13 '12 at 20:11
$begingroup$
Ok, call the indexing set $I = emptyset$. Formally, the condition of the local identity axiom says: for all $i in I$, if $s$ and $s^prime$ get mapped to the same element in $mathcal{F}(U_i)$, then they should agree. But since $I$ is empty, there is nothing to check.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:14
$begingroup$
I agree that there is nothing to check. But then why does that mean that $s,s'$ get mapped to the same element in $mathcal{F}(U_i)$?
$endgroup$
– Manos
Sep 13 '12 at 20:23
1
$begingroup$
Let's denote the statement "$s$ and $s^prime$ agree when restricted to $i$ by $phi(i)$. Then the local identity axiom (for fixed $s$ and $s^prime$ says: $(forall i in I: phi(i)) => (s=s^prime)$. But the condition $(forall i in I: phi(i))$ is a true statement, regardless of what $s$ and $s^prime$ are. It is a general fact that a statement that is quantified over the empty set is always true.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:54
1
$begingroup$
To make this even more precise, let's think about what the statement $(forall i in I: phi(i))$ means. In first-order logic, one would have to write $(forall i) ((i in I) => phi(i))$, and $i in I$ will always evaluate to false, because $I=emptyset$, so $((i in I) => phi(i))$ will always be true (one may deduce anything from a wrong statement ;)). Hope that helps to clarify things.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:55
$begingroup$
What do you mean that "the condition is an empty one"? Why an empty condition is true?
$endgroup$
– Manos
Sep 13 '12 at 20:11
$begingroup$
What do you mean that "the condition is an empty one"? Why an empty condition is true?
$endgroup$
– Manos
Sep 13 '12 at 20:11
$begingroup$
Ok, call the indexing set $I = emptyset$. Formally, the condition of the local identity axiom says: for all $i in I$, if $s$ and $s^prime$ get mapped to the same element in $mathcal{F}(U_i)$, then they should agree. But since $I$ is empty, there is nothing to check.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:14
$begingroup$
Ok, call the indexing set $I = emptyset$. Formally, the condition of the local identity axiom says: for all $i in I$, if $s$ and $s^prime$ get mapped to the same element in $mathcal{F}(U_i)$, then they should agree. But since $I$ is empty, there is nothing to check.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:14
$begingroup$
I agree that there is nothing to check. But then why does that mean that $s,s'$ get mapped to the same element in $mathcal{F}(U_i)$?
$endgroup$
– Manos
Sep 13 '12 at 20:23
$begingroup$
I agree that there is nothing to check. But then why does that mean that $s,s'$ get mapped to the same element in $mathcal{F}(U_i)$?
$endgroup$
– Manos
Sep 13 '12 at 20:23
1
1
$begingroup$
Let's denote the statement "$s$ and $s^prime$ agree when restricted to $i$ by $phi(i)$. Then the local identity axiom (for fixed $s$ and $s^prime$ says: $(forall i in I: phi(i)) => (s=s^prime)$. But the condition $(forall i in I: phi(i))$ is a true statement, regardless of what $s$ and $s^prime$ are. It is a general fact that a statement that is quantified over the empty set is always true.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:54
$begingroup$
Let's denote the statement "$s$ and $s^prime$ agree when restricted to $i$ by $phi(i)$. Then the local identity axiom (for fixed $s$ and $s^prime$ says: $(forall i in I: phi(i)) => (s=s^prime)$. But the condition $(forall i in I: phi(i))$ is a true statement, regardless of what $s$ and $s^prime$ are. It is a general fact that a statement that is quantified over the empty set is always true.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:54
1
1
$begingroup$
To make this even more precise, let's think about what the statement $(forall i in I: phi(i))$ means. In first-order logic, one would have to write $(forall i) ((i in I) => phi(i))$, and $i in I$ will always evaluate to false, because $I=emptyset$, so $((i in I) => phi(i))$ will always be true (one may deduce anything from a wrong statement ;)). Hope that helps to clarify things.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:55
$begingroup$
To make this even more precise, let's think about what the statement $(forall i in I: phi(i))$ means. In first-order logic, one would have to write $(forall i) ((i in I) => phi(i))$, and $i in I$ will always evaluate to false, because $I=emptyset$, so $((i in I) => phi(i))$ will always be true (one may deduce anything from a wrong statement ;)). Hope that helps to clarify things.
$endgroup$
– Alexander Körschgen
Sep 13 '12 at 20:55
|
show 3 more comments
$begingroup$
I think you are taken the wrong definition of the constant presheaf. Since the presheaf $mathcal{F}$ is a functor on the category of open sets of $X$, the constant presheaf means that $mathcal{F}$ is a constant function, i.e., $mathcal{F}(U)= mathcal{F}(V)$ for all $U,V$ open in $X$.
answered Sep 18 '12 at 21:09
CC_AzusaCC_Azusa
1
$endgroup$
$begingroup$
The definition of constant presheaf in the question is correct. Specifically, the OP is considering the constant presheaf where $mathcal(U)=A$ for some fixed set $A$ for all $U$ opin in $mathcal X$.
$endgroup$
– Alex Becker
Sep 19 '12 at 18:44
add a comment |
$begingroup$
I think you are taken the wrong definition of the constant presheaf. Since the presheaf $mathcal{F}$ is a functor on the category of open sets of $X$, the constant presheaf means that $mathcal{F}$ is a constant function, i.e., $mathcal{F}(U)= mathcal{F}(V)$ for all $U,V$ open in $X$.
answered Sep 18 '12 at 21:09
CC_AzusaCC_Azusa
1
$endgroup$
$begingroup$
The definition of constant presheaf in the question is correct. Specifically, the OP is considering the constant presheaf where $mathcal(U)=A$ for some fixed set $A$ for all $U$ opin in $mathcal X$.
$endgroup$
– Alex Becker
Sep 19 '12 at 18:44
add a comment |
$begingroup$
I think you are taken the wrong definition of the constant presheaf. Since the presheaf $mathcal{F}$ is a functor on the category of open sets of $X$, the constant presheaf means that $mathcal{F}$ is a constant function, i.e., $mathcal{F}(U)= mathcal{F}(V)$ for all $U,V$ open in $X$.
answered Sep 18 '12 at 21:09
CC_AzusaCC_Azusa
1
$endgroup$
I think you are taken the wrong definition of the constant presheaf. Since the presheaf $mathcal{F}$ is a functor on the category of open sets of $X$, the constant presheaf means that $mathcal{F}$ is a constant function, i.e., $mathcal{F}(U)= mathcal{F}(V)$ for all $U,V$ open in $X$.
answered Sep 18 '12 at 21:09
CC_AzusaCC_Azusa
1
answered Sep 18 '12 at 21:09
CC_AzusaCC_Azusa
1
answered Sep 18 '12 at 21:09
CC_AzusaCC_Azusa
1
answered Sep 18 '12 at 21:09
CC_AzusaCC_Azusa
1
1
$begingroup$
The definition of constant presheaf in the question is correct. Specifically, the OP is considering the constant presheaf where $mathcal(U)=A$ for some fixed set $A$ for all $U$ opin in $mathcal X$.
$endgroup$
– Alex Becker
Sep 19 '12 at 18:44
add a comment |
$begingroup$
The definition of constant presheaf in the question is correct. Specifically, the OP is considering the constant presheaf where $mathcal(U)=A$ for some fixed set $A$ for all $U$ opin in $mathcal X$.
$endgroup$
– Alex Becker
Sep 19 '12 at 18:44
$begingroup$
The definition of constant presheaf in the question is correct. Specifically, the OP is considering the constant presheaf where $mathcal(U)=A$ for some fixed set $A$ for all $U$ opin in $mathcal X$.
$endgroup$
– Alex Becker
Sep 19 '12 at 18:44
$begingroup$
The definition of constant presheaf in the question is correct. Specifically, the OP is considering the constant presheaf where $mathcal(U)=A$ for some fixed set $A$ for all $U$ opin in $mathcal X$.
$endgroup$
– Alex Becker
Sep 19 '12 at 18:44
add a comment |
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