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Showing that the skyscraper sheaf is a sheaf.
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$DeclareMathOperatorresres$Let $X$ be a topological space, $p in X$ a point, $U subset X$ an open subset covered by $bigcup_i in IU_i$, and $S$ a set (or an abelian group). I'm trying to show that the skyscraper sheaf $i_pS$ given by
$$ i_pS(U) = begincases S & textif , p in U \ e & textelse , endcases $$
is indeed a sheaf. Here's what I've tried so far.
First I want to show the gluing axiom, so I take sections $a_i in i_pS(U_i)$ such that $res^U_i_U_i cap U_j a_i = res^U_j_U_i cap U_j a_j$. We need for there to exist a section $a in i_pS(U)$ such that $res^U_U_ia = a_i$ for all $i in I$. But here's my problem - since the sections are elements of the set (or abelian group) $S$, I'm not sure what the restriction of one of these sections actually is (as opposed to the clear nature of a restriction when we talk about, say, the sheaf of differentiable functions).
I tried to do it case-wise, so say $p in U_i$ and $p notin U_j$ (so $p notin U_i cap U_j$) for example. Then $a_j in e$ so $a_j =e$, and by our assumption, $res^U_i_U_i cap U_j a_i = e$ which means that $a_i$ is $e$ on the part of $U_i$ which overlaps with $U_j$. Does this mean that $a_i$ must be $e$ on the whole of $U_i$? I feel not, because then by this logic I think we would end up with $i_pS(U)= e $. But then if the $a_i$ take different values on different parts of $U$, how does one glue them together to construct a valid $a$, and what does this $a$ even look like?
I haven't yet attempted the other axiom for sheaves, but I think I'd run into similar problems with the above reasoning. Thank you for any help.
general-topology algebraic-geometry sheaf-theory
asked Jul 23 at 22:46
mathphys
961415
add a comment |Â
up vote
2
down vote
favorite
$DeclareMathOperatorresres$Let $X$ be a topological space, $p in X$ a point, $U subset X$ an open subset covered by $bigcup_i in IU_i$, and $S$ a set (or an abelian group). I'm trying to show that the skyscraper sheaf $i_pS$ given by
$$ i_pS(U) = begincases S & textif , p in U \ e & textelse , endcases $$
is indeed a sheaf. Here's what I've tried so far.
First I want to show the gluing axiom, so I take sections $a_i in i_pS(U_i)$ such that $res^U_i_U_i cap U_j a_i = res^U_j_U_i cap U_j a_j$. We need for there to exist a section $a in i_pS(U)$ such that $res^U_U_ia = a_i$ for all $i in I$. But here's my problem - since the sections are elements of the set (or abelian group) $S$, I'm not sure what the restriction of one of these sections actually is (as opposed to the clear nature of a restriction when we talk about, say, the sheaf of differentiable functions).
I tried to do it case-wise, so say $p in U_i$ and $p notin U_j$ (so $p notin U_i cap U_j$) for example. Then $a_j in e$ so $a_j =e$, and by our assumption, $res^U_i_U_i cap U_j a_i = e$ which means that $a_i$ is $e$ on the part of $U_i$ which overlaps with $U_j$. Does this mean that $a_i$ must be $e$ on the whole of $U_i$? I feel not, because then by this logic I think we would end up with $i_pS(U)= e $. But then if the $a_i$ take different values on different parts of $U$, how does one glue them together to construct a valid $a$, and what does this $a$ even look like?
I haven't yet attempted the other axiom for sheaves, but I think I'd run into similar problems with the above reasoning. Thank you for any help.
general-topology algebraic-geometry sheaf-theory
asked Jul 23 at 22:46
mathphys
961415
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$DeclareMathOperatorresres$Let $X$ be a topological space, $p in X$ a point, $U subset X$ an open subset covered by $bigcup_i in IU_i$, and $S$ a set (or an abelian group). I'm trying to show that the skyscraper sheaf $i_pS$ given by
$$ i_pS(U) = begincases S & textif , p in U \ e & textelse , endcases $$
is indeed a sheaf. Here's what I've tried so far.
First I want to show the gluing axiom, so I take sections $a_i in i_pS(U_i)$ such that $res^U_i_U_i cap U_j a_i = res^U_j_U_i cap U_j a_j$. We need for there to exist a section $a in i_pS(U)$ such that $res^U_U_ia = a_i$ for all $i in I$. But here's my problem - since the sections are elements of the set (or abelian group) $S$, I'm not sure what the restriction of one of these sections actually is (as opposed to the clear nature of a restriction when we talk about, say, the sheaf of differentiable functions).
I tried to do it case-wise, so say $p in U_i$ and $p notin U_j$ (so $p notin U_i cap U_j$) for example. Then $a_j in e$ so $a_j =e$, and by our assumption, $res^U_i_U_i cap U_j a_i = e$ which means that $a_i$ is $e$ on the part of $U_i$ which overlaps with $U_j$. Does this mean that $a_i$ must be $e$ on the whole of $U_i$? I feel not, because then by this logic I think we would end up with $i_pS(U)= e $. But then if the $a_i$ take different values on different parts of $U$, how does one glue them together to construct a valid $a$, and what does this $a$ even look like?
I haven't yet attempted the other axiom for sheaves, but I think I'd run into similar problems with the above reasoning. Thank you for any help.
general-topology algebraic-geometry sheaf-theory
asked Jul 23 at 22:46
mathphys
961415
$DeclareMathOperatorresres$Let $X$ be a topological space, $p in X$ a point, $U subset X$ an open subset covered by $bigcup_i in IU_i$, and $S$ a set (or an abelian group). I'm trying to show that the skyscraper sheaf $i_pS$ given by
$$ i_pS(U) = begincases S & textif , p in U \ e & textelse , endcases $$
is indeed a sheaf. Here's what I've tried so far.
First I want to show the gluing axiom, so I take sections $a_i in i_pS(U_i)$ such that $res^U_i_U_i cap U_j a_i = res^U_j_U_i cap U_j a_j$. We need for there to exist a section $a in i_pS(U)$ such that $res^U_U_ia = a_i$ for all $i in I$. But here's my problem - since the sections are elements of the set (or abelian group) $S$, I'm not sure what the restriction of one of these sections actually is (as opposed to the clear nature of a restriction when we talk about, say, the sheaf of differentiable functions).
I tried to do it case-wise, so say $p in U_i$ and $p notin U_j$ (so $p notin U_i cap U_j$) for example. Then $a_j in e$ so $a_j =e$, and by our assumption, $res^U_i_U_i cap U_j a_i = e$ which means that $a_i$ is $e$ on the part of $U_i$ which overlaps with $U_j$. Does this mean that $a_i$ must be $e$ on the whole of $U_i$? I feel not, because then by this logic I think we would end up with $i_pS(U)= e $. But then if the $a_i$ take different values on different parts of $U$, how does one glue them together to construct a valid $a$, and what does this $a$ even look like?
I haven't yet attempted the other axiom for sheaves, but I think I'd run into similar problems with the above reasoning. Thank you for any help.
general-topology algebraic-geometry sheaf-theory
asked Jul 23 at 22:46
mathphys
961415
edited Jul 23 at 23:58
asked Jul 23 at 22:46
mathphys
961415
asked Jul 23 at 22:46
mathphys
961415
asked Jul 23 at 22:46
mathphys
961415
961415
add a comment |Â
add a comment |Â
1 Answer
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For simplicity we look at sheaves of abelian groups.
You can describe $i_pS$ as the sheaf of functions $f : U to S$ with the property that $f(u) = e$ for all $u neq p$. The restriction is the usual restriction of functions.
Now let $f_i : U_i to S$ be a family of sections of $i_pS$, which agree on intersections. Then they glue uniquely to a function $f : X to S$ and obviously $f in i_pS(X)$.
answered Jul 23 at 22:54
Nicolas Hemelsoet
4,855316
Thanks for your answer. So I suppose this way of thinking about it works because these functions "detect" whether $U$ contains $p$ or not. Say $p in U$. Then $f(p) neq e$ and $f(p) in S$. But $f(p)$ only represents one point of $S$; is this not a problem? The definition of the skyscraper sheaf I gave in the question assigns the whole of $S$ rather than just a point of it when $p in U$.
– mathphys
Jul 23 at 23:40
Ah, so I suppose the fact that we can take many possible functions $f$ (all these functions are presumably what makes up the whole sheaf $i_pS(U)$), in particular, for each element of $S$, a function $f$ mapping $p$ to that element of $S$, is what makes this work out?
– mathphys
Jul 23 at 23:47
1
It is of course possible that $f(p) = e$ since $e in S$ and $f(u) = e$ for all $u in U$ is in $i_pS(U)$. Yes, there are exactly $|S|$ sections if $p in U$, and a single section else.
– Nicolas Hemelsoet
Jul 24 at 5:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For simplicity we look at sheaves of abelian groups.
You can describe $i_pS$ as the sheaf of functions $f : U to S$ with the property that $f(u) = e$ for all $u neq p$. The restriction is the usual restriction of functions.
Now let $f_i : U_i to S$ be a family of sections of $i_pS$, which agree on intersections. Then they glue uniquely to a function $f : X to S$ and obviously $f in i_pS(X)$.
answered Jul 23 at 22:54
Nicolas Hemelsoet
4,855316
Thanks for your answer. So I suppose this way of thinking about it works because these functions "detect" whether $U$ contains $p$ or not. Say $p in U$. Then $f(p) neq e$ and $f(p) in S$. But $f(p)$ only represents one point of $S$; is this not a problem? The definition of the skyscraper sheaf I gave in the question assigns the whole of $S$ rather than just a point of it when $p in U$.
– mathphys
Jul 23 at 23:40
Ah, so I suppose the fact that we can take many possible functions $f$ (all these functions are presumably what makes up the whole sheaf $i_pS(U)$), in particular, for each element of $S$, a function $f$ mapping $p$ to that element of $S$, is what makes this work out?
– mathphys
Jul 23 at 23:47
1
It is of course possible that $f(p) = e$ since $e in S$ and $f(u) = e$ for all $u in U$ is in $i_pS(U)$. Yes, there are exactly $|S|$ sections if $p in U$, and a single section else.
– Nicolas Hemelsoet
Jul 24 at 5:04
add a comment |Â
up vote
2
down vote
accepted
For simplicity we look at sheaves of abelian groups.
You can describe $i_pS$ as the sheaf of functions $f : U to S$ with the property that $f(u) = e$ for all $u neq p$. The restriction is the usual restriction of functions.
Now let $f_i : U_i to S$ be a family of sections of $i_pS$, which agree on intersections. Then they glue uniquely to a function $f : X to S$ and obviously $f in i_pS(X)$.
answered Jul 23 at 22:54
Nicolas Hemelsoet
4,855316
Thanks for your answer. So I suppose this way of thinking about it works because these functions "detect" whether $U$ contains $p$ or not. Say $p in U$. Then $f(p) neq e$ and $f(p) in S$. But $f(p)$ only represents one point of $S$; is this not a problem? The definition of the skyscraper sheaf I gave in the question assigns the whole of $S$ rather than just a point of it when $p in U$.
– mathphys
Jul 23 at 23:40
Ah, so I suppose the fact that we can take many possible functions $f$ (all these functions are presumably what makes up the whole sheaf $i_pS(U)$), in particular, for each element of $S$, a function $f$ mapping $p$ to that element of $S$, is what makes this work out?
– mathphys
Jul 23 at 23:47
1
It is of course possible that $f(p) = e$ since $e in S$ and $f(u) = e$ for all $u in U$ is in $i_pS(U)$. Yes, there are exactly $|S|$ sections if $p in U$, and a single section else.
– Nicolas Hemelsoet
Jul 24 at 5:04
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For simplicity we look at sheaves of abelian groups.
You can describe $i_pS$ as the sheaf of functions $f : U to S$ with the property that $f(u) = e$ for all $u neq p$. The restriction is the usual restriction of functions.
Now let $f_i : U_i to S$ be a family of sections of $i_pS$, which agree on intersections. Then they glue uniquely to a function $f : X to S$ and obviously $f in i_pS(X)$.
answered Jul 23 at 22:54
Nicolas Hemelsoet
4,855316
For simplicity we look at sheaves of abelian groups.
You can describe $i_pS$ as the sheaf of functions $f : U to S$ with the property that $f(u) = e$ for all $u neq p$. The restriction is the usual restriction of functions.
Now let $f_i : U_i to S$ be a family of sections of $i_pS$, which agree on intersections. Then they glue uniquely to a function $f : X to S$ and obviously $f in i_pS(X)$.
answered Jul 23 at 22:54
Nicolas Hemelsoet
4,855316
answered Jul 23 at 22:54
Nicolas Hemelsoet
4,855316
answered Jul 23 at 22:54
Nicolas Hemelsoet
4,855316
answered Jul 23 at 22:54
Nicolas Hemelsoet
4,855316
4,855316
Thanks for your answer. So I suppose this way of thinking about it works because these functions "detect" whether $U$ contains $p$ or not. Say $p in U$. Then $f(p) neq e$ and $f(p) in S$. But $f(p)$ only represents one point of $S$; is this not a problem? The definition of the skyscraper sheaf I gave in the question assigns the whole of $S$ rather than just a point of it when $p in U$.
– mathphys
Jul 23 at 23:40
Ah, so I suppose the fact that we can take many possible functions $f$ (all these functions are presumably what makes up the whole sheaf $i_pS(U)$), in particular, for each element of $S$, a function $f$ mapping $p$ to that element of $S$, is what makes this work out?
– mathphys
Jul 23 at 23:47
1
It is of course possible that $f(p) = e$ since $e in S$ and $f(u) = e$ for all $u in U$ is in $i_pS(U)$. Yes, there are exactly $|S|$ sections if $p in U$, and a single section else.
– Nicolas Hemelsoet
Jul 24 at 5:04
add a comment |Â
Thanks for your answer. So I suppose this way of thinking about it works because these functions "detect" whether $U$ contains $p$ or not. Say $p in U$. Then $f(p) neq e$ and $f(p) in S$. But $f(p)$ only represents one point of $S$; is this not a problem? The definition of the skyscraper sheaf I gave in the question assigns the whole of $S$ rather than just a point of it when $p in U$.
– mathphys
Jul 23 at 23:40
Ah, so I suppose the fact that we can take many possible functions $f$ (all these functions are presumably what makes up the whole sheaf $i_pS(U)$), in particular, for each element of $S$, a function $f$ mapping $p$ to that element of $S$, is what makes this work out?
– mathphys
Jul 23 at 23:47
1
It is of course possible that $f(p) = e$ since $e in S$ and $f(u) = e$ for all $u in U$ is in $i_pS(U)$. Yes, there are exactly $|S|$ sections if $p in U$, and a single section else.
– Nicolas Hemelsoet
Jul 24 at 5:04
Thanks for your answer. So I suppose this way of thinking about it works because these functions "detect" whether $U$ contains $p$ or not. Say $p in U$. Then $f(p) neq e$ and $f(p) in S$. But $f(p)$ only represents one point of $S$; is this not a problem? The definition of the skyscraper sheaf I gave in the question assigns the whole of $S$ rather than just a point of it when $p in U$.
– mathphys
Jul 23 at 23:40
Thanks for your answer. So I suppose this way of thinking about it works because these functions "detect" whether $U$ contains $p$ or not. Say $p in U$. Then $f(p) neq e$ and $f(p) in S$. But $f(p)$ only represents one point of $S$; is this not a problem? The definition of the skyscraper sheaf I gave in the question assigns the whole of $S$ rather than just a point of it when $p in U$.
– mathphys
Jul 23 at 23:40
Ah, so I suppose the fact that we can take many possible functions $f$ (all these functions are presumably what makes up the whole sheaf $i_pS(U)$), in particular, for each element of $S$, a function $f$ mapping $p$ to that element of $S$, is what makes this work out?
– mathphys
Jul 23 at 23:47
Ah, so I suppose the fact that we can take many possible functions $f$ (all these functions are presumably what makes up the whole sheaf $i_pS(U)$), in particular, for each element of $S$, a function $f$ mapping $p$ to that element of $S$, is what makes this work out?
– mathphys
Jul 23 at 23:47
1
1
It is of course possible that $f(p) = e$ since $e in S$ and $f(u) = e$ for all $u in U$ is in $i_pS(U)$. Yes, there are exactly $|S|$ sections if $p in U$, and a single section else.
– Nicolas Hemelsoet
Jul 24 at 5:04
It is of course possible that $f(p) = e$ since $e in S$ and $f(u) = e$ for all $u in U$ is in $i_pS(U)$. Yes, there are exactly $|S|$ sections if $p in U$, and a single section else.
– Nicolas Hemelsoet
Jul 24 at 5:04
add a comment |Â
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