Understanding sheaves on a $2$-element set
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I'm working through the Geometry of Schemes and wanted some clarification for an exercise.
Exercise I-5 considers a two-element set $X=0,1$ with the discrete topology and asks the reader to find the relations between the objects of a sheaf (of abelian groups) on $X$.
If we let $mathcalF$ be the sheaf, then it's clear that $mathcalF(emptyset)$ is the trivial group and that we have a commutative diagram of restrictions from $mathcalF(0)$ to $mathcalF(emptyset)$, from $mathcalF(1)$ to $mathcalF(emptyset)$, from $mathcalF(0,1)$ to $mathcalF(0)$, and from $mathcalF(0,1)$ to $mathcalF(emptyset)$.
By the sheaf axiom, it seems like for any $sinmathcalF(0)$ and $tinmathcalF(1)$, $s$ and $t$ restricted to the intersection, $emptyset$, must be the same, so there is some unique section in $mathcalF(0,1)$ that restricts to $s$ and $t$.
What does this say about $mathcalF(0,1)$? I'm guessing it's related to the fiber product, but I'm not particularly well-versed in category theory. Also, how does this generalize to sheaves over different categories?
algebraic-geometry category-theory sheaf-theory
edited Feb 5 at 18:06
Asaf Karagila♦
305k33435766
asked Feb 5 at 16:26
whethamwhetham
16117
$endgroup$
add a comment |
$begingroup$
I'm working through the Geometry of Schemes and wanted some clarification for an exercise.
Exercise I-5 considers a two-element set $X=0,1$ with the discrete topology and asks the reader to find the relations between the objects of a sheaf (of abelian groups) on $X$.
If we let $mathcalF$ be the sheaf, then it's clear that $mathcalF(emptyset)$ is the trivial group and that we have a commutative diagram of restrictions from $mathcalF(0)$ to $mathcalF(emptyset)$, from $mathcalF(1)$ to $mathcalF(emptyset)$, from $mathcalF(0,1)$ to $mathcalF(0)$, and from $mathcalF(0,1)$ to $mathcalF(emptyset)$.
By the sheaf axiom, it seems like for any $sinmathcalF(0)$ and $tinmathcalF(1)$, $s$ and $t$ restricted to the intersection, $emptyset$, must be the same, so there is some unique section in $mathcalF(0,1)$ that restricts to $s$ and $t$.
What does this say about $mathcalF(0,1)$? I'm guessing it's related to the fiber product, but I'm not particularly well-versed in category theory. Also, how does this generalize to sheaves over different categories?
algebraic-geometry category-theory sheaf-theory
edited Feb 5 at 18:06
Asaf Karagila♦
305k33435766
asked Feb 5 at 16:26
whethamwhetham
16117
$endgroup$
1
$begingroup$
If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
$endgroup$
– Derek Elkins
Feb 5 at 21:22
add a comment |
$begingroup$
I'm working through the Geometry of Schemes and wanted some clarification for an exercise.
Exercise I-5 considers a two-element set $X=0,1$ with the discrete topology and asks the reader to find the relations between the objects of a sheaf (of abelian groups) on $X$.
If we let $mathcalF$ be the sheaf, then it's clear that $mathcalF(emptyset)$ is the trivial group and that we have a commutative diagram of restrictions from $mathcalF(0)$ to $mathcalF(emptyset)$, from $mathcalF(1)$ to $mathcalF(emptyset)$, from $mathcalF(0,1)$ to $mathcalF(0)$, and from $mathcalF(0,1)$ to $mathcalF(emptyset)$.
By the sheaf axiom, it seems like for any $sinmathcalF(0)$ and $tinmathcalF(1)$, $s$ and $t$ restricted to the intersection, $emptyset$, must be the same, so there is some unique section in $mathcalF(0,1)$ that restricts to $s$ and $t$.
What does this say about $mathcalF(0,1)$? I'm guessing it's related to the fiber product, but I'm not particularly well-versed in category theory. Also, how does this generalize to sheaves over different categories?
algebraic-geometry category-theory sheaf-theory
edited Feb 5 at 18:06
Asaf Karagila♦
305k33435766
asked Feb 5 at 16:26
whethamwhetham
16117
$endgroup$
I'm working through the Geometry of Schemes and wanted some clarification for an exercise.
Exercise I-5 considers a two-element set $X=0,1$ with the discrete topology and asks the reader to find the relations between the objects of a sheaf (of abelian groups) on $X$.
If we let $mathcalF$ be the sheaf, then it's clear that $mathcalF(emptyset)$ is the trivial group and that we have a commutative diagram of restrictions from $mathcalF(0)$ to $mathcalF(emptyset)$, from $mathcalF(1)$ to $mathcalF(emptyset)$, from $mathcalF(0,1)$ to $mathcalF(0)$, and from $mathcalF(0,1)$ to $mathcalF(emptyset)$.
By the sheaf axiom, it seems like for any $sinmathcalF(0)$ and $tinmathcalF(1)$, $s$ and $t$ restricted to the intersection, $emptyset$, must be the same, so there is some unique section in $mathcalF(0,1)$ that restricts to $s$ and $t$.
What does this say about $mathcalF(0,1)$? I'm guessing it's related to the fiber product, but I'm not particularly well-versed in category theory. Also, how does this generalize to sheaves over different categories?
algebraic-geometry category-theory sheaf-theory
algebraic-geometry category-theory sheaf-theory
edited Feb 5 at 18:06
Asaf Karagila♦
305k33435766
asked Feb 5 at 16:26
whethamwhetham
16117
edited Feb 5 at 18:06
Asaf Karagila♦
305k33435766
asked Feb 5 at 16:26
whethamwhetham
16117
edited Feb 5 at 18:06
Asaf Karagila♦
305k33435766
edited Feb 5 at 18:06
Asaf Karagila♦
305k33435766
edited Feb 5 at 18:06
Asaf Karagila♦
305k33435766
305k33435766
asked Feb 5 at 16:26
whethamwhetham
16117
asked Feb 5 at 16:26
whethamwhetham
16117
asked Feb 5 at 16:26
whethamwhetham
16117
16117
1
$begingroup$
If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
$endgroup$
– Derek Elkins
Feb 5 at 21:22
add a comment |
1
$begingroup$
If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
$endgroup$
– Derek Elkins
Feb 5 at 21:22
1
1
$begingroup$
If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
$endgroup$
– Derek Elkins
Feb 5 at 21:22
$begingroup$
If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
$endgroup$
– Derek Elkins
Feb 5 at 21:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In this case, $mathcalF(0,1)$ is just the direct product of the groups
$mathcalF(0)$ and $mathcalF(1)$.
This commutative diagram has to be a pullback
$requireAMScd$
beginCD
mathcalF(0,1) @>>> mathcalF(1)\
@V V V @VV V\
mathcalF(0) @>>> mathcalF(emptyset)=0
endCD
but as the southeast corner is trivial, the northwest group
is the direct product of the other corners, and the nontrivial
maps are the projections from a direct product to its factors.
answered Feb 5 at 16:29
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
How do we know that $calF(0,1)$ (with the maps downwards and to the right) is universal?
$endgroup$
– whetham
Feb 6 at 16:27
$begingroup$
@whetham The diagram in the answer is essentially a particular instance of the sheaf condition. A pullback can be formulated as an equalizer of products.
$endgroup$
– Derek Elkins
Feb 6 at 19:27
add a comment |
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1 Answer
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oldest
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votes
$begingroup$
In this case, $mathcalF(0,1)$ is just the direct product of the groups
$mathcalF(0)$ and $mathcalF(1)$.
This commutative diagram has to be a pullback
$requireAMScd$
beginCD
mathcalF(0,1) @>>> mathcalF(1)\
@V V V @VV V\
mathcalF(0) @>>> mathcalF(emptyset)=0
endCD
but as the southeast corner is trivial, the northwest group
is the direct product of the other corners, and the nontrivial
maps are the projections from a direct product to its factors.
answered Feb 5 at 16:29
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
How do we know that $calF(0,1)$ (with the maps downwards and to the right) is universal?
$endgroup$
– whetham
Feb 6 at 16:27
$begingroup$
@whetham The diagram in the answer is essentially a particular instance of the sheaf condition. A pullback can be formulated as an equalizer of products.
$endgroup$
– Derek Elkins
Feb 6 at 19:27
add a comment |
$begingroup$
In this case, $mathcalF(0,1)$ is just the direct product of the groups
$mathcalF(0)$ and $mathcalF(1)$.
This commutative diagram has to be a pullback
$requireAMScd$
beginCD
mathcalF(0,1) @>>> mathcalF(1)\
@V V V @VV V\
mathcalF(0) @>>> mathcalF(emptyset)=0
endCD
but as the southeast corner is trivial, the northwest group
is the direct product of the other corners, and the nontrivial
maps are the projections from a direct product to its factors.
answered Feb 5 at 16:29
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
How do we know that $calF(0,1)$ (with the maps downwards and to the right) is universal?
$endgroup$
– whetham
Feb 6 at 16:27
$begingroup$
@whetham The diagram in the answer is essentially a particular instance of the sheaf condition. A pullback can be formulated as an equalizer of products.
$endgroup$
– Derek Elkins
Feb 6 at 19:27
add a comment |
$begingroup$
In this case, $mathcalF(0,1)$ is just the direct product of the groups
$mathcalF(0)$ and $mathcalF(1)$.
This commutative diagram has to be a pullback
$requireAMScd$
beginCD
mathcalF(0,1) @>>> mathcalF(1)\
@V V V @VV V\
mathcalF(0) @>>> mathcalF(emptyset)=0
endCD
but as the southeast corner is trivial, the northwest group
is the direct product of the other corners, and the nontrivial
maps are the projections from a direct product to its factors.
answered Feb 5 at 16:29
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
In this case, $mathcalF(0,1)$ is just the direct product of the groups
$mathcalF(0)$ and $mathcalF(1)$.
This commutative diagram has to be a pullback
$requireAMScd$
beginCD
mathcalF(0,1) @>>> mathcalF(1)\
@V V V @VV V\
mathcalF(0) @>>> mathcalF(emptyset)=0
endCD
but as the southeast corner is trivial, the northwest group
is the direct product of the other corners, and the nontrivial
maps are the projections from a direct product to its factors.
answered Feb 5 at 16:29
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Feb 5 at 16:29
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Feb 5 at 16:29
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Feb 5 at 16:29
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
How do we know that $calF(0,1)$ (with the maps downwards and to the right) is universal?
$endgroup$
– whetham
Feb 6 at 16:27
$begingroup$
@whetham The diagram in the answer is essentially a particular instance of the sheaf condition. A pullback can be formulated as an equalizer of products.
$endgroup$
– Derek Elkins
Feb 6 at 19:27
add a comment |
$begingroup$
How do we know that $calF(0,1)$ (with the maps downwards and to the right) is universal?
$endgroup$
– whetham
Feb 6 at 16:27
$begingroup$
@whetham The diagram in the answer is essentially a particular instance of the sheaf condition. A pullback can be formulated as an equalizer of products.
$endgroup$
– Derek Elkins
Feb 6 at 19:27
$begingroup$
How do we know that $calF(0,1)$ (with the maps downwards and to the right) is universal?
$endgroup$
– whetham
Feb 6 at 16:27
$begingroup$
How do we know that $calF(0,1)$ (with the maps downwards and to the right) is universal?
$endgroup$
– whetham
Feb 6 at 16:27
$begingroup$
@whetham The diagram in the answer is essentially a particular instance of the sheaf condition. A pullback can be formulated as an equalizer of products.
$endgroup$
– Derek Elkins
Feb 6 at 19:27
$begingroup$
@whetham The diagram in the answer is essentially a particular instance of the sheaf condition. A pullback can be formulated as an equalizer of products.
$endgroup$
– Derek Elkins
Feb 6 at 19:27
add a comment |
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$begingroup$
If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
$endgroup$
– Derek Elkins
Feb 5 at 21:22