Generators for category of abelian presheaves
$begingroup$
Let $T$ be a site, and let $mathcal{P}$ denote the category of abelian presheaves on $T$.
Apparently $(Z_U)_{Uin T}$ is a family of generators for $mathcal{P}$ defined by:
$$Z_U(V)=bigoplus_{hom(V,U)}Bbb Z,quad Vin T,$$
Where it is claimed that:
$$F(U)=hom(Bbb{Z},F(U))cong hom(Z_U,F).$$
I can't see why that last isomorphism holds.
It seems that $hom(Z_U,F)$ are presheaf homomorphisms, where if I consider any $V$ I have:
$$hom(Z_U(V),F(V))=hom(bigoplus_{hom(V,U)}Bbb{Z},F(V))cong prod_{hom(V,U)}hom(Bbb Z,F(V))cong F(V)^{hom(V,U)},$$
and I can't see how considering the commuting squares of the natural transformation gives me information allowing me to conclude.
category-theory sheaf-theory
asked Dec 2 '18 at 0:05
user616128user616128
305
$endgroup$
add a comment |
$begingroup$
Let $T$ be a site, and let $mathcal{P}$ denote the category of abelian presheaves on $T$.
Apparently $(Z_U)_{Uin T}$ is a family of generators for $mathcal{P}$ defined by:
$$Z_U(V)=bigoplus_{hom(V,U)}Bbb Z,quad Vin T,$$
Where it is claimed that:
$$F(U)=hom(Bbb{Z},F(U))cong hom(Z_U,F).$$
I can't see why that last isomorphism holds.
It seems that $hom(Z_U,F)$ are presheaf homomorphisms, where if I consider any $V$ I have:
$$hom(Z_U(V),F(V))=hom(bigoplus_{hom(V,U)}Bbb{Z},F(V))cong prod_{hom(V,U)}hom(Bbb Z,F(V))cong F(V)^{hom(V,U)},$$
and I can't see how considering the commuting squares of the natural transformation gives me information allowing me to conclude.
category-theory sheaf-theory
asked Dec 2 '18 at 0:05
user616128user616128
305
$endgroup$
add a comment |
$begingroup$
Let $T$ be a site, and let $mathcal{P}$ denote the category of abelian presheaves on $T$.
Apparently $(Z_U)_{Uin T}$ is a family of generators for $mathcal{P}$ defined by:
$$Z_U(V)=bigoplus_{hom(V,U)}Bbb Z,quad Vin T,$$
Where it is claimed that:
$$F(U)=hom(Bbb{Z},F(U))cong hom(Z_U,F).$$
I can't see why that last isomorphism holds.
It seems that $hom(Z_U,F)$ are presheaf homomorphisms, where if I consider any $V$ I have:
$$hom(Z_U(V),F(V))=hom(bigoplus_{hom(V,U)}Bbb{Z},F(V))cong prod_{hom(V,U)}hom(Bbb Z,F(V))cong F(V)^{hom(V,U)},$$
and I can't see how considering the commuting squares of the natural transformation gives me information allowing me to conclude.
category-theory sheaf-theory
asked Dec 2 '18 at 0:05
user616128user616128
305
$endgroup$
Let $T$ be a site, and let $mathcal{P}$ denote the category of abelian presheaves on $T$.
Apparently $(Z_U)_{Uin T}$ is a family of generators for $mathcal{P}$ defined by:
$$Z_U(V)=bigoplus_{hom(V,U)}Bbb Z,quad Vin T,$$
Where it is claimed that:
$$F(U)=hom(Bbb{Z},F(U))cong hom(Z_U,F).$$
I can't see why that last isomorphism holds.
It seems that $hom(Z_U,F)$ are presheaf homomorphisms, where if I consider any $V$ I have:
$$hom(Z_U(V),F(V))=hom(bigoplus_{hom(V,U)}Bbb{Z},F(V))cong prod_{hom(V,U)}hom(Bbb Z,F(V))cong F(V)^{hom(V,U)},$$
and I can't see how considering the commuting squares of the natural transformation gives me information allowing me to conclude.
category-theory sheaf-theory
category-theory sheaf-theory
asked Dec 2 '18 at 0:05
user616128user616128
305
asked Dec 2 '18 at 0:05
user616128user616128
305
asked Dec 2 '18 at 0:05
user616128user616128
305
asked Dec 2 '18 at 0:05
user616128user616128
305
asked Dec 2 '18 at 0:05
user616128user616128
305
305
add a comment |
add a comment |
1 Answer
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$begingroup$
$Z_U$ is the abelian presheaf freely generated by the set-valued presheaf represented by $U$. Concretely, this says that a map of abelian presheaves out of $Z_U$ is uniquely determined by its action on the standard generators of each direct sum. With this in hand, the desired isomorphism follows from the Yoneda lemma.
Another approach, which I think is a bit nicer, says that an abelian presheaf on $T$ is exactly a presheaf on the preadditive category $mathbb Z[T]$ freely generated by $T$. This is the preadditive category on the same objects as $T$ with homs the free abelian groups generated by the homs or $T$. The presheaves $Z_U$ become representable in their own right when viewed with domain $mathbb Z[T]$, and so the claim is the Yoneda lemma outright, just now the version for preadditive categories.
answered Dec 2 '18 at 4:53
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
$begingroup$
I understand the spirit of the answer, but I'm having trouble making stuff precise. What in general does it mean for me to take an abelian presheaf 'freely generated' by some other set-valued presheaf? Is this like a copower?
$endgroup$
– user616128
Dec 2 '18 at 5:33
$begingroup$
It just means that there's a left adjoint from set-valued to abelian group-valued presheaves (whose right adjoint is faithful), which applies the free abelian group functor levelwise. And $Z_U$ is the image of a representable under this functor.
$endgroup$
– Kevin Carlson
Dec 2 '18 at 8:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$Z_U$ is the abelian presheaf freely generated by the set-valued presheaf represented by $U$. Concretely, this says that a map of abelian presheaves out of $Z_U$ is uniquely determined by its action on the standard generators of each direct sum. With this in hand, the desired isomorphism follows from the Yoneda lemma.
Another approach, which I think is a bit nicer, says that an abelian presheaf on $T$ is exactly a presheaf on the preadditive category $mathbb Z[T]$ freely generated by $T$. This is the preadditive category on the same objects as $T$ with homs the free abelian groups generated by the homs or $T$. The presheaves $Z_U$ become representable in their own right when viewed with domain $mathbb Z[T]$, and so the claim is the Yoneda lemma outright, just now the version for preadditive categories.
answered Dec 2 '18 at 4:53
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
$begingroup$
I understand the spirit of the answer, but I'm having trouble making stuff precise. What in general does it mean for me to take an abelian presheaf 'freely generated' by some other set-valued presheaf? Is this like a copower?
$endgroup$
– user616128
Dec 2 '18 at 5:33
$begingroup$
It just means that there's a left adjoint from set-valued to abelian group-valued presheaves (whose right adjoint is faithful), which applies the free abelian group functor levelwise. And $Z_U$ is the image of a representable under this functor.
$endgroup$
– Kevin Carlson
Dec 2 '18 at 8:31
add a comment |
$begingroup$
$Z_U$ is the abelian presheaf freely generated by the set-valued presheaf represented by $U$. Concretely, this says that a map of abelian presheaves out of $Z_U$ is uniquely determined by its action on the standard generators of each direct sum. With this in hand, the desired isomorphism follows from the Yoneda lemma.
Another approach, which I think is a bit nicer, says that an abelian presheaf on $T$ is exactly a presheaf on the preadditive category $mathbb Z[T]$ freely generated by $T$. This is the preadditive category on the same objects as $T$ with homs the free abelian groups generated by the homs or $T$. The presheaves $Z_U$ become representable in their own right when viewed with domain $mathbb Z[T]$, and so the claim is the Yoneda lemma outright, just now the version for preadditive categories.
answered Dec 2 '18 at 4:53
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
$begingroup$
I understand the spirit of the answer, but I'm having trouble making stuff precise. What in general does it mean for me to take an abelian presheaf 'freely generated' by some other set-valued presheaf? Is this like a copower?
$endgroup$
– user616128
Dec 2 '18 at 5:33
$begingroup$
It just means that there's a left adjoint from set-valued to abelian group-valued presheaves (whose right adjoint is faithful), which applies the free abelian group functor levelwise. And $Z_U$ is the image of a representable under this functor.
$endgroup$
– Kevin Carlson
Dec 2 '18 at 8:31
add a comment |
$begingroup$
$Z_U$ is the abelian presheaf freely generated by the set-valued presheaf represented by $U$. Concretely, this says that a map of abelian presheaves out of $Z_U$ is uniquely determined by its action on the standard generators of each direct sum. With this in hand, the desired isomorphism follows from the Yoneda lemma.
Another approach, which I think is a bit nicer, says that an abelian presheaf on $T$ is exactly a presheaf on the preadditive category $mathbb Z[T]$ freely generated by $T$. This is the preadditive category on the same objects as $T$ with homs the free abelian groups generated by the homs or $T$. The presheaves $Z_U$ become representable in their own right when viewed with domain $mathbb Z[T]$, and so the claim is the Yoneda lemma outright, just now the version for preadditive categories.
answered Dec 2 '18 at 4:53
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
$Z_U$ is the abelian presheaf freely generated by the set-valued presheaf represented by $U$. Concretely, this says that a map of abelian presheaves out of $Z_U$ is uniquely determined by its action on the standard generators of each direct sum. With this in hand, the desired isomorphism follows from the Yoneda lemma.
Another approach, which I think is a bit nicer, says that an abelian presheaf on $T$ is exactly a presheaf on the preadditive category $mathbb Z[T]$ freely generated by $T$. This is the preadditive category on the same objects as $T$ with homs the free abelian groups generated by the homs or $T$. The presheaves $Z_U$ become representable in their own right when viewed with domain $mathbb Z[T]$, and so the claim is the Yoneda lemma outright, just now the version for preadditive categories.
answered Dec 2 '18 at 4:53
Kevin CarlsonKevin Carlson
32.8k23372
answered Dec 2 '18 at 4:53
Kevin CarlsonKevin Carlson
32.8k23372
answered Dec 2 '18 at 4:53
Kevin CarlsonKevin Carlson
32.8k23372
answered Dec 2 '18 at 4:53
Kevin CarlsonKevin Carlson
32.8k23372
32.8k23372
$begingroup$
I understand the spirit of the answer, but I'm having trouble making stuff precise. What in general does it mean for me to take an abelian presheaf 'freely generated' by some other set-valued presheaf? Is this like a copower?
$endgroup$
– user616128
Dec 2 '18 at 5:33
$begingroup$
It just means that there's a left adjoint from set-valued to abelian group-valued presheaves (whose right adjoint is faithful), which applies the free abelian group functor levelwise. And $Z_U$ is the image of a representable under this functor.
$endgroup$
– Kevin Carlson
Dec 2 '18 at 8:31
add a comment |
$begingroup$
I understand the spirit of the answer, but I'm having trouble making stuff precise. What in general does it mean for me to take an abelian presheaf 'freely generated' by some other set-valued presheaf? Is this like a copower?
$endgroup$
– user616128
Dec 2 '18 at 5:33
$begingroup$
It just means that there's a left adjoint from set-valued to abelian group-valued presheaves (whose right adjoint is faithful), which applies the free abelian group functor levelwise. And $Z_U$ is the image of a representable under this functor.
$endgroup$
– Kevin Carlson
Dec 2 '18 at 8:31
$begingroup$
I understand the spirit of the answer, but I'm having trouble making stuff precise. What in general does it mean for me to take an abelian presheaf 'freely generated' by some other set-valued presheaf? Is this like a copower?
$endgroup$
– user616128
Dec 2 '18 at 5:33
$begingroup$
I understand the spirit of the answer, but I'm having trouble making stuff precise. What in general does it mean for me to take an abelian presheaf 'freely generated' by some other set-valued presheaf? Is this like a copower?
$endgroup$
– user616128
Dec 2 '18 at 5:33
$begingroup$
It just means that there's a left adjoint from set-valued to abelian group-valued presheaves (whose right adjoint is faithful), which applies the free abelian group functor levelwise. And $Z_U$ is the image of a representable under this functor.
$endgroup$
– Kevin Carlson
Dec 2 '18 at 8:31
$begingroup$
It just means that there's a left adjoint from set-valued to abelian group-valued presheaves (whose right adjoint is faithful), which applies the free abelian group functor levelwise. And $Z_U$ is the image of a representable under this functor.
$endgroup$
– Kevin Carlson
Dec 2 '18 at 8:31
add a comment |
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