Are sections of an injective sheaf of abelian groups themselves injective abelian groups?
$begingroup$
Let $X$ be a topological space, $mathcal I$ some injective sheaf of abelian groups on $X$ and let $U subset X$ be open. Is $mathcal I(U)$ an injective abelian group?
If not what requirements would we need to make this so (eg. perhaps requirements on the space $X$ or the choice of open set $U$)?
What about if $X$ was a scheme (or just a locally ringed space) and $mathcal I$ an injective sheaf of $mathcal O_X$-modules, is $mathcal I(U)$ an injective $mathcal O_X(U)$-module?
Again if not what requirements would we need to make this so?
(My main question is the one in the title so I don't mind partial answers to any of these slightly separate questions).
algebraic-geometry sheaf-theory schemes
asked Jan 1 at 12:22
FromageFromage
1338
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space, $mathcal I$ some injective sheaf of abelian groups on $X$ and let $U subset X$ be open. Is $mathcal I(U)$ an injective abelian group?
If not what requirements would we need to make this so (eg. perhaps requirements on the space $X$ or the choice of open set $U$)?
What about if $X$ was a scheme (or just a locally ringed space) and $mathcal I$ an injective sheaf of $mathcal O_X$-modules, is $mathcal I(U)$ an injective $mathcal O_X(U)$-module?
Again if not what requirements would we need to make this so?
(My main question is the one in the title so I don't mind partial answers to any of these slightly separate questions).
algebraic-geometry sheaf-theory schemes
asked Jan 1 at 12:22
FromageFromage
1338
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space, $mathcal I$ some injective sheaf of abelian groups on $X$ and let $U subset X$ be open. Is $mathcal I(U)$ an injective abelian group?
If not what requirements would we need to make this so (eg. perhaps requirements on the space $X$ or the choice of open set $U$)?
What about if $X$ was a scheme (or just a locally ringed space) and $mathcal I$ an injective sheaf of $mathcal O_X$-modules, is $mathcal I(U)$ an injective $mathcal O_X(U)$-module?
Again if not what requirements would we need to make this so?
(My main question is the one in the title so I don't mind partial answers to any of these slightly separate questions).
algebraic-geometry sheaf-theory schemes
asked Jan 1 at 12:22
FromageFromage
1338
$endgroup$
Let $X$ be a topological space, $mathcal I$ some injective sheaf of abelian groups on $X$ and let $U subset X$ be open. Is $mathcal I(U)$ an injective abelian group?
If not what requirements would we need to make this so (eg. perhaps requirements on the space $X$ or the choice of open set $U$)?
What about if $X$ was a scheme (or just a locally ringed space) and $mathcal I$ an injective sheaf of $mathcal O_X$-modules, is $mathcal I(U)$ an injective $mathcal O_X(U)$-module?
Again if not what requirements would we need to make this so?
(My main question is the one in the title so I don't mind partial answers to any of these slightly separate questions).
algebraic-geometry sheaf-theory schemes
algebraic-geometry sheaf-theory schemes
asked Jan 1 at 12:22
FromageFromage
1338
asked Jan 1 at 12:22
FromageFromage
1338
asked Jan 1 at 12:22
FromageFromage
1338
asked Jan 1 at 12:22
FromageFromage
1338
asked Jan 1 at 12:22
FromageFromage
1338
1338
add a comment |
add a comment |
1 Answer
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$begingroup$
In general, if $(F,G)$ is a pair of adjoint functor and if $F$ is exact, then $G$ preserve injectives. Indeed, if $I$ is injective, then the functor $Hom(F(.),I)$ is exact as a composite of $F$ (exact by assumption) and $Hom(.,I)$ (exact because $I$ is injective). But $Hom(F(.),I)=Hom(.,G(I))$, so $Hom(.,G(I))$ is exact which means that $G(I)$ is injective.
In your particular case, you consider the functor $Gamma(U,.)$ whose left adjoint is $Amapsto {j_U}_!underline{A}$ where $underline{A}$ is the constant sheaf on $U$ with stalks $A$, and ${j_U}_!$ is the extension by zero outside of $U$. This is an exact functor (as a composite of exact functors). This implies that $Gamma(U,.)$ preserve injectives.
Hence, if $mathcal{I}$ is an injective sheaf, then for all open $Usubset X$, $mathcal{I}(U)$ is injective.
This does not change anything if you consider categories of $mathcal{O}_X$-modules and $mathcal{O}_X(U)$-modules.
answered Jan 1 at 17:14
RolandRoland
7,0291913
$endgroup$
$begingroup$
Great, thanks very much.
$endgroup$
– Fromage
Jan 2 at 9:29
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In general, if $(F,G)$ is a pair of adjoint functor and if $F$ is exact, then $G$ preserve injectives. Indeed, if $I$ is injective, then the functor $Hom(F(.),I)$ is exact as a composite of $F$ (exact by assumption) and $Hom(.,I)$ (exact because $I$ is injective). But $Hom(F(.),I)=Hom(.,G(I))$, so $Hom(.,G(I))$ is exact which means that $G(I)$ is injective.
In your particular case, you consider the functor $Gamma(U,.)$ whose left adjoint is $Amapsto {j_U}_!underline{A}$ where $underline{A}$ is the constant sheaf on $U$ with stalks $A$, and ${j_U}_!$ is the extension by zero outside of $U$. This is an exact functor (as a composite of exact functors). This implies that $Gamma(U,.)$ preserve injectives.
Hence, if $mathcal{I}$ is an injective sheaf, then for all open $Usubset X$, $mathcal{I}(U)$ is injective.
This does not change anything if you consider categories of $mathcal{O}_X$-modules and $mathcal{O}_X(U)$-modules.
answered Jan 1 at 17:14
RolandRoland
7,0291913
$endgroup$
$begingroup$
Great, thanks very much.
$endgroup$
– Fromage
Jan 2 at 9:29
add a comment |
$begingroup$
In general, if $(F,G)$ is a pair of adjoint functor and if $F$ is exact, then $G$ preserve injectives. Indeed, if $I$ is injective, then the functor $Hom(F(.),I)$ is exact as a composite of $F$ (exact by assumption) and $Hom(.,I)$ (exact because $I$ is injective). But $Hom(F(.),I)=Hom(.,G(I))$, so $Hom(.,G(I))$ is exact which means that $G(I)$ is injective.
In your particular case, you consider the functor $Gamma(U,.)$ whose left adjoint is $Amapsto {j_U}_!underline{A}$ where $underline{A}$ is the constant sheaf on $U$ with stalks $A$, and ${j_U}_!$ is the extension by zero outside of $U$. This is an exact functor (as a composite of exact functors). This implies that $Gamma(U,.)$ preserve injectives.
Hence, if $mathcal{I}$ is an injective sheaf, then for all open $Usubset X$, $mathcal{I}(U)$ is injective.
This does not change anything if you consider categories of $mathcal{O}_X$-modules and $mathcal{O}_X(U)$-modules.
answered Jan 1 at 17:14
RolandRoland
7,0291913
$endgroup$
$begingroup$
Great, thanks very much.
$endgroup$
– Fromage
Jan 2 at 9:29
add a comment |
$begingroup$
In general, if $(F,G)$ is a pair of adjoint functor and if $F$ is exact, then $G$ preserve injectives. Indeed, if $I$ is injective, then the functor $Hom(F(.),I)$ is exact as a composite of $F$ (exact by assumption) and $Hom(.,I)$ (exact because $I$ is injective). But $Hom(F(.),I)=Hom(.,G(I))$, so $Hom(.,G(I))$ is exact which means that $G(I)$ is injective.
In your particular case, you consider the functor $Gamma(U,.)$ whose left adjoint is $Amapsto {j_U}_!underline{A}$ where $underline{A}$ is the constant sheaf on $U$ with stalks $A$, and ${j_U}_!$ is the extension by zero outside of $U$. This is an exact functor (as a composite of exact functors). This implies that $Gamma(U,.)$ preserve injectives.
Hence, if $mathcal{I}$ is an injective sheaf, then for all open $Usubset X$, $mathcal{I}(U)$ is injective.
This does not change anything if you consider categories of $mathcal{O}_X$-modules and $mathcal{O}_X(U)$-modules.
answered Jan 1 at 17:14
RolandRoland
7,0291913
$endgroup$
In general, if $(F,G)$ is a pair of adjoint functor and if $F$ is exact, then $G$ preserve injectives. Indeed, if $I$ is injective, then the functor $Hom(F(.),I)$ is exact as a composite of $F$ (exact by assumption) and $Hom(.,I)$ (exact because $I$ is injective). But $Hom(F(.),I)=Hom(.,G(I))$, so $Hom(.,G(I))$ is exact which means that $G(I)$ is injective.
In your particular case, you consider the functor $Gamma(U,.)$ whose left adjoint is $Amapsto {j_U}_!underline{A}$ where $underline{A}$ is the constant sheaf on $U$ with stalks $A$, and ${j_U}_!$ is the extension by zero outside of $U$. This is an exact functor (as a composite of exact functors). This implies that $Gamma(U,.)$ preserve injectives.
Hence, if $mathcal{I}$ is an injective sheaf, then for all open $Usubset X$, $mathcal{I}(U)$ is injective.
This does not change anything if you consider categories of $mathcal{O}_X$-modules and $mathcal{O}_X(U)$-modules.
answered Jan 1 at 17:14
RolandRoland
7,0291913
answered Jan 1 at 17:14
RolandRoland
7,0291913
answered Jan 1 at 17:14
RolandRoland
7,0291913
answered Jan 1 at 17:14
RolandRoland
7,0291913
7,0291913
$begingroup$
Great, thanks very much.
$endgroup$
– Fromage
Jan 2 at 9:29
add a comment |
$begingroup$
Great, thanks very much.
$endgroup$
– Fromage
Jan 2 at 9:29
$begingroup$
Great, thanks very much.
$endgroup$
– Fromage
Jan 2 at 9:29
$begingroup$
Great, thanks very much.
$endgroup$
– Fromage
Jan 2 at 9:29
add a comment |
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