Mixing
Injectivity of locally free sheaf
0
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Let $F$, $G$ be two locally free sheaf on $X$, and let $phi:Frightarrow G$ is injective. Then why is that $phi$ may not be injective on all fibers? Are we still regarding this map as morphism of sheafs, so injective should be equivalent to injective on all fibres?
algebraic-geometry sheaf-theory
asked Jan 5 at 20:07
Peter LiuPeter Liu
315114
$endgroup$
add a comment |
0
$begingroup$
Let $F$, $G$ be two locally free sheaf on $X$, and let $phi:Frightarrow G$ is injective. Then why is that $phi$ may not be injective on all fibers? Are we still regarding this map as morphism of sheafs, so injective should be equivalent to injective on all fibres?
algebraic-geometry sheaf-theory
asked Jan 5 at 20:07
Peter LiuPeter Liu
315114
$endgroup$
2
$begingroup$
There is a difference between stalks and fiber. The injectivity at the level of stalks is indeed equivalent to the injectivity of the sheaf morphism. But an injective morphism may fail to be fiberwise injective.
$endgroup$
– Roland
Jan 5 at 20:10
$begingroup$
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
$endgroup$
– Peter Liu
Jan 5 at 20:12
$begingroup$
I do not know why you think it has to be true.. I am not very comfortable in these topics but for a sheaf $mathcal{F}$, the fiber at a point $xin X$ is $mathcal{F}_xotimes_{mathcal{O}_x}k(x)$ where $mathcal{F}_x$ is stalk of $mathcal{F}$ at $x$, $mathcal{O}_x$ is stalk of the structure sheaf $mathcal{O}$ at $x$ and $k(x)$ is the residue field of $mathcal{O}_x$... Isn't it so?
$endgroup$
– Praphulla Koushik
Jan 5 at 20:57
$begingroup$
@PraphullaKoushik Not really, it should be the pullback along the morphism of $k(x)rightarrow X$ to the point $x$. What you consider is not in general true, but it is true if it is a closed point. I thought it is right is just as Roland's comment, I thought it is the germ. But suppose you take generic point, then it will be totally a different story.
$endgroup$
– Peter Liu
Jan 8 at 23:00
add a comment |
0
0
0
$begingroup$
Let $F$, $G$ be two locally free sheaf on $X$, and let $phi:Frightarrow G$ is injective. Then why is that $phi$ may not be injective on all fibers? Are we still regarding this map as morphism of sheafs, so injective should be equivalent to injective on all fibres?
algebraic-geometry sheaf-theory
asked Jan 5 at 20:07
Peter LiuPeter Liu
315114
$endgroup$
Let $F$, $G$ be two locally free sheaf on $X$, and let $phi:Frightarrow G$ is injective. Then why is that $phi$ may not be injective on all fibers? Are we still regarding this map as morphism of sheafs, so injective should be equivalent to injective on all fibres?
algebraic-geometry sheaf-theory
algebraic-geometry sheaf-theory
asked Jan 5 at 20:07
Peter LiuPeter Liu
315114
asked Jan 5 at 20:07
Peter LiuPeter Liu
315114
asked Jan 5 at 20:07
Peter LiuPeter Liu
315114
asked Jan 5 at 20:07
Peter LiuPeter Liu
315114
asked Jan 5 at 20:07
Peter LiuPeter Liu
315114
315114
2
$begingroup$
There is a difference between stalks and fiber. The injectivity at the level of stalks is indeed equivalent to the injectivity of the sheaf morphism. But an injective morphism may fail to be fiberwise injective.
$endgroup$
– Roland
Jan 5 at 20:10
$begingroup$
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
$endgroup$
– Peter Liu
Jan 5 at 20:12
$begingroup$
I do not know why you think it has to be true.. I am not very comfortable in these topics but for a sheaf $mathcal{F}$, the fiber at a point $xin X$ is $mathcal{F}_xotimes_{mathcal{O}_x}k(x)$ where $mathcal{F}_x$ is stalk of $mathcal{F}$ at $x$, $mathcal{O}_x$ is stalk of the structure sheaf $mathcal{O}$ at $x$ and $k(x)$ is the residue field of $mathcal{O}_x$... Isn't it so?
$endgroup$
– Praphulla Koushik
Jan 5 at 20:57
$begingroup$
@PraphullaKoushik Not really, it should be the pullback along the morphism of $k(x)rightarrow X$ to the point $x$. What you consider is not in general true, but it is true if it is a closed point. I thought it is right is just as Roland's comment, I thought it is the germ. But suppose you take generic point, then it will be totally a different story.
$endgroup$
– Peter Liu
Jan 8 at 23:00
add a comment |
2
$begingroup$
There is a difference between stalks and fiber. The injectivity at the level of stalks is indeed equivalent to the injectivity of the sheaf morphism. But an injective morphism may fail to be fiberwise injective.
$endgroup$
– Roland
Jan 5 at 20:10
$begingroup$
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
$endgroup$
– Peter Liu
Jan 5 at 20:12
$begingroup$
I do not know why you think it has to be true.. I am not very comfortable in these topics but for a sheaf $mathcal{F}$, the fiber at a point $xin X$ is $mathcal{F}_xotimes_{mathcal{O}_x}k(x)$ where $mathcal{F}_x$ is stalk of $mathcal{F}$ at $x$, $mathcal{O}_x$ is stalk of the structure sheaf $mathcal{O}$ at $x$ and $k(x)$ is the residue field of $mathcal{O}_x$... Isn't it so?
$endgroup$
– Praphulla Koushik
Jan 5 at 20:57
$begingroup$
@PraphullaKoushik Not really, it should be the pullback along the morphism of $k(x)rightarrow X$ to the point $x$. What you consider is not in general true, but it is true if it is a closed point. I thought it is right is just as Roland's comment, I thought it is the germ. But suppose you take generic point, then it will be totally a different story.
$endgroup$
– Peter Liu
Jan 8 at 23:00
2
2
$begingroup$
There is a difference between stalks and fiber. The injectivity at the level of stalks is indeed equivalent to the injectivity of the sheaf morphism. But an injective morphism may fail to be fiberwise injective.
$endgroup$
– Roland
Jan 5 at 20:10
$begingroup$
There is a difference between stalks and fiber. The injectivity at the level of stalks is indeed equivalent to the injectivity of the sheaf morphism. But an injective morphism may fail to be fiberwise injective.
$endgroup$
– Roland
Jan 5 at 20:10
$begingroup$
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
$endgroup$
– Peter Liu
Jan 5 at 20:12
$begingroup$
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
$endgroup$
– Peter Liu
Jan 5 at 20:12
$begingroup$
I do not know why you think it has to be true.. I am not very comfortable in these topics but for a sheaf $mathcal{F}$, the fiber at a point $xin X$ is $mathcal{F}_xotimes_{mathcal{O}_x}k(x)$ where $mathcal{F}_x$ is stalk of $mathcal{F}$ at $x$, $mathcal{O}_x$ is stalk of the structure sheaf $mathcal{O}$ at $x$ and $k(x)$ is the residue field of $mathcal{O}_x$... Isn't it so?
$endgroup$
– Praphulla Koushik
Jan 5 at 20:57
$begingroup$
I do not know why you think it has to be true.. I am not very comfortable in these topics but for a sheaf $mathcal{F}$, the fiber at a point $xin X$ is $mathcal{F}_xotimes_{mathcal{O}_x}k(x)$ where $mathcal{F}_x$ is stalk of $mathcal{F}$ at $x$, $mathcal{O}_x$ is stalk of the structure sheaf $mathcal{O}$ at $x$ and $k(x)$ is the residue field of $mathcal{O}_x$... Isn't it so?
$endgroup$
– Praphulla Koushik
Jan 5 at 20:57
$begingroup$
@PraphullaKoushik Not really, it should be the pullback along the morphism of $k(x)rightarrow X$ to the point $x$. What you consider is not in general true, but it is true if it is a closed point. I thought it is right is just as Roland's comment, I thought it is the germ. But suppose you take generic point, then it will be totally a different story.
$endgroup$
– Peter Liu
Jan 8 at 23:00
$begingroup$
@PraphullaKoushik Not really, it should be the pullback along the morphism of $k(x)rightarrow X$ to the point $x$. What you consider is not in general true, but it is true if it is a closed point. I thought it is right is just as Roland's comment, I thought it is the germ. But suppose you take generic point, then it will be totally a different story.
$endgroup$
– Peter Liu
Jan 8 at 23:00
add a comment |
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2
$begingroup$
There is a difference between stalks and fiber. The injectivity at the level of stalks is indeed equivalent to the injectivity of the sheaf morphism. But an injective morphism may fail to be fiberwise injective.
$endgroup$
– Roland
Jan 5 at 20:10
$begingroup$
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
$endgroup$
– Peter Liu
Jan 5 at 20:12
$begingroup$
I do not know why you think it has to be true.. I am not very comfortable in these topics but for a sheaf $mathcal{F}$, the fiber at a point $xin X$ is $mathcal{F}_xotimes_{mathcal{O}_x}k(x)$ where $mathcal{F}_x$ is stalk of $mathcal{F}$ at $x$, $mathcal{O}_x$ is stalk of the structure sheaf $mathcal{O}$ at $x$ and $k(x)$ is the residue field of $mathcal{O}_x$... Isn't it so?
$endgroup$
– Praphulla Koushik
Jan 5 at 20:57
$begingroup$
@PraphullaKoushik Not really, it should be the pullback along the morphism of $k(x)rightarrow X$ to the point $x$. What you consider is not in general true, but it is true if it is a closed point. I thought it is right is just as Roland's comment, I thought it is the germ. But suppose you take generic point, then it will be totally a different story.
$endgroup$
– Peter Liu
Jan 8 at 23:00