Injectivity of locally free sheaf
0
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 2 days ago
Peter LiuPeter Liu
315114
add a comment |
0
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 2 days ago
Peter LiuPeter Liu
315114
2
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.
– Roland
2 days ago
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
– Peter Liu
2 days ago
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?
– Praphulla Koushik
2 days ago
add a comment |
0
0
0
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 2 days ago
Peter LiuPeter Liu
315114
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 2 days ago
Peter LiuPeter Liu
315114
asked 2 days ago
Peter LiuPeter Liu
315114
asked 2 days ago
Peter LiuPeter Liu
315114
asked 2 days ago
Peter LiuPeter Liu
315114
asked 2 days ago
Peter LiuPeter Liu
315114
315114
2
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.
– Roland
2 days ago
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
– Peter Liu
2 days ago
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?
– Praphulla Koushik
2 days ago
add a comment |
2
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.
– Roland
2 days ago
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
– Peter Liu
2 days ago
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?
– Praphulla Koushik
2 days ago
2
2
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.
– Roland
2 days ago
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.
– Roland
2 days ago
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
– Peter Liu
2 days ago
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
– Peter Liu
2 days ago
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?
– Praphulla Koushik
2 days ago
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?
– Praphulla Koushik
2 days ago
add a comment |
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2
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.
– Roland
2 days ago
@Roland Aha, so a fibre means the base extension over $k(x)$, not the germ. I see my problem.
– Peter Liu
2 days ago
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?
– Praphulla Koushik
2 days ago