Mixing
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Neighborhood of fiber contains the full pre-image of an open set
Clash Royale CLAN TAG#URR8PPP
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Let $ f : X to Y $ be a morphism of schemes $ X $ and $ Y $. Suppose that $ f $ is quasi-compact. Is it true that for any $ y in Y $ and any open subset $ U $ of $ X $ which contains $ f^-1 ( y ) $, there is an open subset $ V $ of $ Y $ containing $y$ such that $ f^-1 ( V ) subset U $?
Remark. I have often run into situations where I have to deal with properties of schemes which are local on target and such a result would be helpful in those situations. I don't think it's true in the stated form, but it is true if $ U $ also contains the pre-image of all generizations of $ y $ e.g. when $ y $ is a generic point of $ Y $.
algebraic-geometry schemes
edited Jul 27 at 5:35
Eric Wofsey
162k12188299
asked Jul 27 at 3:06
Hodge-Tate
1187
add a comment |Â
up vote
1
down vote
favorite
Let $ f : X to Y $ be a morphism of schemes $ X $ and $ Y $. Suppose that $ f $ is quasi-compact. Is it true that for any $ y in Y $ and any open subset $ U $ of $ X $ which contains $ f^-1 ( y ) $, there is an open subset $ V $ of $ Y $ containing $y$ such that $ f^-1 ( V ) subset U $?
Remark. I have often run into situations where I have to deal with properties of schemes which are local on target and such a result would be helpful in those situations. I don't think it's true in the stated form, but it is true if $ U $ also contains the pre-image of all generizations of $ y $ e.g. when $ y $ is a generic point of $ Y $.
algebraic-geometry schemes
edited Jul 27 at 5:35
Eric Wofsey
162k12188299
asked Jul 27 at 3:06
Hodge-Tate
1187
I have clarified the question to what I assume was your intended meaning. If your intended meaning was something else feel free to revert.
– Eric Wofsey
Jul 27 at 5:38
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $ f : X to Y $ be a morphism of schemes $ X $ and $ Y $. Suppose that $ f $ is quasi-compact. Is it true that for any $ y in Y $ and any open subset $ U $ of $ X $ which contains $ f^-1 ( y ) $, there is an open subset $ V $ of $ Y $ containing $y$ such that $ f^-1 ( V ) subset U $?
Remark. I have often run into situations where I have to deal with properties of schemes which are local on target and such a result would be helpful in those situations. I don't think it's true in the stated form, but it is true if $ U $ also contains the pre-image of all generizations of $ y $ e.g. when $ y $ is a generic point of $ Y $.
algebraic-geometry schemes
edited Jul 27 at 5:35
Eric Wofsey
162k12188299
asked Jul 27 at 3:06
Hodge-Tate
1187
Let $ f : X to Y $ be a morphism of schemes $ X $ and $ Y $. Suppose that $ f $ is quasi-compact. Is it true that for any $ y in Y $ and any open subset $ U $ of $ X $ which contains $ f^-1 ( y ) $, there is an open subset $ V $ of $ Y $ containing $y$ such that $ f^-1 ( V ) subset U $?
Remark. I have often run into situations where I have to deal with properties of schemes which are local on target and such a result would be helpful in those situations. I don't think it's true in the stated form, but it is true if $ U $ also contains the pre-image of all generizations of $ y $ e.g. when $ y $ is a generic point of $ Y $.
algebraic-geometry schemes
edited Jul 27 at 5:35
Eric Wofsey
162k12188299
asked Jul 27 at 3:06
Hodge-Tate
1187
edited Jul 27 at 5:35
Eric Wofsey
162k12188299
edited Jul 27 at 5:35
Eric Wofsey
162k12188299
edited Jul 27 at 5:35
Eric Wofsey
162k12188299
162k12188299
asked Jul 27 at 3:06
Hodge-Tate
1187
asked Jul 27 at 3:06
Hodge-Tate
1187
asked Jul 27 at 3:06
Hodge-Tate
1187
1187
I have clarified the question to what I assume was your intended meaning. If your intended meaning was something else feel free to revert.
– Eric Wofsey
Jul 27 at 5:38
add a comment |Â
I have clarified the question to what I assume was your intended meaning. If your intended meaning was something else feel free to revert.
– Eric Wofsey
Jul 27 at 5:38
I have clarified the question to what I assume was your intended meaning. If your intended meaning was something else feel free to revert.
– Eric Wofsey
Jul 27 at 5:38
I have clarified the question to what I assume was your intended meaning. If your intended meaning was something else feel free to revert.
– Eric Wofsey
Jul 27 at 5:38
add a comment |Â
1 Answer
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1
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No. For instance, working over your favorite field, let $X=mathbbA^2setminus(0, 0)$ and let $f:XtomathbbA^1$ be the first projection. Then the open set $U=mathbbA^1times(mathbbA^1setminus0)subset X$ contains the entire fiber of $f$ over $0$, but does not contain the preimage of any neighborhood of $0$.
answered Jul 27 at 5:34
Eric Wofsey
162k12188299
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
No. For instance, working over your favorite field, let $X=mathbbA^2setminus(0, 0)$ and let $f:XtomathbbA^1$ be the first projection. Then the open set $U=mathbbA^1times(mathbbA^1setminus0)subset X$ contains the entire fiber of $f$ over $0$, but does not contain the preimage of any neighborhood of $0$.
answered Jul 27 at 5:34
Eric Wofsey
162k12188299
add a comment |Â
up vote
1
down vote
No. For instance, working over your favorite field, let $X=mathbbA^2setminus(0, 0)$ and let $f:XtomathbbA^1$ be the first projection. Then the open set $U=mathbbA^1times(mathbbA^1setminus0)subset X$ contains the entire fiber of $f$ over $0$, but does not contain the preimage of any neighborhood of $0$.
answered Jul 27 at 5:34
Eric Wofsey
162k12188299
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No. For instance, working over your favorite field, let $X=mathbbA^2setminus(0, 0)$ and let $f:XtomathbbA^1$ be the first projection. Then the open set $U=mathbbA^1times(mathbbA^1setminus0)subset X$ contains the entire fiber of $f$ over $0$, but does not contain the preimage of any neighborhood of $0$.
answered Jul 27 at 5:34
Eric Wofsey
162k12188299
No. For instance, working over your favorite field, let $X=mathbbA^2setminus(0, 0)$ and let $f:XtomathbbA^1$ be the first projection. Then the open set $U=mathbbA^1times(mathbbA^1setminus0)subset X$ contains the entire fiber of $f$ over $0$, but does not contain the preimage of any neighborhood of $0$.
answered Jul 27 at 5:34
Eric Wofsey
162k12188299
answered Jul 27 at 5:34
Eric Wofsey
162k12188299
answered Jul 27 at 5:34
Eric Wofsey
162k12188299
answered Jul 27 at 5:34
Eric Wofsey
162k12188299
162k12188299
add a comment |Â
add a comment |Â
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I have clarified the question to what I assume was your intended meaning. If your intended meaning was something else feel free to revert.
– Eric Wofsey
Jul 27 at 5:38