Open Mapping Theorem from $C^n$ to $C^n$
2
$begingroup$
I am looking for an Open Mapping Theorem for a holomorphic function $f: U subset mathbb{C}^n to mathbb{C}^n$ where $U$ is a domain. I believe the following is true:
Let $f: U subset mathbb{C}^n to mathbb{C}^n$ be holomorphic, where $U$ is a domain. Suppose the determinant of the Jacobian of $f$ is not identically zero on $U$. Then $f(U)$ is open.
References and thoughts are welcome.
differential-geometry several-complex-variables
asked Feb 19 '16 at 9:32
wellfedgremlinwellfedgremlin
382119
$endgroup$
add a comment |
2
$begingroup$
I am looking for an Open Mapping Theorem for a holomorphic function $f: U subset mathbb{C}^n to mathbb{C}^n$ where $U$ is a domain. I believe the following is true:
Let $f: U subset mathbb{C}^n to mathbb{C}^n$ be holomorphic, where $U$ is a domain. Suppose the determinant of the Jacobian of $f$ is not identically zero on $U$. Then $f(U)$ is open.
References and thoughts are welcome.
differential-geometry several-complex-variables
asked Feb 19 '16 at 9:32
wellfedgremlinwellfedgremlin
382119
$endgroup$
$begingroup$
@user251257 I'm not sure - I would definitely agree with you if we instead assumed the determinant of the Jacobian is nonvanishing
$endgroup$
– wellfedgremlin
Feb 19 '16 at 9:56
$begingroup$
ah, I misunderstood your statement
$endgroup$
– user251257
Feb 19 '16 at 9:58
add a comment |
2
2
2
3
$begingroup$
I am looking for an Open Mapping Theorem for a holomorphic function $f: U subset mathbb{C}^n to mathbb{C}^n$ where $U$ is a domain. I believe the following is true:
Let $f: U subset mathbb{C}^n to mathbb{C}^n$ be holomorphic, where $U$ is a domain. Suppose the determinant of the Jacobian of $f$ is not identically zero on $U$. Then $f(U)$ is open.
References and thoughts are welcome.
differential-geometry several-complex-variables
asked Feb 19 '16 at 9:32
wellfedgremlinwellfedgremlin
382119
$endgroup$
I am looking for an Open Mapping Theorem for a holomorphic function $f: U subset mathbb{C}^n to mathbb{C}^n$ where $U$ is a domain. I believe the following is true:
Let $f: U subset mathbb{C}^n to mathbb{C}^n$ be holomorphic, where $U$ is a domain. Suppose the determinant of the Jacobian of $f$ is not identically zero on $U$. Then $f(U)$ is open.
References and thoughts are welcome.
differential-geometry several-complex-variables
differential-geometry several-complex-variables
asked Feb 19 '16 at 9:32
wellfedgremlinwellfedgremlin
382119
asked Feb 19 '16 at 9:32
wellfedgremlinwellfedgremlin
382119
edited Feb 19 '16 at 10:01
wellfedgremlin
asked Feb 19 '16 at 9:32
wellfedgremlinwellfedgremlin
382119
asked Feb 19 '16 at 9:32
wellfedgremlinwellfedgremlin
382119
asked Feb 19 '16 at 9:32
wellfedgremlinwellfedgremlin
382119
382119
$begingroup$
@user251257 I'm not sure - I would definitely agree with you if we instead assumed the determinant of the Jacobian is nonvanishing
$endgroup$
– wellfedgremlin
Feb 19 '16 at 9:56
$begingroup$
ah, I misunderstood your statement
$endgroup$
– user251257
Feb 19 '16 at 9:58
add a comment |
$begingroup$
@user251257 I'm not sure - I would definitely agree with you if we instead assumed the determinant of the Jacobian is nonvanishing
$endgroup$
– wellfedgremlin
Feb 19 '16 at 9:56
$begingroup$
ah, I misunderstood your statement
$endgroup$
– user251257
Feb 19 '16 at 9:58
$begingroup$
@user251257 I'm not sure - I would definitely agree with you if we instead assumed the determinant of the Jacobian is nonvanishing
$endgroup$
– wellfedgremlin
Feb 19 '16 at 9:56
$begingroup$
@user251257 I'm not sure - I would definitely agree with you if we instead assumed the determinant of the Jacobian is nonvanishing
$endgroup$
– wellfedgremlin
Feb 19 '16 at 9:56
$begingroup$
ah, I misunderstood your statement
$endgroup$
– user251257
Feb 19 '16 at 9:58
$begingroup$
ah, I misunderstood your statement
$endgroup$
– user251257
Feb 19 '16 at 9:58
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Not true: $(x,y)mapsto(x, xy)$.
edited Feb 19 '16 at 11:07
Daniel Fischer♦
174k16167287
answered Feb 19 '16 at 10:48
user312865user312865
1451
$endgroup$
$begingroup$
The map provided is an open map. This is not a counterexample.
$endgroup$
– wellfedgremlin
Feb 19 '16 at 17:46
$begingroup$
Denote the given map by $f$. I show it is not open at $(0,0)$. Let $r_1<1$. I show that for any $r_2>0$, there exists $z,win B_{r_2}(0)$ such that $(z,w)notin f(B_{r_1}(0)times B_{r_1}(0))$. Indeed, choose $z=r_2r_1, w=r_2$, then the only solution is $x=r_2r_1, y=1/r_1$.
$endgroup$
– Bananach
Feb 26 '16 at 7:19
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
0
$begingroup$
Not true: $(x,y)mapsto(x, xy)$.
edited Feb 19 '16 at 11:07
Daniel Fischer♦
174k16167287
answered Feb 19 '16 at 10:48
user312865user312865
1451
$endgroup$
$begingroup$
The map provided is an open map. This is not a counterexample.
$endgroup$
– wellfedgremlin
Feb 19 '16 at 17:46
$begingroup$
Denote the given map by $f$. I show it is not open at $(0,0)$. Let $r_1<1$. I show that for any $r_2>0$, there exists $z,win B_{r_2}(0)$ such that $(z,w)notin f(B_{r_1}(0)times B_{r_1}(0))$. Indeed, choose $z=r_2r_1, w=r_2$, then the only solution is $x=r_2r_1, y=1/r_1$.
$endgroup$
– Bananach
Feb 26 '16 at 7:19
add a comment |
0
$begingroup$
Not true: $(x,y)mapsto(x, xy)$.
edited Feb 19 '16 at 11:07
Daniel Fischer♦
174k16167287
answered Feb 19 '16 at 10:48
user312865user312865
1451
$endgroup$
$begingroup$
The map provided is an open map. This is not a counterexample.
$endgroup$
– wellfedgremlin
Feb 19 '16 at 17:46
$begingroup$
Denote the given map by $f$. I show it is not open at $(0,0)$. Let $r_1<1$. I show that for any $r_2>0$, there exists $z,win B_{r_2}(0)$ such that $(z,w)notin f(B_{r_1}(0)times B_{r_1}(0))$. Indeed, choose $z=r_2r_1, w=r_2$, then the only solution is $x=r_2r_1, y=1/r_1$.
$endgroup$
– Bananach
Feb 26 '16 at 7:19
add a comment |
0
0
0
$begingroup$
Not true: $(x,y)mapsto(x, xy)$.
edited Feb 19 '16 at 11:07
Daniel Fischer♦
174k16167287
answered Feb 19 '16 at 10:48
user312865user312865
1451
$endgroup$
Not true: $(x,y)mapsto(x, xy)$.
edited Feb 19 '16 at 11:07
Daniel Fischer♦
174k16167287
answered Feb 19 '16 at 10:48
user312865user312865
1451
edited Feb 19 '16 at 11:07
Daniel Fischer♦
174k16167287
edited Feb 19 '16 at 11:07
Daniel Fischer♦
174k16167287
edited Feb 19 '16 at 11:07
Daniel Fischer♦
174k16167287
174k16167287
answered Feb 19 '16 at 10:48
user312865user312865
1451
answered Feb 19 '16 at 10:48
user312865user312865
1451
answered Feb 19 '16 at 10:48
user312865user312865
1451
1451
$begingroup$
The map provided is an open map. This is not a counterexample.
$endgroup$
– wellfedgremlin
Feb 19 '16 at 17:46
$begingroup$
Denote the given map by $f$. I show it is not open at $(0,0)$. Let $r_1<1$. I show that for any $r_2>0$, there exists $z,win B_{r_2}(0)$ such that $(z,w)notin f(B_{r_1}(0)times B_{r_1}(0))$. Indeed, choose $z=r_2r_1, w=r_2$, then the only solution is $x=r_2r_1, y=1/r_1$.
$endgroup$
– Bananach
Feb 26 '16 at 7:19
add a comment |
$begingroup$
The map provided is an open map. This is not a counterexample.
$endgroup$
– wellfedgremlin
Feb 19 '16 at 17:46
$begingroup$
Denote the given map by $f$. I show it is not open at $(0,0)$. Let $r_1<1$. I show that for any $r_2>0$, there exists $z,win B_{r_2}(0)$ such that $(z,w)notin f(B_{r_1}(0)times B_{r_1}(0))$. Indeed, choose $z=r_2r_1, w=r_2$, then the only solution is $x=r_2r_1, y=1/r_1$.
$endgroup$
– Bananach
Feb 26 '16 at 7:19
$begingroup$
The map provided is an open map. This is not a counterexample.
$endgroup$
– wellfedgremlin
Feb 19 '16 at 17:46
$begingroup$
The map provided is an open map. This is not a counterexample.
$endgroup$
– wellfedgremlin
Feb 19 '16 at 17:46
$begingroup$
Denote the given map by $f$. I show it is not open at $(0,0)$. Let $r_1<1$. I show that for any $r_2>0$, there exists $z,win B_{r_2}(0)$ such that $(z,w)notin f(B_{r_1}(0)times B_{r_1}(0))$. Indeed, choose $z=r_2r_1, w=r_2$, then the only solution is $x=r_2r_1, y=1/r_1$.
$endgroup$
– Bananach
Feb 26 '16 at 7:19
$begingroup$
Denote the given map by $f$. I show it is not open at $(0,0)$. Let $r_1<1$. I show that for any $r_2>0$, there exists $z,win B_{r_2}(0)$ such that $(z,w)notin f(B_{r_1}(0)times B_{r_1}(0))$. Indeed, choose $z=r_2r_1, w=r_2$, then the only solution is $x=r_2r_1, y=1/r_1$.
$endgroup$
– Bananach
Feb 26 '16 at 7:19
add a comment |
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$begingroup$
@user251257 I'm not sure - I would definitely agree with you if we instead assumed the determinant of the Jacobian is nonvanishing
$endgroup$
– wellfedgremlin
Feb 19 '16 at 9:56
$begingroup$
ah, I misunderstood your statement
$endgroup$
– user251257
Feb 19 '16 at 9:58