Mixing
If $f: U to mathbb{R}^{n}$ is a $C^{1}$ function, $U subset mathbb{R}^{n}$ open, show that $f$ is an open...
$begingroup$
Let $f: U to mathbb{R}^{n}$ be a $C^{1}$ function with $U subset mathbb{R}^{n}$ open. Show that $f$ is an open map.
If $det Df neq 0$ (maybe it's not so direct), I think I can use the Inverse Function Theorem for show that $f$ is a local diffeomorphism and so, write $f(O)$ as union of open sets for each $O subset U$ open. But without this hypothesis, how can I ensure the same conclusion? I really dont see the way.
real-analysis derivatives
asked Jan 18 at 17:59
Lucas CorrêaLucas Corrêa
1,5771321
$endgroup$
add a comment |
$begingroup$
Let $f: U to mathbb{R}^{n}$ be a $C^{1}$ function with $U subset mathbb{R}^{n}$ open. Show that $f$ is an open map.
If $det Df neq 0$ (maybe it's not so direct), I think I can use the Inverse Function Theorem for show that $f$ is a local diffeomorphism and so, write $f(O)$ as union of open sets for each $O subset U$ open. But without this hypothesis, how can I ensure the same conclusion? I really dont see the way.
real-analysis derivatives
asked Jan 18 at 17:59
Lucas CorrêaLucas Corrêa
1,5771321
$endgroup$
1
$begingroup$
Was there supposed to be an assumption of injectivity or something?
$endgroup$
– Randall
Jan 18 at 18:45
$begingroup$
@Randall no, I know that $f$ need not to be continuously differentiable, but the way it's written, it seemed very strange to me. Fortunately the counterexample is trivial. I think the author forgot to write some more hypothesis.
$endgroup$
– Lucas Corrêa
Jan 18 at 19:44
add a comment |
$begingroup$
Let $f: U to mathbb{R}^{n}$ be a $C^{1}$ function with $U subset mathbb{R}^{n}$ open. Show that $f$ is an open map.
If $det Df neq 0$ (maybe it's not so direct), I think I can use the Inverse Function Theorem for show that $f$ is a local diffeomorphism and so, write $f(O)$ as union of open sets for each $O subset U$ open. But without this hypothesis, how can I ensure the same conclusion? I really dont see the way.
real-analysis derivatives
asked Jan 18 at 17:59
Lucas CorrêaLucas Corrêa
1,5771321
$endgroup$
Let $f: U to mathbb{R}^{n}$ be a $C^{1}$ function with $U subset mathbb{R}^{n}$ open. Show that $f$ is an open map.
If $det Df neq 0$ (maybe it's not so direct), I think I can use the Inverse Function Theorem for show that $f$ is a local diffeomorphism and so, write $f(O)$ as union of open sets for each $O subset U$ open. But without this hypothesis, how can I ensure the same conclusion? I really dont see the way.
real-analysis derivatives
real-analysis derivatives
asked Jan 18 at 17:59
Lucas CorrêaLucas Corrêa
1,5771321
asked Jan 18 at 17:59
Lucas CorrêaLucas Corrêa
1,5771321
asked Jan 18 at 17:59
Lucas CorrêaLucas Corrêa
1,5771321
asked Jan 18 at 17:59
Lucas CorrêaLucas Corrêa
1,5771321
asked Jan 18 at 17:59
Lucas CorrêaLucas Corrêa
1,5771321
1,5771321
1
$begingroup$
Was there supposed to be an assumption of injectivity or something?
$endgroup$
– Randall
Jan 18 at 18:45
$begingroup$
@Randall no, I know that $f$ need not to be continuously differentiable, but the way it's written, it seemed very strange to me. Fortunately the counterexample is trivial. I think the author forgot to write some more hypothesis.
$endgroup$
– Lucas Corrêa
Jan 18 at 19:44
add a comment |
1
$begingroup$
Was there supposed to be an assumption of injectivity or something?
$endgroup$
– Randall
Jan 18 at 18:45
$begingroup$
@Randall no, I know that $f$ need not to be continuously differentiable, but the way it's written, it seemed very strange to me. Fortunately the counterexample is trivial. I think the author forgot to write some more hypothesis.
$endgroup$
– Lucas Corrêa
Jan 18 at 19:44
1
1
$begingroup$
Was there supposed to be an assumption of injectivity or something?
$endgroup$
– Randall
Jan 18 at 18:45
$begingroup$
Was there supposed to be an assumption of injectivity or something?
$endgroup$
– Randall
Jan 18 at 18:45
$begingroup$
@Randall no, I know that $f$ need not to be continuously differentiable, but the way it's written, it seemed very strange to me. Fortunately the counterexample is trivial. I think the author forgot to write some more hypothesis.
$endgroup$
– Lucas Corrêa
Jan 18 at 19:44
$begingroup$
@Randall no, I know that $f$ need not to be continuously differentiable, but the way it's written, it seemed very strange to me. Fortunately the counterexample is trivial. I think the author forgot to write some more hypothesis.
$endgroup$
– Lucas Corrêa
Jan 18 at 19:44
add a comment |
1 Answer
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$begingroup$
You can't prove it, since it is false. Take any constant function, for instance.
answered Jan 18 at 18:09
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
I see! Thank you!
$endgroup$
– Lucas Corrêa
Jan 18 at 18:13
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can't prove it, since it is false. Take any constant function, for instance.
answered Jan 18 at 18:09
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
I see! Thank you!
$endgroup$
– Lucas Corrêa
Jan 18 at 18:13
add a comment |
$begingroup$
You can't prove it, since it is false. Take any constant function, for instance.
answered Jan 18 at 18:09
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
I see! Thank you!
$endgroup$
– Lucas Corrêa
Jan 18 at 18:13
add a comment |
$begingroup$
You can't prove it, since it is false. Take any constant function, for instance.
answered Jan 18 at 18:09
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
You can't prove it, since it is false. Take any constant function, for instance.
answered Jan 18 at 18:09
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 18:09
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 18:09
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 18 at 18:09
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
I see! Thank you!
$endgroup$
– Lucas Corrêa
Jan 18 at 18:13
add a comment |
$begingroup$
I see! Thank you!
$endgroup$
– Lucas Corrêa
Jan 18 at 18:13
$begingroup$
I see! Thank you!
$endgroup$
– Lucas Corrêa
Jan 18 at 18:13
$begingroup$
I see! Thank you!
$endgroup$
– Lucas Corrêa
Jan 18 at 18:13
add a comment |
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1
$begingroup$
Was there supposed to be an assumption of injectivity or something?
$endgroup$
– Randall
Jan 18 at 18:45
$begingroup$
@Randall no, I know that $f$ need not to be continuously differentiable, but the way it's written, it seemed very strange to me. Fortunately the counterexample is trivial. I think the author forgot to write some more hypothesis.
$endgroup$
– Lucas Corrêa
Jan 18 at 19:44