Mixing
Definition of smooth maps between manifolds in $mathbb{R}^n$
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I am currently studying smooth manifold using Jack Lee's introduction to smooth manifold.
In the book, given two manifolds $M in mathbb{R}^n$, $N in mathbb{R}^m$with smooth structures (maximal smooth atlas in which any two charts are compatible), a map between them is called smooth if its coordinate representation is smooth (in the usual sense from an open set in $mathbb{R}^n$).
There's also a definition of smooth map between a subset $A in mathbb{R}^n$, that is, if $f$ is a map from $A$ to $mathbb{R}^m$ such that for every point $x in A$, there exists an open neighborhood $U$ of $x$ and a smooth map from $U$ (in the usual sense) to $mathbb{R}^m$ such that $F|_{U cap A} = f$, then $f$ is said to be a smooth map from $A$ to $mathbb{R}^m$. In fact if I remember correctly this is the definition that Alan Pollack uses in his "differential topology".
Now, I remember my professor made a remark that is along the line of, if a manifold sits in $mathbb{R}^n$, the two notion of smoothness is equivalent. I can see that the second definition implies the first, however I don't see how the 1st definition implies the second. I have tried to come up with such an open neighborhood and an extension of such a map F, I don't think I'm supposed to construct them explicitly, I have been thinking of using inverse function theorem some how but I still haven't quite figure out how, any hint would be appreciated.
thank you!
differential-topology smooth-manifolds
edited Jan 29 at 15:08
J. W. Tanner
4,0611320
asked Jan 29 at 7:05
EcotisticianEcotistician
35718
$endgroup$
add a comment |
$begingroup$
I am currently studying smooth manifold using Jack Lee's introduction to smooth manifold.
In the book, given two manifolds $M in mathbb{R}^n$, $N in mathbb{R}^m$with smooth structures (maximal smooth atlas in which any two charts are compatible), a map between them is called smooth if its coordinate representation is smooth (in the usual sense from an open set in $mathbb{R}^n$).
There's also a definition of smooth map between a subset $A in mathbb{R}^n$, that is, if $f$ is a map from $A$ to $mathbb{R}^m$ such that for every point $x in A$, there exists an open neighborhood $U$ of $x$ and a smooth map from $U$ (in the usual sense) to $mathbb{R}^m$ such that $F|_{U cap A} = f$, then $f$ is said to be a smooth map from $A$ to $mathbb{R}^m$. In fact if I remember correctly this is the definition that Alan Pollack uses in his "differential topology".
Now, I remember my professor made a remark that is along the line of, if a manifold sits in $mathbb{R}^n$, the two notion of smoothness is equivalent. I can see that the second definition implies the first, however I don't see how the 1st definition implies the second. I have tried to come up with such an open neighborhood and an extension of such a map F, I don't think I'm supposed to construct them explicitly, I have been thinking of using inverse function theorem some how but I still haven't quite figure out how, any hint would be appreciated.
thank you!
differential-topology smooth-manifolds
edited Jan 29 at 15:08
J. W. Tanner
4,0611320
asked Jan 29 at 7:05
EcotisticianEcotistician
35718
$endgroup$
add a comment |
$begingroup$
I am currently studying smooth manifold using Jack Lee's introduction to smooth manifold.
In the book, given two manifolds $M in mathbb{R}^n$, $N in mathbb{R}^m$with smooth structures (maximal smooth atlas in which any two charts are compatible), a map between them is called smooth if its coordinate representation is smooth (in the usual sense from an open set in $mathbb{R}^n$).
There's also a definition of smooth map between a subset $A in mathbb{R}^n$, that is, if $f$ is a map from $A$ to $mathbb{R}^m$ such that for every point $x in A$, there exists an open neighborhood $U$ of $x$ and a smooth map from $U$ (in the usual sense) to $mathbb{R}^m$ such that $F|_{U cap A} = f$, then $f$ is said to be a smooth map from $A$ to $mathbb{R}^m$. In fact if I remember correctly this is the definition that Alan Pollack uses in his "differential topology".
Now, I remember my professor made a remark that is along the line of, if a manifold sits in $mathbb{R}^n$, the two notion of smoothness is equivalent. I can see that the second definition implies the first, however I don't see how the 1st definition implies the second. I have tried to come up with such an open neighborhood and an extension of such a map F, I don't think I'm supposed to construct them explicitly, I have been thinking of using inverse function theorem some how but I still haven't quite figure out how, any hint would be appreciated.
thank you!
differential-topology smooth-manifolds
edited Jan 29 at 15:08
J. W. Tanner
4,0611320
asked Jan 29 at 7:05
EcotisticianEcotistician
35718
$endgroup$
I am currently studying smooth manifold using Jack Lee's introduction to smooth manifold.
In the book, given two manifolds $M in mathbb{R}^n$, $N in mathbb{R}^m$with smooth structures (maximal smooth atlas in which any two charts are compatible), a map between them is called smooth if its coordinate representation is smooth (in the usual sense from an open set in $mathbb{R}^n$).
There's also a definition of smooth map between a subset $A in mathbb{R}^n$, that is, if $f$ is a map from $A$ to $mathbb{R}^m$ such that for every point $x in A$, there exists an open neighborhood $U$ of $x$ and a smooth map from $U$ (in the usual sense) to $mathbb{R}^m$ such that $F|_{U cap A} = f$, then $f$ is said to be a smooth map from $A$ to $mathbb{R}^m$. In fact if I remember correctly this is the definition that Alan Pollack uses in his "differential topology".
Now, I remember my professor made a remark that is along the line of, if a manifold sits in $mathbb{R}^n$, the two notion of smoothness is equivalent. I can see that the second definition implies the first, however I don't see how the 1st definition implies the second. I have tried to come up with such an open neighborhood and an extension of such a map F, I don't think I'm supposed to construct them explicitly, I have been thinking of using inverse function theorem some how but I still haven't quite figure out how, any hint would be appreciated.
thank you!
differential-topology smooth-manifolds
differential-topology smooth-manifolds
edited Jan 29 at 15:08
J. W. Tanner
4,0611320
asked Jan 29 at 7:05
EcotisticianEcotistician
35718
edited Jan 29 at 15:08
J. W. Tanner
4,0611320
asked Jan 29 at 7:05
EcotisticianEcotistician
35718
edited Jan 29 at 15:08
J. W. Tanner
4,0611320
edited Jan 29 at 15:08
J. W. Tanner
4,0611320
edited Jan 29 at 15:08
J. W. Tanner
4,0611320
4,0611320
asked Jan 29 at 7:05
EcotisticianEcotistician
35718
asked Jan 29 at 7:05
EcotisticianEcotistician
35718
asked Jan 29 at 7:05
EcotisticianEcotistician
35718
35718
add a comment |
add a comment |
1 Answer
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If your manifold is an embedded submanifold of $mathbb R^n$, then the two notions are equivalent, but you'll probably need to use some submanifold theory to prove it. See Lemma 5.34 (and the paragraph preceding it) in my smooth manifolds book (2nd ed.).
answered Jan 29 at 15:48
Jack LeeJack Lee
27.7k54868
$endgroup$
$begingroup$
Thank you Professor Lee,
$endgroup$
– Ecotistician
Jan 29 at 21:57
$begingroup$
You’re welcome!
$endgroup$
– Jack Lee
Jan 30 at 0:37
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
If your manifold is an embedded submanifold of $mathbb R^n$, then the two notions are equivalent, but you'll probably need to use some submanifold theory to prove it. See Lemma 5.34 (and the paragraph preceding it) in my smooth manifolds book (2nd ed.).
answered Jan 29 at 15:48
Jack LeeJack Lee
27.7k54868
$endgroup$
$begingroup$
Thank you Professor Lee,
$endgroup$
– Ecotistician
Jan 29 at 21:57
$begingroup$
You’re welcome!
$endgroup$
– Jack Lee
Jan 30 at 0:37
add a comment |
$begingroup$
If your manifold is an embedded submanifold of $mathbb R^n$, then the two notions are equivalent, but you'll probably need to use some submanifold theory to prove it. See Lemma 5.34 (and the paragraph preceding it) in my smooth manifolds book (2nd ed.).
answered Jan 29 at 15:48
Jack LeeJack Lee
27.7k54868
$endgroup$
$begingroup$
Thank you Professor Lee,
$endgroup$
– Ecotistician
Jan 29 at 21:57
$begingroup$
You’re welcome!
$endgroup$
– Jack Lee
Jan 30 at 0:37
add a comment |
$begingroup$
If your manifold is an embedded submanifold of $mathbb R^n$, then the two notions are equivalent, but you'll probably need to use some submanifold theory to prove it. See Lemma 5.34 (and the paragraph preceding it) in my smooth manifolds book (2nd ed.).
answered Jan 29 at 15:48
Jack LeeJack Lee
27.7k54868
$endgroup$
If your manifold is an embedded submanifold of $mathbb R^n$, then the two notions are equivalent, but you'll probably need to use some submanifold theory to prove it. See Lemma 5.34 (and the paragraph preceding it) in my smooth manifolds book (2nd ed.).
answered Jan 29 at 15:48
Jack LeeJack Lee
27.7k54868
answered Jan 29 at 15:48
Jack LeeJack Lee
27.7k54868
answered Jan 29 at 15:48
Jack LeeJack Lee
27.7k54868
answered Jan 29 at 15:48
Jack LeeJack Lee
27.7k54868
27.7k54868
$begingroup$
Thank you Professor Lee,
$endgroup$
– Ecotistician
Jan 29 at 21:57
$begingroup$
You’re welcome!
$endgroup$
– Jack Lee
Jan 30 at 0:37
add a comment |
$begingroup$
Thank you Professor Lee,
$endgroup$
– Ecotistician
Jan 29 at 21:57
$begingroup$
You’re welcome!
$endgroup$
– Jack Lee
Jan 30 at 0:37
$begingroup$
Thank you Professor Lee,
$endgroup$
– Ecotistician
Jan 29 at 21:57
$begingroup$
Thank you Professor Lee,
$endgroup$
– Ecotistician
Jan 29 at 21:57
$begingroup$
You’re welcome!
$endgroup$
– Jack Lee
Jan 30 at 0:37
$begingroup$
You’re welcome!
$endgroup$
– Jack Lee
Jan 30 at 0:37
add a comment |
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