Are the two definitions of that a map $F:Ato N$ is smooth on $A$ equivalent?
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In the above pictures, we can see two definitions of that a map $F:Ato N$ is smooth on $A$, are they equivalent?
manifolds smooth-manifolds manifolds-with-boundary
asked Dec 11 '18 at 11:56
Born to be proudBorn to be proud
848510
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add a comment |
0
$begingroup$
In the above pictures, we can see two definitions of that a map $F:Ato N$ is smooth on $A$, are they equivalent?
manifolds smooth-manifolds manifolds-with-boundary
asked Dec 11 '18 at 11:56
Born to be proudBorn to be proud
848510
$endgroup$
$begingroup$
@JackLee Please take a look at my question.
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– Born to be proud
Dec 11 '18 at 13:35
$begingroup$
They're not always equivalent in case $N$ has nonempty boundary. Here's a counterexample: Let $M=mathbb R$, $N = [0,infty)$, $A = [0,infty)subseteq M$, and let $Fcolon Ato N$ be the identity map. Then $F$ is not smooth by the definition in the text, because there is no neighborhood of $0$ in $mathbb R$ on which $F$ has a smooth extension as a map into $N$. But the coordinate representation of $F$ using standard coordinates in domain and codomain (which is essentially the same map) has an extension to an open subset of $mathbb R$ as a map into $mathbb R$.
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– Jack Lee
Dec 11 '18 at 23:15
$begingroup$
See also this question.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:17
$begingroup$
@JackLee Thank you very much!
$endgroup$
– Born to be proud
Dec 11 '18 at 23:54
add a comment |
0
0
0
$begingroup$
In the above pictures, we can see two definitions of that a map $F:Ato N$ is smooth on $A$, are they equivalent?
manifolds smooth-manifolds manifolds-with-boundary
asked Dec 11 '18 at 11:56
Born to be proudBorn to be proud
848510
$endgroup$
In the above pictures, we can see two definitions of that a map $F:Ato N$ is smooth on $A$, are they equivalent?
manifolds smooth-manifolds manifolds-with-boundary
manifolds smooth-manifolds manifolds-with-boundary
asked Dec 11 '18 at 11:56
Born to be proudBorn to be proud
848510
asked Dec 11 '18 at 11:56
Born to be proudBorn to be proud
848510
asked Dec 11 '18 at 11:56
Born to be proudBorn to be proud
848510
asked Dec 11 '18 at 11:56
Born to be proudBorn to be proud
848510
asked Dec 11 '18 at 11:56
Born to be proudBorn to be proud
848510
848510
$begingroup$
@JackLee Please take a look at my question.
$endgroup$
– Born to be proud
Dec 11 '18 at 13:35
$begingroup$
They're not always equivalent in case $N$ has nonempty boundary. Here's a counterexample: Let $M=mathbb R$, $N = [0,infty)$, $A = [0,infty)subseteq M$, and let $Fcolon Ato N$ be the identity map. Then $F$ is not smooth by the definition in the text, because there is no neighborhood of $0$ in $mathbb R$ on which $F$ has a smooth extension as a map into $N$. But the coordinate representation of $F$ using standard coordinates in domain and codomain (which is essentially the same map) has an extension to an open subset of $mathbb R$ as a map into $mathbb R$.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:15
$begingroup$
See also this question.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:17
$begingroup$
@JackLee Thank you very much!
$endgroup$
– Born to be proud
Dec 11 '18 at 23:54
add a comment |
$begingroup$
@JackLee Please take a look at my question.
$endgroup$
– Born to be proud
Dec 11 '18 at 13:35
$begingroup$
They're not always equivalent in case $N$ has nonempty boundary. Here's a counterexample: Let $M=mathbb R$, $N = [0,infty)$, $A = [0,infty)subseteq M$, and let $Fcolon Ato N$ be the identity map. Then $F$ is not smooth by the definition in the text, because there is no neighborhood of $0$ in $mathbb R$ on which $F$ has a smooth extension as a map into $N$. But the coordinate representation of $F$ using standard coordinates in domain and codomain (which is essentially the same map) has an extension to an open subset of $mathbb R$ as a map into $mathbb R$.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:15
$begingroup$
See also this question.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:17
$begingroup$
@JackLee Thank you very much!
$endgroup$
– Born to be proud
Dec 11 '18 at 23:54
$begingroup$
@JackLee Please take a look at my question.
$endgroup$
– Born to be proud
Dec 11 '18 at 13:35
$begingroup$
@JackLee Please take a look at my question.
$endgroup$
– Born to be proud
Dec 11 '18 at 13:35
$begingroup$
They're not always equivalent in case $N$ has nonempty boundary. Here's a counterexample: Let $M=mathbb R$, $N = [0,infty)$, $A = [0,infty)subseteq M$, and let $Fcolon Ato N$ be the identity map. Then $F$ is not smooth by the definition in the text, because there is no neighborhood of $0$ in $mathbb R$ on which $F$ has a smooth extension as a map into $N$. But the coordinate representation of $F$ using standard coordinates in domain and codomain (which is essentially the same map) has an extension to an open subset of $mathbb R$ as a map into $mathbb R$.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:15
$begingroup$
They're not always equivalent in case $N$ has nonempty boundary. Here's a counterexample: Let $M=mathbb R$, $N = [0,infty)$, $A = [0,infty)subseteq M$, and let $Fcolon Ato N$ be the identity map. Then $F$ is not smooth by the definition in the text, because there is no neighborhood of $0$ in $mathbb R$ on which $F$ has a smooth extension as a map into $N$. But the coordinate representation of $F$ using standard coordinates in domain and codomain (which is essentially the same map) has an extension to an open subset of $mathbb R$ as a map into $mathbb R$.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:15
$begingroup$
See also this question.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:17
$begingroup$
See also this question.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:17
$begingroup$
@JackLee Thank you very much!
$endgroup$
– Born to be proud
Dec 11 '18 at 23:54
$begingroup$
@JackLee Thank you very much!
$endgroup$
– Born to be proud
Dec 11 '18 at 23:54
add a comment |
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$begingroup$
@JackLee Please take a look at my question.
$endgroup$
– Born to be proud
Dec 11 '18 at 13:35
$begingroup$
They're not always equivalent in case $N$ has nonempty boundary. Here's a counterexample: Let $M=mathbb R$, $N = [0,infty)$, $A = [0,infty)subseteq M$, and let $Fcolon Ato N$ be the identity map. Then $F$ is not smooth by the definition in the text, because there is no neighborhood of $0$ in $mathbb R$ on which $F$ has a smooth extension as a map into $N$. But the coordinate representation of $F$ using standard coordinates in domain and codomain (which is essentially the same map) has an extension to an open subset of $mathbb R$ as a map into $mathbb R$.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:15
$begingroup$
See also this question.
$endgroup$
– Jack Lee
Dec 11 '18 at 23:17
$begingroup$
@JackLee Thank you very much!
$endgroup$
– Born to be proud
Dec 11 '18 at 23:54