Extension of a smooth map from the boundary of a smooth manifold
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1
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Suppose $f: partial M to S^n$ is a smooth map, is it always possible to extend it to a map on $M$?
If $dim M > n$ it is not always possible for sure. Just consider $overline{B}^m$. What about $dim M leq n$? The usual partition of unity argument does not work because one cannot guarantee that the extension stays in $S^n$.
differential-geometry
asked Nov 16 at 7:05
z.z
34017
add a comment |
up vote
1
down vote
favorite
Suppose $f: partial M to S^n$ is a smooth map, is it always possible to extend it to a map on $M$?
If $dim M > n$ it is not always possible for sure. Just consider $overline{B}^m$. What about $dim M leq n$? The usual partition of unity argument does not work because one cannot guarantee that the extension stays in $S^n$.
differential-geometry
asked Nov 16 at 7:05
z.z
34017
Do you mean "what about $dim Mleq n$"?
– Amitai Yuval
Nov 16 at 9:52
Because if you do, then the answer to your question is "yes".
– Amitai Yuval
Nov 16 at 12:48
Oh yes! I edited the question. Can you explain why? @AmitaiYuval
– z.z
Nov 16 at 14:58
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $f: partial M to S^n$ is a smooth map, is it always possible to extend it to a map on $M$?
If $dim M > n$ it is not always possible for sure. Just consider $overline{B}^m$. What about $dim M leq n$? The usual partition of unity argument does not work because one cannot guarantee that the extension stays in $S^n$.
differential-geometry
asked Nov 16 at 7:05
z.z
34017
Suppose $f: partial M to S^n$ is a smooth map, is it always possible to extend it to a map on $M$?
If $dim M > n$ it is not always possible for sure. Just consider $overline{B}^m$. What about $dim M leq n$? The usual partition of unity argument does not work because one cannot guarantee that the extension stays in $S^n$.
differential-geometry
differential-geometry
asked Nov 16 at 7:05
z.z
34017
asked Nov 16 at 7:05
z.z
34017
edited Nov 16 at 14:58
asked Nov 16 at 7:05
z.z
34017
asked Nov 16 at 7:05
z.z
34017
asked Nov 16 at 7:05
z.z
34017
34017
Do you mean "what about $dim Mleq n$"?
– Amitai Yuval
Nov 16 at 9:52
Because if you do, then the answer to your question is "yes".
– Amitai Yuval
Nov 16 at 12:48
Oh yes! I edited the question. Can you explain why? @AmitaiYuval
– z.z
Nov 16 at 14:58
add a comment |
Do you mean "what about $dim Mleq n$"?
– Amitai Yuval
Nov 16 at 9:52
Because if you do, then the answer to your question is "yes".
– Amitai Yuval
Nov 16 at 12:48
Oh yes! I edited the question. Can you explain why? @AmitaiYuval
– z.z
Nov 16 at 14:58
Do you mean "what about $dim Mleq n$"?
– Amitai Yuval
Nov 16 at 9:52
Do you mean "what about $dim Mleq n$"?
– Amitai Yuval
Nov 16 at 9:52
Because if you do, then the answer to your question is "yes".
– Amitai Yuval
Nov 16 at 12:48
Because if you do, then the answer to your question is "yes".
– Amitai Yuval
Nov 16 at 12:48
Oh yes! I edited the question. Can you explain why? @AmitaiYuval
– z.z
Nov 16 at 14:58
Oh yes! I edited the question. Can you explain why? @AmitaiYuval
– z.z
Nov 16 at 14:58
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
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accepted
If $dim Mleq n,$ then $dimpartial M<n=dim S^n,$ and this implies that $f$ is not surjective. So, you can think of $f$ as a map to $mathbb{R}^n$, and as such, it can be extended from $partial M$ to $M$.
answered Nov 16 at 16:03
Amitai Yuval
14.9k11026
Great! But just to be more specific: there is always a chart of $S^n$ covering $S^n {some point}$, then your argument works. Is my interpretation correct?
– z.z
Nov 16 at 17:00
@z.z Yes, this is what I'm saying.
– Amitai Yuval
Nov 16 at 18:25
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $dim Mleq n,$ then $dimpartial M<n=dim S^n,$ and this implies that $f$ is not surjective. So, you can think of $f$ as a map to $mathbb{R}^n$, and as such, it can be extended from $partial M$ to $M$.
answered Nov 16 at 16:03
Amitai Yuval
14.9k11026
Great! But just to be more specific: there is always a chart of $S^n$ covering $S^n {some point}$, then your argument works. Is my interpretation correct?
– z.z
Nov 16 at 17:00
@z.z Yes, this is what I'm saying.
– Amitai Yuval
Nov 16 at 18:25
add a comment |
up vote
1
down vote
accepted
If $dim Mleq n,$ then $dimpartial M<n=dim S^n,$ and this implies that $f$ is not surjective. So, you can think of $f$ as a map to $mathbb{R}^n$, and as such, it can be extended from $partial M$ to $M$.
answered Nov 16 at 16:03
Amitai Yuval
14.9k11026
Great! But just to be more specific: there is always a chart of $S^n$ covering $S^n {some point}$, then your argument works. Is my interpretation correct?
– z.z
Nov 16 at 17:00
@z.z Yes, this is what I'm saying.
– Amitai Yuval
Nov 16 at 18:25
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $dim Mleq n,$ then $dimpartial M<n=dim S^n,$ and this implies that $f$ is not surjective. So, you can think of $f$ as a map to $mathbb{R}^n$, and as such, it can be extended from $partial M$ to $M$.
answered Nov 16 at 16:03
Amitai Yuval
14.9k11026
If $dim Mleq n,$ then $dimpartial M<n=dim S^n,$ and this implies that $f$ is not surjective. So, you can think of $f$ as a map to $mathbb{R}^n$, and as such, it can be extended from $partial M$ to $M$.
answered Nov 16 at 16:03
Amitai Yuval
14.9k11026
answered Nov 16 at 16:03
Amitai Yuval
14.9k11026
answered Nov 16 at 16:03
Amitai Yuval
14.9k11026
answered Nov 16 at 16:03
Amitai Yuval
14.9k11026
14.9k11026
Great! But just to be more specific: there is always a chart of $S^n$ covering $S^n {some point}$, then your argument works. Is my interpretation correct?
– z.z
Nov 16 at 17:00
@z.z Yes, this is what I'm saying.
– Amitai Yuval
Nov 16 at 18:25
add a comment |
Great! But just to be more specific: there is always a chart of $S^n$ covering $S^n {some point}$, then your argument works. Is my interpretation correct?
– z.z
Nov 16 at 17:00
@z.z Yes, this is what I'm saying.
– Amitai Yuval
Nov 16 at 18:25
Great! But just to be more specific: there is always a chart of $S^n$ covering $S^n {some point}$, then your argument works. Is my interpretation correct?
– z.z
Nov 16 at 17:00
Great! But just to be more specific: there is always a chart of $S^n$ covering $S^n {some point}$, then your argument works. Is my interpretation correct?
– z.z
Nov 16 at 17:00
@z.z Yes, this is what I'm saying.
– Amitai Yuval
Nov 16 at 18:25
@z.z Yes, this is what I'm saying.
– Amitai Yuval
Nov 16 at 18:25
add a comment |
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Do you mean "what about $dim Mleq n$"?
– Amitai Yuval
Nov 16 at 9:52
Because if you do, then the answer to your question is "yes".
– Amitai Yuval
Nov 16 at 12:48
Oh yes! I edited the question. Can you explain why? @AmitaiYuval
– z.z
Nov 16 at 14:58