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0

$begingroup$

1. Suppose $M$ is a manifold, $N$ is a compact submanifold(with non-empty boundary) of same dimension, $f:Nrightarrowmathbb R^{n}$ is an embedding into sufficiently-high-dimensional Euclidean space. Under what extra assumptions can we find an extension of $f$ to the whole $M$?




2. A possibly relative question: with $M$ and $N$ same as above, suppose in the following diagram $i$ is inclusion, $f$ and $g$ are embeddings. When is there an embedding $j$ that makes the diagram commute?

$require{AMScd}$
begin{CD}
N @>displaystyle i>> M\
@V displaystyle f V V# @VV displaystyle g V\
mathbb R^{n} @>>displaystyle j> mathbb R^{n}
end{CD}

manifolds differential-topology compact-manifolds

share|cite|improve this question

edited Mar 13 at 16:49

1830rbc03

asked Mar 13 at 15:17

1830rbc031830rbc03

41048

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assume $N$ is a manifold with boundary? Otherwise the only cases of a submanifold of the same dimension are $M=N$ or $M$ is not connected and $N$ is one or several of the connected components.
  $endgroup$
  – quarague
  Mar 13 at 15:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's so vague question. I mean everything can always locally embedded inside $R^n$. But unless you specified what manifold M and N you are interested and what kind of embedding we are talking its next to impossible to have an answer. For example take an nbd of arbitrat complicated knot in M and embed it as an unknot...
  $endgroup$
  – Anubhav Mukherjee
  Mar 13 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anubhav Mukherjee I know it is vague, so I am looking for advice for "extra assumptions"......You may assume M is also compact or something about the boundary if you like. Please forgive my ignorance, do you mean in your example the embedding cannot be extended?
  $endgroup$
  – 1830rbc03
  Mar 13 at 19:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this is an interesting question, even if it might be difficult to answer. For example the unit disk embeds in $mathbb{R}^3$. The unit disk can be taken to be N for any 2-manifold M. So let's pick something that won't work: the Klein bottle.
  $endgroup$
  – Prototank
  Mar 14 at 13:09
  

  
  
  
  

add a comment | 

0

$begingroup$

1. Suppose $M$ is a manifold, $N$ is a compact submanifold(with non-empty boundary) of same dimension, $f:Nrightarrowmathbb R^{n}$ is an embedding into sufficiently-high-dimensional Euclidean space. Under what extra assumptions can we find an extension of $f$ to the whole $M$?




2. A possibly relative question: with $M$ and $N$ same as above, suppose in the following diagram $i$ is inclusion, $f$ and $g$ are embeddings. When is there an embedding $j$ that makes the diagram commute?

$require{AMScd}$
begin{CD}
N @>displaystyle i>> M\
@V displaystyle f V V# @VV displaystyle g V\
mathbb R^{n} @>>displaystyle j> mathbb R^{n}
end{CD}

manifolds differential-topology compact-manifolds

share|cite|improve this question

edited Mar 13 at 16:49

1830rbc03

asked Mar 13 at 15:17

1830rbc031830rbc03

41048

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assume $N$ is a manifold with boundary? Otherwise the only cases of a submanifold of the same dimension are $M=N$ or $M$ is not connected and $N$ is one or several of the connected components.
  $endgroup$
  – quarague
  Mar 13 at 15:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's so vague question. I mean everything can always locally embedded inside $R^n$. But unless you specified what manifold M and N you are interested and what kind of embedding we are talking its next to impossible to have an answer. For example take an nbd of arbitrat complicated knot in M and embed it as an unknot...
  $endgroup$
  – Anubhav Mukherjee
  Mar 13 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anubhav Mukherjee I know it is vague, so I am looking for advice for "extra assumptions"......You may assume M is also compact or something about the boundary if you like. Please forgive my ignorance, do you mean in your example the embedding cannot be extended?
  $endgroup$
  – 1830rbc03
  Mar 13 at 19:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this is an interesting question, even if it might be difficult to answer. For example the unit disk embeds in $mathbb{R}^3$. The unit disk can be taken to be N for any 2-manifold M. So let's pick something that won't work: the Klein bottle.
  $endgroup$
  – Prototank
  Mar 14 at 13:09
  

  
  
  
  

add a comment | 

$begingroup$

1. Suppose $M$ is a manifold, $N$ is a compact submanifold(with non-empty boundary) of same dimension, $f:Nrightarrowmathbb R^{n}$ is an embedding into sufficiently-high-dimensional Euclidean space. Under what extra assumptions can we find an extension of $f$ to the whole $M$?




2. A possibly relative question: with $M$ and $N$ same as above, suppose in the following diagram $i$ is inclusion, $f$ and $g$ are embeddings. When is there an embedding $j$ that makes the diagram commute?

$require{AMScd}$
begin{CD}
N @>displaystyle i>> M\
@V displaystyle f V V# @VV displaystyle g V\
mathbb R^{n} @>>displaystyle j> mathbb R^{n}
end{CD}

manifolds differential-topology compact-manifolds

share|cite|improve this question

edited Mar 13 at 16:49

1830rbc03

asked Mar 13 at 15:17

1830rbc031830rbc03

41048

$endgroup$

share|cite|improve this question

edited Mar 13 at 16:49

1830rbc03

asked Mar 13 at 15:17

1830rbc031830rbc03

41048

share|cite|improve this question

edited Mar 13 at 16:49

1830rbc03

asked Mar 13 at 15:17

1830rbc031830rbc03

41048

asked Mar 13 at 15:17

1830rbc031830rbc03

41048

asked Mar 13 at 15:17

1830rbc031830rbc03

41048