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Is $pi: mathcalC^infty (M,N) to mathcalC^infty (S,N)$, $pi(f) = left. fright|_S$ a quocient map in the $mathcal C^1$ topology?

3

$begingroup$

Let $M, N$ be smooth manifolds (without boundary), so we can put a topology in the space $C^infty(M, N)$ using $mathcalC^1$ Whitney Topology.

Now, consider $Ssubset M$ a submanifold of $M$ with boundary such that $textdimS=textdimM$, using the same process we can put a topology in $C^infty(S,N)$ using the $mathcalC^1$ Whitney Topology. There is a natural projection of $C^infty(M, N)$ to $C^infty(S,N)$, definide by

beginalign*
pi: C^infty(M, N) &to C^infty(S,N)\
f&mapsto left.fright|_S.
endalign*

My Question: Is $pi$ an open map or at least a quocient map? (If necessary we can suppose that $M$ is a compact and connected manifold)

functions differential-topology

share|cite|improve this question

edited Mar 24 at 18:30

Matheus Manzatto

asked Mar 24 at 18:08

Matheus ManzattoMatheus Manzatto

1,2991626

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is naturally identifiable with the quotient map for the equivalence relation on $C^infty(M,N)$ given by $f sim g$ if $f|_S = g|_S$.
  $endgroup$
  – Paul Sinclair
  Mar 25 at 3:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried this identification. however, I was not able to show that these two topologies are the same.
  $endgroup$
  – Matheus Manzatto
  Mar 25 at 3:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is hard to show that $pi^-1(pi (A) $ is open if $A $ is open. It require the existence of a weird extension property.
  $endgroup$
  – Matheus Manzatto
  Mar 25 at 15:14
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $M, N$ be smooth manifolds (without boundary), so we can put a topology in the space $C^infty(M, N)$ using $mathcalC^1$ Whitney Topology.

Now, consider $Ssubset M$ a submanifold of $M$ with boundary such that $textdimS=textdimM$, using the same process we can put a topology in $C^infty(S,N)$ using the $mathcalC^1$ Whitney Topology. There is a natural projection of $C^infty(M, N)$ to $C^infty(S,N)$, definide by

beginalign*
pi: C^infty(M, N) &to C^infty(S,N)\
f&mapsto left.fright|_S.
endalign*

My Question: Is $pi$ an open map or at least a quocient map? (If necessary we can suppose that $M$ is a compact and connected manifold)

functions differential-topology

share|cite|improve this question

edited Mar 24 at 18:30

Matheus Manzatto

asked Mar 24 at 18:08

Matheus ManzattoMatheus Manzatto

1,2991626

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is naturally identifiable with the quotient map for the equivalence relation on $C^infty(M,N)$ given by $f sim g$ if $f|_S = g|_S$.
  $endgroup$
  – Paul Sinclair
  Mar 25 at 3:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried this identification. however, I was not able to show that these two topologies are the same.
  $endgroup$
  – Matheus Manzatto
  Mar 25 at 3:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is hard to show that $pi^-1(pi (A) $ is open if $A $ is open. It require the existence of a weird extension property.
  $endgroup$
  – Matheus Manzatto
  Mar 25 at 15:14
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $M, N$ be smooth manifolds (without boundary), so we can put a topology in the space $C^infty(M, N)$ using $mathcalC^1$ Whitney Topology.

Now, consider $Ssubset M$ a submanifold of $M$ with boundary such that $textdimS=textdimM$, using the same process we can put a topology in $C^infty(S,N)$ using the $mathcalC^1$ Whitney Topology. There is a natural projection of $C^infty(M, N)$ to $C^infty(S,N)$, definide by

beginalign*
pi: C^infty(M, N) &to C^infty(S,N)\
f&mapsto left.fright|_S.
endalign*

My Question: Is $pi$ an open map or at least a quocient map? (If necessary we can suppose that $M$ is a compact and connected manifold)

functions differential-topology

share|cite|improve this question

edited Mar 24 at 18:30

Matheus Manzatto

asked Mar 24 at 18:08

Matheus ManzattoMatheus Manzatto

1,2991626

$endgroup$

share|cite|improve this question

edited Mar 24 at 18:30

Matheus Manzatto

asked Mar 24 at 18:08

Matheus ManzattoMatheus Manzatto

1,2991626

share|cite|improve this question

edited Mar 24 at 18:30

Matheus Manzatto

asked Mar 24 at 18:08

Matheus ManzattoMatheus Manzatto

1,2991626

asked Mar 24 at 18:08

Matheus ManzattoMatheus Manzatto

1,2991626

asked Mar 24 at 18:08

Matheus ManzattoMatheus Manzatto

1,2991626