Mixing
Proving the No Retraction Theorem
$begingroup$
I try to understand a proof of the No Retraction Theorem which states that for any compact smooth manifold $M$ with boundary $partial M neq emptyset$, there is no smooth map
$$
f : M to partial M , f|_{partial M } = id.
$$
For the proof we suppose that such a map exists. Then by Sard's theorem we find a regular value $r in partial M$ and the preimage of $r$ under $f$ is a 1-dimensional manifold. The author now continues to say that the boundary of this preimage has to be a subset of $partial M$. My questions is: Why do we know that?
general-topology functions differential-topology
edited Jan 27 at 20:00
Saucy O'Path
6,1841627
asked Jan 27 at 18:23
MuziMuzi
429320
$endgroup$
add a comment |
$begingroup$
I try to understand a proof of the No Retraction Theorem which states that for any compact smooth manifold $M$ with boundary $partial M neq emptyset$, there is no smooth map
$$
f : M to partial M , f|_{partial M } = id.
$$
For the proof we suppose that such a map exists. Then by Sard's theorem we find a regular value $r in partial M$ and the preimage of $r$ under $f$ is a 1-dimensional manifold. The author now continues to say that the boundary of this preimage has to be a subset of $partial M$. My questions is: Why do we know that?
general-topology functions differential-topology
edited Jan 27 at 20:00
Saucy O'Path
6,1841627
asked Jan 27 at 18:23
MuziMuzi
429320
$endgroup$
add a comment |
$begingroup$
I try to understand a proof of the No Retraction Theorem which states that for any compact smooth manifold $M$ with boundary $partial M neq emptyset$, there is no smooth map
$$
f : M to partial M , f|_{partial M } = id.
$$
For the proof we suppose that such a map exists. Then by Sard's theorem we find a regular value $r in partial M$ and the preimage of $r$ under $f$ is a 1-dimensional manifold. The author now continues to say that the boundary of this preimage has to be a subset of $partial M$. My questions is: Why do we know that?
general-topology functions differential-topology
edited Jan 27 at 20:00
Saucy O'Path
6,1841627
asked Jan 27 at 18:23
MuziMuzi
429320
$endgroup$
I try to understand a proof of the No Retraction Theorem which states that for any compact smooth manifold $M$ with boundary $partial M neq emptyset$, there is no smooth map
$$
f : M to partial M , f|_{partial M } = id.
$$
For the proof we suppose that such a map exists. Then by Sard's theorem we find a regular value $r in partial M$ and the preimage of $r$ under $f$ is a 1-dimensional manifold. The author now continues to say that the boundary of this preimage has to be a subset of $partial M$. My questions is: Why do we know that?
general-topology functions differential-topology
general-topology functions differential-topology
edited Jan 27 at 20:00
Saucy O'Path
6,1841627
asked Jan 27 at 18:23
MuziMuzi
429320
edited Jan 27 at 20:00
Saucy O'Path
6,1841627
asked Jan 27 at 18:23
MuziMuzi
429320
edited Jan 27 at 20:00
Saucy O'Path
6,1841627
edited Jan 27 at 20:00
Saucy O'Path
6,1841627
edited Jan 27 at 20:00
Saucy O'Path
6,1841627
6,1841627
asked Jan 27 at 18:23
MuziMuzi
429320
asked Jan 27 at 18:23
MuziMuzi
429320
asked Jan 27 at 18:23
MuziMuzi
429320
429320
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is due to the form that the regular value theorem takes for manifolds with boundary:
Let $f:Mto N$ be a smooth map between manifolds with boundary and let $partial N=emptyset$. Let $cin N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $dim f^{-1}(c)=dim M-dim N$ and $partial f^{-1}(c)= f^{-1}(c)cap partial M$.
answered Jan 27 at 18:35
Saucy O'PathSaucy O'Path
6,1841627
$endgroup$
$begingroup$
That helps a lot. I didn't know this version of the regular value theorem
$endgroup$
– Muzi
Jan 27 at 18:43
add a comment |
$begingroup$
The conclusion of the Sard's Theorem argument is a bit stronger than what you say.
The real conclusion is that the map $f : M to partial M$ has a regular value $r in partial M$.
It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary.
Now suppose that you take $x in f^{-1}(r) cap text{interior}(M)$. The derivative map $D_x f : T_x M to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $partial M$.
One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $partial M cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc ${(x,y) mid x^2 + y^2 < 0, |y| ge 1}$ with the line $x=0$.
answered Jan 27 at 18:38
Lee MosherLee Mosher
51.2k33889
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is due to the form that the regular value theorem takes for manifolds with boundary:
Let $f:Mto N$ be a smooth map between manifolds with boundary and let $partial N=emptyset$. Let $cin N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $dim f^{-1}(c)=dim M-dim N$ and $partial f^{-1}(c)= f^{-1}(c)cap partial M$.
answered Jan 27 at 18:35
Saucy O'PathSaucy O'Path
6,1841627
$endgroup$
$begingroup$
That helps a lot. I didn't know this version of the regular value theorem
$endgroup$
– Muzi
Jan 27 at 18:43
add a comment |
$begingroup$
This is due to the form that the regular value theorem takes for manifolds with boundary:
Let $f:Mto N$ be a smooth map between manifolds with boundary and let $partial N=emptyset$. Let $cin N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $dim f^{-1}(c)=dim M-dim N$ and $partial f^{-1}(c)= f^{-1}(c)cap partial M$.
answered Jan 27 at 18:35
Saucy O'PathSaucy O'Path
6,1841627
$endgroup$
$begingroup$
That helps a lot. I didn't know this version of the regular value theorem
$endgroup$
– Muzi
Jan 27 at 18:43
add a comment |
$begingroup$
This is due to the form that the regular value theorem takes for manifolds with boundary:
Let $f:Mto N$ be a smooth map between manifolds with boundary and let $partial N=emptyset$. Let $cin N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $dim f^{-1}(c)=dim M-dim N$ and $partial f^{-1}(c)= f^{-1}(c)cap partial M$.
answered Jan 27 at 18:35
Saucy O'PathSaucy O'Path
6,1841627
$endgroup$
This is due to the form that the regular value theorem takes for manifolds with boundary:
Let $f:Mto N$ be a smooth map between manifolds with boundary and let $partial N=emptyset$. Let $cin N$ be a regular value for $f$. Then, $f^{-1}(c)$ is a submanifold of $M$ such that $dim f^{-1}(c)=dim M-dim N$ and $partial f^{-1}(c)= f^{-1}(c)cap partial M$.
answered Jan 27 at 18:35
Saucy O'PathSaucy O'Path
6,1841627
answered Jan 27 at 18:35
Saucy O'PathSaucy O'Path
6,1841627
answered Jan 27 at 18:35
Saucy O'PathSaucy O'Path
6,1841627
answered Jan 27 at 18:35
Saucy O'PathSaucy O'Path
6,1841627
6,1841627
$begingroup$
That helps a lot. I didn't know this version of the regular value theorem
$endgroup$
– Muzi
Jan 27 at 18:43
add a comment |
$begingroup$
That helps a lot. I didn't know this version of the regular value theorem
$endgroup$
– Muzi
Jan 27 at 18:43
$begingroup$
That helps a lot. I didn't know this version of the regular value theorem
$endgroup$
– Muzi
Jan 27 at 18:43
$begingroup$
That helps a lot. I didn't know this version of the regular value theorem
$endgroup$
– Muzi
Jan 27 at 18:43
add a comment |
$begingroup$
The conclusion of the Sard's Theorem argument is a bit stronger than what you say.
The real conclusion is that the map $f : M to partial M$ has a regular value $r in partial M$.
It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary.
Now suppose that you take $x in f^{-1}(r) cap text{interior}(M)$. The derivative map $D_x f : T_x M to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $partial M$.
One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $partial M cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc ${(x,y) mid x^2 + y^2 < 0, |y| ge 1}$ with the line $x=0$.
answered Jan 27 at 18:38
Lee MosherLee Mosher
51.2k33889
$endgroup$
add a comment |
$begingroup$
The conclusion of the Sard's Theorem argument is a bit stronger than what you say.
The real conclusion is that the map $f : M to partial M$ has a regular value $r in partial M$.
It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary.
Now suppose that you take $x in f^{-1}(r) cap text{interior}(M)$. The derivative map $D_x f : T_x M to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $partial M$.
One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $partial M cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc ${(x,y) mid x^2 + y^2 < 0, |y| ge 1}$ with the line $x=0$.
answered Jan 27 at 18:38
Lee MosherLee Mosher
51.2k33889
$endgroup$
add a comment |
$begingroup$
The conclusion of the Sard's Theorem argument is a bit stronger than what you say.
The real conclusion is that the map $f : M to partial M$ has a regular value $r in partial M$.
It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary.
Now suppose that you take $x in f^{-1}(r) cap text{interior}(M)$. The derivative map $D_x f : T_x M to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $partial M$.
One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $partial M cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc ${(x,y) mid x^2 + y^2 < 0, |y| ge 1}$ with the line $x=0$.
answered Jan 27 at 18:38
Lee MosherLee Mosher
51.2k33889
$endgroup$
The conclusion of the Sard's Theorem argument is a bit stronger than what you say.
The real conclusion is that the map $f : M to partial M$ has a regular value $r in partial M$.
It follows, first, that $f^{-1}(r)$ is a 1-manifold with boundary.
Now suppose that you take $x in f^{-1}(r) cap text{interior}(M)$. The derivative map $D_x f : T_x M to T_r M$ has rank $1$, so its kernel is a 1-dimensional subspace of $M$. Applying the implicit function theorem, and using that $x$ is in the interior of $M$, it follows that $f^{-1}(r)$ intersected with a small neighborhood of $x$ is a 1-manifold without boundary containing $r$. Thus, $x$ cannot be a boundary point of the 1-manifold $f^{-1}(r)$, therefore all of the boundary points of $f^{-1}(r)$ are contained in $partial M$.
One can make a still stronger conclusion, namely that $f^{-1}(r)$ is a "properly embedded" submanifold of $M$, which means that each point on $partial M cap f^{-1}(r)$ has a neighborhood in which the intersection of that neighborhood with $f^{-1}(r)$ looks like the intersection of the half-disc ${(x,y) mid x^2 + y^2 < 0, |y| ge 1}$ with the line $x=0$.
answered Jan 27 at 18:38
Lee MosherLee Mosher
51.2k33889
answered Jan 27 at 18:38
Lee MosherLee Mosher
51.2k33889
answered Jan 27 at 18:38
Lee MosherLee Mosher
51.2k33889
answered Jan 27 at 18:38
Lee MosherLee Mosher
51.2k33889
51.2k33889
add a comment |
add a comment |
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