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Pre-image of $mathbbR^+$ is a smooth manifold with boundary if $0$ is regular value.
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I want to prove that: Given $M$ smooth $m$ dimensional manifold and $f: M to mathbbR$ smooth map such that $0$ is a regular value, then $x in M : f(x) geq 0$ is a smooth manifold with boundary $f^-1(0)$.
So, I know $f^-1(0)$ is a embedded submanifold of codimension $1$. If $H:=x in M : f(x) geq 0$ is a manifold with boundary $f^-1(0)$, then $H$ must have dimension $2$.
With this I have my strategy, it is to find local harts to $mathbbR^2$. With $f^-1(0)$ being an embedded submanifold of $M$, I know that for every $p$ in $f^-1(0)$, there exists $(U,phi)$ local charts of $p$ such that $phi(U cap f^-1(0))= V times y simeq V subset mathbbR$.
My idea is to work around this function: $x mapsto (phi(x),f(x))$ (This is an abuse of notation since $phi(x)$ lies in $mathbbR^m$, but I see it as a real number by composing $phi$ with an homeomorphism $V times y to V$).
This is well defined in $U cap f^-1(0)$ (The idea is that I can do this for any of those local charts $(U,phi)$ and define these ''new local charts'' then try to extend them to $Ucap H$ by a map $x mapsto (g(x),f(x))$, where $g$ is an extension of $phi restriction_Ucap f^-1(0)$ to $U cap H$, so that $xmapsto (g(x),f(x))$ is an homeomorphism. Doing this, I can assure local charts for any $p in f^-1(0) subset H$.
But I cannot seem to get anywhere with this, since I do not know how the local charts work in $H-f^-1(0)$.
Any help would be apreciated, thanks in advanced.
differential-topology smooth-manifolds
asked Jul 18 at 22:07
Bajo Fondo
376213
add a comment |Â
up vote
1
down vote
favorite
I want to prove that: Given $M$ smooth $m$ dimensional manifold and $f: M to mathbbR$ smooth map such that $0$ is a regular value, then $x in M : f(x) geq 0$ is a smooth manifold with boundary $f^-1(0)$.
So, I know $f^-1(0)$ is a embedded submanifold of codimension $1$. If $H:=x in M : f(x) geq 0$ is a manifold with boundary $f^-1(0)$, then $H$ must have dimension $2$.
With this I have my strategy, it is to find local harts to $mathbbR^2$. With $f^-1(0)$ being an embedded submanifold of $M$, I know that for every $p$ in $f^-1(0)$, there exists $(U,phi)$ local charts of $p$ such that $phi(U cap f^-1(0))= V times y simeq V subset mathbbR$.
My idea is to work around this function: $x mapsto (phi(x),f(x))$ (This is an abuse of notation since $phi(x)$ lies in $mathbbR^m$, but I see it as a real number by composing $phi$ with an homeomorphism $V times y to V$).
This is well defined in $U cap f^-1(0)$ (The idea is that I can do this for any of those local charts $(U,phi)$ and define these ''new local charts'' then try to extend them to $Ucap H$ by a map $x mapsto (g(x),f(x))$, where $g$ is an extension of $phi restriction_Ucap f^-1(0)$ to $U cap H$, so that $xmapsto (g(x),f(x))$ is an homeomorphism. Doing this, I can assure local charts for any $p in f^-1(0) subset H$.
But I cannot seem to get anywhere with this, since I do not know how the local charts work in $H-f^-1(0)$.
Any help would be apreciated, thanks in advanced.
differential-topology smooth-manifolds
asked Jul 18 at 22:07
Bajo Fondo
376213
You should either mention that $M$ is a two-dimensional manifold or consider the more general case of an $n$-dimensional manifold, the proof being identical with that in two dimensions.
– Paul Frost
Jul 18 at 22:48
Ok.. gonna edit it.
– Bajo Fondo
Jul 18 at 22:49
Because $f^-1(0)$ is an embedded submanifold, there are "slice charts" covering it (these are charts of $M$). The other piece of your preimage is an open subset of $M$, so has the same charts as $M$. This should make it easy to make sure they agree where they are supposed to.
– Steve D
Jul 18 at 23:06
Note that $H-f^-1(0)$ is an open subset of $M$.
– Ted Shifrin
Jul 18 at 23:33
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to prove that: Given $M$ smooth $m$ dimensional manifold and $f: M to mathbbR$ smooth map such that $0$ is a regular value, then $x in M : f(x) geq 0$ is a smooth manifold with boundary $f^-1(0)$.
So, I know $f^-1(0)$ is a embedded submanifold of codimension $1$. If $H:=x in M : f(x) geq 0$ is a manifold with boundary $f^-1(0)$, then $H$ must have dimension $2$.
With this I have my strategy, it is to find local harts to $mathbbR^2$. With $f^-1(0)$ being an embedded submanifold of $M$, I know that for every $p$ in $f^-1(0)$, there exists $(U,phi)$ local charts of $p$ such that $phi(U cap f^-1(0))= V times y simeq V subset mathbbR$.
My idea is to work around this function: $x mapsto (phi(x),f(x))$ (This is an abuse of notation since $phi(x)$ lies in $mathbbR^m$, but I see it as a real number by composing $phi$ with an homeomorphism $V times y to V$).
This is well defined in $U cap f^-1(0)$ (The idea is that I can do this for any of those local charts $(U,phi)$ and define these ''new local charts'' then try to extend them to $Ucap H$ by a map $x mapsto (g(x),f(x))$, where $g$ is an extension of $phi restriction_Ucap f^-1(0)$ to $U cap H$, so that $xmapsto (g(x),f(x))$ is an homeomorphism. Doing this, I can assure local charts for any $p in f^-1(0) subset H$.
But I cannot seem to get anywhere with this, since I do not know how the local charts work in $H-f^-1(0)$.
Any help would be apreciated, thanks in advanced.
differential-topology smooth-manifolds
asked Jul 18 at 22:07
Bajo Fondo
376213
I want to prove that: Given $M$ smooth $m$ dimensional manifold and $f: M to mathbbR$ smooth map such that $0$ is a regular value, then $x in M : f(x) geq 0$ is a smooth manifold with boundary $f^-1(0)$.
So, I know $f^-1(0)$ is a embedded submanifold of codimension $1$. If $H:=x in M : f(x) geq 0$ is a manifold with boundary $f^-1(0)$, then $H$ must have dimension $2$.
With this I have my strategy, it is to find local harts to $mathbbR^2$. With $f^-1(0)$ being an embedded submanifold of $M$, I know that for every $p$ in $f^-1(0)$, there exists $(U,phi)$ local charts of $p$ such that $phi(U cap f^-1(0))= V times y simeq V subset mathbbR$.
My idea is to work around this function: $x mapsto (phi(x),f(x))$ (This is an abuse of notation since $phi(x)$ lies in $mathbbR^m$, but I see it as a real number by composing $phi$ with an homeomorphism $V times y to V$).
This is well defined in $U cap f^-1(0)$ (The idea is that I can do this for any of those local charts $(U,phi)$ and define these ''new local charts'' then try to extend them to $Ucap H$ by a map $x mapsto (g(x),f(x))$, where $g$ is an extension of $phi restriction_Ucap f^-1(0)$ to $U cap H$, so that $xmapsto (g(x),f(x))$ is an homeomorphism. Doing this, I can assure local charts for any $p in f^-1(0) subset H$.
But I cannot seem to get anywhere with this, since I do not know how the local charts work in $H-f^-1(0)$.
Any help would be apreciated, thanks in advanced.
differential-topology smooth-manifolds
asked Jul 18 at 22:07
Bajo Fondo
376213
edited Jul 18 at 22:50
asked Jul 18 at 22:07
Bajo Fondo
376213
asked Jul 18 at 22:07
Bajo Fondo
376213
asked Jul 18 at 22:07
Bajo Fondo
376213
376213
You should either mention that $M$ is a two-dimensional manifold or consider the more general case of an $n$-dimensional manifold, the proof being identical with that in two dimensions.
– Paul Frost
Jul 18 at 22:48
Ok.. gonna edit it.
– Bajo Fondo
Jul 18 at 22:49
Because $f^-1(0)$ is an embedded submanifold, there are "slice charts" covering it (these are charts of $M$). The other piece of your preimage is an open subset of $M$, so has the same charts as $M$. This should make it easy to make sure they agree where they are supposed to.
– Steve D
Jul 18 at 23:06
Note that $H-f^-1(0)$ is an open subset of $M$.
– Ted Shifrin
Jul 18 at 23:33
add a comment |Â
You should either mention that $M$ is a two-dimensional manifold or consider the more general case of an $n$-dimensional manifold, the proof being identical with that in two dimensions.
– Paul Frost
Jul 18 at 22:48
Ok.. gonna edit it.
– Bajo Fondo
Jul 18 at 22:49
Because $f^-1(0)$ is an embedded submanifold, there are "slice charts" covering it (these are charts of $M$). The other piece of your preimage is an open subset of $M$, so has the same charts as $M$. This should make it easy to make sure they agree where they are supposed to.
– Steve D
Jul 18 at 23:06
Note that $H-f^-1(0)$ is an open subset of $M$.
– Ted Shifrin
Jul 18 at 23:33
You should either mention that $M$ is a two-dimensional manifold or consider the more general case of an $n$-dimensional manifold, the proof being identical with that in two dimensions.
– Paul Frost
Jul 18 at 22:48
You should either mention that $M$ is a two-dimensional manifold or consider the more general case of an $n$-dimensional manifold, the proof being identical with that in two dimensions.
– Paul Frost
Jul 18 at 22:48
Ok.. gonna edit it.
– Bajo Fondo
Jul 18 at 22:49
Ok.. gonna edit it.
– Bajo Fondo
Jul 18 at 22:49
Because $f^-1(0)$ is an embedded submanifold, there are "slice charts" covering it (these are charts of $M$). The other piece of your preimage is an open subset of $M$, so has the same charts as $M$. This should make it easy to make sure they agree where they are supposed to.
– Steve D
Jul 18 at 23:06
Because $f^-1(0)$ is an embedded submanifold, there are "slice charts" covering it (these are charts of $M$). The other piece of your preimage is an open subset of $M$, so has the same charts as $M$. This should make it easy to make sure they agree where they are supposed to.
– Steve D
Jul 18 at 23:06
Note that $H-f^-1(0)$ is an open subset of $M$.
– Ted Shifrin
Jul 18 at 23:33
Note that $H-f^-1(0)$ is an open subset of $M$.
– Ted Shifrin
Jul 18 at 23:33
add a comment |Â
1 Answer
1
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1
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You have to show the following:
(1) Each $x in B = f^-1(0)$ is contained in a boundary chart, i.e. there exists a homeomorphism $h : U to W$ with $U subset H$ an open neighborhood of $x$, $W subset mathbbR^n_+ = mathbbR^n-1 times [0,infty)$ open and $h(x) in mathbbR^n-1 times 0 $.
(2) For any two boundary charts $h_i : U_i to W_i$, the transitition $h_12 : h_1(U_1 cap U_2) to h_2(U_1 cap U_2)$ is a diffeomorphism between open subsets of subsets of $mathbbR^n_+$ which means that it is the restriction of a diffeomorphism between open subsets of $mathbbR^n$.
(1) Since $0$ is a regular value, each $x in B$ has a chart $h' : U' to W'$ in $M$ such that $h'(U' cap B) = W' cap mathbbR^n-1 times 0 $ and $f (h')^-1(x_1,...,x_n) = x_n$ for $(x_1,...,x_n) in W'$. Then you see that $h : U = U' cap H stackrelh'rightarrow W = W' cap mathbbR^n_+$ is a boundary chart.
(2) This is shown completely similar.
answered Jul 18 at 23:32
Paul Frost
3,678420
My problem was with the difference between dimension and co-dimension. Thanks
– Bajo Fondo
Aug 6 at 17:29
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have to show the following:
(1) Each $x in B = f^-1(0)$ is contained in a boundary chart, i.e. there exists a homeomorphism $h : U to W$ with $U subset H$ an open neighborhood of $x$, $W subset mathbbR^n_+ = mathbbR^n-1 times [0,infty)$ open and $h(x) in mathbbR^n-1 times 0 $.
(2) For any two boundary charts $h_i : U_i to W_i$, the transitition $h_12 : h_1(U_1 cap U_2) to h_2(U_1 cap U_2)$ is a diffeomorphism between open subsets of subsets of $mathbbR^n_+$ which means that it is the restriction of a diffeomorphism between open subsets of $mathbbR^n$.
(1) Since $0$ is a regular value, each $x in B$ has a chart $h' : U' to W'$ in $M$ such that $h'(U' cap B) = W' cap mathbbR^n-1 times 0 $ and $f (h')^-1(x_1,...,x_n) = x_n$ for $(x_1,...,x_n) in W'$. Then you see that $h : U = U' cap H stackrelh'rightarrow W = W' cap mathbbR^n_+$ is a boundary chart.
(2) This is shown completely similar.
answered Jul 18 at 23:32
Paul Frost
3,678420
My problem was with the difference between dimension and co-dimension. Thanks
– Bajo Fondo
Aug 6 at 17:29
add a comment |Â
up vote
1
down vote
accepted
You have to show the following:
(1) Each $x in B = f^-1(0)$ is contained in a boundary chart, i.e. there exists a homeomorphism $h : U to W$ with $U subset H$ an open neighborhood of $x$, $W subset mathbbR^n_+ = mathbbR^n-1 times [0,infty)$ open and $h(x) in mathbbR^n-1 times 0 $.
(2) For any two boundary charts $h_i : U_i to W_i$, the transitition $h_12 : h_1(U_1 cap U_2) to h_2(U_1 cap U_2)$ is a diffeomorphism between open subsets of subsets of $mathbbR^n_+$ which means that it is the restriction of a diffeomorphism between open subsets of $mathbbR^n$.
(1) Since $0$ is a regular value, each $x in B$ has a chart $h' : U' to W'$ in $M$ such that $h'(U' cap B) = W' cap mathbbR^n-1 times 0 $ and $f (h')^-1(x_1,...,x_n) = x_n$ for $(x_1,...,x_n) in W'$. Then you see that $h : U = U' cap H stackrelh'rightarrow W = W' cap mathbbR^n_+$ is a boundary chart.
(2) This is shown completely similar.
answered Jul 18 at 23:32
Paul Frost
3,678420
My problem was with the difference between dimension and co-dimension. Thanks
– Bajo Fondo
Aug 6 at 17:29
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have to show the following:
(1) Each $x in B = f^-1(0)$ is contained in a boundary chart, i.e. there exists a homeomorphism $h : U to W$ with $U subset H$ an open neighborhood of $x$, $W subset mathbbR^n_+ = mathbbR^n-1 times [0,infty)$ open and $h(x) in mathbbR^n-1 times 0 $.
(2) For any two boundary charts $h_i : U_i to W_i$, the transitition $h_12 : h_1(U_1 cap U_2) to h_2(U_1 cap U_2)$ is a diffeomorphism between open subsets of subsets of $mathbbR^n_+$ which means that it is the restriction of a diffeomorphism between open subsets of $mathbbR^n$.
(1) Since $0$ is a regular value, each $x in B$ has a chart $h' : U' to W'$ in $M$ such that $h'(U' cap B) = W' cap mathbbR^n-1 times 0 $ and $f (h')^-1(x_1,...,x_n) = x_n$ for $(x_1,...,x_n) in W'$. Then you see that $h : U = U' cap H stackrelh'rightarrow W = W' cap mathbbR^n_+$ is a boundary chart.
(2) This is shown completely similar.
answered Jul 18 at 23:32
Paul Frost
3,678420
You have to show the following:
(1) Each $x in B = f^-1(0)$ is contained in a boundary chart, i.e. there exists a homeomorphism $h : U to W$ with $U subset H$ an open neighborhood of $x$, $W subset mathbbR^n_+ = mathbbR^n-1 times [0,infty)$ open and $h(x) in mathbbR^n-1 times 0 $.
(2) For any two boundary charts $h_i : U_i to W_i$, the transitition $h_12 : h_1(U_1 cap U_2) to h_2(U_1 cap U_2)$ is a diffeomorphism between open subsets of subsets of $mathbbR^n_+$ which means that it is the restriction of a diffeomorphism between open subsets of $mathbbR^n$.
(1) Since $0$ is a regular value, each $x in B$ has a chart $h' : U' to W'$ in $M$ such that $h'(U' cap B) = W' cap mathbbR^n-1 times 0 $ and $f (h')^-1(x_1,...,x_n) = x_n$ for $(x_1,...,x_n) in W'$. Then you see that $h : U = U' cap H stackrelh'rightarrow W = W' cap mathbbR^n_+$ is a boundary chart.
(2) This is shown completely similar.
answered Jul 18 at 23:32
Paul Frost
3,678420
answered Jul 18 at 23:32
Paul Frost
3,678420
answered Jul 18 at 23:32
Paul Frost
3,678420
answered Jul 18 at 23:32
Paul Frost
3,678420
3,678420
My problem was with the difference between dimension and co-dimension. Thanks
– Bajo Fondo
Aug 6 at 17:29
add a comment |Â
My problem was with the difference between dimension and co-dimension. Thanks
– Bajo Fondo
Aug 6 at 17:29
My problem was with the difference between dimension and co-dimension. Thanks
– Bajo Fondo
Aug 6 at 17:29
My problem was with the difference between dimension and co-dimension. Thanks
– Bajo Fondo
Aug 6 at 17:29
add a comment |Â
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You should either mention that $M$ is a two-dimensional manifold or consider the more general case of an $n$-dimensional manifold, the proof being identical with that in two dimensions.
– Paul Frost
Jul 18 at 22:48
Ok.. gonna edit it.
– Bajo Fondo
Jul 18 at 22:49
Because $f^-1(0)$ is an embedded submanifold, there are "slice charts" covering it (these are charts of $M$). The other piece of your preimage is an open subset of $M$, so has the same charts as $M$. This should make it easy to make sure they agree where they are supposed to.
– Steve D
Jul 18 at 23:06
Note that $H-f^-1(0)$ is an open subset of $M$.
– Ted Shifrin
Jul 18 at 23:33