Expansion of homeomorphism outside a disk
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The following is an exercise in Bloch's Intro to Geometric Topology
Let $B subseteq Bbb R^2$ be a set homeomorphic to the closed unit disk and $h :partial B to partial B$, a homeomorphism. By Schonflies we can find a homeomorphism $F$ of $Bbb R^2$ that is $F(D^2)=B$ and $F$ is the identity outside a disk. Then we can expand $F^{-1}circ hcirc F$ to homeomorphism $g$ of the unit disk. Then $F circ g circ F^{-1}$ will give us a homeomorphism of $B$ that is $h$ on the boundary.
My question is if there is a way to expand $F circ g circ F^{-1}$ in all $Bbb R^2$?
geometric-topology
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The following is an exercise in Bloch's Intro to Geometric Topology
Let $B subseteq Bbb R^2$ be a set homeomorphic to the closed unit disk and $h :partial B to partial B$, a homeomorphism. By Schonflies we can find a homeomorphism $F$ of $Bbb R^2$ that is $F(D^2)=B$ and $F$ is the identity outside a disk. Then we can expand $F^{-1}circ hcirc F$ to homeomorphism $g$ of the unit disk. Then $F circ g circ F^{-1}$ will give us a homeomorphism of $B$ that is $h$ on the boundary.
My question is if there is a way to expand $F circ g circ F^{-1}$ in all $Bbb R^2$?
geometric-topology
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The following is an exercise in Bloch's Intro to Geometric Topology
Let $B subseteq Bbb R^2$ be a set homeomorphic to the closed unit disk and $h :partial B to partial B$, a homeomorphism. By Schonflies we can find a homeomorphism $F$ of $Bbb R^2$ that is $F(D^2)=B$ and $F$ is the identity outside a disk. Then we can expand $F^{-1}circ hcirc F$ to homeomorphism $g$ of the unit disk. Then $F circ g circ F^{-1}$ will give us a homeomorphism of $B$ that is $h$ on the boundary.
My question is if there is a way to expand $F circ g circ F^{-1}$ in all $Bbb R^2$?
geometric-topology
The following is an exercise in Bloch's Intro to Geometric Topology
Let $B subseteq Bbb R^2$ be a set homeomorphic to the closed unit disk and $h :partial B to partial B$, a homeomorphism. By Schonflies we can find a homeomorphism $F$ of $Bbb R^2$ that is $F(D^2)=B$ and $F$ is the identity outside a disk. Then we can expand $F^{-1}circ hcirc F$ to homeomorphism $g$ of the unit disk. Then $F circ g circ F^{-1}$ will give us a homeomorphism of $B$ that is $h$ on the boundary.
My question is if there is a way to expand $F circ g circ F^{-1}$ in all $Bbb R^2$?
geometric-topology
geometric-topology
asked yesterday
Amontillado
422313
422313
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Assuming that you do actually know how to extend $F^{-1} circ h circ F$ to a homeomorphism of the unit disc, and assuming you know how to do this so that the origin $mathcal O$ is fixed, then this is possible.
Simply work in the one-point compactification $mathbb R^2 cup {infty}$, and use the "inversion" homeomorphism
$$g : mathbb R^2 cup {infty} to mathbb R^2 cup {infty}, qquad g(x) = begin{cases}
frac{x}{|x|^2} & quadtext{if $x notin {0,infty}$} \
mathcal O &quad text{if $x = infty$} \
infty &quad text{if $x=mathcal O$}
end{cases}
$$
You can then restrict $g^{-1} circ (F^{-1} circ h circ F) circ g$ to $partial B$, next you can extend that to a homeomorphism $k : B to B$ which fixes $mathcal O$, and then the map $g circ k circ g^{-1}$, suitably restricted, is the extension that you want.
Thank you very much
– Amontillado
1 hour ago
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Any continuous map $f : S^1 to S^1$ extends to a continuous map $e(f) : mathbb{R^2} to mathbb{R^2}$ by defining
$$e(f)(x) =
begin{cases}
0 & x = 0 \
lVert x rVert f(frac{x}{lVert x rVert }) & x ne 0
end{cases}
$$
Note that $e(g circ f) = e(g) circ e(f)$. Thus, if $h : S^1 to S^1$ is a homeomorphism, then $e(h)$ is a homeomorphism. In fact, $e(h^{-1})$ is the inverse homeomorphism.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Assuming that you do actually know how to extend $F^{-1} circ h circ F$ to a homeomorphism of the unit disc, and assuming you know how to do this so that the origin $mathcal O$ is fixed, then this is possible.
Simply work in the one-point compactification $mathbb R^2 cup {infty}$, and use the "inversion" homeomorphism
$$g : mathbb R^2 cup {infty} to mathbb R^2 cup {infty}, qquad g(x) = begin{cases}
frac{x}{|x|^2} & quadtext{if $x notin {0,infty}$} \
mathcal O &quad text{if $x = infty$} \
infty &quad text{if $x=mathcal O$}
end{cases}
$$
You can then restrict $g^{-1} circ (F^{-1} circ h circ F) circ g$ to $partial B$, next you can extend that to a homeomorphism $k : B to B$ which fixes $mathcal O$, and then the map $g circ k circ g^{-1}$, suitably restricted, is the extension that you want.
Thank you very much
– Amontillado
1 hour ago
add a comment |
up vote
1
down vote
accepted
Assuming that you do actually know how to extend $F^{-1} circ h circ F$ to a homeomorphism of the unit disc, and assuming you know how to do this so that the origin $mathcal O$ is fixed, then this is possible.
Simply work in the one-point compactification $mathbb R^2 cup {infty}$, and use the "inversion" homeomorphism
$$g : mathbb R^2 cup {infty} to mathbb R^2 cup {infty}, qquad g(x) = begin{cases}
frac{x}{|x|^2} & quadtext{if $x notin {0,infty}$} \
mathcal O &quad text{if $x = infty$} \
infty &quad text{if $x=mathcal O$}
end{cases}
$$
You can then restrict $g^{-1} circ (F^{-1} circ h circ F) circ g$ to $partial B$, next you can extend that to a homeomorphism $k : B to B$ which fixes $mathcal O$, and then the map $g circ k circ g^{-1}$, suitably restricted, is the extension that you want.
Thank you very much
– Amontillado
1 hour ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Assuming that you do actually know how to extend $F^{-1} circ h circ F$ to a homeomorphism of the unit disc, and assuming you know how to do this so that the origin $mathcal O$ is fixed, then this is possible.
Simply work in the one-point compactification $mathbb R^2 cup {infty}$, and use the "inversion" homeomorphism
$$g : mathbb R^2 cup {infty} to mathbb R^2 cup {infty}, qquad g(x) = begin{cases}
frac{x}{|x|^2} & quadtext{if $x notin {0,infty}$} \
mathcal O &quad text{if $x = infty$} \
infty &quad text{if $x=mathcal O$}
end{cases}
$$
You can then restrict $g^{-1} circ (F^{-1} circ h circ F) circ g$ to $partial B$, next you can extend that to a homeomorphism $k : B to B$ which fixes $mathcal O$, and then the map $g circ k circ g^{-1}$, suitably restricted, is the extension that you want.
Assuming that you do actually know how to extend $F^{-1} circ h circ F$ to a homeomorphism of the unit disc, and assuming you know how to do this so that the origin $mathcal O$ is fixed, then this is possible.
Simply work in the one-point compactification $mathbb R^2 cup {infty}$, and use the "inversion" homeomorphism
$$g : mathbb R^2 cup {infty} to mathbb R^2 cup {infty}, qquad g(x) = begin{cases}
frac{x}{|x|^2} & quadtext{if $x notin {0,infty}$} \
mathcal O &quad text{if $x = infty$} \
infty &quad text{if $x=mathcal O$}
end{cases}
$$
You can then restrict $g^{-1} circ (F^{-1} circ h circ F) circ g$ to $partial B$, next you can extend that to a homeomorphism $k : B to B$ which fixes $mathcal O$, and then the map $g circ k circ g^{-1}$, suitably restricted, is the extension that you want.
answered 14 hours ago
Lee Mosher
47.2k33681
47.2k33681
Thank you very much
– Amontillado
1 hour ago
add a comment |
Thank you very much
– Amontillado
1 hour ago
Thank you very much
– Amontillado
1 hour ago
Thank you very much
– Amontillado
1 hour ago
add a comment |
up vote
0
down vote
Any continuous map $f : S^1 to S^1$ extends to a continuous map $e(f) : mathbb{R^2} to mathbb{R^2}$ by defining
$$e(f)(x) =
begin{cases}
0 & x = 0 \
lVert x rVert f(frac{x}{lVert x rVert }) & x ne 0
end{cases}
$$
Note that $e(g circ f) = e(g) circ e(f)$. Thus, if $h : S^1 to S^1$ is a homeomorphism, then $e(h)$ is a homeomorphism. In fact, $e(h^{-1})$ is the inverse homeomorphism.
add a comment |
up vote
0
down vote
Any continuous map $f : S^1 to S^1$ extends to a continuous map $e(f) : mathbb{R^2} to mathbb{R^2}$ by defining
$$e(f)(x) =
begin{cases}
0 & x = 0 \
lVert x rVert f(frac{x}{lVert x rVert }) & x ne 0
end{cases}
$$
Note that $e(g circ f) = e(g) circ e(f)$. Thus, if $h : S^1 to S^1$ is a homeomorphism, then $e(h)$ is a homeomorphism. In fact, $e(h^{-1})$ is the inverse homeomorphism.
add a comment |
up vote
0
down vote
up vote
0
down vote
Any continuous map $f : S^1 to S^1$ extends to a continuous map $e(f) : mathbb{R^2} to mathbb{R^2}$ by defining
$$e(f)(x) =
begin{cases}
0 & x = 0 \
lVert x rVert f(frac{x}{lVert x rVert }) & x ne 0
end{cases}
$$
Note that $e(g circ f) = e(g) circ e(f)$. Thus, if $h : S^1 to S^1$ is a homeomorphism, then $e(h)$ is a homeomorphism. In fact, $e(h^{-1})$ is the inverse homeomorphism.
Any continuous map $f : S^1 to S^1$ extends to a continuous map $e(f) : mathbb{R^2} to mathbb{R^2}$ by defining
$$e(f)(x) =
begin{cases}
0 & x = 0 \
lVert x rVert f(frac{x}{lVert x rVert }) & x ne 0
end{cases}
$$
Note that $e(g circ f) = e(g) circ e(f)$. Thus, if $h : S^1 to S^1$ is a homeomorphism, then $e(h)$ is a homeomorphism. In fact, $e(h^{-1})$ is the inverse homeomorphism.
answered 11 hours ago
Paul Frost
7,0311526
7,0311526
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