Mixing
Existence of a special homeomorphism on $mathbb{T}^2$.
$begingroup$
Let $A, B$ be closed topological subspaces of $mathbb{T}^2$. Suppose that $A$ and $B$ are homeomorphic as topological spaces.
My Question: Is it possible to construct a homeomorphism $h: mathbb{T}^2 to
mathbb{T}^2$, such that $h(A) = B?$
If necessary, we can assume $A$ and $B$ as $mathcal{C}^0$- manifold with boundary (topological manifold with boundary).
I've been stuck in this problem for a long time; everything I tried did not come close to achieving the result. Can anyone help me?
general-topology manifolds manifolds-with-boundary compact-manifolds
edited Feb 1 at 0:45
J. W. Tanner
4,6291320
asked Feb 1 at 0:32
Matheus ManzattoMatheus Manzatto
1,2991626
$endgroup$
add a comment |
$begingroup$
Let $A, B$ be closed topological subspaces of $mathbb{T}^2$. Suppose that $A$ and $B$ are homeomorphic as topological spaces.
My Question: Is it possible to construct a homeomorphism $h: mathbb{T}^2 to
mathbb{T}^2$, such that $h(A) = B?$
If necessary, we can assume $A$ and $B$ as $mathcal{C}^0$- manifold with boundary (topological manifold with boundary).
I've been stuck in this problem for a long time; everything I tried did not come close to achieving the result. Can anyone help me?
general-topology manifolds manifolds-with-boundary compact-manifolds
edited Feb 1 at 0:45
J. W. Tanner
4,6291320
asked Feb 1 at 0:32
Matheus ManzattoMatheus Manzatto
1,2991626
$endgroup$
1
$begingroup$
It is impossible in general, when $A, B$ are topological circles. When each of them is a disjoint union of compact arcs, such a homeomorphism does exist, but it is not completely trivial. One needs Schoenflies theorem for topological arcs.
$endgroup$
– Moishe Kohan
Feb 1 at 0:44
add a comment |
$begingroup$
Let $A, B$ be closed topological subspaces of $mathbb{T}^2$. Suppose that $A$ and $B$ are homeomorphic as topological spaces.
My Question: Is it possible to construct a homeomorphism $h: mathbb{T}^2 to
mathbb{T}^2$, such that $h(A) = B?$
If necessary, we can assume $A$ and $B$ as $mathcal{C}^0$- manifold with boundary (topological manifold with boundary).
I've been stuck in this problem for a long time; everything I tried did not come close to achieving the result. Can anyone help me?
general-topology manifolds manifolds-with-boundary compact-manifolds
edited Feb 1 at 0:45
J. W. Tanner
4,6291320
asked Feb 1 at 0:32
Matheus ManzattoMatheus Manzatto
1,2991626
$endgroup$
Let $A, B$ be closed topological subspaces of $mathbb{T}^2$. Suppose that $A$ and $B$ are homeomorphic as topological spaces.
My Question: Is it possible to construct a homeomorphism $h: mathbb{T}^2 to
mathbb{T}^2$, such that $h(A) = B?$
If necessary, we can assume $A$ and $B$ as $mathcal{C}^0$- manifold with boundary (topological manifold with boundary).
I've been stuck in this problem for a long time; everything I tried did not come close to achieving the result. Can anyone help me?
general-topology manifolds manifolds-with-boundary compact-manifolds
general-topology manifolds manifolds-with-boundary compact-manifolds
edited Feb 1 at 0:45
J. W. Tanner
4,6291320
asked Feb 1 at 0:32
Matheus ManzattoMatheus Manzatto
1,2991626
edited Feb 1 at 0:45
J. W. Tanner
4,6291320
asked Feb 1 at 0:32
Matheus ManzattoMatheus Manzatto
1,2991626
edited Feb 1 at 0:45
J. W. Tanner
4,6291320
edited Feb 1 at 0:45
J. W. Tanner
4,6291320
edited Feb 1 at 0:45
J. W. Tanner
4,6291320
4,6291320
asked Feb 1 at 0:32
Matheus ManzattoMatheus Manzatto
1,2991626
asked Feb 1 at 0:32
Matheus ManzattoMatheus Manzatto
1,2991626
asked Feb 1 at 0:32
Matheus ManzattoMatheus Manzatto
1,2991626
1,2991626
1
$begingroup$
It is impossible in general, when $A, B$ are topological circles. When each of them is a disjoint union of compact arcs, such a homeomorphism does exist, but it is not completely trivial. One needs Schoenflies theorem for topological arcs.
$endgroup$
– Moishe Kohan
Feb 1 at 0:44
add a comment |
1
$begingroup$
It is impossible in general, when $A, B$ are topological circles. When each of them is a disjoint union of compact arcs, such a homeomorphism does exist, but it is not completely trivial. One needs Schoenflies theorem for topological arcs.
$endgroup$
– Moishe Kohan
Feb 1 at 0:44
1
1
$begingroup$
It is impossible in general, when $A, B$ are topological circles. When each of them is a disjoint union of compact arcs, such a homeomorphism does exist, but it is not completely trivial. One needs Schoenflies theorem for topological arcs.
$endgroup$
– Moishe Kohan
Feb 1 at 0:44
$begingroup$
It is impossible in general, when $A, B$ are topological circles. When each of them is a disjoint union of compact arcs, such a homeomorphism does exist, but it is not completely trivial. One needs Schoenflies theorem for topological arcs.
$endgroup$
– Moishe Kohan
Feb 1 at 0:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general it's not possible. Consider A = circle that you can colapse in a point and B = circle that "cuts" the torus(One generator of the fundamental group). So $A^c$ retracts to a 8 shapped figure and $B^c$ is a cilinder.
answered Feb 1 at 1:32
Ygor ArthurYgor Arthur
462
$endgroup$
$begingroup$
Perfect counter-exemple.
$endgroup$
– Matheus Manzatto
Feb 1 at 1:35
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
In general it's not possible. Consider A = circle that you can colapse in a point and B = circle that "cuts" the torus(One generator of the fundamental group). So $A^c$ retracts to a 8 shapped figure and $B^c$ is a cilinder.
answered Feb 1 at 1:32
Ygor ArthurYgor Arthur
462
$endgroup$
$begingroup$
Perfect counter-exemple.
$endgroup$
– Matheus Manzatto
Feb 1 at 1:35
add a comment |
$begingroup$
In general it's not possible. Consider A = circle that you can colapse in a point and B = circle that "cuts" the torus(One generator of the fundamental group). So $A^c$ retracts to a 8 shapped figure and $B^c$ is a cilinder.
answered Feb 1 at 1:32
Ygor ArthurYgor Arthur
462
$endgroup$
$begingroup$
Perfect counter-exemple.
$endgroup$
– Matheus Manzatto
Feb 1 at 1:35
add a comment |
$begingroup$
In general it's not possible. Consider A = circle that you can colapse in a point and B = circle that "cuts" the torus(One generator of the fundamental group). So $A^c$ retracts to a 8 shapped figure and $B^c$ is a cilinder.
answered Feb 1 at 1:32
Ygor ArthurYgor Arthur
462
$endgroup$
In general it's not possible. Consider A = circle that you can colapse in a point and B = circle that "cuts" the torus(One generator of the fundamental group). So $A^c$ retracts to a 8 shapped figure and $B^c$ is a cilinder.
answered Feb 1 at 1:32
Ygor ArthurYgor Arthur
462
answered Feb 1 at 1:32
Ygor ArthurYgor Arthur
462
answered Feb 1 at 1:32
Ygor ArthurYgor Arthur
462
answered Feb 1 at 1:32
Ygor ArthurYgor Arthur
462
462
$begingroup$
Perfect counter-exemple.
$endgroup$
– Matheus Manzatto
Feb 1 at 1:35
add a comment |
$begingroup$
Perfect counter-exemple.
$endgroup$
– Matheus Manzatto
Feb 1 at 1:35
$begingroup$
Perfect counter-exemple.
$endgroup$
– Matheus Manzatto
Feb 1 at 1:35
$begingroup$
Perfect counter-exemple.
$endgroup$
– Matheus Manzatto
Feb 1 at 1:35
add a comment |
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$begingroup$
It is impossible in general, when $A, B$ are topological circles. When each of them is a disjoint union of compact arcs, such a homeomorphism does exist, but it is not completely trivial. One needs Schoenflies theorem for topological arcs.
$endgroup$
– Moishe Kohan
Feb 1 at 0:44