Annulus without ambient space?
$begingroup$
An annulus $A = {(x,y) in mathbb{R}^2: 1<x^2+y^2 < 2}$ is a topological subspace of 2-D Euclidean space $mathbb{R}^2$ and has a boundary composed of two components. But if we forget $mathbb{R}^2 - A$ completely (so the whole universe is $A$ itself), then it clearly has no boundary. This space should be topologically different from $A$. However, in terms of loops (I just started studying elementary algebraic topology), the two spaces seems to be equivalent. Intuitively the two spaces have the same classes of continuously deformable loops. With what algebraic topological tool can I distinguish the two spaces?
general-topology algebraic-topology
asked Dec 26 '18 at 8:08
BalbadakBalbadak
755
$endgroup$
|
show 6 more comments
$begingroup$
An annulus $A = {(x,y) in mathbb{R}^2: 1<x^2+y^2 < 2}$ is a topological subspace of 2-D Euclidean space $mathbb{R}^2$ and has a boundary composed of two components. But if we forget $mathbb{R}^2 - A$ completely (so the whole universe is $A$ itself), then it clearly has no boundary. This space should be topologically different from $A$. However, in terms of loops (I just started studying elementary algebraic topology), the two spaces seems to be equivalent. Intuitively the two spaces have the same classes of continuously deformable loops. With what algebraic topological tool can I distinguish the two spaces?
general-topology algebraic-topology
asked Dec 26 '18 at 8:08
BalbadakBalbadak
755
$endgroup$
$begingroup$
What is the boundary of $A$ as a subsapce of $mathbb{R}^2$?
$endgroup$
– ODF
Dec 26 '18 at 8:12
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@ODF The circles of radius 1 and $sqrt{2}$ form the boundary. So?
$endgroup$
– Balbadak
Dec 26 '18 at 8:22
2
$begingroup$
No. The boundary of a set depends on the embedding (i.e. it depends on the surrounding space), not on the topology of the set itself. So the annulus does in fact have the same topology no matter what space it's embedded in.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:01
1
$begingroup$
Other spaces that have the same topology as the annulus include the plane with a point removed, the sphere with two points removed, and the curved face of a cylinder.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:07
1
$begingroup$
Usually when topologists informally speak about two topological spaces being the "same" or "different", they are referring to the equivalence relation of "homeomorphism". You have said that "This space should be topologically different from $A$", and yet the two spaces are homeomorphic. Do you have some other equivalence relation in mind than homeomorphism?
$endgroup$
– Lee Mosher
Dec 26 '18 at 18:35
|
show 6 more comments
$begingroup$
An annulus $A = {(x,y) in mathbb{R}^2: 1<x^2+y^2 < 2}$ is a topological subspace of 2-D Euclidean space $mathbb{R}^2$ and has a boundary composed of two components. But if we forget $mathbb{R}^2 - A$ completely (so the whole universe is $A$ itself), then it clearly has no boundary. This space should be topologically different from $A$. However, in terms of loops (I just started studying elementary algebraic topology), the two spaces seems to be equivalent. Intuitively the two spaces have the same classes of continuously deformable loops. With what algebraic topological tool can I distinguish the two spaces?
general-topology algebraic-topology
asked Dec 26 '18 at 8:08
BalbadakBalbadak
755
$endgroup$
An annulus $A = {(x,y) in mathbb{R}^2: 1<x^2+y^2 < 2}$ is a topological subspace of 2-D Euclidean space $mathbb{R}^2$ and has a boundary composed of two components. But if we forget $mathbb{R}^2 - A$ completely (so the whole universe is $A$ itself), then it clearly has no boundary. This space should be topologically different from $A$. However, in terms of loops (I just started studying elementary algebraic topology), the two spaces seems to be equivalent. Intuitively the two spaces have the same classes of continuously deformable loops. With what algebraic topological tool can I distinguish the two spaces?
general-topology algebraic-topology
general-topology algebraic-topology
asked Dec 26 '18 at 8:08
BalbadakBalbadak
755
asked Dec 26 '18 at 8:08
BalbadakBalbadak
755
asked Dec 26 '18 at 8:08
BalbadakBalbadak
755
asked Dec 26 '18 at 8:08
BalbadakBalbadak
755
asked Dec 26 '18 at 8:08
BalbadakBalbadak
755
755
$begingroup$
What is the boundary of $A$ as a subsapce of $mathbb{R}^2$?
$endgroup$
– ODF
Dec 26 '18 at 8:12
$begingroup$
@ODF The circles of radius 1 and $sqrt{2}$ form the boundary. So?
$endgroup$
– Balbadak
Dec 26 '18 at 8:22
2
$begingroup$
No. The boundary of a set depends on the embedding (i.e. it depends on the surrounding space), not on the topology of the set itself. So the annulus does in fact have the same topology no matter what space it's embedded in.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:01
1
$begingroup$
Other spaces that have the same topology as the annulus include the plane with a point removed, the sphere with two points removed, and the curved face of a cylinder.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:07
1
$begingroup$
Usually when topologists informally speak about two topological spaces being the "same" or "different", they are referring to the equivalence relation of "homeomorphism". You have said that "This space should be topologically different from $A$", and yet the two spaces are homeomorphic. Do you have some other equivalence relation in mind than homeomorphism?
$endgroup$
– Lee Mosher
Dec 26 '18 at 18:35
|
show 6 more comments
$begingroup$
What is the boundary of $A$ as a subsapce of $mathbb{R}^2$?
$endgroup$
– ODF
Dec 26 '18 at 8:12
$begingroup$
@ODF The circles of radius 1 and $sqrt{2}$ form the boundary. So?
$endgroup$
– Balbadak
Dec 26 '18 at 8:22
2
$begingroup$
No. The boundary of a set depends on the embedding (i.e. it depends on the surrounding space), not on the topology of the set itself. So the annulus does in fact have the same topology no matter what space it's embedded in.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:01
1
$begingroup$
Other spaces that have the same topology as the annulus include the plane with a point removed, the sphere with two points removed, and the curved face of a cylinder.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:07
1
$begingroup$
Usually when topologists informally speak about two topological spaces being the "same" or "different", they are referring to the equivalence relation of "homeomorphism". You have said that "This space should be topologically different from $A$", and yet the two spaces are homeomorphic. Do you have some other equivalence relation in mind than homeomorphism?
$endgroup$
– Lee Mosher
Dec 26 '18 at 18:35
$begingroup$
What is the boundary of $A$ as a subsapce of $mathbb{R}^2$?
$endgroup$
– ODF
Dec 26 '18 at 8:12
$begingroup$
What is the boundary of $A$ as a subsapce of $mathbb{R}^2$?
$endgroup$
– ODF
Dec 26 '18 at 8:12
$begingroup$
@ODF The circles of radius 1 and $sqrt{2}$ form the boundary. So?
$endgroup$
– Balbadak
Dec 26 '18 at 8:22
$begingroup$
@ODF The circles of radius 1 and $sqrt{2}$ form the boundary. So?
$endgroup$
– Balbadak
Dec 26 '18 at 8:22
2
2
$begingroup$
No. The boundary of a set depends on the embedding (i.e. it depends on the surrounding space), not on the topology of the set itself. So the annulus does in fact have the same topology no matter what space it's embedded in.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:01
$begingroup$
No. The boundary of a set depends on the embedding (i.e. it depends on the surrounding space), not on the topology of the set itself. So the annulus does in fact have the same topology no matter what space it's embedded in.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:01
1
1
$begingroup$
Other spaces that have the same topology as the annulus include the plane with a point removed, the sphere with two points removed, and the curved face of a cylinder.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:07
$begingroup$
Other spaces that have the same topology as the annulus include the plane with a point removed, the sphere with two points removed, and the curved face of a cylinder.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:07
1
1
$begingroup$
Usually when topologists informally speak about two topological spaces being the "same" or "different", they are referring to the equivalence relation of "homeomorphism". You have said that "This space should be topologically different from $A$", and yet the two spaces are homeomorphic. Do you have some other equivalence relation in mind than homeomorphism?
$endgroup$
– Lee Mosher
Dec 26 '18 at 18:35
$begingroup$
Usually when topologists informally speak about two topological spaces being the "same" or "different", they are referring to the equivalence relation of "homeomorphism". You have said that "This space should be topologically different from $A$", and yet the two spaces are homeomorphic. Do you have some other equivalence relation in mind than homeomorphism?
$endgroup$
– Lee Mosher
Dec 26 '18 at 18:35
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
First of all, "loops" will not detect something as subtle as you mean. However, it is also true that taking the closure of an open subset and considering the boundary of the resulting space depends on how you embed a space.
For example the subsets ${(x,y,z) mid z neq pm 1} subset S^2$ is topologically a cylinder $S^1 times mathbb R$ which is homeomorphic to the open annulus. Likewise, you can consider ${(x,y,z) mid |z|<1/2}$ is also an annulus. The "boundary" of the first is two points and the boundary of the latter is $S^1 coprod S^1$, despite them being homeomorphic.
On the other hand "loops" will only detect up to homotopy. For example, as a space unto itself, $S^1$ and $S^1 times mathbb R$ will have the same fundamental group.
answered Dec 26 '18 at 18:04
Andres MejiaAndres Mejia
16.2k21549
$endgroup$
add a comment |
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$begingroup$
First of all, "loops" will not detect something as subtle as you mean. However, it is also true that taking the closure of an open subset and considering the boundary of the resulting space depends on how you embed a space.
For example the subsets ${(x,y,z) mid z neq pm 1} subset S^2$ is topologically a cylinder $S^1 times mathbb R$ which is homeomorphic to the open annulus. Likewise, you can consider ${(x,y,z) mid |z|<1/2}$ is also an annulus. The "boundary" of the first is two points and the boundary of the latter is $S^1 coprod S^1$, despite them being homeomorphic.
On the other hand "loops" will only detect up to homotopy. For example, as a space unto itself, $S^1$ and $S^1 times mathbb R$ will have the same fundamental group.
answered Dec 26 '18 at 18:04
Andres MejiaAndres Mejia
16.2k21549
$endgroup$
add a comment |
$begingroup$
First of all, "loops" will not detect something as subtle as you mean. However, it is also true that taking the closure of an open subset and considering the boundary of the resulting space depends on how you embed a space.
For example the subsets ${(x,y,z) mid z neq pm 1} subset S^2$ is topologically a cylinder $S^1 times mathbb R$ which is homeomorphic to the open annulus. Likewise, you can consider ${(x,y,z) mid |z|<1/2}$ is also an annulus. The "boundary" of the first is two points and the boundary of the latter is $S^1 coprod S^1$, despite them being homeomorphic.
On the other hand "loops" will only detect up to homotopy. For example, as a space unto itself, $S^1$ and $S^1 times mathbb R$ will have the same fundamental group.
answered Dec 26 '18 at 18:04
Andres MejiaAndres Mejia
16.2k21549
$endgroup$
add a comment |
$begingroup$
First of all, "loops" will not detect something as subtle as you mean. However, it is also true that taking the closure of an open subset and considering the boundary of the resulting space depends on how you embed a space.
For example the subsets ${(x,y,z) mid z neq pm 1} subset S^2$ is topologically a cylinder $S^1 times mathbb R$ which is homeomorphic to the open annulus. Likewise, you can consider ${(x,y,z) mid |z|<1/2}$ is also an annulus. The "boundary" of the first is two points and the boundary of the latter is $S^1 coprod S^1$, despite them being homeomorphic.
On the other hand "loops" will only detect up to homotopy. For example, as a space unto itself, $S^1$ and $S^1 times mathbb R$ will have the same fundamental group.
answered Dec 26 '18 at 18:04
Andres MejiaAndres Mejia
16.2k21549
$endgroup$
First of all, "loops" will not detect something as subtle as you mean. However, it is also true that taking the closure of an open subset and considering the boundary of the resulting space depends on how you embed a space.
For example the subsets ${(x,y,z) mid z neq pm 1} subset S^2$ is topologically a cylinder $S^1 times mathbb R$ which is homeomorphic to the open annulus. Likewise, you can consider ${(x,y,z) mid |z|<1/2}$ is also an annulus. The "boundary" of the first is two points and the boundary of the latter is $S^1 coprod S^1$, despite them being homeomorphic.
On the other hand "loops" will only detect up to homotopy. For example, as a space unto itself, $S^1$ and $S^1 times mathbb R$ will have the same fundamental group.
answered Dec 26 '18 at 18:04
Andres MejiaAndres Mejia
16.2k21549
answered Dec 26 '18 at 18:04
Andres MejiaAndres Mejia
16.2k21549
answered Dec 26 '18 at 18:04
Andres MejiaAndres Mejia
16.2k21549
answered Dec 26 '18 at 18:04
Andres MejiaAndres Mejia
16.2k21549
16.2k21549
add a comment |
add a comment |
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$begingroup$
What is the boundary of $A$ as a subsapce of $mathbb{R}^2$?
$endgroup$
– ODF
Dec 26 '18 at 8:12
$begingroup$
@ODF The circles of radius 1 and $sqrt{2}$ form the boundary. So?
$endgroup$
– Balbadak
Dec 26 '18 at 8:22
2
$begingroup$
No. The boundary of a set depends on the embedding (i.e. it depends on the surrounding space), not on the topology of the set itself. So the annulus does in fact have the same topology no matter what space it's embedded in.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:01
1
$begingroup$
Other spaces that have the same topology as the annulus include the plane with a point removed, the sphere with two points removed, and the curved face of a cylinder.
$endgroup$
– Akiva Weinberger
Dec 26 '18 at 9:07
1
$begingroup$
Usually when topologists informally speak about two topological spaces being the "same" or "different", they are referring to the equivalence relation of "homeomorphism". You have said that "This space should be topologically different from $A$", and yet the two spaces are homeomorphic. Do you have some other equivalence relation in mind than homeomorphism?
$endgroup$
– Lee Mosher
Dec 26 '18 at 18:35