How do I prove this using van-Kampen theorem informally ? (2)
$begingroup$
The second( and the last) problem is this
Could someone please help me how to calculate $pi_1(X)$?
algebraic-topology
asked Dec 17 '14 at 18:15
RubertosRubertos
5,6802824
$endgroup$
add a comment |
$begingroup$
The second( and the last) problem is this
Could someone please help me how to calculate $pi_1(X)$?
algebraic-topology
asked Dec 17 '14 at 18:15
RubertosRubertos
5,6802824
$endgroup$
$begingroup$
@DanielRust I want to believe it..
$endgroup$
– Rubertos
Dec 17 '14 at 18:22
$begingroup$
Is that a link, @Rubertos?
$endgroup$
– Balarka Sen
Dec 17 '14 at 18:37
$begingroup$
@BalarkaSen No, I can just upload a picture for you. It's just one page
$endgroup$
– Rubertos
Dec 17 '14 at 18:39
add a comment |
$begingroup$
The second( and the last) problem is this
Could someone please help me how to calculate $pi_1(X)$?
algebraic-topology
asked Dec 17 '14 at 18:15
RubertosRubertos
5,6802824
$endgroup$
The second( and the last) problem is this
Could someone please help me how to calculate $pi_1(X)$?
algebraic-topology
algebraic-topology
asked Dec 17 '14 at 18:15
RubertosRubertos
5,6802824
asked Dec 17 '14 at 18:15
RubertosRubertos
5,6802824
edited Dec 19 '15 at 1:17
Rubertos
asked Dec 17 '14 at 18:15
RubertosRubertos
5,6802824
asked Dec 17 '14 at 18:15
RubertosRubertos
5,6802824
asked Dec 17 '14 at 18:15
RubertosRubertos
5,6802824
5,6802824
$begingroup$
@DanielRust I want to believe it..
$endgroup$
– Rubertos
Dec 17 '14 at 18:22
$begingroup$
Is that a link, @Rubertos?
$endgroup$
– Balarka Sen
Dec 17 '14 at 18:37
$begingroup$
@BalarkaSen No, I can just upload a picture for you. It's just one page
$endgroup$
– Rubertos
Dec 17 '14 at 18:39
add a comment |
$begingroup$
@DanielRust I want to believe it..
$endgroup$
– Rubertos
Dec 17 '14 at 18:22
$begingroup$
Is that a link, @Rubertos?
$endgroup$
– Balarka Sen
Dec 17 '14 at 18:37
$begingroup$
@BalarkaSen No, I can just upload a picture for you. It's just one page
$endgroup$
– Rubertos
Dec 17 '14 at 18:39
$begingroup$
@DanielRust I want to believe it..
$endgroup$
– Rubertos
Dec 17 '14 at 18:22
$begingroup$
@DanielRust I want to believe it..
$endgroup$
– Rubertos
Dec 17 '14 at 18:22
$begingroup$
Is that a link, @Rubertos?
$endgroup$
– Balarka Sen
Dec 17 '14 at 18:37
$begingroup$
Is that a link, @Rubertos?
$endgroup$
– Balarka Sen
Dec 17 '14 at 18:37
$begingroup$
@BalarkaSen No, I can just upload a picture for you. It's just one page
$endgroup$
– Rubertos
Dec 17 '14 at 18:39
$begingroup$
@BalarkaSen No, I can just upload a picture for you. It's just one page
$endgroup$
– Rubertos
Dec 17 '14 at 18:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An informal calculation might go as follows. First, let's "push" one of the circles to infinity so that we're instead removing a copy of $S^1$ from $mathbb{R}^3$ and a copy of $mathbb{R}$ which 'goes through' the circle and goes off towards infinity along the $z$-coordinate. You should hopefully be able to see that this space is a kind of 'maximally fattened up torus' in $3$-space. That is, if we just start expanding the torus as much as we can in all direction in $mathbb{R}^3$, the only bits which we would not be able to 'fill in' by this fattening process would be a circle inside the complement of the torus, and a line going through the 'hole' in our torus. So our space should have the same fundamental group as the torus, namely $mathbb{Z}^2$.
To prove this formally, one would need to use Van-Kampen's theorem, to show that $pi_1(S^3setminus L)cong pi_1(mathbb{R}^3setminus L)$ where we view $S^3$ as the one-point compactification of $mathbb{R}^3$. This makes the 'pushing the circle to infinity' part of the above actually work. The 'fattening up' process is really just saying that this new space deformaiton retracts onto a torus.
answered Dec 17 '14 at 18:30
Dan RustDan Rust
22.8k114884
$endgroup$
$begingroup$
(Sorry I have to leave so won't be able to answer any questions about this, admittedly hard to visualise, answer - hopefully someone else can help if it's needed)
$endgroup$
– Dan Rust
Dec 17 '14 at 18:33
$begingroup$
I get it. Thank you so much!!
$endgroup$
– Rubertos
Dec 17 '14 at 18:50
add a comment |
$begingroup$
You also need to understand the intuition of a relation at a crossing, as follows:
I have demonstrated this to children with a copper tubing pentoil and a nice length of rope. For the connection with the van Kampen theorem, see my book Topology and Groupoids, p. 349.
answered Dec 17 '14 at 18:55
Ronnie BrownRonnie Brown
12k12938
$endgroup$
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
$begingroup$
An informal calculation might go as follows. First, let's "push" one of the circles to infinity so that we're instead removing a copy of $S^1$ from $mathbb{R}^3$ and a copy of $mathbb{R}$ which 'goes through' the circle and goes off towards infinity along the $z$-coordinate. You should hopefully be able to see that this space is a kind of 'maximally fattened up torus' in $3$-space. That is, if we just start expanding the torus as much as we can in all direction in $mathbb{R}^3$, the only bits which we would not be able to 'fill in' by this fattening process would be a circle inside the complement of the torus, and a line going through the 'hole' in our torus. So our space should have the same fundamental group as the torus, namely $mathbb{Z}^2$.
To prove this formally, one would need to use Van-Kampen's theorem, to show that $pi_1(S^3setminus L)cong pi_1(mathbb{R}^3setminus L)$ where we view $S^3$ as the one-point compactification of $mathbb{R}^3$. This makes the 'pushing the circle to infinity' part of the above actually work. The 'fattening up' process is really just saying that this new space deformaiton retracts onto a torus.
answered Dec 17 '14 at 18:30
Dan RustDan Rust
22.8k114884
$endgroup$
$begingroup$
(Sorry I have to leave so won't be able to answer any questions about this, admittedly hard to visualise, answer - hopefully someone else can help if it's needed)
$endgroup$
– Dan Rust
Dec 17 '14 at 18:33
$begingroup$
I get it. Thank you so much!!
$endgroup$
– Rubertos
Dec 17 '14 at 18:50
add a comment |
$begingroup$
An informal calculation might go as follows. First, let's "push" one of the circles to infinity so that we're instead removing a copy of $S^1$ from $mathbb{R}^3$ and a copy of $mathbb{R}$ which 'goes through' the circle and goes off towards infinity along the $z$-coordinate. You should hopefully be able to see that this space is a kind of 'maximally fattened up torus' in $3$-space. That is, if we just start expanding the torus as much as we can in all direction in $mathbb{R}^3$, the only bits which we would not be able to 'fill in' by this fattening process would be a circle inside the complement of the torus, and a line going through the 'hole' in our torus. So our space should have the same fundamental group as the torus, namely $mathbb{Z}^2$.
To prove this formally, one would need to use Van-Kampen's theorem, to show that $pi_1(S^3setminus L)cong pi_1(mathbb{R}^3setminus L)$ where we view $S^3$ as the one-point compactification of $mathbb{R}^3$. This makes the 'pushing the circle to infinity' part of the above actually work. The 'fattening up' process is really just saying that this new space deformaiton retracts onto a torus.
answered Dec 17 '14 at 18:30
Dan RustDan Rust
22.8k114884
$endgroup$
$begingroup$
(Sorry I have to leave so won't be able to answer any questions about this, admittedly hard to visualise, answer - hopefully someone else can help if it's needed)
$endgroup$
– Dan Rust
Dec 17 '14 at 18:33
$begingroup$
I get it. Thank you so much!!
$endgroup$
– Rubertos
Dec 17 '14 at 18:50
add a comment |
$begingroup$
An informal calculation might go as follows. First, let's "push" one of the circles to infinity so that we're instead removing a copy of $S^1$ from $mathbb{R}^3$ and a copy of $mathbb{R}$ which 'goes through' the circle and goes off towards infinity along the $z$-coordinate. You should hopefully be able to see that this space is a kind of 'maximally fattened up torus' in $3$-space. That is, if we just start expanding the torus as much as we can in all direction in $mathbb{R}^3$, the only bits which we would not be able to 'fill in' by this fattening process would be a circle inside the complement of the torus, and a line going through the 'hole' in our torus. So our space should have the same fundamental group as the torus, namely $mathbb{Z}^2$.
To prove this formally, one would need to use Van-Kampen's theorem, to show that $pi_1(S^3setminus L)cong pi_1(mathbb{R}^3setminus L)$ where we view $S^3$ as the one-point compactification of $mathbb{R}^3$. This makes the 'pushing the circle to infinity' part of the above actually work. The 'fattening up' process is really just saying that this new space deformaiton retracts onto a torus.
answered Dec 17 '14 at 18:30
Dan RustDan Rust
22.8k114884
$endgroup$
An informal calculation might go as follows. First, let's "push" one of the circles to infinity so that we're instead removing a copy of $S^1$ from $mathbb{R}^3$ and a copy of $mathbb{R}$ which 'goes through' the circle and goes off towards infinity along the $z$-coordinate. You should hopefully be able to see that this space is a kind of 'maximally fattened up torus' in $3$-space. That is, if we just start expanding the torus as much as we can in all direction in $mathbb{R}^3$, the only bits which we would not be able to 'fill in' by this fattening process would be a circle inside the complement of the torus, and a line going through the 'hole' in our torus. So our space should have the same fundamental group as the torus, namely $mathbb{Z}^2$.
To prove this formally, one would need to use Van-Kampen's theorem, to show that $pi_1(S^3setminus L)cong pi_1(mathbb{R}^3setminus L)$ where we view $S^3$ as the one-point compactification of $mathbb{R}^3$. This makes the 'pushing the circle to infinity' part of the above actually work. The 'fattening up' process is really just saying that this new space deformaiton retracts onto a torus.
answered Dec 17 '14 at 18:30
Dan RustDan Rust
22.8k114884
answered Dec 17 '14 at 18:30
Dan RustDan Rust
22.8k114884
answered Dec 17 '14 at 18:30
Dan RustDan Rust
22.8k114884
answered Dec 17 '14 at 18:30
Dan RustDan Rust
22.8k114884
22.8k114884
$begingroup$
(Sorry I have to leave so won't be able to answer any questions about this, admittedly hard to visualise, answer - hopefully someone else can help if it's needed)
$endgroup$
– Dan Rust
Dec 17 '14 at 18:33
$begingroup$
I get it. Thank you so much!!
$endgroup$
– Rubertos
Dec 17 '14 at 18:50
add a comment |
$begingroup$
(Sorry I have to leave so won't be able to answer any questions about this, admittedly hard to visualise, answer - hopefully someone else can help if it's needed)
$endgroup$
– Dan Rust
Dec 17 '14 at 18:33
$begingroup$
I get it. Thank you so much!!
$endgroup$
– Rubertos
Dec 17 '14 at 18:50
$begingroup$
(Sorry I have to leave so won't be able to answer any questions about this, admittedly hard to visualise, answer - hopefully someone else can help if it's needed)
$endgroup$
– Dan Rust
Dec 17 '14 at 18:33
$begingroup$
(Sorry I have to leave so won't be able to answer any questions about this, admittedly hard to visualise, answer - hopefully someone else can help if it's needed)
$endgroup$
– Dan Rust
Dec 17 '14 at 18:33
$begingroup$
I get it. Thank you so much!!
$endgroup$
– Rubertos
Dec 17 '14 at 18:50
$begingroup$
I get it. Thank you so much!!
$endgroup$
– Rubertos
Dec 17 '14 at 18:50
add a comment |
$begingroup$
You also need to understand the intuition of a relation at a crossing, as follows:
I have demonstrated this to children with a copper tubing pentoil and a nice length of rope. For the connection with the van Kampen theorem, see my book Topology and Groupoids, p. 349.
answered Dec 17 '14 at 18:55
Ronnie BrownRonnie Brown
12k12938
$endgroup$
add a comment |
$begingroup$
You also need to understand the intuition of a relation at a crossing, as follows:
I have demonstrated this to children with a copper tubing pentoil and a nice length of rope. For the connection with the van Kampen theorem, see my book Topology and Groupoids, p. 349.
answered Dec 17 '14 at 18:55
Ronnie BrownRonnie Brown
12k12938
$endgroup$
add a comment |
$begingroup$
You also need to understand the intuition of a relation at a crossing, as follows:
I have demonstrated this to children with a copper tubing pentoil and a nice length of rope. For the connection with the van Kampen theorem, see my book Topology and Groupoids, p. 349.
answered Dec 17 '14 at 18:55
Ronnie BrownRonnie Brown
12k12938
$endgroup$
You also need to understand the intuition of a relation at a crossing, as follows:
I have demonstrated this to children with a copper tubing pentoil and a nice length of rope. For the connection with the van Kampen theorem, see my book Topology and Groupoids, p. 349.
answered Dec 17 '14 at 18:55
Ronnie BrownRonnie Brown
12k12938
edited Jan 14 at 15:06
answered Dec 17 '14 at 18:55
Ronnie BrownRonnie Brown
12k12938
answered Dec 17 '14 at 18:55
Ronnie BrownRonnie Brown
12k12938
answered Dec 17 '14 at 18:55
Ronnie BrownRonnie Brown
12k12938
12k12938
add a comment |
add a comment |
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$begingroup$
@DanielRust I want to believe it..
$endgroup$
– Rubertos
Dec 17 '14 at 18:22
$begingroup$
Is that a link, @Rubertos?
$endgroup$
– Balarka Sen
Dec 17 '14 at 18:37
$begingroup$
@BalarkaSen No, I can just upload a picture for you. It's just one page
$endgroup$
– Rubertos
Dec 17 '14 at 18:39