First fundamental group $pi_1(X)$ of $mathbb{R^3} setminus {$a circumference $cup$ a line tangent to a point...
$begingroup$
I have to find the fundamental group $pi_1(X)$ of $X=mathbb{R^3} setminus {$a line $cup$ a circumference$}$.
Line passing through the centre of the circle, we have: $pi_1(X) congleft(mathbb{R^3}setminus{0}right)times S^1 cong mathbb{Z}timesmathbb{Z}$
Line "out" from the circumference, using the Van Kampen theorem, $X=Acup B$ where $A=mathbb{R^3} setminus S^1$, $B= mathbb{R^3} setminus {$the line$}$ and $Acap B=mathbb{R^3}$, we have $pi_1(X) cong cfrac{mathbb{Z}*mathbb{Z}}{{e}} congmathbb{Z}*mathbb{Z}$
What about the $pi_1(X)$ when the line tangent to a point on the circle? I have no ideas
algebraic-topology fundamental-groups
asked Dec 12 '18 at 21:57
F.incF.inc
415210
$endgroup$
add a comment |
$begingroup$
I have to find the fundamental group $pi_1(X)$ of $X=mathbb{R^3} setminus {$a line $cup$ a circumference$}$.
Line passing through the centre of the circle, we have: $pi_1(X) congleft(mathbb{R^3}setminus{0}right)times S^1 cong mathbb{Z}timesmathbb{Z}$
Line "out" from the circumference, using the Van Kampen theorem, $X=Acup B$ where $A=mathbb{R^3} setminus S^1$, $B= mathbb{R^3} setminus {$the line$}$ and $Acap B=mathbb{R^3}$, we have $pi_1(X) cong cfrac{mathbb{Z}*mathbb{Z}}{{e}} congmathbb{Z}*mathbb{Z}$
What about the $pi_1(X)$ when the line tangent to a point on the circle? I have no ideas
algebraic-topology fundamental-groups
asked Dec 12 '18 at 21:57
F.incF.inc
415210
$endgroup$
3
$begingroup$
Your last $X$ is homotopy equivalent to "$8$" (bouquet of two circles), so you get a free group with two generators.
$endgroup$
– user8268
Dec 12 '18 at 22:23
2
$begingroup$
Very similar to math.stackexchange.com/q/3026468.
$endgroup$
– Paul Frost
Dec 12 '18 at 22:33
1
$begingroup$
@user8268 how to see that? Because at first glance I would have said $mathbb{Z}timesmathbb{Z}$: thinking at the loop which "passes" around the intersection gives me the idea that the generators can commute.
$endgroup$
– F.inc
Dec 13 '18 at 16:47
1
$begingroup$
I'll try to explain: $Bbb R^3$ minus a circle deformation retracts onto a sphere with a diameter. In place of a circle and a tangent line, let me rather remove the circle $x^2+y^2=1$, $z=0$, and the two half-lines $xleq-1$, $xgeq 1$ ($y=z=0$). This thing now retracts to a sphere with a diameter, but with two holes in the sphere (that's where the half-lines cross it), which in turn retracts to a "$ominus$" shape, and that is homotopy equivalent to "$8$"
$endgroup$
– user8268
Dec 13 '18 at 21:09
add a comment |
$begingroup$
I have to find the fundamental group $pi_1(X)$ of $X=mathbb{R^3} setminus {$a line $cup$ a circumference$}$.
Line passing through the centre of the circle, we have: $pi_1(X) congleft(mathbb{R^3}setminus{0}right)times S^1 cong mathbb{Z}timesmathbb{Z}$
Line "out" from the circumference, using the Van Kampen theorem, $X=Acup B$ where $A=mathbb{R^3} setminus S^1$, $B= mathbb{R^3} setminus {$the line$}$ and $Acap B=mathbb{R^3}$, we have $pi_1(X) cong cfrac{mathbb{Z}*mathbb{Z}}{{e}} congmathbb{Z}*mathbb{Z}$
What about the $pi_1(X)$ when the line tangent to a point on the circle? I have no ideas
algebraic-topology fundamental-groups
asked Dec 12 '18 at 21:57
F.incF.inc
415210
$endgroup$
I have to find the fundamental group $pi_1(X)$ of $X=mathbb{R^3} setminus {$a line $cup$ a circumference$}$.
Line passing through the centre of the circle, we have: $pi_1(X) congleft(mathbb{R^3}setminus{0}right)times S^1 cong mathbb{Z}timesmathbb{Z}$
Line "out" from the circumference, using the Van Kampen theorem, $X=Acup B$ where $A=mathbb{R^3} setminus S^1$, $B= mathbb{R^3} setminus {$the line$}$ and $Acap B=mathbb{R^3}$, we have $pi_1(X) cong cfrac{mathbb{Z}*mathbb{Z}}{{e}} congmathbb{Z}*mathbb{Z}$
What about the $pi_1(X)$ when the line tangent to a point on the circle? I have no ideas
algebraic-topology fundamental-groups
algebraic-topology fundamental-groups
asked Dec 12 '18 at 21:57
F.incF.inc
415210
asked Dec 12 '18 at 21:57
F.incF.inc
415210
asked Dec 12 '18 at 21:57
F.incF.inc
415210
asked Dec 12 '18 at 21:57
F.incF.inc
415210
asked Dec 12 '18 at 21:57
F.incF.inc
415210
415210
3
$begingroup$
Your last $X$ is homotopy equivalent to "$8$" (bouquet of two circles), so you get a free group with two generators.
$endgroup$
– user8268
Dec 12 '18 at 22:23
2
$begingroup$
Very similar to math.stackexchange.com/q/3026468.
$endgroup$
– Paul Frost
Dec 12 '18 at 22:33
1
$begingroup$
@user8268 how to see that? Because at first glance I would have said $mathbb{Z}timesmathbb{Z}$: thinking at the loop which "passes" around the intersection gives me the idea that the generators can commute.
$endgroup$
– F.inc
Dec 13 '18 at 16:47
1
$begingroup$
I'll try to explain: $Bbb R^3$ minus a circle deformation retracts onto a sphere with a diameter. In place of a circle and a tangent line, let me rather remove the circle $x^2+y^2=1$, $z=0$, and the two half-lines $xleq-1$, $xgeq 1$ ($y=z=0$). This thing now retracts to a sphere with a diameter, but with two holes in the sphere (that's where the half-lines cross it), which in turn retracts to a "$ominus$" shape, and that is homotopy equivalent to "$8$"
$endgroup$
– user8268
Dec 13 '18 at 21:09
add a comment |
3
$begingroup$
Your last $X$ is homotopy equivalent to "$8$" (bouquet of two circles), so you get a free group with two generators.
$endgroup$
– user8268
Dec 12 '18 at 22:23
2
$begingroup$
Very similar to math.stackexchange.com/q/3026468.
$endgroup$
– Paul Frost
Dec 12 '18 at 22:33
1
$begingroup$
@user8268 how to see that? Because at first glance I would have said $mathbb{Z}timesmathbb{Z}$: thinking at the loop which "passes" around the intersection gives me the idea that the generators can commute.
$endgroup$
– F.inc
Dec 13 '18 at 16:47
1
$begingroup$
I'll try to explain: $Bbb R^3$ minus a circle deformation retracts onto a sphere with a diameter. In place of a circle and a tangent line, let me rather remove the circle $x^2+y^2=1$, $z=0$, and the two half-lines $xleq-1$, $xgeq 1$ ($y=z=0$). This thing now retracts to a sphere with a diameter, but with two holes in the sphere (that's where the half-lines cross it), which in turn retracts to a "$ominus$" shape, and that is homotopy equivalent to "$8$"
$endgroup$
– user8268
Dec 13 '18 at 21:09
3
3
$begingroup$
Your last $X$ is homotopy equivalent to "$8$" (bouquet of two circles), so you get a free group with two generators.
$endgroup$
– user8268
Dec 12 '18 at 22:23
$begingroup$
Your last $X$ is homotopy equivalent to "$8$" (bouquet of two circles), so you get a free group with two generators.
$endgroup$
– user8268
Dec 12 '18 at 22:23
2
2
$begingroup$
Very similar to math.stackexchange.com/q/3026468.
$endgroup$
– Paul Frost
Dec 12 '18 at 22:33
$begingroup$
Very similar to math.stackexchange.com/q/3026468.
$endgroup$
– Paul Frost
Dec 12 '18 at 22:33
1
1
$begingroup$
@user8268 how to see that? Because at first glance I would have said $mathbb{Z}timesmathbb{Z}$: thinking at the loop which "passes" around the intersection gives me the idea that the generators can commute.
$endgroup$
– F.inc
Dec 13 '18 at 16:47
$begingroup$
@user8268 how to see that? Because at first glance I would have said $mathbb{Z}timesmathbb{Z}$: thinking at the loop which "passes" around the intersection gives me the idea that the generators can commute.
$endgroup$
– F.inc
Dec 13 '18 at 16:47
1
1
$begingroup$
I'll try to explain: $Bbb R^3$ minus a circle deformation retracts onto a sphere with a diameter. In place of a circle and a tangent line, let me rather remove the circle $x^2+y^2=1$, $z=0$, and the two half-lines $xleq-1$, $xgeq 1$ ($y=z=0$). This thing now retracts to a sphere with a diameter, but with two holes in the sphere (that's where the half-lines cross it), which in turn retracts to a "$ominus$" shape, and that is homotopy equivalent to "$8$"
$endgroup$
– user8268
Dec 13 '18 at 21:09
$begingroup$
I'll try to explain: $Bbb R^3$ minus a circle deformation retracts onto a sphere with a diameter. In place of a circle and a tangent line, let me rather remove the circle $x^2+y^2=1$, $z=0$, and the two half-lines $xleq-1$, $xgeq 1$ ($y=z=0$). This thing now retracts to a sphere with a diameter, but with two holes in the sphere (that's where the half-lines cross it), which in turn retracts to a "$ominus$" shape, and that is homotopy equivalent to "$8$"
$endgroup$
– user8268
Dec 13 '18 at 21:09
add a comment |
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$begingroup$
Your last $X$ is homotopy equivalent to "$8$" (bouquet of two circles), so you get a free group with two generators.
$endgroup$
– user8268
Dec 12 '18 at 22:23
2
$begingroup$
Very similar to math.stackexchange.com/q/3026468.
$endgroup$
– Paul Frost
Dec 12 '18 at 22:33
1
$begingroup$
@user8268 how to see that? Because at first glance I would have said $mathbb{Z}timesmathbb{Z}$: thinking at the loop which "passes" around the intersection gives me the idea that the generators can commute.
$endgroup$
– F.inc
Dec 13 '18 at 16:47
1
$begingroup$
I'll try to explain: $Bbb R^3$ minus a circle deformation retracts onto a sphere with a diameter. In place of a circle and a tangent line, let me rather remove the circle $x^2+y^2=1$, $z=0$, and the two half-lines $xleq-1$, $xgeq 1$ ($y=z=0$). This thing now retracts to a sphere with a diameter, but with two holes in the sphere (that's where the half-lines cross it), which in turn retracts to a "$ominus$" shape, and that is homotopy equivalent to "$8$"
$endgroup$
– user8268
Dec 13 '18 at 21:09