First fundamental group $pi_1(X)$ of $mathbb{R^3} setminus {$a circumference $cup$ a line tangent to a point...












5












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















5












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














5












5








5


2



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










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