Fundamental group of $mathbb{R}^3setminus$ line+2 circles











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I am trying to find the fundamental group of the space $X=mathbb{R}^3setminus S$ where $S$ is the union of the $z$-axis, and two circles of radius 1 on the $xy$-plane, centered at $(0,0,1)$ and $(0,0,-1)$.



My approach:
We know that the fundamental group of $mathbb{R}^3$ minus circle+$z$-axis is that of a torus, which is $mathbb{Z}_2$. Split the given space into two, namely, into $U_1=Xcap {(x,y,z-epsilon):(x,y,z)intext{upper half plane}, 0<epsilon<1}$ and $U_2=Xcap {(x,y,z+epsilon):(x,y,z)intext{lower half plane}, 0<epsilon<1}$.
Then $U_1cup U_2=X$, and $U_1$, $U_2$ are panth connected and open. The intersection then deformation retracts to $mathbb{R}^2-{0,0}$, whose fundamental group is $mathbb{Z}$.



We now have to figure out the induced homomorphism $i^*:mathbb{Z}tomathbb{Z}_2$. This is where I am most uncertain. I thought, since the generating loop of $U_1cap U_2$ simply gets included to one of the generating loops of the torus (the one going around the center hole), we get the presentation
begin{equation}
pi(X) cong <a,b,c,d:ab=ba,cd=dc,a=c>?
end{equation}
(What is this group, by the way?)



Is what I did correct?



Many thanks!






algebraic-topology fundamental-groups







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  • Two circles of radius $1$ on the xy-plane that's centered at (0,0,1) and (0,0,-1)? Do you mean for the 1 and -1 to be the y coordinates?
    – Yunus Syed
    Dec 5 at 1:39















up vote
1
down vote

favorite
1












I am trying to find the fundamental group of the space $X=mathbb{R}^3setminus S$ where $S$ is the union of the $z$-axis, and two circles of radius 1 on the $xy$-plane, centered at $(0,0,1)$ and $(0,0,-1)$.



My approach:
We know that the fundamental group of $mathbb{R}^3$ minus circle+$z$-axis is that of a torus, which is $mathbb{Z}_2$. Split the given space into two, namely, into $U_1=Xcap {(x,y,z-epsilon):(x,y,z)intext{upper half plane}, 0<epsilon<1}$ and $U_2=Xcap {(x,y,z+epsilon):(x,y,z)intext{lower half plane}, 0<epsilon<1}$.
Then $U_1cup U_2=X$, and $U_1$, $U_2$ are panth connected and open. The intersection then deformation retracts to $mathbb{R}^2-{0,0}$, whose fundamental group is $mathbb{Z}$.



We now have to figure out the induced homomorphism $i^*:mathbb{Z}tomathbb{Z}_2$. This is where I am most uncertain. I thought, since the generating loop of $U_1cap U_2$ simply gets included to one of the generating loops of the torus (the one going around the center hole), we get the presentation
begin{equation}
pi(X) cong <a,b,c,d:ab=ba,cd=dc,a=c>?
end{equation}
(What is this group, by the way?)



Is what I did correct?



Many thanks!






algebraic-topology fundamental-groups







share|cite|improve this question






















  • Two circles of radius $1$ on the xy-plane that's centered at (0,0,1) and (0,0,-1)? Do you mean for the 1 and -1 to be the y coordinates?
    – Yunus Syed
    Dec 5 at 1:39













up vote
1
down vote

favorite
1









up vote
1
down vote

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1





I am trying to find the fundamental group of the space $X=mathbb{R}^3setminus S$ where $S$ is the union of the $z$-axis, and two circles of radius 1 on the $xy$-plane, centered at $(0,0,1)$ and $(0,0,-1)$.



My approach:
We know that the fundamental group of $mathbb{R}^3$ minus circle+$z$-axis is that of a torus, which is $mathbb{Z}_2$. Split the given space into two, namely, into $U_1=Xcap {(x,y,z-epsilon):(x,y,z)intext{upper half plane}, 0<epsilon<1}$ and $U_2=Xcap {(x,y,z+epsilon):(x,y,z)intext{lower half plane}, 0<epsilon<1}$.
Then $U_1cup U_2=X$, and $U_1$, $U_2$ are panth connected and open. The intersection then deformation retracts to $mathbb{R}^2-{0,0}$, whose fundamental group is $mathbb{Z}$.



We now have to figure out the induced homomorphism $i^*:mathbb{Z}tomathbb{Z}_2$. This is where I am most uncertain. I thought, since the generating loop of $U_1cap U_2$ simply gets included to one of the generating loops of the torus (the one going around the center hole), we get the presentation
begin{equation}
pi(X) cong <a,b,c,d:ab=ba,cd=dc,a=c>?
end{equation}
(What is this group, by the way?)



Is what I did correct?



Many thanks!






algebraic-topology fundamental-groups







share|cite|improve this question













I am trying to find the fundamental group of the space $X=mathbb{R}^3setminus S$ where $S$ is the union of the $z$-axis, and two circles of radius 1 on the $xy$-plane, centered at $(0,0,1)$ and $(0,0,-1)$.



My approach:
We know that the fundamental group of $mathbb{R}^3$ minus circle+$z$-axis is that of a torus, which is $mathbb{Z}_2$. Split the given space into two, namely, into $U_1=Xcap {(x,y,z-epsilon):(x,y,z)intext{upper half plane}, 0<epsilon<1}$ and $U_2=Xcap {(x,y,z+epsilon):(x,y,z)intext{lower half plane}, 0<epsilon<1}$.
Then $U_1cup U_2=X$, and $U_1$, $U_2$ are panth connected and open. The intersection then deformation retracts to $mathbb{R}^2-{0,0}$, whose fundamental group is $mathbb{Z}$.



We now have to figure out the induced homomorphism $i^*:mathbb{Z}tomathbb{Z}_2$. This is where I am most uncertain. I thought, since the generating loop of $U_1cap U_2$ simply gets included to one of the generating loops of the torus (the one going around the center hole), we get the presentation
begin{equation}
pi(X) cong <a,b,c,d:ab=ba,cd=dc,a=c>?
end{equation}
(What is this group, by the way?)



Is what I did correct?



Many thanks!






algebraic-topology fundamental-groups




algebraic-topology fundamental-groups






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asked Dec 5 at 1:21









user134070

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  • Two circles of radius $1$ on the xy-plane that's centered at (0,0,1) and (0,0,-1)? Do you mean for the 1 and -1 to be the y coordinates?
    – Yunus Syed
    Dec 5 at 1:39


















  • Two circles of radius $1$ on the xy-plane that's centered at (0,0,1) and (0,0,-1)? Do you mean for the 1 and -1 to be the y coordinates?
    – Yunus Syed
    Dec 5 at 1:39
















Two circles of radius $1$ on the xy-plane that's centered at (0,0,1) and (0,0,-1)? Do you mean for the 1 and -1 to be the y coordinates?
– Yunus Syed
Dec 5 at 1:39




Two circles of radius $1$ on the xy-plane that's centered at (0,0,1) and (0,0,-1)? Do you mean for the 1 and -1 to be the y coordinates?
– Yunus Syed
Dec 5 at 1:39










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Maybe it's easier to think of $X$ as a plane with two holes that rotates around the $z$ axis. If the plane with two holes is $Y$,then $X$ should be homeomorphic to $Y times S^1$. And thus the fundamental group is $(mathbb{Z}*mathbb{Z})timesmathbb{Z}$. I hope it helps.



And, by the way, the fundamental group of the torus is $mathbb{Z}timesmathbb{Z}$, how did you get $mathbb{Z_{2}}$?






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    Maybe it's easier to think of $X$ as a plane with two holes that rotates around the $z$ axis. If the plane with two holes is $Y$,then $X$ should be homeomorphic to $Y times S^1$. And thus the fundamental group is $(mathbb{Z}*mathbb{Z})timesmathbb{Z}$. I hope it helps.



    And, by the way, the fundamental group of the torus is $mathbb{Z}timesmathbb{Z}$, how did you get $mathbb{Z_{2}}$?






    share|cite|improve this answer



























      up vote
      1
      down vote













      Maybe it's easier to think of $X$ as a plane with two holes that rotates around the $z$ axis. If the plane with two holes is $Y$,then $X$ should be homeomorphic to $Y times S^1$. And thus the fundamental group is $(mathbb{Z}*mathbb{Z})timesmathbb{Z}$. I hope it helps.



      And, by the way, the fundamental group of the torus is $mathbb{Z}timesmathbb{Z}$, how did you get $mathbb{Z_{2}}$?






      share|cite|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









        Maybe it's easier to think of $X$ as a plane with two holes that rotates around the $z$ axis. If the plane with two holes is $Y$,then $X$ should be homeomorphic to $Y times S^1$. And thus the fundamental group is $(mathbb{Z}*mathbb{Z})timesmathbb{Z}$. I hope it helps.



        And, by the way, the fundamental group of the torus is $mathbb{Z}timesmathbb{Z}$, how did you get $mathbb{Z_{2}}$?






        share|cite|improve this answer














        Maybe it's easier to think of $X$ as a plane with two holes that rotates around the $z$ axis. If the plane with two holes is $Y$,then $X$ should be homeomorphic to $Y times S^1$. And thus the fundamental group is $(mathbb{Z}*mathbb{Z})timesmathbb{Z}$. I hope it helps.



        And, by the way, the fundamental group of the torus is $mathbb{Z}timesmathbb{Z}$, how did you get $mathbb{Z_{2}}$?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 5 at 15:25

























        answered Dec 5 at 11:07









        Sendobren

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