Fundamental group and the universal covering space for $X$ which is obtained by attaching a Mobius band to a...












1












$begingroup$


Let $X$ denote the space which is obtained by attaching a Möbius band via a homeomorphism from the boundary circle of the Möbius band to the circle $S^1×{x_0}$ in a torus $S^1times S^1$.



I need to find the fundamental group and the universal covering space $tilde{X}$ for $X$.



I found http://qcpages.qc.cuny.edu/~jterilla/topology/ps4b_answers.pdf on google. For the fundamental group, I am still confused why can we have $alpha$ is a loop in $A_2$. I think in $A_2$, $alpha$ is just half of the circle and the intersection of $A_1,A_2$ is $alpha^2$ and we have $alpha^2=delta$. What is wrong here?
And I am still stuck on finding the universal covering space. May I please ask how to construct it? Thanks in advance!






algebraic-topology covering-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    @reuns I do not think this is the shape of the space. If it is exact the fact. May I please ask for some explaination?
    $endgroup$
    – PropositionX
    Aug 23 '17 at 23:16










  • $begingroup$
    Yes you are right. About the covering space, there is a group of transformations of $(x,y) in mathbb{R}^2$ whose presentation is $langle beta, delta | beta delta^2 = delta^2 beta rangle$ with $delta(x,y) = (x+2,y)$ and $beta(x,y) = (x+1,-y)$
    $endgroup$
    – reuns
    Aug 24 '17 at 0:06












  • $begingroup$
    @reuns Sorry I cannot understand what do you mean by the universal covering space construction. May I please ask for some explaintion? What does it look like?
    $endgroup$
    – PropositionX
    Aug 26 '17 at 9:36
















1












$begingroup$


Let $X$ denote the space which is obtained by attaching a Möbius band via a homeomorphism from the boundary circle of the Möbius band to the circle $S^1×{x_0}$ in a torus $S^1times S^1$.



I need to find the fundamental group and the universal covering space $tilde{X}$ for $X$.



I found http://qcpages.qc.cuny.edu/~jterilla/topology/ps4b_answers.pdf on google. For the fundamental group, I am still confused why can we have $alpha$ is a loop in $A_2$. I think in $A_2$, $alpha$ is just half of the circle and the intersection of $A_1,A_2$ is $alpha^2$ and we have $alpha^2=delta$. What is wrong here?
And I am still stuck on finding the universal covering space. May I please ask how to construct it? Thanks in advance!






algebraic-topology covering-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    @reuns I do not think this is the shape of the space. If it is exact the fact. May I please ask for some explaination?
    $endgroup$
    – PropositionX
    Aug 23 '17 at 23:16










  • $begingroup$
    Yes you are right. About the covering space, there is a group of transformations of $(x,y) in mathbb{R}^2$ whose presentation is $langle beta, delta | beta delta^2 = delta^2 beta rangle$ with $delta(x,y) = (x+2,y)$ and $beta(x,y) = (x+1,-y)$
    $endgroup$
    – reuns
    Aug 24 '17 at 0:06












  • $begingroup$
    @reuns Sorry I cannot understand what do you mean by the universal covering space construction. May I please ask for some explaintion? What does it look like?
    $endgroup$
    – PropositionX
    Aug 26 '17 at 9:36














1












1








1





$begingroup$


Let $X$ denote the space which is obtained by attaching a Möbius band via a homeomorphism from the boundary circle of the Möbius band to the circle $S^1×{x_0}$ in a torus $S^1times S^1$.



I need to find the fundamental group and the universal covering space $tilde{X}$ for $X$.



I found http://qcpages.qc.cuny.edu/~jterilla/topology/ps4b_answers.pdf on google. For the fundamental group, I am still confused why can we have $alpha$ is a loop in $A_2$. I think in $A_2$, $alpha$ is just half of the circle and the intersection of $A_1,A_2$ is $alpha^2$ and we have $alpha^2=delta$. What is wrong here?
And I am still stuck on finding the universal covering space. May I please ask how to construct it? Thanks in advance!






algebraic-topology covering-spaces







share|cite|improve this question











$endgroup$




Let $X$ denote the space which is obtained by attaching a Möbius band via a homeomorphism from the boundary circle of the Möbius band to the circle $S^1×{x_0}$ in a torus $S^1times S^1$.



I need to find the fundamental group and the universal covering space $tilde{X}$ for $X$.



I found http://qcpages.qc.cuny.edu/~jterilla/topology/ps4b_answers.pdf on google. For the fundamental group, I am still confused why can we have $alpha$ is a loop in $A_2$. I think in $A_2$, $alpha$ is just half of the circle and the intersection of $A_1,A_2$ is $alpha^2$ and we have $alpha^2=delta$. What is wrong here?
And I am still stuck on finding the universal covering space. May I please ask how to construct it? Thanks in advance!






algebraic-topology covering-spaces




algebraic-topology covering-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 24 '17 at 6:40









Daniel Bernoulli

1,369620




1,369620










asked Aug 23 '17 at 22:27









PropositionXPropositionX

1,645620




1,645620












  • $begingroup$
    @reuns I do not think this is the shape of the space. If it is exact the fact. May I please ask for some explaination?
    $endgroup$
    – PropositionX
    Aug 23 '17 at 23:16










  • $begingroup$
    Yes you are right. About the covering space, there is a group of transformations of $(x,y) in mathbb{R}^2$ whose presentation is $langle beta, delta | beta delta^2 = delta^2 beta rangle$ with $delta(x,y) = (x+2,y)$ and $beta(x,y) = (x+1,-y)$
    $endgroup$
    – reuns
    Aug 24 '17 at 0:06












  • $begingroup$
    @reuns Sorry I cannot understand what do you mean by the universal covering space construction. May I please ask for some explaintion? What does it look like?
    $endgroup$
    – PropositionX
    Aug 26 '17 at 9:36


















  • $begingroup$
    @reuns I do not think this is the shape of the space. If it is exact the fact. May I please ask for some explaination?
    $endgroup$
    – PropositionX
    Aug 23 '17 at 23:16










  • $begingroup$
    Yes you are right. About the covering space, there is a group of transformations of $(x,y) in mathbb{R}^2$ whose presentation is $langle beta, delta | beta delta^2 = delta^2 beta rangle$ with $delta(x,y) = (x+2,y)$ and $beta(x,y) = (x+1,-y)$
    $endgroup$
    – reuns
    Aug 24 '17 at 0:06












  • $begingroup$
    @reuns Sorry I cannot understand what do you mean by the universal covering space construction. May I please ask for some explaintion? What does it look like?
    $endgroup$
    – PropositionX
    Aug 26 '17 at 9:36
















$begingroup$
@reuns I do not think this is the shape of the space. If it is exact the fact. May I please ask for some explaination?
$endgroup$
– PropositionX
Aug 23 '17 at 23:16




$begingroup$
@reuns I do not think this is the shape of the space. If it is exact the fact. May I please ask for some explaination?
$endgroup$
– PropositionX
Aug 23 '17 at 23:16












$begingroup$
Yes you are right. About the covering space, there is a group of transformations of $(x,y) in mathbb{R}^2$ whose presentation is $langle beta, delta | beta delta^2 = delta^2 beta rangle$ with $delta(x,y) = (x+2,y)$ and $beta(x,y) = (x+1,-y)$
$endgroup$
– reuns
Aug 24 '17 at 0:06






$begingroup$
Yes you are right. About the covering space, there is a group of transformations of $(x,y) in mathbb{R}^2$ whose presentation is $langle beta, delta | beta delta^2 = delta^2 beta rangle$ with $delta(x,y) = (x+2,y)$ and $beta(x,y) = (x+1,-y)$
$endgroup$
– reuns
Aug 24 '17 at 0:06














$begingroup$
@reuns Sorry I cannot understand what do you mean by the universal covering space construction. May I please ask for some explaintion? What does it look like?
$endgroup$
– PropositionX
Aug 26 '17 at 9:36




$begingroup$
@reuns Sorry I cannot understand what do you mean by the universal covering space construction. May I please ask for some explaintion? What does it look like?
$endgroup$
– PropositionX
Aug 26 '17 at 9:36










1 Answer
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oldest

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1












$begingroup$

Use Seifert - van Kampen to calculate the fundamental group. For $U$ take the torus (plus a little bit, s.t. it is open). For $V$ take the Möbius band (plus a little bit, s.t. it is open). Let $X$ be your space.



You might know the fundamental group of the torus, which is $pi_1(U)=Bbb ZtimesBbb Z=langlealpha,beta|alphabeta=betaalpharangle$ with $alpha$ and $beta$ the generators. We can deformation retract the Möbius band to $S^1$. Therefore $pi_1(V)=Bbb Z$ with generator $gamma$. The deformation retract of the intersection $Ucap V$ is a $1$-sphere $S^1$. Hence $pi_1(Ucap V)=Bbb Z$ with generator $delta$.



Consider the inclusion $iota_Ucolon Ucap Vhookrightarrow U$ and $iota_Vcolon Ucap Vhookrightarrow V$ which induce homomorphisms $iota_U^astcolon pi_1(Ucap V)hookrightarrow pi_1(U)$ and $iota_V^astcolon pi_1(Ucap V)hookrightarrow pi_1(V)$. The kernel of the surjection $phicolon pi_1(U)astpi_1(V)to pi_1(X)$ is generated by elements of the form $iota_U^ast(delta)iota_V^ast(delta)^{-1}=alphagamma^{-2}$. This is clear by looking at the inclusions.



So we have
$$pi_1(X)=pi_1(U)ast_{pi_1(Ucap V)}pi_1(V)=langlealpha,beta,gamma|alphabeta=betaalpha,alpha=gamma^2rangle=langlebeta,gamma|gamma^2beta=betagamma^2rangle.$$



For the other part of your question: see comments.






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    1 Answer
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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    Use Seifert - van Kampen to calculate the fundamental group. For $U$ take the torus (plus a little bit, s.t. it is open). For $V$ take the Möbius band (plus a little bit, s.t. it is open). Let $X$ be your space.



    You might know the fundamental group of the torus, which is $pi_1(U)=Bbb ZtimesBbb Z=langlealpha,beta|alphabeta=betaalpharangle$ with $alpha$ and $beta$ the generators. We can deformation retract the Möbius band to $S^1$. Therefore $pi_1(V)=Bbb Z$ with generator $gamma$. The deformation retract of the intersection $Ucap V$ is a $1$-sphere $S^1$. Hence $pi_1(Ucap V)=Bbb Z$ with generator $delta$.



    Consider the inclusion $iota_Ucolon Ucap Vhookrightarrow U$ and $iota_Vcolon Ucap Vhookrightarrow V$ which induce homomorphisms $iota_U^astcolon pi_1(Ucap V)hookrightarrow pi_1(U)$ and $iota_V^astcolon pi_1(Ucap V)hookrightarrow pi_1(V)$. The kernel of the surjection $phicolon pi_1(U)astpi_1(V)to pi_1(X)$ is generated by elements of the form $iota_U^ast(delta)iota_V^ast(delta)^{-1}=alphagamma^{-2}$. This is clear by looking at the inclusions.



    So we have
    $$pi_1(X)=pi_1(U)ast_{pi_1(Ucap V)}pi_1(V)=langlealpha,beta,gamma|alphabeta=betaalpha,alpha=gamma^2rangle=langlebeta,gamma|gamma^2beta=betagamma^2rangle.$$



    For the other part of your question: see comments.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Use Seifert - van Kampen to calculate the fundamental group. For $U$ take the torus (plus a little bit, s.t. it is open). For $V$ take the Möbius band (plus a little bit, s.t. it is open). Let $X$ be your space.



      You might know the fundamental group of the torus, which is $pi_1(U)=Bbb ZtimesBbb Z=langlealpha,beta|alphabeta=betaalpharangle$ with $alpha$ and $beta$ the generators. We can deformation retract the Möbius band to $S^1$. Therefore $pi_1(V)=Bbb Z$ with generator $gamma$. The deformation retract of the intersection $Ucap V$ is a $1$-sphere $S^1$. Hence $pi_1(Ucap V)=Bbb Z$ with generator $delta$.



      Consider the inclusion $iota_Ucolon Ucap Vhookrightarrow U$ and $iota_Vcolon Ucap Vhookrightarrow V$ which induce homomorphisms $iota_U^astcolon pi_1(Ucap V)hookrightarrow pi_1(U)$ and $iota_V^astcolon pi_1(Ucap V)hookrightarrow pi_1(V)$. The kernel of the surjection $phicolon pi_1(U)astpi_1(V)to pi_1(X)$ is generated by elements of the form $iota_U^ast(delta)iota_V^ast(delta)^{-1}=alphagamma^{-2}$. This is clear by looking at the inclusions.



      So we have
      $$pi_1(X)=pi_1(U)ast_{pi_1(Ucap V)}pi_1(V)=langlealpha,beta,gamma|alphabeta=betaalpha,alpha=gamma^2rangle=langlebeta,gamma|gamma^2beta=betagamma^2rangle.$$



      For the other part of your question: see comments.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Use Seifert - van Kampen to calculate the fundamental group. For $U$ take the torus (plus a little bit, s.t. it is open). For $V$ take the Möbius band (plus a little bit, s.t. it is open). Let $X$ be your space.



        You might know the fundamental group of the torus, which is $pi_1(U)=Bbb ZtimesBbb Z=langlealpha,beta|alphabeta=betaalpharangle$ with $alpha$ and $beta$ the generators. We can deformation retract the Möbius band to $S^1$. Therefore $pi_1(V)=Bbb Z$ with generator $gamma$. The deformation retract of the intersection $Ucap V$ is a $1$-sphere $S^1$. Hence $pi_1(Ucap V)=Bbb Z$ with generator $delta$.



        Consider the inclusion $iota_Ucolon Ucap Vhookrightarrow U$ and $iota_Vcolon Ucap Vhookrightarrow V$ which induce homomorphisms $iota_U^astcolon pi_1(Ucap V)hookrightarrow pi_1(U)$ and $iota_V^astcolon pi_1(Ucap V)hookrightarrow pi_1(V)$. The kernel of the surjection $phicolon pi_1(U)astpi_1(V)to pi_1(X)$ is generated by elements of the form $iota_U^ast(delta)iota_V^ast(delta)^{-1}=alphagamma^{-2}$. This is clear by looking at the inclusions.



        So we have
        $$pi_1(X)=pi_1(U)ast_{pi_1(Ucap V)}pi_1(V)=langlealpha,beta,gamma|alphabeta=betaalpha,alpha=gamma^2rangle=langlebeta,gamma|gamma^2beta=betagamma^2rangle.$$



        For the other part of your question: see comments.






        share|cite|improve this answer











        $endgroup$



        Use Seifert - van Kampen to calculate the fundamental group. For $U$ take the torus (plus a little bit, s.t. it is open). For $V$ take the Möbius band (plus a little bit, s.t. it is open). Let $X$ be your space.



        You might know the fundamental group of the torus, which is $pi_1(U)=Bbb ZtimesBbb Z=langlealpha,beta|alphabeta=betaalpharangle$ with $alpha$ and $beta$ the generators. We can deformation retract the Möbius band to $S^1$. Therefore $pi_1(V)=Bbb Z$ with generator $gamma$. The deformation retract of the intersection $Ucap V$ is a $1$-sphere $S^1$. Hence $pi_1(Ucap V)=Bbb Z$ with generator $delta$.



        Consider the inclusion $iota_Ucolon Ucap Vhookrightarrow U$ and $iota_Vcolon Ucap Vhookrightarrow V$ which induce homomorphisms $iota_U^astcolon pi_1(Ucap V)hookrightarrow pi_1(U)$ and $iota_V^astcolon pi_1(Ucap V)hookrightarrow pi_1(V)$. The kernel of the surjection $phicolon pi_1(U)astpi_1(V)to pi_1(X)$ is generated by elements of the form $iota_U^ast(delta)iota_V^ast(delta)^{-1}=alphagamma^{-2}$. This is clear by looking at the inclusions.



        So we have
        $$pi_1(X)=pi_1(U)ast_{pi_1(Ucap V)}pi_1(V)=langlealpha,beta,gamma|alphabeta=betaalpha,alpha=gamma^2rangle=langlebeta,gamma|gamma^2beta=betagamma^2rangle.$$



        For the other part of your question: see comments.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 '18 at 16:38









        jawheele

        677




        677










        answered Aug 24 '17 at 6:46









        Daniel BernoulliDaniel Bernoulli

        1,369620




        1,369620






























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