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 $S^1\times S^1$.

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!

  • $\begingroup$ @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 think there is something wrong there. Notice that $\beta^{2}=\delta$ $\endgroup$ – user135520 Jun 21 '19 at 21:14
  • $\begingroup$ math.stackexchange.com/questions/3478166/… $\endgroup$ – user135520 Dec 17 '19 at 17:27

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 $\pi_1(U)=\Bbb Z\times\Bbb Z=\langle\alpha,\beta|\alpha\beta=\beta\alpha\rangle$ 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 $U\cap V$ is a $1$-sphere $S^1$. Hence $\pi_1(U\cap V)=\Bbb Z$ with generator $\delta$.

Consider the inclusion $\iota_U\colon U\cap V\hookrightarrow U$ and $\iota_V\colon U\cap V\hookrightarrow V$ which induce homomorphisms $\iota_U^\ast\colon \pi_1(U\cap V)\hookrightarrow \pi_1(U)$ and $\iota_V^\ast\colon \pi_1(U\cap V)\hookrightarrow \pi_1(V)$. The kernel of the surjection $\phi\colon \pi_1(U)\ast\pi_1(V)\to \pi_1(X)$ is generated by elements of the form $\iota_U^\ast(\delta)\iota_V^\ast(\delta)^{-1}=\alpha\gamma^{-2}$. This is clear by looking at the inclusions.

So we have $$\pi_1(X)=\pi_1(U)\ast_{\pi_1(U\cap V)}\pi_1(V)=\langle\alpha,\beta,\gamma|\alpha\beta=\beta\alpha,\alpha=\gamma^2\rangle=\langle\beta,\gamma|\gamma^2\beta=\beta\gamma^2\rangle.$$

For the other part of your question: see comments.


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