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Consider the usual cell structure on $\mathbb R P^2$, with one 1-cell and one 2-cell attached via a map of degree 2. Consider the space $X$ obtained by gluing a Möbius band along the 1-cell via a homeomorphism with its boundary circle. Using van Kampen's theorem, I believe I can show that the fundamental group of this space is $\mathbb Z/2 * \mathbb Z/2$. However, I would like to know what its universal cover is, and what the deck transformations are. I have tried gluing spheres and Möbius strips in various configurations, but have so far not been successful. Any suggestions?

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Let $M$ be the Möbius band and let $D$ be the $2$-cell of $RP^2$. Then $X$ is the result of gluing $M$ to $D$ along a map $\partial D\rightarrow \partial M$ of degree $2$. Hence $\pi_1(X)$ has a single generator $\gamma$, that comes from the inclusion $M\rightarrow X$, and the attached disc $D$ kills $\gamma^4$, hence $\pi_1(X)\cong {\mathbb Z}/4$. Alternatively, take the homotopy $M\times I\rightarrow M$ that shrinks the Möbius band to its central circle $S\subset M$; this gives a homotopy from $X = M\cup D$ to the space ${\mathbb S}^1\cup_f D$, where $f\colon \partial D\rightarrow {\mathbb S}^1$ is a map of degree $4$.

The universal cover of the space ${\mathbb S}^1\cup_f D$ is described in Hatcher's Example 1.47, and it is the homeomorphic to the quotient of $D\times\{1,2,3,4\}$ by the relation $(x,i)\sim (y,j)$ iff $x=y$ and $x\in \partial D\times \{i\}$.

Now let $D$ denote de closed unit disc in ${\mathbb R}^2$. The universal cover of the space $X$ is homeomorphic to the quotient of $D\times \{a,b,c,d\}\cup S^1\times [-1,1]$ by the relations

  • $(x,a)\sim (x,b)\sim (x,1)$ for all $x\in S^1 = \partial D$
  • $(x,c)\sim(x,d)\sim (x,-1)$ for all $x\in S^1=\partial D$

and $\pi_1(X)\cong {\mathbb Z}/4$ acts as follows:

  • $(re^{2\pi i\theta},a)\to (re^{2\pi i(\theta + 1/4)}, c)\to (re^{2\pi i(\theta+1/2)},b)\to (re^{2\pi i (\theta + 3/4)},d)\rightarrow (re^{2\pi i\theta},a)$ for the points in the discs $D\times \{a,b,c,d\}$
  • $(e^{2\pi i \theta},t)\to (e^{2\pi i (\theta+1/4)},-t)\to (e^{2\pi i (\theta + 1/2)},t)\to (e^{2\pi i(\theta + 3/4)},-t)\rightarrow (e^{2\pi i \theta},t)$ for the points in $S^1\times [-1,1]$.

and the map $\tilde{X}\rightarrow X$ sends the four discs to the disc and the cylinder to the Möbius band.

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