Universal Cover of projective plane glued to Möbius band

This is the second part of exercise 1.3.21 in page 81 of Hatcher's book Algebraic topology, and the first part is answered here.

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.

Compute $\pi_1(X)$, describe the universal cover of $X$ and the action of $\pi_1(X)$ on the universal cover.

Using van Kampen's theorem, $\pi_1(X)=\langle x,y \mid x^2=1, x=y^2\rangle=\langle y\mid y^4=1\rangle=\mathbb Z_4$.

I have tried gluing spheres and Möbius strips in various configurations, but have so far not been successful. Any suggestions?

Solution 1:

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.