Fundamental group of two Moebius bands identified on boundary circles

Let $M_1, M_2$ be two copies of the M\"obius band, and let $\partial > M_i$ its boundary circle for $i=1,2$. Let $f:M_1\to M_2$ be an homeomorphism, and consider the topological space $$X=(M_1\bigsqcup > M_2)/\sim$$ where $p\sim f(p), p\in\partial M_1$. Let $G=\pi_1(X)$. a) Give a presentation of $G$ with generators and relations. b) Show that there exists a surjective homomorphism $\phi:G\to \mathbb Z/2*\mathbb > Z/2$. c) Say whether the universal covering of $X$ is compact or not.

My solution

a) Let $X_1=M_1\cup V\subseteq X$, where $V$ is an open neighbourhood of $\partial M_2$ in $M_2$ which deformation retracts onto $\partial M_2$, and $X_2=U\cup M_2\subseteq X$ (where $U$ is defined similarly). Then $X_1$ retracts on $M_1$, $X_2$ retracts on $M_2$ and $X_1\cap X_2$ retracts on the identified circle. So we have two generators, $a$ and $b$ (one for each $X_i$) and the relation $a^2=b^2$, since two loops of the generator of the $\pi_1$ of the Moebius band correspond to a loop in the boundary circle.

b) Defining $$\phi(a^{e_1}b^{f_1}\dots a^{e_n}b^{f_n})=\bar{a}^{e_1}\bar b^{f_1}\dots \bar a^{e_n} \bar b^{e_n}$$ we obtain an homomorphism, since the relation $a^2=b^2$ is respected: $\bar a^2=1=\bar b^2$. The relation allows us to say also that $\phi$ is surjective (one should be more detailed in saying why?).

c) The fundamental group action on the universal covering is free and properly discontinuous. Therefore, if $\tilde X$ is compact, $\pi_1(X)$ it must be finite, but $G$ has a surjection onto $\mathbb Z/2*\mathbb Z/2$, wich is not finite (all elements $\bar a\bar b$, $\bar a\bar b\bar a$, $\bar a\bar b\bar a\bar b$, etc. are distinct, where $\bar a,\bar b$ are the two generators).

Is this correct?

For part b, I'd be explicit on why it's surjective: for a given element $g$ of the target group, identify an element of the domain group that is sent to $g$. For example, for the element $\bar{a}\bar{b}\bar{a}$, the element $aba$ is sent to it.
You might ask yourself the question "just what surface is $X$?" Once you do so, it might be easier to see what $pi_1(X)$ looks like.
• Thank you. I'll think about your suggestion! Do you confirm that part (a) is correct, in particular that the structure of $G$ does not depend on the homeomorphism $f$? – W. Rether Aug 3 '16 at 13:23
• Yes, that's right. A homeomorphism from $S^1 \times I$ to itself always takes a path that traverses the generator $S^1 \times \{0\}$ to something homotopic to a path that traverses the corresponding generator in the image. The only possibility is that the orientation of that path is reversed, but this can be fixed by renaming $b$ to be $-b$. :) – John Hughes Aug 3 '16 at 16:25