1
$\begingroup$

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?

Thank you in advance.

$\endgroup$
2
$\begingroup$

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.

For part c, what you've said looks OK to me.

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.

For instance, if your mobius band are generated by taking a long strip and gluing together the ends with a half-twist, you might consider undoing that gluing, doing the gluing suggested in the problem (leaving the four original "ends" not completely glued up) and then trying to see how those four end up glue to each other, etc. You may be surprised at the resulting surface.

$\endgroup$
  • $\begingroup$ 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 $f$? $\endgroup$ – W. Rether Aug 3 '16 at 13:23
  • 1
    $\begingroup$ 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 $-b$. :) $\endgroup$ – John Hughes Aug 3 '16 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.