4

I want to calculate by two different way the fundamental group of the Klein Bottle.

First one:

I want to use that the Klein Bottle is can be decomposed in two Mobiüs Band as the following picture shows:

Quotient that gives the Klein bottle from two Mobiüs band

Then I have to choose two open set $U$ and $V$. And want to use the Mobiüs band as the open sets, but they are not open. Therefore I define $U$ and $V$ as the picture shows below

Open sets U and V

I think that those sets are homotopic to the Mobiüs band. Know the first question: What is $U\cap V$? I think that it could be a cilinder, a Mobiüs band or a Mobiüs band with double twist. Anyway I think that the group of any of this is isomorphic to $\mathbb{Z}$. Am I right?

With this method I pretend to get two generators $\alpha, \beta$ satisfying $\alpha^2 \beta^2=1$

Second one

This is the classic way to calculate the fundamental group of the Klein bottle. Let $q$ be the quotient map from $I^2=[-1,1]^2$ to the Klein bottle. We choose $U$ to be $q(I^2-\partial I^2)$ this set is simply connected and $U$ is $q(I^2-{0})$. The set $U\cap V$ has a fundamental group isomorphic to $\mathbb{Z}$.

My question: I know that $V$ has a fundamental group homeomorphic to $S^1 \wedge S^1$ but I don't know how to prove.

With this method I pretend to get two generators $\alpha, \beta$ satisfying $\alpha\beta\alpha\beta^{-1}=1$

I want to prove this without using the concept of surface or polygonal representation of surface

Remark: I have seen a lot of question similar to this, but at least the first method it is not in MS!

Thanks for help!

EQJ
  • 4,369
  • 1
    Regarding your first question: You have chosen as an open cover two thickened intersecting Mobius strips $U$ and $V$. First note that each Mobius strip deformation retracts onto a circle, namely the circle in the "middle circle" of the Mobius strip. $U\cap V$ also deformation retracts onto a circle, namely the boundary circle of each Mobius strip. The generating loop of $\pi_1(U\cap V)$ can be written as twice the generating loop of $\pi_1(U)$, and twice the generating loop of $\pi_1(V)$. Using this, try to apply Van Kampen's theorem. – iwriteonbananas Jun 16 '15 at 07:23

0 Answers0