Exercise 74.3

Question

The \emph{	extbf{Klein bottle}} $K$ is the space obtained form a square by means of the labeling scheme $aba^{-1}b$. Figure $74.11$ indicates how $K$ can be pictured as an immersed surface in $\mathbb{R}^3$.
\begin{enumerate}[label=(\alph*)]
\item Find a presentation for the fundamental group of $K$.
\item Find a double covering map $p : T 	o K$, where $T$ is the torus. Describe the induced homomorphism of fundamental groups.
\end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$]
  $\pi_1(K) = \braket{a,b \mid aba^{-1}b = 1}$ by Theorem $74.2$.
\end{proof}
\begin{proof}[Proof of $(b)$]
  We consider $T$ as $[0,1] 	imes [0,1]$ with the relations $(0,y)\sim(1,y)$,
  $(x,0) \sim (x,1)$, and $K$ as $[0,1] 	imes [0,1]$ with the relations $(0,y)
  \sim (1,1-y)$, $(x,0) \sim (x,1)$. Then, define $p\colon T 	o K$ by
  \begin{equation*}
    p(x,y) = \begin{cases}
      (2x,y) & 	ext{if}~x \in [0,1/2],\
      (2x-1,1-y) & 	ext{if}~x \in [1/2,1].
    \end{cases}
  \end{equation*}
  This is continuous in each region, and agrees on the boundary since $p(1/2,y) = (2 \cdot 1/2,y) = (1,y) = (0,1-y) = (2 \cdot 1/2 - 1,1-y)$ and $p(1,1-y) = (1,1-y) = (0,y) = p(0,y)$.
  \par Now recall that $\pi_1(T) = \braket{\alpha,\beta \mid \alpha\beta\alpha^{-1}\beta^{-1} = 1}$. Looking at Figures $74.4$ and $74.11$, we see $p_*(\alpha) = a^2,p_*(\beta) = b$. Since $aba^{-1}b = 1 \iff bab = a$, we see that $p_*(\alpha)p_*(\beta) = a^2b = babab = ba^2 = p_*(\beta)p_*(\alpha)$.
\end{proof}

Exercise 74.3

Answers

Comments

Exercise 74.3

Answers

Comments

Add answer