Exercise 0.20

Question

Show that the subspace $X \subset \mathbb{R}^3$ formed by a Klein bottle intersecting itself in a circle, as shown in the figure, is homotopy equivalent to $S^1 \vee S^1 \vee S^2$.

Amit Awasthi · Accepted Answer

\documentclass[12pt]{article} \usepackage{geometry} \usepackage[utf8]{inputenc} \usepackage[unicode]{hyperref} \usepackage{amsmath,amsthm,amssymb} \usepackage{mathtools} \usepackage{tikz} \usetikzlibrary{decorations.markings,calc,shapes,snakes} \usepackage{pgflibraryarrows} \usepackage{tikz-cd} ikzset{->-/.style={decoration={ markings, mark=at position #1 with {\arrow[scale=2,draw=black]{>}}},postaction={decorate}}} ikzset{-<-/.style={decoration={ markings, mark=at position #1 with {\arrow[scale=2,draw=black]{<}}},postaction={decorate}}} ikzset{->--/.style={decoration={ markings, mark=at position #1 with {\arrow[scale=1.2]{>}}},postaction={decorate}}} ikzset{label/.style={% postaction={ decorate,transform shape, decoration={ markings, mark=at position .55 with ode #1;}}}} ikzset{decorate sep/.style 2 args= {decorate,decoration={shape backgrounds,shape=circle,shape size=#1,shape sep=#2}}} \begin{document} \begin{proof} Using the characterization of the Klein bottle as a quotient of a square, we have that $X$ is the quotient of the following square: \begin{center} \begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=1.0cm,y=1.0cm] \fill [gray] (0,0) rectangle (2,2); \draw (0,1) node[anchor=east] {$a$}; \draw (2,1) node[anchor=west] {$a$}; \draw (1,0) node[anchor=north] {$b$}; \draw (1,2) node[anchor=south] {$b$}; \draw [->-=0.55] (0,0) -- (0,2); \draw [->-=0.55] (2,2) -- (0,2); \draw [->-=0.55] (2,0) -- (2,2); \draw [->-=0.55] (0,0) -- (2,0); \draw (1,1) node[anchor=north] {$b$}; \draw [decoration={markings, mark=at position 0.77 with {\arrow[scale=2]{>}}}, postaction={decorate}] (1,1) circle(0.5); \end{tikzpicture} \end{center} We first deformation retract the disc in the center to a point: \begin{center} \begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=1.0cm,y=1.0cm] \fill [gray] (0,0) rectangle (2,2); \draw (0,1) node[anchor=east] {$a$}; \draw (2,1) node[anchor=west] {$a$}; \draw (1,0) node[anchor=north] {$b$}; \draw (1,2) node[anchor=south] {$b$}; \draw [->-=0.55] (0,0) -- (0,2); \draw [->-=0.55] (2,2) -- (0,2); \draw [->-=0.55] (2,0) -- (2,2); \draw [->-=0.55] (0,0) -- (2,0); \draw (1,1) node[anchor=north] {$b$}; \fill (1,1) circle (0.08); \end{tikzpicture} \end{center} But then, since the two horizontal edges are identified with this center point, we can contract them to points in our diagram to get the space \begin{center} \begin{tikzpicture}[line cap=round,line join=round,>=stealth,x=1.0cm,y=1.0cm] \fill [gray] (0,0) circle (1); \draw[decoration={markings, mark=at position 0.52 with {\arrow[scale=2]{<}}}, postaction={decorate}] (0,0) circle (1); \draw[decoration={markings, mark=at position 0.01 with {\arrow[scale=2]{>}}}, postaction={decorate}] (0,0) circle (1); \draw (-1,0) node[anchor=east] {$a$}; \draw (1,0) node[anchor=west] {$a$}; \fill (0,1) circle (0.08); \fill (0,0) circle (0.08); \fill (0,-1) circle (0.08); %\draw (0,1) node[anchor=east] {$a$}; %\draw (2,1) node[anchor=west] {$a$}; %\draw (1,0) node[anchor=north] {$b$}; %\draw (1,2) node[anchor=south] {$b$}; %\draw [->-=0.55] (0,0) -- (0,2); %\draw [->-=0.55] (2,2) -- (0,2); %\draw [->-=0.55] (2,0) -- (2,2); %\draw [->-=0.55] (0,0) -- (2,0); %\draw (1,1) node[anchor=north] {$b$}; %\draw [decoration={markings, mark=at position 0.77 with {\arrow[scale=2]{>}}}, postaction={decorate}] (1,1) circle(0.5); \end{tikzpicture} \end{center} where the three labelled points are identified. This is the same as a sphere with three points identified, and so, attaching $1$-cells as in Example $0.8$, we see that $X$ is homotopy equivalent to $S^1 \vee S^1 \vee S^2$. \end{proof} \end{document}

Exercise 0.20

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Exercise 0.20

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