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The Klein bottle $K$ is the surface obtained from a square by gluing top and bottom sides by translation, and gluing left and right sides with a "twist". I can show that a presentation for the fundamental group is given by$$\pi_1(K, p) \cong \langle a, b \mid aba^{-1}b\rangle.$$What is another cell decomposition of $K$ with one polygon and one vertex that gives another presentation for the fundamental group as$$\pi_1(K, p) \cong \langle a, b \mid a^2 b^2\rangle?$$

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  • $\begingroup$ It should be $ \pi_1(K, p) \cong \langle a, b \mid a^2b^2\rangle$ $\endgroup$ – iwriteonbananas Oct 24 '16 at 7:56
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You can consider a square with all sides oriented in a counterclockwise fashion, identified as $\ a,a,b,b. \ $ Of course the loop of the perimeter is $\ a^2 b^2 \ $ and goes to zero. To see this is the Klein bottle, just cut this square along the diagonal which connects the endpoint of the first $\ b \ $ to the endpoint of the first $\ a. \ $ Call the diagonal $\ c: \ $ pasting along $\ a \ $ gives your presentation as $\ bc^{-1}b^{-1}c^{-1}: \ $

To see this drawn: my presentation

$$ \begin{array}{lcr}\ \ * \xleftarrow{ \qquad b \qquad } * \\ b\Bigg\downarrow \qquad \searrow c \ \quad \Bigg\uparrow a \\ \ \ * \xrightarrow[ \qquad a \qquad ]\ * \end{array} $$

goes to (glue lateral $\ a \ $ to bottom $\ a \ $ )

$$ \begin{array}{lcr}\ \ * \xrightarrow{ \qquad c \qquad } * \\ b\Bigg\downarrow \qquad \nearrow a \ \quad \Bigg\downarrow b \\ \ \ * \xleftarrow[ \qquad c \qquad ]\ * \end{array} $$

that is just classic Klein surface.

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