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The problem asks me to show that the fundamental group of the Klein bottle is generated by "latitudinal" loops $a$ and "longitudinal" loops $b$ where $a$ and $b$ obey the relation $aba^{-1} = b^{-1}$.

The problem is that I don't understand what I'm supposed to do, since the problem is stated a bit imprecisely. I haven't covered the Seifert-Van Kampen theorem either.

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Draw a short cartoon that shows the loop $aba^{-1}$ homotoping into $b^{-1}$, in the squares-with-edges-identified model.

For clarity, let the loop cross the corners one at a time, and be sure to indicate the direction of every visible segment of the loop in each frame.

(Writing down the homotopy symbolically, with coordinates, would probably be overkill).

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Hint: you can obtain the Klein bottle from the unit square identifying its edges accordingly.

The blue one is $a$ and the red one is $b$.

enter image description here

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Yes, I know that the Klein bottle is the unit square with opposite vertical edges identified in the same direction, and opposite horizontal edges identified in the opposite direction. – user61642 Feb 10 '13 at 0:05

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