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What is the fundamental group of the Möbius strip?
Is it given by $\{-1,1\}$ as the lemma of Synge supposes, or am I wrong and it does not apply there?

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The moebius strip is homotopy-equivalent to the circle, so has the same fundamental group which is $\mathbb Z$.

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Thank you so so far.So why does the lemma of Synge not apply here? It says that a manifold in even dimension has fundamental group {-1,1}, if it is not orientable. – AlexisZorbas Oct 27 '12 at 15:35
Synge only applies if you have a compact manifold with no boundary and positive curvature. The Mobius band has a boundary (or is not compact), and has no (complete) metric of positive curvature. – Jason DeVito Oct 27 '12 at 15:37
Thank you! That helped a lot! – AlexisZorbas Oct 27 '12 at 15:39

It is $\mathbb{Z}$. You can prove it via seeing the Möbius strip as a quotient of a square , with sides identified properly. Draw a diagonal dividing this square, and show that the Möbius strip deformation retracts onto this circle .

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Instead of the diagonal, you could use the line through the center of the square and parallel to the unidentified edges. – Andreas Blass May 24 '13 at 17:12

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