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?


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

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    $\begingroup$ 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. $\endgroup$ – AlexisZorbas Oct 27 '12 at 15:35
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    $\begingroup$ 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. $\endgroup$ – Jason DeVito Oct 27 '12 at 15:37
  • $\begingroup$ Thank you! That helped a lot! $\endgroup$ – 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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    $\begingroup$ Instead of the diagonal, you could use the line through the center of the square and parallel to the unidentified edges. $\endgroup$ – Andreas Blass May 24 '13 at 17:12

The fundamental group of the moebius strip is $\{a,b|a^2=b^2\}$. Cf. http://www2.math.ou.edu/~forester/5863S14/fsol.pdf

  • $\begingroup$ That is actually the fundamental group of the Klein bottle, although it can be constructed using Mobius bands. $\endgroup$ – D. Zack Garza Dec 9 '17 at 21:45

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