I am having trouble seeing why for the the regular cylinder $C=\{(x,y,z)|x^2+y^2=1,|z|\le 1\}$, $C/\mathbb{Z}_2$ is homeomorphic to the Möbius band ($x_0 \in C$). Can someone explain?
2026-03-25 14:28:44.1774448924
Möbius Strip Cylinder
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Add on the top and on the bottom of C unit disks. The result is topological sphere (a closed barrel) with antipodal $Z/2$ action. The quotient is projective plane. Now, remove the image of these disks from the quotient. As you know, projective plane minus disk is Moebius band.