# Möbius Strip Cylinder

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?

• You mean not "with" but "quotioned by". It also depends what the action is, since the quotient can also be homeomorphic to the plane. You also need to specify which definition of Moebius strip you know. – Moishe Kohan Feb 13 '14 at 4:10
• The action is $\pm 1 \cdot x = \pm x$. The definition I know is the one where we identify a pair of opposite sides of a square that are oriented in opposite directions. – Jonny Feb 13 '14 at 4:19
• What is x in your last formula? Same as in the post? Please, correct your post to make things clear. Also, do you know what is the relation of Moebius band and projective plane and the definition of projective plane as the quotient of sphere by antipodal map? – Moishe Kohan Feb 13 '14 at 4:25
• Yes I know of both of those. The mobius band can be sewn into the disk to obtain the projective plane. – Jonny Feb 13 '14 at 4:32
• Given the notation now in the problem statement, the action is $\pm 1 \cdot x_0 = \pm x_0$. – Jonny Feb 13 '14 at 4:39

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.