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As picture below ,it is a Mobius band with a cylinder crossing it .Let it be $\Omega$ . Obviously , $\partial \Omega$ is a circle. Now , what is $\Omega/\partial \Omega$ ( I mean glue the boundary to a point )? And how to show it ?

enter image description here

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  • $\begingroup$ @ArtW How do you feel it ? $\endgroup$
    – lanse7pty
    Jun 21, 2016 at 13:11
  • $\begingroup$ Gluing the boundary of a Mobius band we find a real projective plane: en.wikipedia.org/wiki/Real_projective_plane. So I suppose the in your case we have a real projective plane with a cylinder crossing it. $\endgroup$ Jun 21, 2016 at 13:22

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This is $\mathbb{R}P^2\#\mathbb{R}P^2\#\mathbb{R}P^2\simeq\mathbb{R}P^2\#T^2\simeq \mathbb{R}P^2\#K$, where as usual $T^2$ is the torus, $\mathbb{R}P^2$ the real projective plane and $K$ the klein bottle.

I will only give an intuitive argument for this. Consider the following sketch:$\mathbb{R}P^2\#\mathbb{R}P^2\#\mathbb{R}P^2$

To see this represents your manifold, note that this is a Möbius strip (left and right side of the strip are identified in the opposite direction) with 2 holes in it that are to be identified, which is your cylinder. Finally, if we contract the boundary of your modified Möbius strip, we identify the upper and lower edge of my drawing in the same direction (the red arrows).

The fact that both diagrams give the same manifold follows if we consider one of the circular holes and move it through the vertical boundary in the diagram. (fun exercise: show these two surfaces have the same homology and Euler characteristic)

Observe how the left side represents $\mathbb{R}P^2\#K$ and the right side $\mathbb{R}P^2\#T^2$. Indeed, the two circles that are cut out in the center can be considered a klein bottle on the left and a torus on the right.
The fact that $\mathbb{R}P^2\#T^2\simeq \mathbb{R}P^2\#\mathbb{R}P^2\#\mathbb{R}P^2$ is classic (and one of the key steps in the classification of 2-manifolds), so I won't show this here. This gives us the third description.

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    $\begingroup$ Where is the $RP^2$ from ? Gluing the rectangle yield Klein .And It is not #, it is cutting Klein and torus ,then glue they to Klein . $\endgroup$
    – lanse7pty
    Jun 23, 2016 at 3:51
  • $\begingroup$ You are totally right - I messed up one of the arrows in my drawing. I'll edit it soon. $\endgroup$
    – ArtW
    Jun 23, 2016 at 6:04

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