What is this manifold?

Question

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 ?

Answer 1

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:

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.