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Let $I\times I/(0,t){\sim}(1,1-t)$ be the Möbius band and let $S=\{(x,y): (x,y)\in M, 1/4<y<3/4\}$ be its middle part. How can I show that $M-S$ is connected? I tried to write a continuous surjective map from a connected space to $M-S$ and it all got messy, is there another way?

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  • $\begingroup$ Take a narrow strip of paper and draw the center line of the soon to be mobius strip , tape it together , cut it down the center with a pair scissors. You now have some intuition into what is going on. Also, look at a standard parametrization of the mobius strip, now remove the section you need to remove. This might make a cool sketch, I might try it myself, thanks! $\endgroup$ – Alan May 14 '14 at 18:10
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    $\begingroup$ Actually this is what I did so I can write up a continuous map $\endgroup$ – cartman May 14 '14 at 18:13
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the upper and lower parts are obviously connected (separately). now the point $(1,1)$ is the same as $(0,0)$ in your topology. hence these two connected sets have a common point which means their union is connected as well (that's a basic fact)

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This is a standard parametrization of a mobius strip with a center strip removed. As you can see , it does remain connected.

$$x(u,v)= \left(1+\frac{v}{2} \cos \frac{u}{2}\right)\cos u$$

$$y(u,v)= \left(1+\frac{v}{2} \cos\frac{u}{2}\right)\sin u$$

$$z(u,v)= \frac{v}{2}\sin \frac{u}{2}$$

Where $ 0 ≤ u < 2\pi $ and importantly $-1 ≤ v ≤ -\frac{1}{2}$ and $ \frac{1}{2} ≤ v ≤ 1 $ are the two sections of the strip with the center removed.

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If you make the Möbius band from a long narrow rectangle, you start with two short edges and two long edges.

Making the band you give a single twist and tape the two short edges together so they disappear. You have also joined the two long edges so they are now a single edge. I.e. the Möbius band has a single edge.

When you remove or cut the central part of the Möbius band, you do not touch that edge, and what remains of the band is connected to that edge, so is connected overall.

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