Hey everyone,

I have to show that the cylinder minus a point is homotopy equivalent to the torus minus a point. What is the quickest way to do it?

I would really appreciate any help you can give me. Thanks.

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- November 10th 2009, 11:26 PMDJDorianGrayCylinder and Torus minus a point
Hey everyone,

I have to show that the cylinder minus a point is homotopy equivalent to the torus minus a point. What is the quickest way to do it?

I would really appreciate any help you can give me. Thanks. - November 11th 2009, 01:36 AMHallsofIvy
Well, look at the closed paths not homotopic to a point. On a cylinder minus a point, we can have paths that go around the cylinder, paths that do not go around the cylinder but do go around the missing point, and paths that go around either the cylinder or the point. Do you see that that gives three distinct homotopy classes? Do the same thing for the torus minus a point.

- November 11th 2009, 02:22 AMtonio
- November 11th 2009, 10:16 AMcribby
- November 11th 2009, 11:07 AMtonio
- November 11th 2009, 11:35 AMaliceinwonderland
My suggestion is to use polygons with identifications (link).

http://upload.wikimedia.org/wikipedi...are.svg/313px-

If we pairwise identify both As and Bs, then the resulting space is a torus. Now, WLOG, let's make a hole in the center. Then it deformation retracts to its boundary. When we pairwise identify both As and Bs, the resulting space is a wedge sum of two circles, a.k.a, figure 8 space.

For a space , we can use the above figure with a slight modification. We don't pairwise identify either As or Bs. Assume we pairwise identify Bs only. Actually, B is the real line, so we cannot draw using a polygon. This is just for illustration purpose. The resulting space having a hole at the center after deformation retract is "O-O", where "-" is a real line. A real line can be deformation retracted to a point, "O-O" is homotopy equivalent to "OO", which is a figure 8 space.

Thus we conclude that both spaces are homotopy equivalent.