# Cylinder and Torus minus a point

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• November 10th 2009, 11:26 PM
DJDorianGray
Cylinder and Torus minus a point
Hey everyone,
I have to show that the cylinder $S^1 \times \mathbb{R}$ minus a point is homotopy equivalent to the torus $T^2$ 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 AM
HallsofIvy
Quote:

Originally Posted by DJDorianGray
Hey everyone,
I have to show that the cylinder $S^1 \times \mathbb{R}$ minus a point is homotopy equivalent to the torus $T^2$ minus a point. What is the quickest way to do it?
I would really appreciate any help you can give me. Thanks.

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 AM
tonio
Quote:

Originally Posted by HallsofIvy
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.

I guess that three lines before the end it should be "...paths that go NEITHER around the..."

Tonio
• November 11th 2009, 10:16 AM
cribby
Quote:

Originally Posted by tonio
I guess that three lines before the end it should be "...paths that go NEITHER around the..."

Tonio

Paths that go neither around the cylinder nor the removed point are homotopic to a point, right? So you really only need concern yourself with the other two types of paths.
• November 11th 2009, 11:07 AM
tonio
Quote:

Originally Posted by cribby
Paths that go neither around the cylinder nor the removed point are homotopic to a point, right? So you really only need concern yourself with the other two types of paths.

Well, a null-homotopic path is ALSO a path, but yes: the interesting ones are the other two kinds.

Tonio
• November 11th 2009, 11:35 AM
aliceinwonderland
Quote:

Originally Posted by DJDorianGray
Hey everyone,
I have to show that the cylinder $S^1 \times \mathbb{R}$ minus a point is homotopy equivalent to the torus $T^2$ minus a point. What is the quickest way to do it?
I would really appreciate any help you can give me. Thanks.

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 $S^1 \times R$, 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.