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This is problem 11 (b) from the first chapter of "Basic Topology" by M.A. Armstrong. The author hasn't had time to develop many theorems or mathematical machinery, so this problem should be able to be solved by just picturing a series of intermediate steps. It goes

Imagine all the spaces shown in Fig. 1.23 to be made of rubber. For each pair of spaces X, Y, convince yourself that X can be continuously deformed into Y.

I'm having trouble with one of the pairs of spaces (the other examples in the problem are unrelated, so I neglected to draw them). The two spaces which I can't seem to think of a continuous deformation for are

X = punctured torus, Y = Two cylinders glued together over a square patch

The caption for the first picture reads "X = punctured torus", while the caption for the second picture is "Y = Two cylinders glued together over a square patch". I'm trying to think of some intermediate steps in the problem. Working backwards, I can see how each of the cylinders in the second picture could be deformed to spheres with two punctures each, but I'm having trouble seeing how the "handle" on the torus is created.

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    $\begingroup$ The correct spelling is Torus. Taurus is an ancient word for bull, Greek or Latin or something, I believe also a sign of the Zodiac. I have no way to supply intermediate pictures. Note that this is the reason a punctured torus can be successfully inverted. I would recommend a bicycle tube but real rubber is not stretchy enough to do this. $\endgroup$
    – Will Jagy
    Aug 27, 2012 at 20:41
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    $\begingroup$ @WillJagy: Thanks, I can't believe I did that. I'm crediting that particular error to the type of car that I drive (it's a good car, but I guess it hasn't helped my math spelling)... $\endgroup$
    – Andrew
    Aug 27, 2012 at 20:47
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    $\begingroup$ At the risk of self-promotion, this is relevant: youtube.com/watch?v=j2HxBUaoaPU $\endgroup$
    – geometrian
    Dec 1, 2013 at 20:24

4 Answers 4

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To give you an alternative way to see this, you certainly know how to obtain a torus by taking a square of paper and gluing the edges together in couples. Now the punctured torus can be obtained the same way by making a hole in the square of paper. If you deform the hole enough and leave only a small strip around the edges, when you glue them together you'll get your figure.

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    $\begingroup$ Ah, this was helpful. +1, thanks. $\endgroup$ Jun 28, 2015 at 11:40
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This is probably clear, given your answer, but just in case a verbal description is helpful for you or others:

I like to think of this the following way: put your hands in the puncture, one on either side, and begin to stretch the puncture around the torus; once you do this, you can imagine that the torus is mostly puncture, with just two small "ribs" left, as in the picture in your answer.

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The following intermediate picture, which is taken from a step in a video I found on YouTube uploaded by user esterdalvit, was sufficient to help me see that there is a continuous deformation between spaces X and Y:

Intermediate deformatoin

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    $\begingroup$ Good. As I mentioned before, if you paint the outside of the original punctured torus green, you can see that this step is so symmetric, you can continue on to a punctured torus with the green color on the inside. $\endgroup$
    – Will Jagy
    Aug 27, 2012 at 22:42
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    $\begingroup$ A second way of seeing it is by looking at the torus as a square with edges identified appropriately. If you puncture the middle of the square, you can retract to a neighborhood of the edges. After you identify the edges, you have the glued cylinders. $\endgroup$
    – Neal
    Aug 27, 2012 at 22:48
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    $\begingroup$ The pther spelling thing to watch for is that the plural is tori, while torii is a sort of Japanese gate: see en.wikipedia.org/wiki/Torus and then en.wikipedia.org/wiki/Torii Finally, from a recent crossword puzzle, umami is the name for a fifth basic taste en.wikipedia.org/wiki/Umami $\endgroup$
    – Will Jagy
    Aug 27, 2012 at 23:05
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Let me share a portable network graphics' handcrafted, which one has to see it as a drawing but in ${\Bbb{R}}^3$,

enter image description here

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