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The problem is in this video at 18:00.

If you have a 2-holed 2D torus with a line ($\mathbb{A}^1$) going through one of its holes. How do you deform it to make the line go through both holes.

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The two-holed "torus" is not $S^1\times S^1$. It could be argued to be (homotopy) equivalent to the wedge sum $S^1\wedge S^1$ for the purpose of this exercise, though. –  Henning Makholm Feb 10 '12 at 13:01
    
Whoops.. you are right.. I edited that out! Thanks. –  aelguindy Feb 10 '12 at 13:16

1 Answer 1

up vote 7 down vote accepted

This would be easiest to describe with a series of pictures but I don't have the time to draw them at the moment. Imagine your 2-holed torus is actually a 1-holed torus with another handle attached to one side along two disks. Now move one of these disks so that it circumambulates the hole of the central torus, moving around the perimeter until it comes back to near where it started. Now both of the torus's holes are penetrated by the line, kind of like two concentric circles which are tangent at a point. If you want you can push one of these circles up a bit so that the line goes through each hole at a different height. I hope this helps!

Update: here is a hand-drawn sequence of pictures which I took a photo of with my cellphone.

enter image description here

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If I understand correctly, what you do involves passing a part of the surface through itself. That's not allowed. Did I misunderstand you? –  aelguindy Feb 10 '12 at 19:50
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@aelguindy: nope I'm not moving the surface through itself. I am sliding the end of one handle along the surface of the central torus. This is an isotopy. Which part of what I said looks like moving the surface through itself? –  Grumpy Parsnip Feb 10 '12 at 22:39
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I'll draw a picture. –  Grumpy Parsnip Feb 10 '12 at 22:43
    
Awesome! Thank yo very much! –  aelguindy Feb 11 '12 at 0:17

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