I've been watching Norman Wildberger's lectures on Algebraic Topology and one of his problems really got me stuck. The question is to show how a double-holed torus with a line of infinite length passing through one of the holes is homeomorphic to a the same double-holed torus but with the line passing through both holes. I know it's been answered before, I'm just having a hard time finding it on this or any other site.
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3One such homeomorphism takes the line in the first situation to the line in the second, and the surface in the first situation to the surface in the second. Of course, it's not defined on the rest of 3-space, but it doesn't have to be. But I suspect you're asking something different, like "is there an isotopy of 3-space carrying one to the other?" Are you sure you've phrased it correctly? – John Hughes Jun 08 '16 at 22:54
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You are correct. I understand the fact that they are homeomorphic, I have trouble seeing the transformation that leads from one to the other – Connor James Jun 08 '16 at 23:04
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Can you post a link to the lecture and cite the time in which he states the problem? – Mnifldz Jun 08 '16 at 23:08
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1Yeah @Mnifldz, I've been trying to track it down for a while so when I find it I will post it here. EDIT: Found it! https://www.youtube.com/watch?v=4U9XzZjxMFI starting at roughly 16 minutes. – Connor James Jun 08 '16 at 23:14
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1Ah...so the question is not whether the two things are homeomorphic, but whether there's an ambient isotopy carrying one to the other. By the way, there's no ambient isotopy that leaves the two-holed torus fixed. So the real question is whether there's an ambient isotopy that takes the first 2-holed torus to a new position in which the line is linked with two holes in the result...but they're not necessarily the same two holes, so to speak. – John Hughes Jun 09 '16 at 00:42
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Here's an answer in pictures. The top line shows the "after" and "before" situations, with the sole difference that I've drawn the line in a different color (red) from the 2-holed torus (blue), and made the line curved into a sort of "U" shape.
Below those two top drawings are two schematic drawings -- the red line remains the same, but the blue torus has been replaced by a "core" -- a collection of curves. If you take all points a little distance from the curves, you'll get back a 2-holed torus. Below this second line is a deformation of that "core"; if you fatten up the core in each of these, you get a deformation of the 2-holed torus. Yay!
Now I want to briefly address your comment: "Yes, I understand that they are homeomorphic, I have trouble seeing the transformation that leads from one to another."
Situation A: Consider two unit circles: one, $A_1$ in the $xz$ plane, centered at the origin, the other, $A_2$ in the $xy$ plane, centered at $(1,0,0)$.
Situation B: Consider two unit circles: one, $B_1$, in the $xz$ plane, centered at the origin, the other, $B_2$, in the $xy$ plane, centered at $(3,0,0)$.
The sets of two circles in these situations are homeomorphic, with a homeomorphism being given by
$$
f(x, y, z) = \begin{cases}
(x,y, z) & (x, y, z) \in A_1 \\
(x+2, y, z) & (x, y, z) \in A_2
\end{cases}.
$$
On the other hand, there's no ambient isotopy from the first situation to the second, because the linking number of circles $A_1$ and $A_2$ is 1, but for the $B$ circles it's $0$.
There is a real difference between homeomorphism and ambient isotopy, and it's worth understanding. The initial assertion in your question that we have a line and a two-holed torus embedded in two different ways, and you want to see that they're homeomorphic has, as an answer, that they have to be homoemorphic, since they're embeddings of the same union of manifolds. So that wasn't actually the question you wanted to ask at all. It's quite difficult to phrase questions like this carefully, especially in a field as "intuitive" as elementary topology. It's also worth the time (I believe) to do so.
John Hughes
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