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I've seen this example given numerous times, but have never seen a real proof in a textbook.

InquilineKea
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    What's your definition of "coffee mug"? ;-) – Hans Lundmark Aug 07 '11 at 08:43
  • ... or "donut" for that matter. – lhf Aug 07 '11 at 11:30
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    This isn't really the kind of statement one can attach a rigorous proof to until the meaning of the terms "topological equivalence," "donut," and "coffee mug" are fixed. In my experience the first term can mean homotopy, ambient isotopy, or homeomorphism and it is never really clear which one is meant (sometimes intentionally so). "Donut" could refer either to a torus or a solid torus. And a coffee mug... is it thickened? Do you care about the surface? Etc. – Qiaochu Yuan Aug 07 '11 at 14:47
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    As far as I'm concerned you don't need a more rigorous proof of this kind of statement (which is clearly meant more as an easily-grasped intuitive example than anything else) than "if you made a coffee mug out of clay, you could reshape it into a donut without breaking or attaching anything." – Qiaochu Yuan Aug 07 '11 at 14:49
  • why "donut" and not a "bagel"? – PA6OTA May 22 '14 at 16:47
  • So what's the answer? Suppose I had a topological definition of a coffee mug. I want to show that the mug is homeomorphic to a doughnut (torus). In order to do that, I need to PROVE (if not explicitly give the formula) the existence of a function that's a homeomorphism between those two topological spaces. "It intuitively exists" is not a valid argument in mathematics in my opinion. Or is it in topology? – user4205580 Feb 13 '15 at 22:37
  • I think, by the classification theorem of closed surfaces, the matters are to give the definition of the cup and to prove the cup is a closed surface and orientable and its Euler characteristic is 0 (where "donuts":=topological torus).

    Maybe this is a minor problem (but I don't know the details since I almost haven't studied geometries, sorry).

     – user682141 Aug 30 '21 at 08:37

3 Answers3

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The torus and coffee mug are homeomorphic by the identity map $\operatorname{Id}_{\mathbb{T}^2}:\mathbb{T}^2\to \mathbb{T}^2$ ;-).

spinosarus123
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They are both homeomorphic to a sphere with one handle attached. This is quite clear for the coffee mug (where the handle is precisely the handle.....) and it is easily obtained for the donut (a.k.a. torus) by playing a bit with its representation as a square with opposite edges identified.

Andrea Mori
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    Actually, I wouldn't want to eat one of your two-dimensional donuts :) My coffee mugs are rather three-dimensional as well... – t.b. Aug 07 '11 at 10:59
  • Not even if it's filled with cream? :) – Andrea Mori Aug 07 '11 at 11:04
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    If it has a cream filling, it certainly cannot be two-dimensional anymore... :D – J. M. ain't a mathematician Aug 07 '11 at 11:10
  • The point is that the topological equivalence will not preserve any cream filling, as that will flow all over the place unless thoroughly beaten. – Tim Porter Aug 07 '11 at 11:33
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    The problem with the question is that neither the coffee cup nor the doughnut has been made that precise. In this case it is very difficult to give and explicit homeomorphism between them. On the other hand if you start with a mug without handle, specified as ever you like then it should be possible to produce a flat disc (thickness that of the mug) and then add a handle. (Is the torus a solid torus or just the surface... ditto for the mug?) – Tim Porter Aug 07 '11 at 11:39
  • As far as I ever understood, the doghnut/coffe mug examples made implicit reference to their surfaces. Said that, it is also clear that real doughnuts and coffe mugs are not $2$-dimensional (like any real object, as a matter of fact) – Andrea Mori Aug 07 '11 at 11:52
  • Ditto, my understanding of the equivalence is that they're both genus-1 surfaces... – J. M. ain't a mathematician Aug 07 '11 at 11:54
  • So what's the answer? Suppose I had a topological definition of a coffee mug. I want to show that the mug is homeomorphic to a doughnut (torus). In order to do that, I need to PROVE (if not explicitly give the formula) the existence of a function that's a homeomorphism between those two topological spaces. "It intuitively exists" is not a valid argument in mathematics in my opinion. Or is it in topology? – user4205580 Feb 14 '15 at 16:47
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I am assuming that the questioner knows that the question is about either the surfaces or the 3-manifolds in question. A rigorous, yet diagrammatic proof is in our book Knotted Surfaces and Their Diagrams.

Scott Carter
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