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Suppose I have a cartwheel-like structure minus the rim where all the spokes are made up of infinite chains of tori, say my structure has n such spokes, how many "ends" (in the topological sense) does this 3D surface have?

Is such a surface homeomorphic to all surfaces made up of infinitely many tori with that number of ends? If not, how might I count the number of 3D surfaces up to homeomorphism that are made up of gluing together infinitely many tori?

Thank you.

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"this 3D surface"? The surface is 2-dimensional. If you're considering not homeomorphisms of those surfaces but homeomorphisms of an $\mathbb{R}^3$ with the surfaces embedded therein, then the answer is obviously no (consider a non-trivially knotted chain of tori). – leftaroundabout Aug 19 '11 at 22:36
@leftroundabout: Thanks, that is very true. The last question, though, still stands... I am pretty sure I came across some material earlier today that said that $R^3$ surfaces composed of infinitely many tori, up to homeomorphism, are countable. – brian Aug 19 '11 at 22:44
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The surface you've constructed is an orientable infinite genus surface with $n$ ends. If you make sure that every neighborhood of an end is not an annulus, then all such surfaces are homeomorphic. (A trivial way of making an end is to cut out a disk, but we don't want to count those.) This result is due to Ian Richards.

Compare Benoit Kloeckner's MO answer.

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"If you make sure that every neighborhood of an end is not an annulus" well, any neighbourhood of an end is still an infinite chain of tori, isn't it? – leftaroundabout Aug 19 '11 at 22:57
Exactly, that's what it is in your example. I'm just saying that if you say "infinite genus plus n-ends" it's not enough to classify. You also need the additional condition you just quoted. – Grumpy Parsnip Aug 19 '11 at 23:06

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