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The boundary of a disk or of a Möbius band is a circle.

Which other manifolds share that property?

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Doesn't the boundary of the Möbius strip have two components? – Mercy King Jul 4 '12 at 17:56
@Mercy : No. Just one. – Michael Hardy Jul 4 '12 at 17:57
Would it just be every surface without a boundary, since you could just puncture it? – Michael Hardy Jul 4 '12 at 17:58
Yes I see, but it seems it isn't really a circle, meaning that it's not contained in a plane. – Mercy King Jul 4 '12 at 18:01
@Mercy : It's topologically a circle. That's what I had in mind. I.e. it's homeomorphic to the ordinary circle in the plane. – Michael Hardy Jul 4 '12 at 18:03
up vote 6 down vote accepted

For the compact case, I believe the answer is, as you said in the comments above, any closed surface with a single puncture, i.e. a disk removed. I claim that this completely classifies compact surfaces with boundary $S^1$. This is because you can glue a disk to the surface along its boundary to obtain a closed surface, and there is a unique way to do this (see, for example, Example 4.1.4(c) in Gompf and Stipsicz's 4-Manifolds and Kirby Calculus). So by the classification of surfaces, there should then be a unique surface with boundary $S^1$ corresponding to each closed surface.

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So if there were more than one way to glue the disk back on to the punctured surface, then two different surfaces when punctured would result in the same surface with the circle as its boundary. And you're saying that does not happen. So maybe my question was not as trivial as I thought just after I posted it. Or maybe the proof of this uniqueness claim is still trivial. – Michael Hardy Jul 4 '12 at 22:28
In this case, the different ways to attach a disk ($2$-handle) to a surface with boundary $S^1$ are in one-to-one correspondence with elements of $\pi_1(\mathrm{O}(0)) = 0$. See the following answer regarding the relevant part of Gompf and Stipsicz on this topic: – Henry T. Horton Jul 4 '12 at 23:58

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