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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?2012-07-04
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    @Mercy : No. Just one.2012-07-04
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    Would it just be every surface without a boundary, since you could just puncture it?2012-07-04
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    Yes I see, but it seems it isn't really a circle, meaning that it's not contained in a plane.2012-07-04
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    @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.2012-07-04
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    I wonder if what I've done here is asked a question with a completely trivial answer, when I should have asked a different question with a less trivial answer?2012-07-04

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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.2012-07-04
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    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: http://math.stackexchange.com/a/131947/249342012-07-04