Dear all,

I am thinking about a problem as follows: Suppose a simply-connected 2-dimensional manifold has an $S^1$ boundary, is it homeomorphic to the open disk $D^2$? In fact, I would like to understand the general higher-dimensional case, i.e., how to decide the homeomorphism type of a (connected) manifold $M$ via its boundary submanifold $N$? Of course we need to put some condition on $M$ such like contractible or else, otherwise it could be arbitrary. In particular, when could we know that $M$ is the trivial fill-in manifold of $N$? For example, $M$ is the solid $k$-torus whereas $N=T^k=S^1\times\cdots\times S^1$.

I have no idea whether this is a research problem or just an exercise. I am sorry about my few knowledge on topology. If it is quite standard, could you please suggest me some reference books? Thanks a lot.


Yes, assuming the manifold compact. One way to see this is this: if we glue a disk along the boundary, we get a manifold which is simply-connected manifold by Seifert-van Kampen, and closed, hence a sphere by the classification theorem.

The same argument plus the Whitehead theorem and the topological Poincare conjecture shows that any compact contractible manifold that bounds a sphere must be homeomorphic (but not necessarily diffeomorphic) to the unit disk.

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    $\begingroup$ ... and in dimension $n \geq 3$ there are $n$-manifolds which are not spheres that bound contractible manifolds. Google "Mazur manifold" for details. $\endgroup$ – Ryan Budney Nov 1 '11 at 4:08
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    $\begingroup$ But if an $n$-manifold bounds a contractible manifold, by Poincare Duality it must be a homology sphere -- meaning having the same homology has $S^n$. $\endgroup$ – Ryan Budney Nov 1 '11 at 4:12
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    $\begingroup$ Dear Chih-Wei -- a torus $T^k$ inside some manifold $M$ does not necessarily split $M$: take $M=T^k\times T^1$. And yes, no contractible manifold has $T^k$ as boundary unless $k=1$. $\endgroup$ – algori Nov 1 '11 at 4:37
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    $\begingroup$ A compact contractible smooth manifold bounded by a sphere $S^{n-1}$ is actually also diffeomorphic to a disc $D^n$ provided that $n>4$: there are exotic spheres, but there are no exotic balls as far as I remember. In fact, every known exotic sphere is the union of two standard smooth balls glued via a nonstandard smooth map along their boundaries. So if you remove a ball from an exotic sphere you get a standard ball. The dimension-4 case is of course still open and of completely different flavour. $\endgroup$ – Bruno Martelli Nov 9 '11 at 21:02
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    $\begingroup$ Bruno -- re "there are no exotic balls as far as I remember": indeed, by the h-cobordism theorem. $\endgroup$ – algori Nov 10 '11 at 12:54

I'm wondering whether there isn't a much easier argument for 2-dimensional case, without using Seifert-van Kampen or the classification of surfaces (which is a middling hard theorem).

Fix the boundary of a collar $f\colon\, S^1\to M$ of the $S^1$-boundary of your simply-connected 2-manifold M, and contract it to a point via a homotopy $H\colon\, S^1\times [0,1]\to M$ with $H(S^1\times\{0\})=f$ to $H(S^1\times\{1\})$ a constant function. To keep things simple, this might as well be free homotopy- the loops have no basepoint. This homotopy H gives rise to a codimension 1 foliation of a submanifold N of M, whose leaves are the loops $H(S^1\times\{t\}$ with $t\in [0,1]$. The submanifold N has to be a disc, because the homotopy induces a Morse function ($t$ is "the height") which has one critical point at $t=1$ (maybe after an eentsy weentsy perturbation- so Sard's theorem is implicit at this step, which is also a middling difficult theorem, but, I would argue, easier than the classification of closed surfaces), and a 2-manifold with a Morse function with one boundary component and one critical point can only be a disc. Now you argue that because M is a 2-manifold, there can't be more than one choice of disc N, otherwise in the neighbourhood of their intersection would not localy look like $\mathbb{R}^2$. So M must be N (plus the collar), and you're finished.
Does this make sense?

  • $\begingroup$ Thanks Daniel, it seems to be a nice proof, however I do not yet have enough ability to say it works or not. I'll try to pick up related knowledge. Thanks a lot. $\endgroup$ – Chih-Wei Chen Nov 17 '11 at 2:20

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