# Decide a manifold via its boundary

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.

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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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... and in dimension $n \geq 3$ there are $n$-manifolds which are not spheres that bound contractible manifolds. Google "Mazur manifold" for details. –  Ryan Budney Nov 1 '11 at 4:08
But if an $n$-manifold bounds a contractible manifold, by Poincare Duality it must be a homology sphere -- meaning having the same homology has $S^n$. –  Ryan Budney Nov 1 '11 at 4:12
Dear algori and Ryan, thanks for your answers. I'd like to learn more about this. Suppose we have a complete non-compact contractible manifold $M$ with a $T^k\times [-1,1]$ neck inside, can we say that the $T^k$ must divide $M$ into two parts and one of them is compact and contractible (so that $T^k$ shrinks to a point inside)? (or such neck cannot exist in a contractible manifold, by Ryan's second comment?) Thank you a lot. –  Chih-Wei Chen Nov 1 '11 at 4:28
In the comment above, $k$ is less than the dimension $n$ of $M$. And I forget to put the power $n-k$ on the interval $[-1,1]$, sorry. In case that $k<n-1$, it seems quite different and hard to our discussion before. –  Chih-Wei Chen Nov 1 '11 at 4:35
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 $k=1$. –  algori Nov 1 '11 at 4:37
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.