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.
 A: 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.
A: 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?
