# Compact contractible topological manifold with boundary=sphere is a ball

(this question is joint with Steven Karp and Thomas Lam) We need to use the following fact in our paper:

Theorem 1. Let $$M^n$$ be a compact contractible topological manifold with boundary, such that the boundary $$\partial M$$ is homeomorphic to a sphere $$S^{n-1}$$. Then $$M$$ is homeomorphic to a closed ball.

Q1. Is there a reference where this is stated with a proof? One reference that we found is the (slightly stronger) Theorem 10.3.3(ii) in Davis's book. He states this for all $$n$$, but only gives a proof sketch for $$n\geq 6$$.

Q2. What is the simplest way to prove Theorem 1? Here is an argument that we gathered from various MO posts: The boundary of $$M$$ is collared by Brown's theorem. Thus we can glue an $$n$$-ball to $$M$$, and by van Kampen and Mayer–Vietoris, it follows that the resulting space is a simply connected homology sphere. Thus it is a sphere by the Poincaré conjecture, therefore $$M$$ is a closed ball by Brown's Schoenflies theorem.

Note that we do not require the interior of $$M$$ to be an open ball (and instead we require that the boundary is a sphere), which is why this question is not a duplicate of this question.

This is a consequence of the h-cobordism theorem. You can find proofs in the smooth (respectively PL) categories in Milnor's Lectures on the h-Cobordism Theorem and Rourke and Sanderson, Introduction to Piecewise-Linear Topology. These are based on manipulations of handlebody decompositions (or Morse functions) and require n at least 6; there is a special argument for $$n=5$$ due independently to Barden and Smale.

Your question states the theorem in the topological category, and so the answer is more elaborate. You need to know that all of the tools of handlebody theory work in the stated dimensions, for topological manifolds. This is due to Kirby and Siebenmann, and is explained in their book, Foundational Essays on Topological Manifolds.

Finally, the argument you suggest (add a ball and apply the solution to the Poincaré conjecture) is sort of circular. That's because one usually proves the Poincaré conjecture by puncturing the homotopy sphere in two places and applying the h-cobordism theorem.

• Thanks! Actually, do you know if the topological h-Cobordism Theorem is even known in all dimensions? Specifically, we suspect it is open for n=4. So our "circular" argument is the only short proof we know that works for all n. – Pavel Galashin Feb 27 '19 at 3:27
• The topological h-cobordism theorem is known for n=4, by Freedman's work.The proof is by the argument you give; first show that a homotopy 4-sphere is homeomorphic to S^4, then remove a ball. But the key point in that statement is that a homotopy 4-sphere is h-cobordant to S^4, so you are using the 5-dimensional h-cobordism theorem. – Danny Ruberman Feb 27 '19 at 12:07
• As you explain, Freedman shows that every 5-dimensional h-cobordism between 4-manifolds is trivial (in the topological category), and Freedman+Perelman shows that every 4-dimensional h-cobordism between a 3-manifold and a 3-sphere is trivial (by gluing balls on both sides). But which reference shows that 4-dimensional h-cobordisms between general 3-manifolds are trivial? – Pavel Galashin Feb 28 '19 at 4:07
• @PavelGalashin By Perel'man there is only one simply connected 3-manifold, $S^3$, so Freedman's result covers the simply connected case. For non-simply connected manifolds, you need to have an additional hypothesis about Whitehead torsion; when it vanishes you have an s-cobordism which is known to be a product in high dimensions. (In dimension 5 at present you need a hypothesis on $\pi_1$.) But there are 4-dimensional s-cobordisms that are not homeomorphic to products. See work of Cappell-Shaneson and Kwasik-Schultz. None of these are known to be smoothable. – Danny Ruberman Feb 28 '19 at 12:57