# 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.
• @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 at 12:57