TITLE: When a compact topological manifold with boundary is a ball?
QUESTION [15 upvotes]: Let $X$ be a compact topological manifold with boundary. Suppose its interior is homeomorphic to  $\mathbb{R}^n$. Is $X$ homeomorphic to a (closed) ball?
Context: I want to show that a certain compactification of moduli of convex $n$-polygons (up to scaling and rotations) is a $(2n-4)$-cell. I can degenerate all sides (but $2$, of course), keeping records of the slopes, and all angles to $\pi$ (again except $2$). If more structure is needed, like smoothness on the interior, PL at the boundary - you have it.

REPLY [7 votes]: Igor's answer is spot-on, avoiding the h-cobordism theorem, which is fragile in low dimensions. Here is a proposal to tweak his answer (which works in all dimensions), so as to use the Poincare Conjecture only once, just where it is required.
Step 1. Boundary X is a homotopy sphere (noted already), and hence a sphere, by the PC. (Big machinery here, which varies with the dimension, but there's no way to avoid it. If you know in advance that boundary X is a sphere, so much the better.)
The next two steps use two classic theorems of Morton Brown (both mentioned already), and they work uniformly in all dimensions.
Step 2. Boundary X is collared in X (Brown, beautifully redone by Connelly). And so, by pulling X inward into its interior, we can assume that X lies in R^n, hence in S^n.
Step 3. Now Brown's Schoenflies Theorem (pure magic) implies that boundary X bounds a closed ball. Done.