When a compact topological manifold with boundary is a ball? 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. 
 A: I wish to address the case $n=4$. This would follow if any $4$-dimensional h-cobordism between $3$-spheres were trivial but at the moment I am not sure how to prove this. There is however a different argument. 
Brown proved in [A proof of the generalized Schoenflies theorem, Bull. Amer. Math. Soc. 66 (1960), 74-76],  http://www.ams.org/journals/bull/1960-66-02/S0002-9904-1960-10400-4/home.html that any topological embedding of $S^{n-1}\times [-1,1]$ into $S^n$, then the closures of either component of the complement of $S^{n-1}\times \{0\}$ is a topological disk. 
Now let's show that $X$ is homeomorphic to $D^4$. Since $\partial X$ is a simply-connected $3$-manifold it is homeomorphic to $S^3$ (by Perelman). By the topological collar theorem (also due to Brown) there is a collar neighborhood $C$ of $\partial X$ in $X$. Attaching a $4$-disk along $\partial X$ gives a $4$-sphere. Thus $C$ is a copy of $S^3\times I$ in $S^4$, hence by Brown the closure of $X-C$ is homeomorphic to $D^4$, and hence the same is true about $X$.
Perhaps a variation of the above shows that any h-coborsims between 3-spheres is topologically a product but I do not see this.
A: 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.
A: For $n \geq 6$, this follows from the topological h-cobordism theorem (due to Kirby-Siebenmann). Indeed, removing a topological ball from the interiour of $X$ yields an h-cobordism. If this is trivial, i.e., homeomorphic to $S^{n-1} \times [0,1]$ relative boundary, you glue the ball back in to see that $X$ must be a topological ball.
