# Is every open, bounded and simply connected subset of $\mathbb{R}^n$ essentially a ball?

Let $\Omega\subset \mathbb{R}^n$ be open, bounded and simply connected. I wonder if the answer to the following question is known:

Is there a homeomorphism $\Omega\to \operatorname{B}_1(0)$, where $\operatorname{B}_R(0)$ is the open ball (in the topology induced by the metric) with radius $R$ around $0\in \mathbb{R}^n$?

The Poincare conjecture comes to mind, but it only concerns manifolds without border, as far as I understand.

Thanks for any hints to literature, theorems or counterexamples etc... :)

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$B_2(0)\setminus\bar{B}_1(0)$ is open, bounded, simply connected subset of $\mathbb{R}^3$ but not a ball. – achille hui Feb 13 '13 at 15:08
in dimension 2 this is true : it is the riemann mapping theorem – Glougloubarbaki Feb 13 '13 at 15:09
Those are both answers, not comments! – Ben Millwood Feb 13 '13 at 15:12
@achille The set you considered is punctured. – Student Feb 13 '13 at 15:17
Might be worth looking at the higher homotopy groups. Every path in $B_2(0) \setminus \overline B_1(0)$ contracts to a point, but there is a surface which doesn't, so the second homotopy group isn't trivial. In a convex set like the open ball, every homotopy group is trivial. – Ben Millwood Feb 13 '13 at 15:22

$B_1(0)\setminus\{0\}$ is a counterexample when $n>2$.
In $\mathbb R^2$, yes, every open, bounded and simply connected set is homeomorphic to $B_1(0)$.
Another link for the $n=2$ case, from MathOverflow, with the requirement of not using the Riemann mapping theorem: mathoverflow.net/questions/66048/…. Boundedness is unnecessary. (A trivial remark is that nonemptiness is necessary unless that is already part of the definition of simply connected.) – Jonas Meyer Feb 13 '13 at 15:20