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

Question

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... :)

$B_2(0)\setminus\bar{B}_1(0)$ 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 — Albert, Feb 13 '13 at 15:09
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
See also: necessary and sufficient conditions to be homeomorphic to $\mathbb R^m$ — , Feb 13 '13 at 16:44

score 14 · Accepted Answer · answered Feb 13 '13 at 15:09

14

$B_1(0)\setminus\{0\}$ is a counterexample when $n>2$.

answered Feb 13 '13 at 15:09

Jonas Meyer

53,602

Of coarse! Thank you :) – Sh4pe Feb 13 '13 at 15:15

score 3 · Answer 2 · edited Apr 13 '17 at 12:19

3

In $\mathbb R^2$, yes, every open, bounded and simply connected set is homeomorphic to $B_1(0)$.

Look at Are simply connected open sets in $\mathbb{R}^2$ homeomorphic to an open ball? and Proof that convex open sets in $\mathbb{R}^n$ are homeomorphic?

edited Apr 13 '17 at 12:19

Community

1

answered Feb 13 '13 at 15:14

superAnnoyingUser

4,379

Yes, indeed. I appologise. – superAnnoyingUser Feb 13 '13 at 15:20
Another link for the $n=2$ case, from MathOverflow, with the requirement of not using the Riemann mapping theorem: http://mathoverflow.net/questions/66048/riemann-mapping-theorem-for-homeomorphisms. 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
I edited to make the restriction to two dimensions more obvious. – Ben Millwood Feb 13 '13 at 15:35

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

2 Answers2