Connectedness of the complement of a homeomorph of a ball Let $n\geq 2$. My question is in two parts:


*

*If $B\subseteq\mathbb{R}^n$ is bounded and homeomorphic to the open unit ball in $\mathbb{R}^n$, is $\mathbb{R}^n\setminus B$ connected? Does the answer depend on the dimension $n$?

*If $B\subseteq\mathbb{R}^n$ is homeomorphic to the closed unit ball in $\mathbb{R}^n$, is $\mathbb{R}^n\setminus B$ connected? Does the answer depend on the dimension $n$?${}$

 A: Answers to both questions are positive, they are applications of Poincare-Alexander duality. 


*

*Let $U$ be a bounded open subset of $R^n$ and  $n\ge 2$, then connectivity of $R^n\setminus U$ is equivalent to connectivity of $C=S^n\setminus U$, where $S^n$ is the 1-point compactification of $R^n$. Now, $C$ is compact and we can apply duality:
$$
{H}_{n-1}(U)\cong {\check{H}}^0(C)\ne 0,
$$
if $C$ is not connected. (Here the cohomology is Chech.) Hence, $U$ cannot be contractible if $C$ is not connected. Thus, if $U$ is homeomorphic to open ball then $C$ is connected. 

*Suppose that $B$ is a compact subset of $R^n$ homeomorphic to a closed ball (it suffices to assume contractibility). By duality, you get
$$
\tilde{H}_0(R^n \setminus B)\cong H^n(B)=0. 
$$ 
Hence, $R^n \setminus B$ is connected. 
A: Question 2 can also be answered using Brouwer's topological degree. More precisely, it can be shown that:

Theorem: Let $\Omega_1, \Omega_2 \subset \mathbb{R}^n$ be two homeomorphic compact subspaces. Then $\mathbb{R} \backslash \Omega_1$ and $\mathbb{R} \backslash \Omega_2$ have the same number of connected components.

See Brouwer’s Topological Degree (IV): Jordan Curve Theorem.
