how to show that a open set of $S^3$ is simply connected?

Question

Let $B$ be the union of the compactification point and $(\mathbb R^3-X)$ in the one-point compactification of $\mathbb R^3$. (Here $X$ is a closed ball in $\mathbb R^3$.)

Then I think $B$ is somehow simply connected..but not able to prove it. Can anyone help me with this?

Answer 1

We may take $X$ to be $\{x\in\mathbb R^3:\|x\|\le 1\}$.

The mapping $f:X^\circ\to B$ given by $$ f(x) = \begin{cases} x/\|x\|^2 & \text{if }\|x\|>0, \\[6pt] \text{the compactification point} & \text{if }x=0, \end{cases} $$ is a homeomorphism from $X^\circ$ to $B$ (where $X^\circ$ is the interior of $X$). Therefore if $X^\circ$ is simply connected, then so is $B$.

Answer 2

The space is homeomorphic an open ball in $\mathbb{R^3}$.

I would look first at the analogous case : $X$ closed ball in $\mathbb{R^2}$ imbedded in the sphere $S^2$ by a stereographic projection from the North pole.