Q:

Deformation retraction of $ \mathbb{R^3}\setminus S^2.$

I was wondering that if $ \mathbb{R^3}{\setminus}S^1$    retracts to $S^2 \vee S^1 $ ( it becomes $S^2$ more a dianeter I think).
Then does $ \mathbb{R^3}{\setminus}{S^2}$ retract to $S^2 \vee S^2 $ ? It can be generalized ? Maybe $\mathbb{R^n}{\setminus}{S^m}$ $(n > m)$ retracts to $S^{n-1} \vee S^m$ ? Is that true ?
Thank you.

A:

$\mathbb{R}^3 \setminus  S^2$ is not connected. If you allow retractions for disconnected spaces it retracts to $S^2 \sqcup \{ pt\}$ (the disjoint union of $S^2$ and one point).
For the general statement, $\mathbb{R}^n \setminus S^m$ with $m < n-1$ retracts on $\mathbb{R}^{m+2} \setminus S^m$; I am not sure if there is a nice way to retract it more.

