Homology and homotopy groups of $\mathbb{R}^n\setminus D^n$ Let $D^n\subset \mathbb{R}^n$ be a closed ball. What are the homology and homotopy groups of $\mathbb{R}^n\setminus D^n$? 
I suspect that $\pi_n(\mathbb{R}^n\setminus D^n)\neq 0$ since a sphere $S^n$ placed around the disc $D^n$ cannot be contracted to a point. 
By contracting $D^n$ to a point, is it enough to calculate $\mathbb{R}^n\setminus \{\mathrm{pt}\}$? Since $\mathbb{R}^n\setminus \{\mathrm{pt}\}$ is homotopic to $S^{n-1}$, the problem reduces to calculating the homology and homotopy groups of spheres. Is this reasoning correct?
 A: If you just want $\pi_n$ you can get an explicit deformation retract to $S^{n-1}$
by retracting to $\partial D^n$.
In particular,  take the straight line homotopy $f(x,t)=tx+(1-t)x/\|x\|$ which is well defined since $\|x\| \neq 0$.
Now, $\pi_n(S^n)$ can be computed in a few ways.
The easiest is the Hurewicz theorem which says that if a topological space is $n-1$ connected, then there is an isomorphism $\phi:\pi_n\to H_n$. So, all you need to do is calculate $H_n(S^n)$. You can do this in many ways. Cellular homology is easy using the standard  structure of $S^n$ with one $0$-cell and $1$ $n$-cell.
Or, mayer vietoris tells that 
$$0 \to H_n(S^n) \to H_{n-1}(S^{n-1}) \to 0$$
is exact. In either case, one can deduce that $H_n(S^n) \cong \mathbb Z$.
or use the fact that $S^n$ is an orientable manifold here
In any case, that is doable. 
The lower homotopy groups are easy as well  (they are all $0$) which can be obtained via cellular approximation.
For $\pi_k(S^n)$ with $k>n$, there are partial results, but the problem is widely open.
