Homology and homotopy groups of $\mathbb{R}^n\setminus D^n$

Question

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?

Answer 1

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.