Computing homology groups of complements of $S^2$ I have to compute the homology groups of $X=\mathbb{R}^3-S^2$ and $Y=\mathbb{R}^4-S^2$.
In the first case I thought that, since $X$ is not connected, its homology groups are sum of the two connected componets that are the internal part and the external part of the sphere. Since the first one is contractible, I reduced to compute the homology groups of $\mathbb{R}^3-B^3$, which I think it is homotopic equivalent to $S^2$, but I'm not sure of this.
For $Y$ I'm not able to see what to do, even in finding open sets in order to apply Mayer-Vietoris in a useful way.
Could someone give a hint? Thanks!
 A: Hint:
$\mathbb{R}^3 - B^3$ is indeed homotopy equivalent to $S^2$. To show this, you want to find a map $\mathbb{R}^3 - B^3 \rightarrow S^2$. Remember that $S^2$ is the space of points of $\mathbb{R}^3$ which are of unit distance from the origin. Can you think of a map which takes a point of $\mathbb{R^3}$ and sends it to some sort of "corresponding" point on the unit sphere?
For the second one, you're right to be using Mayer Vietoris. My hint would be to compare it to the space $\mathbb{R}^3 - S^1$. Perhaps use this as a toy example - since here your intuition can guide you because you can visualise, you might be able to pick your open sets more easily - and complete the exercise, and then see how this can apply to your space.
A: Let us compute, $\Bbb R^N-\Bbb S^n$. Note that $N=n+1$ implies $\Bbb R^N-\Bbb S^n=\Bbb R^N-\Bbb S^{N-1}$ which has two components, one of which is an $N$-ball, and hence contractible, while the second one is homotopically equivalent to $\Bbb S^{N-1}$.
Next, if $N>n+1$, then, $\Bbb R^N-\Bbb S^n$ is homotopically equivalent to $\Bbb S^{N-1}\lor\Bbb S^{N-n-1}$.
Now, $\widetilde H_k(\Bbb S^i\lor \Bbb S^j)=\widetilde H_k(\Bbb S^i)\oplus \widetilde H_k(\Bbb S^j)$. And, $$H_k(\Bbb S^i)=\begin{cases}\Bbb Z&\text{ if }k=0,i\\0&\text{ otherwise.} \end{cases}$$
