Singular homology about $\mathbb{R}^n$ without coordinates axis

Question

I've tried to compute singular homology of $\mathbb{R}^n$ without his $n$ coordinates axis.

I think that Mayer Vietoris is a good idea but I dont know what are the better open.

Thanks

Answer 1

Following the idea of @Omnomnomnom (and fixing it) the space deformation retracts onto $S^{n-1}\backslash\{\pm e_1,\ldots, \pm e_n\}$ where each $e_i=(0,\ldots,0,1,0,\ldots,0)$ is a unit vector. So all together it is a sphere without $2n$ points. The deformation is induced by normalization $v\mapsto\frac{1}{\lVert v\rVert}v$.

A sphere without a point is homeomorphic to the Euclidean space of the same dimension (via stereographic projection) and so we obtain that your space is homotopy equivalent to $\mathbb{R}^{n-1}\backslash\{v_i\}_{i=1}^{2n-1}$. Note that we "lost" one point due to homeomorphism.

This space on the other hand can be shown to be homotopy equivalent to the wedge sum of $2n-1$ spheres each of dimension ${n-2}$. And thus the final (reduced) homology is

$$\tilde H_k(X)=\bigoplus_{i=1}^{2n-1} \tilde H_k(S^{n-2})=\begin{cases}\mathbb{Z}^{2n-1} &\text{if }k=n-2 \\ 0 &\text{otherwise} \end{cases}$$

The cases $n=0,1$ have to be treated separately but are rather trivial if valid.