$H_k (R^m, R^m \setminus 0) \cong H_k (S^m, S^m \setminus x)$?

Question

Why does (for singular homology) for all $k\in Z$ $H_k (R^m, R^m \setminus 0) \cong H_k (S^m, S^m \setminus x)$ for some point $x \in S^m$ hold? I thought that this could follow from the fact that $R^m$ ist hty equivalent to $S^{m-1}$ or that $R^m $ is homeomorphic to $S^m$ without a point.

Answer 1

The homology group $H_*(\mathbb R^m,\mathbb R^m\smallsetminus 0)$ fits into a long exact sequence of reduced homology, and the connecting morphism $H_*(\mathbb R^m,\mathbb R^m\smallsetminus 0)\to H_{*-1}(\mathbb R^m\smallsetminus 0)$ is an isomorphism, since $\mathbb R^m$ is contractible.

The group $H_*(S^m,S^m\smallsetminus \ast)$ also fits in a LES, and this time $S^m\smallsetminus \ast$ is contractible so you get an isomorphism with $H_*(S^m)$.

If you put all of this information together, you should be able to conclude.