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

Question

Why does for all $k\in Z$ $H_k (S^m, S^m \setminus x) \cong H_k (S^m, pt)$ (for $x\neq pt)$ hold? The inclusion $(S^m, pt) \rightarrow (S^m, S^m \setminus x)$ might induce an isomorphism on homology, but I am not sure about that.

Answer 1

The identity on $S^m$ yields a map of pairs $i : (S^m,\{ pt \}) \to (S^m, S^m \setminus \{ x \})$. We have $i_{abs} = id : S^m \to S^m$ and $i_{rel} = j = \text{inclusion } \{ pt \} \to S^m \setminus \{ x \}$ which is a homotopy equivalence because $S^m \setminus \{ x \}$ is homeomorphic to $\mathbb{R}^m$ which is contractible.

Now consider the long exact sequences of the pairs $(S^m,\{ pt \})$ and $(S^m, S^m \setminus \{ x \})$ which are connected by the homomorphisms $j_*: H_*(\{ pt \}) \to H_*(S^m \setminus \{ x \})$, $id_* = id : H_*(S^m) \to H_*(S^m)$ and $i_* : H_*(S^m,\{ pt \}) \to H_* (S^m, S^m \setminus \{ x \})$. $id_*$ and $j_*$ are isomorphisms, therefore the five lemma shows that also $i_*$ is one.