Example 38.1:

Let $X = S^n$ and $A = S^n - {\bf e}_{n+1}$. We take $U$ to be the complement of the polar ice cap namely the set of all ${\bf x} \in S^n$ such that $x_{n+1} \leq 2/3$ (reader is invited to draw a picture). Applying the excision theorem, and denoting the polar ice cap by $D$,

$$H_n(S^n, S^n - {\bf e}_{n+1}) \cong H_n(S^n - U,\; D - {\bf e}_{n+1}) = H_n(D,\; D - {\bf e}_{n+1}).$$

Theorem (37.2) gives $H_n(S^n, S^n - {\bf e}_{n+1}) \cong H_n(S^n)$ and $H_n(D, D - {\bf e}_{n+1}) \cong H_{n-1}(D - {\bf e}_{n+1})$. Since the polar ice cap is homeomorphic to an open ball,

$$H_n(S^n) \cong H_{n-1}(S^{n-1}),\quad n \geq 2.$$

Using theorem (32.1) we conclude that $H_n(S^n) \cong \mathbb{Z}$ for $n\geq 1$.