Proof:

The result is clear if $k = 0$. We proceed by induction on $k$ and assume the result with $k-1$ in place of $k$. Let $A = A^{+}\cup A^{-}$ where $A^{+}$ and $A^{-}$ are each homeomorphic to $S^{k-1}$ and $A^{+}\cap A^{-}$ is homeomorphic to $S^{k-1}$. The Mayer Vietoris sequence may be applied to the open cover $\{S^n - A^{+},\; S^n - A^{-}\}$ of $S^n - A$ and the reader ought to verify that

$$H_{j+1}(S^n - A^{+}\cap A^{-}) \cong H_j(S^n - A),\quad j > 0.$$

By induction hypothesis we get (41.5) for the case $j > 0$. Let us now consider the case $j = 0$. The tail end of the Mayer Vietoris sequence gives

$$\begin{CD} 0@> >> H_1(S^n - S^{k-1})@> >> H_0(S^n-S^k) @> r >> \mathbb{Z}\oplus\mathbb{Z} @> q >>$$   img$$\;q @> >> 0\\ \end{CD}$$

Since the image of $q$ is isomorphic to $\mathbb{Z}$, we see that the kernel of $q$ must also be isomorphic to $\mathbb{Z}$ giving a short exact sequence

$$\begin{CD} 0@> >> H_1(S^n - S^{k-1})@> >> H_0(S^n-S^k) @> r >>$$   img$$\;r @> >>0. \\ \end{CD} \eqno(41.6)$$

Since the image of $r$ is free of rank one, (41.6) splits and we have

$$H_0(S^n-S^k) = H_1(S^n - S^{k-1})\oplus \mathbb{Z}.$$

If $k = n-1$ then $1 = n-(k-1)-1$ and so the induction hypothesis gives $H_1(S^n - S^{k-1}) = \mathbb{Z}$ whereas if $k \leq n-2$ then $H_1(S^n - S^{k-1}) = 0$. $\square$