In class today we showed that $H_i(S^n - h(D^k))=0$ where $h(D^k)$ is the embedding of the k-disc into $S^n$. The proof was very technical and used induction. I am wondering why the following argument doesn't work:
Since $D^k \cong h(D^k)$ and we know that $D^k$ is contractible we know that $h(D^k)$ is contractible. Thus $H_i(S^n - h(D^k))=H_i(S^n - \{p\})$ and since $S^n - {p}$ is contracible we get $H_i(S^n - h(D^k)=0$ $\forall i>0$
Perhaps we do not know that $h(D^k)$ is contractible because the ambient space may add complications and so we do not know for sure that $S^n - h(D^k) \cong S^n - D^k$ or something like that. Does anyone have any insight, clarification or general remarks on the matter? Thanks!
It seems that you argue that $S^n \setminus h(D^k)$ is homeomorphic (or at least homotopy equivalent) to $S^n \setminus \{ p\}$. However, in general $S^n \setminus h(D^k)$ is not even simply connected. An example is the Alexander horned ball (see https://en.wikipedia.org/wiki/Alexander_horned_sphere).