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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!

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    How do you know that $H_i(S^n-h(D^k))=H_i(S^n-{p})$ using the fact that $h(D^k)$ is contractible ? – Adam Chalumeau Apr 24 '19 at 21:22
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    Also be carefull $S^n-D^k$ doesn't really make sens – Adam Chalumeau Apr 24 '19 at 21:24
  • By contracting $h(D^K)$ to ${p}$ was the idea –  Apr 24 '19 at 21:32
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    Yes but you are using the fact that the contraction of $h(D^k)$ onto ${p}$ implies a contraction of $S^n-p$ onto $S^n-h(D^k)$ right ? – Adam Chalumeau Apr 24 '19 at 21:50
  • Yes, this is indeed true and something that I was assuming. I mean, couldn't we like take the inverse of the contraction or something? What makes this complicated to do? –  Apr 24 '19 at 22:10
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    I don't no if it is complicated to do but I don't know how to do it. Do you have a simple proof ? – Adam Chalumeau Apr 24 '19 at 22:11
  • Let $C: h(D^k) \times I \rightarrow h(D^k)$ be a contraction of $h(D^k)$ to a point. Let $R$ be the mapping s.t. it is $C$ except it sends $t \in I$ to $(1-t) \in I$ or something like that. Reverse the contraction ya diG? Oh and it starts with a point and goes to $h(D^k)$ idk something like that hard to type in comments. –  Apr 24 '19 at 22:14
  • Let us continue this discussion in chat. – Adam Chalumeau Apr 24 '19 at 22:15
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    The image of a contractible space doesn’t have to be contractible. A line is contractible but can wrap around a circle. – Connor Malin Apr 25 '19 at 00:03
  • but if we wrap a line around a circle that's not a homeomorphism because we are identifying end points. The case I stated is special because it's an embedding. –  Apr 25 '19 at 00:47
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    Ok yes I agree with that. – Connor Malin Apr 25 '19 at 00:52
  • so my embedded disc is contractible you think? hmm –  Apr 25 '19 at 11:40
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    In general you don't know if the complement of a contractible space has the same homology type as the complement of a point. For example take $X= \Bbb R ^3$ and consider the complement of $1$ line that passes through the origin and of $2$ lines that pass through the origin. – hedphelym Apr 26 '19 at 02:20
  • thank you! leo lerena! –  Apr 26 '19 at 09:44
  • @LeoLerena Your example is not compact as $h(D^k)$ is. – Paul Frost Jul 17 '19 at 23:51

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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).

Paul Frost
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