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For those who have the book, in corollary 19.6 page 234 of Bredon, the Jordan-Brouwer separation theorem has just been proven and so we know that if S is homeomorphic to S^(n-1), then S^n\S has two connected components, which are open in S^n and their boundary is S.
So we let V be one of the components of S^n\S and let x be in V. Then Bredon writes that excision implies the isomorphism
[tex]H_{i+1}(V,V\backslash \{x\})\cong H_{i+1}(\mathbb{D}^n,\mathbb{D}^n\backslash\{0\})[/tex]
(where D^n is the closed n-disk).
It's like he said "V u S is homeomorphic to D^n (with x being sent to 0), so excising the boundary [tex]\partial\mathbb{D}^n\cong S[/tex] gives the above isomorphism".
But for n>2, it is not true in general that V u S is homeomorphic to D^n as the (counter-) example of the Alexander horned disk shows...
So what is it that he's doing in that step?P.S. Don't bother with Google Book, page 234 is missing from the preview. :(
So we let V be one of the components of S^n\S and let x be in V. Then Bredon writes that excision implies the isomorphism
[tex]H_{i+1}(V,V\backslash \{x\})\cong H_{i+1}(\mathbb{D}^n,\mathbb{D}^n\backslash\{0\})[/tex]
(where D^n is the closed n-disk).
It's like he said "V u S is homeomorphic to D^n (with x being sent to 0), so excising the boundary [tex]\partial\mathbb{D}^n\cong S[/tex] gives the above isomorphism".
But for n>2, it is not true in general that V u S is homeomorphic to D^n as the (counter-) example of the Alexander horned disk shows...
So what is it that he's doing in that step?P.S. Don't bother with Google Book, page 234 is missing from the preview. :(