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This is a Theorem in Vick's Book about Singular homology:

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I understand this proof, but what I don't get is how to prove for the case when $j=0$, since we won't have an isomorphism, Vick did only the case in which $j>0$. Any ideas are welcome, thanks!

Endov
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  • $S^n-B$ has two path components, so $H_0(S^n-B)=\mathbb Z\oplus \mathbb Z.$ – Matematleta Sep 26 '18 at 15:58
  • @Driver8 But the point is proving this statment. – Endov Sep 26 '18 at 18:46
  • If $X$ is a toplogical space, then $H_0(X)$ is a direct sum of $\mathbb Z$ ,one for each path component of $X$. I thought you knew this already because Mayer-Vietoris is usually presented after (at least, Rotman and Munkres do it that way). – Matematleta Sep 26 '18 at 22:13
  • @Driver8 Yes, the fact that $H_0(X)$ is the direct sum of $\mathbb{Z}$ I know, but to state that $S^n-B$ has two path components you use what argument? Because I am trying to use this theorem to prove it. From what you are saying you go the opposite way, you know that $S^n-B$ has two path components and therefore you conclude that $H_0(S^n-B)=\mathbb{Z}\oplus \mathbb{Z}$, but I want to show that $H_0(S^n-B)=\mathbb{Z}\oplus\mathbb{Z}$ to conclude that it has two path components. (Notice that this is valid for $k=n-1$ in the theorem, in the other cases we will have $H_0(S^n-B)=\mathbb{Z}$. – Endov Sep 27 '18 at 13:49
  • OK, I thought that you wanted a proof of the corollary for j=0. – Matematleta Sep 27 '18 at 18:12

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