- #1
- 4,807
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Hi everyone.
Take the open sets A=S^1 - N and B=S^1 - S, that is, the circle minus the north and south pole resp.. Noting that AnB=S^0 and that A and B are contractible, the Mayer-Vietoris sequence in reduced homology for this decomposition writes,
[tex]\ldots \rightarrow \widetilde{H}_n(\mathbb{S}^0)\rightarrow 0\oplus 0\rightarrow \widetilde{H}_n(\mathbb{S}^1)\rightarrow\widetilde{H}_{n-1}(\mathbb{S}^0)\rightarrow 0\oplus 0\rightarrow\ldots[/tex]
But in reduced homology, [tex]\widetilde{H}_n(\mathbb{S}^0)=0[/tex] in all degree, so we conclude that [tex]\widetilde{H}_n(\mathbb{S}^1)=0[/tex] in all degrees.
But this is not so because [tex]\widetilde{H}_1(\mathbb{S}^1)=\mathbb{Z}[/tex].
So where am I mistaken in the above?
Take the open sets A=S^1 - N and B=S^1 - S, that is, the circle minus the north and south pole resp.. Noting that AnB=S^0 and that A and B are contractible, the Mayer-Vietoris sequence in reduced homology for this decomposition writes,
[tex]\ldots \rightarrow \widetilde{H}_n(\mathbb{S}^0)\rightarrow 0\oplus 0\rightarrow \widetilde{H}_n(\mathbb{S}^1)\rightarrow\widetilde{H}_{n-1}(\mathbb{S}^0)\rightarrow 0\oplus 0\rightarrow\ldots[/tex]
But in reduced homology, [tex]\widetilde{H}_n(\mathbb{S}^0)=0[/tex] in all degree, so we conclude that [tex]\widetilde{H}_n(\mathbb{S}^1)=0[/tex] in all degrees.
But this is not so because [tex]\widetilde{H}_1(\mathbb{S}^1)=\mathbb{Z}[/tex].
So where am I mistaken in the above?