Homology of spheres from the M-V sequence

[quasar987]

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,

\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

But in reduced homology, \widetilde{H}_n(\mathbb{S}^0)=0 in all degree, so we conclude that \widetilde{H}_n(\mathbb{S}^1)=0 in all degrees.

But this is not so because \widetilde{H}_1(\mathbb{S}^1)=\mathbb{Z}.

So where am I mistaken in the above?

 

[yyat]

Hi quasar987!

Remember that S^0 consists of two points (connected components), so its reduced 0-th homology is non-trivial.

 

[quasar987]

But's isn't the reduced homology of a space equal to the the direct sum of the reduced homology of its connected components, so that \widetilde{H}_n(\mathbb{S}^0)=\widetilde{H}_n({-1})\oplus\widetilde{H}_n({1})=0\oplus 0=0??

 

[yyat]

No, that is only true for unreduced homology (for reduced homology it's true if you take the wedge sum instead of the disjoint sum).
Unreduced homology is exactly like the reduced homology but with an additional Z summand in the degree 0 term. Since H_0(S^0)=\mathbb{Z}^2, one must have \widetilde{H}_0(S^0)=\mathbb{Z}.

 

[quasar987]

Thanks a lot for point that out yyat. I had convinced myself that the formula H_*(X)=\bigoplus_{\alpha} H_*(X_{\alpha}) held for reduced homology as well. I am right in thinking it holds for relative homology though? Namely, H_*(X,A)=\bigoplus_{\alpha} H_*(X_{\alpha},X_{\alpha}\cap A) ?

 

[yyat]

Thanks a lot for point that out yyat. I had convinced myself that the formula H_*(X)=\bigoplus_{\alpha} H_*(X_{\alpha}) held for reduced homology as well. I am right in thinking it holds for relative homology though? Namely, H_*(X,A)=\bigoplus_{\alpha} H_*(X_{\alpha},X_{\alpha}\cap A) ?

Yes. It also follows from the relative Mayer-Vietoris sequence in the case of finite disjoint unions.