Homology of spheres from the M-V sequence

1. Mar 10, 2009

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?

2. Mar 10, 2009

yyat

Hi quasar987!

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

3. Mar 10, 2009

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 $$\widetilde{H}_n(\mathbb{S}^0)=\widetilde{H}_n({-1})\oplus\widetilde{H}_n({1})=0\oplus 0=0$$??

4. Mar 10, 2009

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 $$\widetilde{H}_0(S^0)=\mathbb{Z}$$.

5. Mar 10, 2009

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, $$H_*(X,A)=\bigoplus_{\alpha} H_*(X_{\alpha},X_{\alpha}\cap A)$$ ?

6. Mar 10, 2009

yyat

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