Use the Mayer–Vietoris sequence to calculate the homology groups of \(S^n\).
Solution
1. For \(n\ge1\), choose open sets \(U,V\subset S^n\) obtained by slightly enlarging the northern and southern hemispheres.Then\[ U\simeq *, \qquad V\simeq *, \qquad U\cap V\simeq S^{n-1}. \]
Each enlarged hemisphere deformation retracts to a point. Their overlap is an open equatorial band, which deformation retracts onto the equator \(S^{n-1}\).
2. The reduced Mayer–Vietoris sequence gives natural isomorphisms\[ \widetilde H_k(S^n;\mathbb Z) \cong \widetilde H_{k-1}(S^{n-1};\mathbb Z) \]for every \(k\).
The reduced Mayer–Vietoris sequence for \[ S^n=U\cup V \] contains \[ \widetilde H_k(U)\oplus\widetilde H_k(V) \longrightarrow \widetilde H_k(S^n) \xrightarrow{\partial} \widetilde H_{k-1}(U\cap V) \longrightarrow \widetilde H_{k-1}(U)\oplus\widetilde H_{k-1}(V). \] By 1, \(U\) and \(V\) are contractible, so both outside terms vanish. Exactness therefore makes the connecting map an isomorphism \[ \widetilde H_k(S^n) \xrightarrow{\cong} \widetilde H_{k-1}(U\cap V) \cong \widetilde H_{k-1}(S^{n-1}). \]
3. The reduced homology of \(S^0\) is\[ \widetilde H_k(S^0;\mathbb Z) \cong \begin{cases} \mathbb Z,&k=0,\\ 0,&k\ne0. \end{cases} \]
The space \(S^0\) consists of two points, so \[ H_0(S^0;\mathbb Z)\cong\mathbb Z^2. \] Passing to reduced homology removes the diagonal copy of \(\mathbb Z\), leaving \[ \widetilde H_0(S^0;\mathbb Z)\cong\mathbb Z. \] There are no positive-dimensional cells, so all higher homology groups vanish.
4. Iterating 2 and using 3 gives\[ \widetilde H_k(S^n;\mathbb Z) \cong \begin{cases} \mathbb Z,&k=n,\\ 0,&k\ne n \end{cases} \qquad(n\ge1). \]
Repeatedly applying 2 yields \[ \widetilde H_k(S^n) \cong \widetilde H_{k-1}(S^{n-1}) \cong\cdots\cong \widetilde H_{k-n}(S^0). \] By 3, the last group is \(\mathbb Z\) exactly when \(k-n=0\), and is zero otherwise.
5. Therefore, for \(n\ge1\),\[ \boxed{ H_k(S^n;\mathbb Z) \cong \begin{cases} \mathbb Z,&k=0,n,\\ 0,&\text{otherwise}. \end{cases}} \]For \(n=0\), one has \(H_0(S^0;\mathbb Z)\cong\mathbb Z^2\) and all other groups vanish.
For \(n\ge1\), the sphere \(S^n\) is path-connected, so \[ H_0(S^n;\mathbb Z)\cong\mathbb Z. \] The positive-degree groups agree with reduced homology, so 4 gives the displayed answer. The \(n=0\) case is 3 before reduction.