Show that \(H_i(X \times S^n) \approx H_i(X) \oplus H_{i-n}(X)\) for all \(i\) and \(n\), where \(H_i = 0\) for \(i < 0\) by definition. Namely, show \(H_i(X \times S^n) \approx H_i(X) \oplus H_i(X \times S^n, X \times \{x_0\})\) and \(H_i(X \times S^n, X \times \{x_0\}) \approx H_{i-1}(X \times S^{n-1}, X \times \{x_0\})\). [For the latter isomorphism the relative Mayer–Vietoris sequence yields an easy proof.]
Solution
Fix \(x_0\in S^n\) and write \[ P_n=X\times S^n, \qquad A_n=X\times\{x_0\}. \]
1. The long exact sequence of \((P_n,A_n)\) splits and gives\[ H_i(P_n)\cong H_i(X)\oplus H_i(P_n,A_n). \]
The inclusion \[ A_n\cong X\hookrightarrow X\times S^n \] has the projection \(X\times S^n\to X\) as a retraction. Hence its induced map on homology is injective and split. The long exact sequence of the pair therefore breaks into split short exact sequences \[ 0\to H_i(X)\to H_i(P_n)\to H_i(P_n,A_n)\to0. \]
2. For \(n\ge1\) there are natural isomorphisms\[ H_i(P_n,A_n)\cong H_{i-1}(P_{n-1},A_{n-1}). \]
Decompose \(S^n\) into upper and lower hemispheres \(D_+^n\) and \(D_-^n\), meeting in \(S^{n-1}\), and choose \(x_0\) on the equator. Apply relative Mayer–Vietoris to the pair \[ (X\times S^n,\,X\times\{x_0\}). \] The two hemisphere pairs \[ (X\times D_\pm^n,\,X\times\{x_0\}) \] deformation retract to \((X,X)\) and hence have zero relative homology. Their intersection pair is \[ (X\times S^{n-1},\,X\times\{x_0\}). \] Thus the connecting homomorphism in the relative Mayer–Vietoris sequence is an isomorphism shifting degree by one.
3. One has\[ H_i(P_n,A_n)\cong H_{i-n}(X). \]
Iterate 2 until \(n=0\). Since \(S^0\) consists of two points and \(A_0\) is one copy of \(X\), \[ H_j(X\times S^0,\,X\times\{x_0\})\cong H_j(X). \] Therefore after \(n\) shifts, \[ H_i(P_n,A_n)\cong H_{i-n}(X), \] where the latter is zero for negative indices.
Combining 1 and 3 gives \[ \boxed{H_i(X\times S^n)\cong H_i(X)\oplus H_{i-n}(X).} \]