Compute \(H_*(X\cross S^n)\) in terms of \(H_*(X)\)
- Compute \(H_*(T^n)\)
Solution
1. For every \(k\),\[ \boxed{H_k(X\times S^n;\mathbb Z)\cong H_k(X;\mathbb Z)\oplus H_{k-n}(X;\mathbb Z),} \]where \(H_j(X)=0\) for \(j<0\).
The homology of \(S^n\) is free and nonzero only in degrees \(0\) and \(n\), where it is \(\mathbb Z\). Hence the Künneth theorem has no Tor terms and gives \[ H_k(X\times S^n) \cong H_k(X)\otimes H_0(S^n) \oplus H_{k-n}(X)\otimes H_n(S^n), \] which is the displayed direct sum.
2. For the \(n\)-torus \(T^n=(S^1)^n\),\[ \boxed{H_k(T^n;\mathbb Z)\cong\mathbb Z^{\binom nk}} \]for \(0\le k\le n\), and \(H_k(T^n)=0\) otherwise.
Proceed by induction using \(T^n=T^{n-1}\times S^1\) and 1: \[ H_k(T^n)\cong H_k(T^{n-1})\oplus H_{k-1}(T^{n-1}). \] If the ranks are \(\binom{n-1}{k}\) and \(\binom{n-1}{k-1}\), Pascal’s identity gives \[ \binom{n-1}{k}+\binom{n-1}{k-1}=\binom nk. \] The base case \(T^1=S^1\) is immediate.