Show that if \(X\) is a CW complex then \(H_n(X^n)\) is free by identifying it with the kernel of the cellular boundary map \(H_n(X^n, X^{n-1}) \to H_{n-1}(X^{n-1}, X^{n-2})\).
Solution
Consider the cellular boundary map \[ d_n: H_n(X^n,X^{n-1}) \longrightarrow H_{n-1}(X^{n-1},X^{n-2}). \] It is the composite of the connecting map for \((X^n,X^{n-1})\) with the natural map into the relative group: \[ H_n(X^n,X^{n-1}) \xrightarrow{\partial} H_{n-1}(X^{n-1}) \xrightarrow{j} H_{n-1}(X^{n-1},X^{n-2}). \]
1. The map\[ j:H_{n-1}(X^{n-1})\to H_{n-1}(X^{n-1},X^{n-2}) \]is injective.
The long exact sequence of the pair \((X^{n-1},X^{n-2})\) contains \[ H_{n-1}(X^{n-2}) \longrightarrow H_{n-1}(X^{n-1}) \xrightarrow{j}H_{n-1}(X^{n-1},X^{n-2}). \] Since \(X^{n-2}\) has dimension at most \(n-2\), \[ H_{n-1}(X^{n-2})=0. \] Thus \(j\) is injective.
2. The natural map\[ H_n(X^n)\to H_n(X^n,X^{n-1}) \]is injective and identifies \(H_n(X^n)\) with \(\ker d_n\).
The long exact sequence of \((X^n,X^{n-1})\) begins in this range as \[ 0=H_n(X^{n-1}) \longrightarrow H_n(X^n) \longrightarrow H_n(X^n,X^{n-1}) \xrightarrow{\partial}H_{n-1}(X^{n-1}). \] Hence \(H_n(X^n)\) injects into the relative group with image \(\ker\partial\). By 1, \(j\) is injective, so \[ \ker d_n=\ker(j\partial)=\ker\partial. \] Therefore \[ \boxed{H_n(X^n)\cong\ker d_n.} \]
3. The group \(H_n(X^n)\) is free abelian.
The relative group \[ H_n(X^n,X^{n-1}) \] is free abelian with basis the \(n\)-cells of \(X\). By 2, \(H_n(X^n)\) is a subgroup of this free abelian group. Every subgroup of a free abelian group is free abelian. Hence \(H_n(X^n)\) is free.