Show that a chain complex of free abelian groups \(C_n\) splits as a direct sum of subcomplexes \(0 \to L_{n+1} \to K_n \to 0\) with at most two nonzero terms. [Show the short exact sequence \(0 \to \ker\partial \to C_n \to \operatorname{Im}\partial \to 0\) splits and take \(K_n = \ker\partial\).]
In case the groups \(C_n\) are finitely generated, show there is a further splitting into summands \(0 \to \mathbb{Z} \to 0\) and \(0 \to \mathbb{Z} \xrightarrow{m} \mathbb{Z} \to 0\). [Reduce the matrix of the boundary map \(L_{n+1} \to K_n\) to echelon form by elementary row and column operations.]
Deduce that if \(X\) is a CW complex with finitely many cells in each dimension, then \(H_n(X; G)\) is the direct sum of the following groups:
a copy of \(G\) for each \(\mathbb{Z}\) summand of \(H_n(X)\)
a copy of \(G/mG\) for each \(\mathbb{Z}_m\) summand of \(H_n(X)\)
a copy of the kernel of \(G \xrightarrow{m} G\) for each \(\mathbb{Z}_m\) summand of \(H_{n-1}(X)\)
Solution
Let \[ \cdots\xrightarrow{\partial_{n+1}}C_n\xrightarrow{\partial_n}C_{n-1}\xrightarrow{}\cdots \] be a chain complex of free abelian groups.
1. Put\[ K_n=\ker\partial_n. \]Then the short exact sequence\[ 0\to K_n\to C_n\xrightarrow{\partial_n}\operatorname{im}\partial_n\to0 \]splits.
The subgroup \(\operatorname{im}\partial_n\) of the free abelian group \(C_{n-1}\) is free abelian. A short exact sequence with free quotient splits, so choose a complement \(L_n\subset C_n\) such that \[ C_n=K_n\oplus L_n. \] The restriction \[ \partial_n|_{L_n}:L_n\xrightarrow{\cong}\operatorname{im}\partial_n \] is an isomorphism.
2. The original chain complex is the direct sum, over \(n\), of two-term subcomplexes\[ 0\to L_{n+1}\xrightarrow{\partial_{n+1}}K_n\to0. \]
At degree \(n\) the decomposition is \[ C_n=K_n\oplus L_n. \] The differential vanishes on \(K_n\) and maps \(L_n\) into \(K_{n-1}\). Thus each \(L_{n+1}\) interacts only with \(K_n\), and the direct sum of these two-term pieces recovers every chain group and every differential.
Assume now that all \(C_n\) are finitely generated.
3. Each map\[ L_{n+1}\to K_n \]can be diagonalized by Smith normal form, so the chain complex further decomposes into summands\[ 0\to\mathbb Z\to0 \]and\[ 0\to\mathbb Z\xrightarrow{m}\mathbb Z\to0 \qquad(m>0). \]
The groups \(L_{n+1}\) and \(K_n\) are finitely generated free abelian. Smith normal form gives bases in which the injective map \(L_{n+1}\to K_n\) is diagonal with nonzero diagonal entries \(m_1,\dots,m_r\). Each diagonal entry yields a two-term summand \[ \mathbb Z\xrightarrow{m_j}\mathbb Z, \] while any unused basis vectors of \(K_n\) yield one-term summands \(\mathbb Z\).
Now let \(X\) be a CW complex with finitely many cells in each dimension and tensor its cellular chain complex with an abelian group \(G\).
4. A one-term summand \(\mathbb Z\) in degree \(n\) contributes one copy of \(G\) to \(H_n(X;G)\).
After tensoring with \(G\), the summand becomes a one-term complex with \(G\) in degree \(n\) and zero differential. Such summands correspond exactly to free \(\mathbb Z\) summands of \(H_n(X)\).
5. A two-term summand\[ 0\to\mathbb Z\xrightarrow{m}\mathbb Z\to0 \]in degrees \(n+1,n\) contributes\[ G/mG \]to \(H_n(X;G)\) and\[ \ker(m:G\to G) \]to \(H_{n+1}(X;G)\).
Tensoring gives \[ 0\to G\xrightarrow{m}G\to0. \] Its homology in the lower degree is the cokernel \(G/mG\), while its homology in the upper degree is the kernel of multiplication by \(m\).
6. Therefore\[ H_n(X;G) \]is the direct sum of:
- one copy of \(G\) for each \(\mathbb Z\) summand of \(H_n(X)\);
- one copy of \(G/mG\) for each \(\mathbb Z_m\) summand of \(H_n(X)\);
- one copy of \(\ker(m:G\to G)\) for each \(\mathbb Z_m\) summand of \(H_{n-1}(X)\).
A torsion summand \(\mathbb Z_m\) in integral homology arises as the cokernel of one of the multiplication-by-\(m\) two-term pieces. Apply <1>4–<1>5 and collect all direct summands contributing in degree \(n\).