The short exact sequences \(0 \to C_n(A) \to C_n(X) \to C_n(X, A) \to 0\) always split, but why does this not always yield splittings \(H_n(X) \approx H_n(A) \oplus H_n(X, A)\)?
Solution
The issue is that a splitting of the short exact sequence \[ 0\to C_n(A)\to C_n(X)\to C_n(X,A)\to0 \] for each individual degree \(n\) need not be compatible with the boundary operators.
1. Degreewise splittings need not assemble to a splitting of chain complexes.
A group-theoretic section \[ s_n:C_n(X,A)\to C_n(X) \] only satisfies that the quotient map composed with \(s_n\) is the identity. There is no reason for \[ \partial s_n=s_{n-1}\partial. \] Without this chain-map condition, \(s_n\) need not take relative cycles to absolute cycles, nor relative boundaries to absolute boundaries. Hence it does not in general induce a map on homology.
2. The connecting homomorphism in the long exact sequence measures this failure.
The short exact sequence of chain complexes yields \[ \cdots\to H_n(A)\to H_n(X)\to H_n(X,A) \xrightarrow{\partial}H_{n-1}(A)\to\cdots. \] If the chain sequence split as chain complexes, this connecting map would be zero and the homology sequence would split. In general the connecting map can be nonzero, precisely because a degreewise section fails to commute with boundaries.
3. For example, take\[ (X,A)=(D^n,S^{n-1}), \qquad n\ge2. \]Then no decomposition\[ H_n(X)\cong H_n(A)\oplus H_n(X,A) \]is possible.
Here \[ H_n(D^n)=0, \qquad H_n(S^{n-1})=0, \qquad H_n(D^n,S^{n-1})\cong\mathbb Z. \] Thus the proposed splitting would assert \[ 0\cong0\oplus\mathbb Z, \] a contradiction. Indeed the connecting map \[ H_n(D^n,S^{n-1})\xrightarrow{\cong}H_{n-1}(S^{n-1}) \] is an isomorphism.
Therefore degreewise splittings of singular chain groups do not imply splittings on homology.