For $A \subset X$, $C_n(X) \cong C_{n}(A) \oplus C_n(X,A)$ doesn't imply $H_n(X) \cong H_{n}(A) \oplus H_n(X,A)$ in general. This is exercise 2.2.27 in Hatcher's Algebraic Topology

The short exact sequence $0 \to C_{n}(A) \to C_{n}(X) \to C_{n}(X, A) \to 0$ always splits, but why does this not always yield splittings $H_n(X) \cong H_{n}(A) \oplus H_n(X,A)$? 

My understandings are moved to the answer.
Are there better ideas? Thanks for your time and effort.
 A: There are some incorrect statements in the existing answer. 
1) It is true that $C_*(X,A)$ is isomorphic, as a graded free abelian group, to the space of simplices not contained in $A$. What is not true is that this isomorphism preserves the differential; in fact, the boundary of a simplex not contained in $A$ may have faces in $A$, so this subspace of $C_*(X)$ isn't even a chain complex. 
The boundary operator on $C_*(X,A)$, thinking of this as being free on the simplices not contained in $A$, is the usual boundary operator, except you send any boundary face contained in $A$ to zero. 
2) The reason that $C_*(X,A)$ is a free abelian group on those generators is very general. If you have a free abelian group with a specified basis (for instance, $C_*(X)$), and you have a subgroup spanned by a subset of that basis ($C_*(A)$, spanned by simplices contained in $A$), then the quotient is free on the remaining basis elements. This is just a definition chase.
3) The correct statement is that the sequence $C_*(A) \to C_*(X) \to C_*(X,A)$ does not admit a splitting as chain complexes, that is, there is no section $C_*(X,A) \to C_*(X)$ which preserves the boundary operator. The obvious section described in (1) does not preserve the boundary operator. 
4) It seems claimed in the other answer that for any exact sequence $$0 \to A \to B \to C \to 0$$ with $A,B$ free abelian, this sequence splits. That is false. In fact there is such a short exact sequence for any abelian group $C$ whatsoever. It should be easy to see how to construct one with $C = \Bbb Z/n$. 
