Let $C_n(X)$ be the free Abelian group generated by all the singular $n$-simplices of the topological space $X$. I'm reading Bredon's Topology and Geometry and there's a statement which says:
The group $C_n(X,A) = C_n(X)/C_n(A)$ is free Abelian. This gives a splitting $C_*(X,A) \rightarrow C_*(X)$. However, this splitting is not a chain map and so it does not induce a map in homology.
By splitting, does it mean $C_*(X) \cong C_*(A) \oplus C_*(X,A)$? What does it mean for a splitting to be a chain map? (I know the definition of a chain map, but for a splitting?) Also, we do have an exact homology long sequence of the pair $(X,A)$, so what is the induced map in homology in the statement referring to?