Show that if $A$ is a retract of $X$ then for all $n \ge 0$ $$H_n(X) \simeq H_n(A) \oplus H_n(X,A)$$
So we have a retraction $r:X \to A$, which is surjective.
Consider the long exact sequence
$$\cdots \to H_n(A) \to H_n(X) \to H_n(X,A) \to H_{n-1}(A) \cdots$$
As $r$ is surjective we have that $H_n(X) \to H_n(X,A)$ is surjective, and hence $H_n(A) \to H_n(X)$ is injective. Thus we have a short exact sequence
$$0 \to H_n(A) \to H_n(X) \to H_n(X,A) \to 0$$
I am unsure how to go from the fact this is exact, to the result (assuming the above is correct!)
Best Answer
If you can find a homomorphism $H_n(X,A) \to H_n(X)$ which 'splits' the quotient map $H_n(X) \to H_n(X,A)$ then you can use the splitting lemma to give you your direct sum.