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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!)

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    Well, don't you have a map $H_n(X) \to H_n(A) \oplus H_n(X,A)$ ? One factor comes from the retract, the other from the long exact sequence.2011-03-07

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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.

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    thanks. I think I've got it now. $r:A \to X$ is surjective, $i:X \to A$ (the inclusion map) is injective. Use these to show that the short exact sequence splits; we have two maps $r_*:H_n(X) \to H_n(X,A)$ and $i_*:H_n(X,A) \to H_n(X)$, and then apply the splitting lemma2011-03-07