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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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I assume in the first line $H_n(x)$ should be $H_n(X)$. If you can show that the short exact sequence splits, then you get the result; presumably, you'd want to use the retraction $r\colon X\to A$ to get a retraction $\overline{r}\colon H_n(X)\to H_n(A)$... – Arturo Magidin Mar 7 '11 at 4:02
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. – Ryan Budney Mar 7 '11 at 4:03
up vote 2 down vote accepted

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 lemma – Juan S Mar 7 '11 at 5:17

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