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Let $A$ be a retract of $X$. Show that if $i:A\to X$ is the inclusion operator , then $i_*:H_n(A) \to H_n(X)$ is an isomorphism

(It's easy to see that it's a monomorphism. But why is it an epimorphism?)

Thanks!

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it's not. Every $X$ retracts to a point. You might mean deformation retract. –  user8268 Feb 16 '12 at 11:58

2 Answers 2

Let $i:A \to X$ be the inclusion and $r:X \to A$ the retraction. By definition $r \circ i = \text{id}_A$. Then use the fact homology is a functor.

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Yes, but this implies i is injective, but why is it surjective? –  joshua Feb 16 '12 at 15:01
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As someone anonymous pointed out, the statement is not true. –  Dylan Wilson Feb 16 '12 at 17:12
    
can you give me a counterexample? –  joshua Feb 16 '12 at 17:44
    
@Jack user8268's comment provided a universal counterexample. Here is another: Take $X$ to be the wedge of two $S^1$s and let $A$ be one of the two. Then $H_1(X)=\mathbb{Z}^2$ and $H_1(A)=\mathbb{Z}$. However, $A$ is a retract of $X$ by the map which collapses the second $S^1$ of $X$. –  Adam Feb 16 '12 at 18:02

As others have mentioned, this statement is not true without a stronger condition on $r$: that it is a deformation retraction. That is, you also want $i\circ r$ to be homotopic to the identity on $X$. Then $(i\circ r)_*=(\text{id}_X)_*$ is the identity so $i_*$ is surjective. Combine this with the injectivity of $i_*$ to get that it is an isomorphism.

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Thanks! !!!!!!! –  joshua Feb 17 '12 at 8:11

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