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I asked this question here. Can someone tell me if this is right:

claim: There are no retractions $r:X \rightarrow A$

proof: (by contradiction)

(i) If $f:X \rightarrow Y$ is a homotopy equivalence then the induced homomorphism $f_* : \pi_1(X, x_0) \rightarrow \pi_1(Y,f(x_0))$ is an isomorphism.

(ii) If $X$ deformation retracts onto $A \subset X$ then $r$, the retraction from $X$ to $A$, is a homotopy equivalence.

Assume there was a retraction. Then by proposition 1.17. (Hatcher p. 36) the homomorphism induced by the inclusion $i_* : \pi_1(A, x_0) \rightarrow \pi_1(X,x_0)$ would be injective.

But $A$ deformation retracts to a point in $X$ so by (i) $i_*(\pi_1(A, x_0))$ is isomorphic to $\{ e \}$, the trivial group. Therefore $i_*$ cannot be injective. Contradiction. There are no retractions $r: X \rightarrow A$.

Many thanks!!

Edit I've just read this again and I think it's wrong, $i_\ast: 0 \rightarrow \mathbb{Z}$ is actually injective! What am I missing? Thanks for your help!

Edit 2 If $r$ is a retraction and $i$ the inclusion then $(r \circ i)_\ast = id$ so $(r \circ i)_\ast$ is an isomorphism which is a contradiction to $(r \circ i)_\ast = 0$?

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    @Ryan: Yes thanks, I think that's a good suggestion that I'll stick to.2011-09-03

1 Answers 1

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Looks good to me!!!!!!!!!!!!!!!!!!!!

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    (my original answer was short of the min. 30 characters)2011-05-05