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 $(r \circ i)_\ast = 0$?

  • $\begingroup$ Looks good to me. $\endgroup$ – Aaron Mazel-Gee May 5 '11 at 8:39
  • $\begingroup$ @Aaron: thank you! Would you make that comment into an answer? Then I can accept it and this question is answered. $\endgroup$ – Rudy the Reindeer May 5 '11 at 9:01
  • $\begingroup$ In your edit, you make the claim that $i_* : 0 \to \mathbb Z$ is injective. But the domain of $i_*$ is $\pi_1(A,x_0)$, which is the integers, because $A$ is (homeomorphic to) a circle. $\endgroup$ – Ryan Budney Aug 31 '11 at 16:00
  • $\begingroup$ I think somewhere you've mixed-up the domain of $i_*$ and its image in $\pi_1(X,x_0)$. $\endgroup$ – Ryan Budney Aug 31 '11 at 16:05
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    $\begingroup$ One of the most fundamental theorems about the circle is that its fundamental group is infinite cyclic, so it can't be contractible. IMO you're using misleading terminology. That you can contract $A$ to a point in $X$ is not "contractible". Being "contractible" means you can contract $A$ to a point in $A$. You'll get confused less often if you say "$A$ is null-homotopic in $X$" rather than "contractible". $\endgroup$ – Ryan Budney Sep 2 '11 at 16:27

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

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    $\begingroup$ (my original answer was short of the min. 30 characters) $\endgroup$ – Aaron Mazel-Gee May 5 '11 at 9:28

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