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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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Rudy the Reindeer
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  • Looks good to me. – Aaron Mazel-Gee May 05 '11 at 08:39
  • @Aaron: thank you! Would you make that comment into an answer? Then I can accept it and this question is answered. – Rudy the Reindeer May 05 '11 at 09:01
  • 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. – Ryan Budney Aug 31 '11 at 16:00
  • I think somewhere you've mixed-up the domain of $i_*$ and its image in $\pi_1(X,x_0)$. – Ryan Budney Aug 31 '11 at 16:05
  • I thought $\pi_1(A, x_0) = 0$ because $A$ is contractible. So there is a contradiction: it's homeomorphic to a circle and homotopy equivalent to a point. Both yielding the same fundamental groups? – Rudy the Reindeer Sep 01 '11 at 05:46
  • OK, that is only after embedding $A$ in $X$ via $i$! – Rudy the Reindeer Sep 01 '11 at 06:48
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    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". – Ryan Budney Sep 02 '11 at 16:27
  • @Ryan: Yes thanks, I think that's a good suggestion that I'll stick to. – Rudy the Reindeer Sep 03 '11 at 09:43

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