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$X = S^1 \times D^2$ and $A$ the circle shown in the figure, Show that there are no retractions $r \colon X \to A$.

enter image description here

Assume for contradiction that there is a retraction $r \colon X \to A$, then that means $$i_*\colon \pi_1(A, x_0) \to \pi_1 (X, x_0)$$ is injective.

$$\pi_1(S^1 \times D^2) \simeq \pi_1(S^1) \times \pi_1(D^2) \simeq \mathbb{Z},$$ and $$\pi_1(A) \simeq ?$$

So I know that $A$ is a non-trivial knot on the torus, which can't be homotopic to $S^1$ as the space $X$. But I don't know any further. May I request for some suggestions to proceed?

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  • $\begingroup$ $\pi_1(A) \sim \mathbb Z$ also, but it is not a problem of having an injection from a group to another which is impossible because $\mathrm{Hom}(\pi_1(A), \pi_1(X)) = 0$. The problem here is that $r_*$ must be a non-trivial homomorphism that maps $1$ to $0$ (see my answer). $\endgroup$ Sep 21, 2013 at 14:45

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The biggest problem with the existence of a retraction is because if you parametrize $A$ via $\varphi : S^1 \to S^1 \times D^2$ with $\varphi(S^1) = A$, then $\varphi$ is homotopic to a constant map (there is no problem with self-crossing in this context because the loop lives in $S^1 \times D^2$). If we had a retraction $r : S^1 \times D^2 \to A$, what can you say about $r \circ \varphi$?

If you need more details, ask.

Hope that helps,

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  • $\begingroup$ Oh, $r \circ \varphi$ is also a constant map.. $\endgroup$ Sep 16, 2013 at 0:27
  • $\begingroup$ Well, it is homotopic to one. :P The problem is that the retraction has implications on the induced map $r_*$ on the fundamental groups of $S^1 \times D^2$ and $A$. Did you find it? $\endgroup$ Sep 16, 2013 at 11:10
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    $\begingroup$ @Jellyfish : To be fair, I have already given quite some thought about this question here : math.stackexchange.com/questions/477055/… $\endgroup$ Sep 16, 2013 at 19:06

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