# Retraction from torus to linked circle [duplicate]

Possible Duplicate:
Why is this entangled circle not a retract of the solid torus?

I am stuck with exercise 16 (c), pag.39 of Hatcher's Algebraic Topology: prove that there is no retraction from $S^1\times D^2$ onto the set $A$, which is described by an image in the book, and which you can see here below.

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## marked as duplicate by Rudy the Reindeer, Steve D, TMM, Henry T. Horton, Brett FrankelDec 29 '12 at 19:54

1) Please show us your work and where you get stuck. 2) that should be $S^1\times D^2$ the solid torus. – Chris Gerig Dec 29 '12 at 19:03
Uhm, 2) you're obviously right, edited. 1) There's not much work...I supposed such a retraction existed and tried to reason about the induced homomorphism $\pi(X)\rightarrow \pi(A)$, where $X$ is the solid torus, which is an isomorphism. – random Dec 29 '12 at 19:10

Hint: View the subspace "$A$" as a path in the space $A$, and then as a path in the space $S^1\times D^2$. Then see what that means about the desired retraction and inclusion maps.

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oh now I can see this, I just kept on thinking that $A$ as a path in $S^1\times D^2$ is homotopic to the circle, but it's actually nullhomotopic! Thank you! – random Dec 29 '12 at 19:53
Dear deleted answer, I really thought that you added a lot and were much nicer than what is in the other thread. Given the comment above it seems that OP understands the exercise and can solve it. Hence I would see no harm in undeleting. – Rudy the Reindeer Dec 29 '12 at 20:06

This is indeed a duplicate. Just note that the map $S^1 \cong A \to S^1 \times D^2$ induces a map

$\pi_1 S^1 \to \pi_1 (S^1 \times D^2)$

which is the zero map, $0: \mathbb{Z} \to \mathbb{Z}$. This is because $A$ can be shrunk to a point in $S^1 \times D^2$. (You're allowed to homotope $A$ through itself, meaning you can `unlink' it from itself.)

A retraction $S^1 \times D^2 \to A$, however, would induce an isomorphism $\pi_1 A \cong \pi_1 A$ via the composition

$$A \to S^1 \times D^2 \to A$$

which is impossible because the first map induces the zero map on $\pi_1$.

• I would leave this as a comment but I don't have enough reputation. Matt, it is not true that $\pi_1 A \cong \pi_1(\ast)$. You mean to say that the inclusion $A \to S^1 \times D^2$ induces the zero map, but this does not mean that $\pi_1 A$ is trivial.
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Excellent, thank you for pointing this out. – Rudy the Reindeer Dec 29 '12 at 19:44
...wow, thanks for not giving the OP a chance to solve it for himself. The point of this thread is to help the OP; if he wanted to, he could go to that other thread which already contains the answer. – Chris Gerig Dec 29 '12 at 19:48
Chris, you're right. I'll take down the answer. I've clicked the delete button. – user54535 Dec 29 '12 at 19:50