Retraction from torus to linked circle

Question

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.

enter image description here

Uhm, 2) you're obviously right, edited.

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

score 3 · Answer 1 · answered Dec 29 '12 at 19:37

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.

...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. — , Dec 29 '12 at 19:50

score 2 · Accepted Answer · answered Dec 29 '12 at 19:19

2

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.

answered Dec 29 '12 at 19:19

Chris Gerig

2,001

1

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
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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

Retraction from torus to linked circle

2 Answers2

Linked