2
$\begingroup$

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

  • 1
    1) Please show us your work and where you get stuck. 2) that should be $S^1\times D^2$ the *solid* torus.2012-12-29
  • 1
    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.2012-12-29

2 Answers 2

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.

  • 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!2012-12-29
  • 1
    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.2012-12-29
2

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.
  • 0
    Excellent, thank you for pointing this out.2012-12-29
  • 0
    ...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*.2012-12-29
  • 0
    Chris, you're right. I'll take down the answer. I've clicked the delete button.2012-12-29