Retraction from torus to linked circle 
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.

 A: 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.

A: 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.
