Why is this entangled circle not a retract of the solid torus? I'm doing exercise 16 on page 39 in Hatcher:


*Show that there are no retractions $r: X \rightarrow A$ in the following cases:
(a) $X = \mathbb{R}^3$ with $A$ any
subspace homeomorphic to $S^1$
(b) $X = S^1 \times D^2$ with $A$ its
boundary torus $S^1 \times S^1$
(c) $X = S^1 \times D^2$ and $A$ the circle shown in the figure.

I've done (a) and (b) using proposition 1.17. i.e. I assumed there was a retraction, then the map between the fundamental groups has got to be injective, therefore contradiction.
Now I'm stuck with (c) because according to my understanding the circle on the picture has the same fundamental group as $S^1$ which means it also has the same fundamental group as the solid torus.
What is a different way of proving that there is no retraction (not using prop. 1.17.)? Many thanks for your help!
 A: Proposition 1.17 shows that there is no retraction $r : X \to A$.
Unlike the previous examples, as abstract groups you can have an injective map $\pi_1(A) \to \pi_1(A)$, because they are both isomorphic to $\mathbb{Z}$. So you are not done yet.
If $r : X \to A$ is a retraction, then $r_* : \pi_1(A) \to \pi_1(X)$ sends the loop in $A$ on a path in $X$ that is null-homotopic, so $r_*$ is not injective.
Inside $X$, you can attach a disk around $A$ (the disk crosses itself but it's not important). Taking the image of this disk by a retraction would give you a disk around $A$ inside $A$, which is not possible.
A: (i) If $f:X \rightarrow Y$ is a homotopy equivalence then the induced homomorphism $f_* : \pi_1(X, x_0) \rightarrow \pi_1(Y,f(x_0))$ is an isomorphism.
(ii) If $X$ deformation retracts onto $A \subset X$ then $r$, the retraction from $X$ to $A$, is a homotopy equivalence.
claim: There are no retractions $r:X \rightarrow A$
proof: (by contradiction)
Assume there was a retraction. Then by proposition 1.17. (Hatcher p. 36) the homomorphism induced by the inclusion $i_* : \pi_1(A, x_0) \rightarrow \pi_1(X,x_0)$ would be injective. 
But $A$ deformation retracts to a point in $X$ so by (i) $i_*(\pi_1(A, x_0))$ is isomorphic to $\{ e \}$, the trivial group. Therefore $i_*$ cannot be injective. Contradiction. There are no retractions $r: X \rightarrow A$.
Can someone tell me if I got it right?
