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

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.

(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?