$X = S^1 \times D^2$ and $A$ the circle shown in the figure, Show that there are no retractions $r \colon X \to A$.
Assume for contradiction that there is a retraction $r \colon X \to A$, then that means $$i_*\colon \pi_1(A, x_0) \to \pi_1 (X, x_0)$$ is injective.
$$\pi_1(S^1 \times D^2) \simeq \pi_1(S^1) \times \pi_1(D^2) \simeq \mathbb{Z},$$ and $$\pi_1(A) \simeq ?$$
So I know that $A$ is a non-trivial knot on the torus, which can't be homotopic to $S^1$ as the space $X$. But I don't know any further. May I request for some suggestions to proceed?