Show that there are not retractions $r:X \to A$ when $X=S^1 \times D^2$ and $A$ is the circle shown in the following picture. 
Show that there are not retractions $r:X \to A$ when $X=S^1 \times D^2$ and $A$ is the circle shown in the following picture 

I'm reading a solution for this which I don't really understand. The parts I'm not figuring out are first why is $A \approx S^1$? The second part is that if I accept the fact that $A \approx S^1$, then $\pi_1(S^1 \times D^2)=\pi_1(A)=\Bbb Z$ and so if we have a retraction the map $\iota_*:\pi_1(A) \to \pi_1(S^1 \times D^2)$ is an injection by Hatcher's proposition $1.17$. Now it's stated that

However, the map $\iota_*:\pi_1(A) \to \pi_1(S^1 \times D^2)$ is actually trivial, not injective. This is because $\iota_*$ maps a loop that goes around $A$ once to a loop that goes around the $S^1$ part of $X$ once and then goes around the $S^1$ part of $X$ in the opposite direction, which is homotopic to not looping around anything. So the generator of $\pi_1(A)$ gets sent to $0$, and $\iota_*$ can’t possibly be injective.

What is meant by a trivial map here? Evertyhing gets send to $0$? And where is this conclusion drawn that

$\iota_*$ maps a loop that goes around $A$ once to a loop that goes around the $S^1$ part of $X$ once and then goes around the $S^1$ part of $X$ in the opposite direction.

 A: There are a few ways to show this, first lies in the fact that your "circle" $A$ can be contracted by pulling the two ends of $A$ through each other and so every loop of $\pi_1(A)$ under $\iota_*$ gets mapped to a loop homotopic to the trivial loop. Another one in the fact that if you take a generating loop of $\pi_1(A)$ and map it under $\iota_*$ it will first go around the $S^1$ factor of $X$ and then it'll go around the same factor, but in the opposite direction i.e. it's homotopic to the loop that "doesn't do anything".

Suppose that there exists a retraction $r :X \to A$, then the induced map $\iota_*:\pi_1(A) \to \pi_1(X)$ is injective. As $S^1 \times D^2$ is path-connected we have that $\pi_1(S^1 \times D^2)=\pi_1(S^1) \times \pi_1(D^2)$ and since $D^2$ is contractible you'll end up with $\pi_1(X) \cong \Bbb Z$. Also as $A \approx S^1$ you'll have that $\pi_1(A) \cong \Bbb Z$.
Now as assumed the map $\iota_*$ should be injective, but since it maps the homotopy class of any generating loop in $\pi_1(A)$ to a homotopy class of loop in $\pi_1(X)$ that is homotopic to a trivial loop the map $\iota_*$ is actually the trivial map thusfore not injective implying that no retraction $r:X \to A$ exists.

"And where is this conclusion drawn that"

$\iota_*$ maps a loop that goes around $A$ once to a loop that goes around the $S^1$ part of $X$ once and then goes around the $S^1$ part of $X$ in the opposite direction.

I don't have the reputation to post images here, but here is a link to a sketch which should give you the intuition. Imagine your loops starts at the red point on $A$ and traverses up to the blue point. After the blue point the loop will start to traverse in the reverse direction and "undoes" what it previously did.
