How to see that this circle is contractible on $S^1\times D^2$?

Question

In Hatcher 1.16(c), the circle is shown in the graph:

I found an answer that says that the inclusion of this circle is null-homotopic. But how to see this fact? Could some one explain? Better with a picture how the deformation process.

Answer 1

To see that $A$ is nullhomotopic in $X=S^1\times D^2$ imagine $X$ deformation retracted to a circle $S^1$. The induced map $S^1\cong A\to S^1$ will have winding number zero. To see this, the image of $A$ will pass through most points on $S^1$ twice, in opposite directions.

Good luck with getting a picture though; you won't get one from me!