How is this circle in $S^1 \times D^2$ null-homotopic?

Question

This picture is from Hatcher's AT:

I have been told that this circle is null-homotopic, but I can't see why. I know $S^1 \times D^2$ is a solid torus, but $A$ is linked with itself. How are we to unwind $A$?

Answer 1

$A$ is allowed to cross itself in the homotopy.

For example, "homotopic to a point" means that at the end, all of $A$ is at a single point (which is obviously not injective).

If you believe that $\pi_1(\mathbf R^3) = \{e\}$ then you believe that every knot in $\mathbf R^3$ is null-homotopic. This is the same thing.

This is also why knots are not defined up to "homotopy" but rather a different relation called "ambient isotopy."