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 $A$?
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$A$ 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."
Trevor Gunn
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Why/How is $A$ allowed to cross itself during the homotopy? – user5826 Feb 12 '19 at 04:11
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1@AlJebr There's no rule that says that it can't. You need to allow paths to cross if you want to consider the constant map as a path. – Trevor Gunn Feb 12 '19 at 04:13
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@AlJebr, you may be confusing "homotopy" with "isotopy." – Cheerful Parsnip Feb 12 '19 at 05:15
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@TrevorGunn Is a space crossing itself during a homotopy a continuous function? – user5826 Mar 08 '19 at 05:49