Map of $\mathbb{R}^3-Knot \to S^1$

Question

Reading Bachman's "A Geometric Approach to Differential Forms", in section 7.8.1 about the Lining Number invariant, I have stumbled upon the following assertion.

Let the knot $K$ be defined as a (continuous and differentiable) map from unit circle $S^1$ to 3d Eucledian space $\mathbb{R}^3$, so:

$K: S^1 \to \mathbb{R}^3$

i.e. $K$ is a closed curve in 3d space parametrized by a single real variable.

Let $U=\mathbb{R}^3-K$, i.e. it is the 3d space without the points associated with the knot (curve).

Then there always exists a map: $A:U\to S^1$.

Unfortunatelly author then says that the proof is outside the scope of the book, and I am not sufficiently knowledgeable to know where to find the proof. Can anyone please suggest a good reference for a keen amateur?

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Paragraph from the book:

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The linking number of a two-component link is precisely the analogous measure that you get when you treat one of the components as the z-axis and the other as the 1-chain over which we integrate. Given any knot $K$ in $\mathbb{R}^3$, there is a function $A: \mathbb{R}^3 − K \to S^1$. (The existence of such a function is, unfortunately, beyond the scope of this book.) If we think of A as a 0-form, then we can differentiate it to get a 1-form on $\mathbb{R}^3 − K$. This 1-form is precisely what we can integrate over a second knot to measure how many times it “links” with $K$. Interestingly, there is also a point $p \in S^1$ such that $A^{-1}(p)$ is a surface whose boundary is $K$. Just as before, the linking number can also be computed just by appropriately counting the number of intersections with this surface.

"""

Answer 1

The standard reason is homological (Hatcher's book is the usual reference). Alexander duality relates the homology of $K$ to the cohomology of $\mathbb{R}^3-K$. In particular, $H^1(\mathbb{R}^3-K;\mathbb{Z})$ is isomorphic to $H_1(K;\mathbb{Z})$ is isomorphic to $\mathbb{Z}$. First cohomology with $\mathbb{Z}$ coefficients is the same as homotopy classes of maps $\mathbb{R}^3-K \to S^1$. If you take the generator $1\in\mathbb{Z}$ and run through the isomorphisms, you get a map $A:\mathbb{R}^3-K\to S^1$, which we can assume is smooth. Supposing $S^1$ is parameterized by $\theta$, then $d\theta$ is its $1$-form. The map $A^*$ is the pullback for $1$-forms, giving a closed $1$-form $A^*d\theta$ on $\mathbb{R}^3-K$. (I'm a little uncomfortable calling $dA$ a $1$-form, but it seems $dA=A^*d\theta$.)

Given an oriented curve $C\subset\mathbb{R}^3-K$, the linking number of $C$ with $K$ is $$\int_C A^*d\theta.$$

Unfortunately, I don't know of an elementary way to see such an $A$ must exist! (Perhaps you could integrate the form in the Gauss linking integral that N. Owad mentions in the comments. The idea would be you fix one knot $K$ permanently, then vary the second knot. Fix a point $x_0\in \mathbb{R}^3-K$ and define $A(x)$ to be the integral along any arc from $x_0$ to $x$ in $\mathbb{R}^3-K$. I haven't checked, but the $2$-form ought to be closed, so $A(x)$ is well-defined modulo $1$, hence $A$ can be thought of as a map to a circle.)

A possibly less-mysterious way to get an $A$ is through a Seifert surface. There is a triangulation of $\mathbb{R}^3-K$ where the Seifert surface is a subcomplex, and then you can define a function to $S^1$ by choosing where all the edges in the triangulation go---paths from one side of the Seifert surface to the other should go around $S^1$ exactly once---and then there is a way to fill in the triangles and tetrahedra. After this, you have to find a smooth approximation, which is easy in theory.