Linking Number $0$ When One Knot is a Boundary

Question

I am given two smooth knots $K$ and $L$ embedded in $\mathbb{R}^3$. I am also given that $K$ is the boundary of an oriented, compact surface disjoint from $L$, call it $S$. I must prove that the linking number of two such knots is $0$ where my definition of the linking number is the degree of the map $$ T: K \times L \longrightarrow S^2 $$ $$ T(x,y) = \frac{x-y}{||x-y||}. $$ My thought is that $K \times L$ is the boundary of $S \times L$. If I can show that $T$ extends to $S \times L$, I'll be done. But, I am uncertain why that should be.

Answer 1

Your intuition was good. Because $S$ is disjoint from $L$, you can simply extend $T$ by the same formula: $$\begin{array}{cccc} \widehat T :& S \times L & \to & S^2 \\ & (x,y) & \mapsto & \dfrac{x-y}{\|x-y\|},\end{array}$$ and your proof works!