Hi I don't have that many resources to learn this module and my exam is this Saturday. I'm having trouble proving the following:
(a) Let $L$ be an oriented 2-component link of components $L_{1}, L_{2}$. Prove that if $L$ is weakly split then $lk(L_{1},L_{2})=0$, where $lk$ is the linking number. An oriented link $L$ is weakly split if it admits a disconnected spanning surface without closed components.
(b) Let $F,G\subset\mathbb{R^3}$ be disjoint compact oriented surfaces. Prove that $lk(\partial F,\partial G)=0$.
If anyone could go through them I'd be so grateful.
(a) The linking number can be computed in many ways, and a useful one here is that $\operatorname{lk}(L_1,L_2)=\Sigma_1\cdot L_2$, where $\Sigma_1$ is a Seifert surface (a connected oriented spanning surface) for $L_1$, and where $\Sigma_1\cdot L_2$ is the algebraic intersection number of the two manifolds. Assuming weakly split was supposed to include "oriented," then we have $\Sigma_1$ that is disjoint from $L_2$, so the linking number is $0$.
Maybe that is a little too magical of a proof for the definition of linking number you have on hand. Something else you can use is that linking numbers are invariant under performing homological operations to each link, holding the other constant. That is, while holding one link constant, you can let the other link do pass-through moves and saddle moves with itself (in the complement of the first link). This is illustrated by the following equations, where the usual "everything outside the portrayed piece of the link is the same in both cases" convention is being used:
Every connected surface with nonempty boundary has a handle decomposition of a particularly simple kind: it can be decomposed as a disk with strips attached to its boundary. For example, a disk with two strips as a decomposition of a punctured torus:
If $\Sigma_1$ is a Seifert surface for $L_1$, then we can give $\Sigma_1$ such a decomposition. Since $L_2$ is disjoint from $\Sigma_1$, there is no obstruction to our using the saddle moves. Notice that along a strip, the two boundaries have opposite orientations, so the saddle move does apply. Thus, we can "cut" all the strips to modify $L_1$ without changing the linking number. What is left is that $L_1$ is an unknot (since it bounds a disk), and then we can move $L_1$ away from $L_2$, making it obvious that their linking number is $0$.
(b) We only need that $F$ is disjoint from $\partial G$. Assume $F$ has no closed components by throwing away any such components. Now, for each connected component of $F$ we can do the same handle decomposition argument, reducing the problem to the case of $F$ being a collection of disks disjoint from $\partial G$.
(Intuitively speaking, $F$ obstructs $\partial G$ from "linking through" $\partial F$.)