Irreducible link complements in $\mathbb S^3$ Let $L$ be an oriented link in the 3-sphere $\mathbb S^3$, consisting of two knot components, $\gamma_1$ and $\gamma_2$. I wonder now if the following is true:
If the linking number $N(L)$ of $L$ is not $0$, then $S^3 \setminus L$ is an irreducible 3 manifold.
This is equivalent to showing:
If there is a smoothly embedded 2-sphere $\Sigma \subset \mathbb S^3$, with $\Sigma \cap L = \emptyset$, such that $\gamma_1$ is contained in a different component of $\mathbb S^3 \setminus \Sigma$ than $\gamma_2$, then $N(L) = 0$.
I had the following idea: In the closed 3-ball $B_1$, whose interior is the component of $\mathbb S^3 \setminus \Sigma$ containing $\gamma_1$, perform an orientation-preserving self-homeomorphism that is the identity on $\Sigma$. This can be naturally extended to a self-homeomorphism on all of $\mathbb S^3$ by the identitiy on $\mathbb S^3 \setminus B_1$. Call the resulting map $h$ and let $L'$ be $h(L)$. 
I want to find $h$ such that there exists a regular projection of $h(L)$ such that $h(\gamma_1)$ and $h(\gamma_2)$ will have disjoint images under that projection. So far this hasn't worked out for me. Can anyone give me a hint ?
Thanks in advance!
 A: It might be easier to show that having a sphere in the complement that separates the two link components allows you to construct disjoint Seifert surfaces. Start with a Seifert surface for one component. It will hit the sphere in a union of circles which can be surgered along inner most disks to make the Seifert surface disjoint from the sphere. Now do the same procedure for the other link component. Once you have disjoint Seifert surfaces, the linking number is trivial, which can be seen by using the fact that the linking number is the algebraic intersection number of the seifert surface of one component with the other component.
A: Here's an argument based on another (standard) definition of linking number: Recall that the first homology of a knot complement is freely generated by a meridian $\mu$ for the knot (i.e. a curve bounding a small disk transverse to $K$). More precisely, there's an isomorphism $H_1(S^3 \setminus K)\cong \mathbb{Z}$ given by $[\mu]\mapsto 1$. 
Definition. If $K'\subset S^3$ is a knot disjoint from $K$, define the linking number of $K'$ and $K$ to be the integer $n$ such that 
$$[K']= n \cdot [\mu]$$
as classes in $H_1(S^3 \setminus K)$.
Now we want to show that if $K$ and $K'$ are separated by an embedded 2-sphere $\Sigma \subset S^3$, then their linking number is zero. As you wrote in your original post, $\Sigma$ separates $S^3$ into a pair of 3-balls. Letting $B$ denote the ball containing $K'$, we see that the inclusion $i:S^3 \setminus K \hookrightarrow (S^3 \setminus K,B)$ induces an isomorphism $H_1(S^3 \setminus K) \cong H_1(S^3 \setminus K,B)$ using the LES associated to the pair $(S^3 \setminus K,B)$. We also note that the homology class of $K'\subset B$ vanishes in $H_1(S^3 \setminus K,B)$. Thus, since $i_*([K'])=0$ and $i_*$ is an isomorphism, we conclude that $[K']=0$ in $H_1(S^3 \setminus K)$ as desired. 

Edit: 
For the first statement, see The first homology group $ H_1(E(K); Z) $ of a knot exterior is an infinite cyclic group which is generated by the class of the meridian.
For the equivalence of the homological and diagrammatic definitions of linking number, consider any regular diagram for $K \cup K'$. Define a new link $K''$ as follows: At each crossing where $K'$ passes over $K$, replace the overstrand with an understrand and add a small meridional loop around $K$. Here's the local picture at a positive crossing:

The local picture for a negative crossing is similar, with the orientation on $K$ reversed. It is easy to check that $lk(K',K)=lk(K'',K)$ using the diagrammatic definition. Note that each meridional loop contributes either +1 or -1 to $lk(K'',K)$ based on whether it came from a positive or negative crossing. It is also straightforward to convince oneself  that $K'$ and $K''$ are homologous in $S^3 \setminus K$. (There's an obvious annulus spanned by pushing $K'$ straight down, and we get the meridional loops by cutting out holes for $K$ to pass through. The resulting surface lies in $S^3 \setminus K$ and has boundary $K' \sqcup -K''$.) Now $K''$ consists of a bunch of meridional loops and a disjoint component which lies entirely underneath $K$. We can modify the diagram by pulling the latter component out from under $K$, so that it clearly does not contribute to the diagrammatic linking number. This means that the diagrammatic linking number of $K''$ and $K$ is a signed count of the meridional loops. Similarly, since the "large" component of $K''$ bounds a Seifert surface disjoint from the rest of the components and $K$, it does not contribute to the homological linking number. This means that this latter linking number is also a signed count of the meridional loops, and it's clear that the two signed counts agree (assuming $\mu$ is a positive meridional loop). Since $K''$ is homologous to $K'$ in $S^3 \setminus K$, we conclude that the two definitions of linking number are the same.
