Let $L$ be a non-splittable link in $S^3$. Non-splittable means that there is no smooth embedding $s:S^2\to S^3\setminus L$ which splits $L$, i. e. such that both connected components of $S^3\setminus\operatorname{im}s$ intersect $L$.


The [Hopf link](https://en.wikipedia.org/wiki/Hopf_link) is an example.



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> Proof. Let $l\_1, l\_2:S^1 \to S^3$ be two smooth embeddings with images $L\_1$ and $L\_2$, forming the Hopf link $L=L\_1\sqcup L\_2$, $s:S^2\to S^3\setminus L$ a smooth embedding which splits $L$, $B$ a connected component of $S^3\setminus \operatorname{im}s$. Suppose that $L\_1\subset B$ and $L\_2\cap B=\varnothing$. By Alexander's theorem, $B$ is homeomorphic to the 3-ball, so $l\_1$ is contractible as a map to $B=B\setminus L\_2$, hence contractible as a map to $S^3\setminus L\_2$. This is a contradiction (for example, one can construct a retraction of $S^3\setminus L\_2$ onto $L\_1$, which is homeomorphic to $S^1$, so non-contractible).
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Can there still be a *continuous* embedding $s:S^2\to S^3\setminus L$ which splits $L$? The analogue of Alexander's theorem is false for continuous embeddings, so the above proof doesn't work.



Can a non-splittable link be split by a wild sphere?
Let $L\_1 \cup L\_2$ be a nonsplittable link in $S^3$, and let $\phi : S^2 \to S^3 \setminus (L\_1 \cup L\_2)$ be an embedding of a $2$-sphere. We want to show that the $S^2$ does not separate $L\_1$ from $L\_2$.


Choose a points $p\_1$ and $p\_2$ on $L\_1$ and $L\_2$, so $S^3 \setminus \{ p\_1, p\_2 \} \cong S^2 \times \mathbb{R}$ and $H\_2(S^3 \setminus \{ p\_1, p\_2 \} ) \cong \mathbb{Z}$. The sphere $\phi(S^2)$ will separate $L\_1$ from $L\_2$ if and only if $\phi\_{\ast}[S^2]$ is nontrivial in $H\_2(S^3 \setminus \{ p\_1, p\_2 \} )$.


The OP points me to a [paper of Papakyriakopoulos](https://www.maths.ed.ac.uk/%7Ev1ranick/papers/papa.pdf) which proves that $\pi\_2(S^3 \setminus (L\_1 \cup L\_2))$ is trivial. Therefore, there is a homotopy between $\phi$ and a constant map $S^2 \to S^3 \setminus (L\_1 \cup L\_2)$, with the homotopy staying within $S^3 \setminus (L\_1 \cup L\_2)$. But then, in particular, the homotopy stays within $S^3 \setminus \{ p\_1, p\_2 \}$, so the two homotopic maps induce that same map $H\_2(S^2) \to H\_2(S^3 \setminus \{ p\_1, p\_2 \} )$. So $\phi\_{\ast}$ is zero on $H\_2$, and thus $\phi(S^2)$ does not separate $p\_1$ from $p\_2$.


