Can a non-splittable link be split by a wild sphere? 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 is an example.

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).

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.
 A: 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 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 $p_2$.
