Embedding of $T^{2}$ on $S^{1}\times S^{2}$.

Question

Let $i:T^{2}\rightarrow S^{1}\times S^{2}$ be a embedding map and $i_{*}:\pi(T^{2})\rightarrow \pi(S^{1}\times S^{2})$ be the homomorphism induced by $i$. If $i(T^{2})$ is separable in $S^{1}\times S^{2}$ and $Im i_{*}=0$, then $$S^{1}\times S^{2}-i(T^{2})\cong K^{c}\cup (S^{1}\times D^{2})\sharp (S^{1}\times D^{2})$$ or

$$S^{1}\times S^{2}-i(T^{2})\cong (S^{1}\times D^{2})\cup(S^{1}\times D^{2})\sharp K^{c}$$

Here $S^{1}\times D^{2}$ is an open solid torus and $K^{c}$ compliment knot on $S^{3}$.

Answer 1

I remember this chestnut. Taken together, your two equations say that the torus $i(T)$ is contained in a 3-ball $B$ embedded in $S^2 \times S^1$: the first equation says that $i(T)$ is obtained up to isotopy by drilling a knotted hole through $B$ and taking the resulting boundary; the second equation says that $i(T)$ is the boundary of a thickened knot in $B$ (at least, I think that's what your notation means...). However, these are not the only possibilities. There are separating embeddings of $T$ in $S^2 \times S^1$ which are trivial on $\pi_1$ but are not contained in a ball: take the boundary of Whitehead's link in $S^1 \times D^2$ (described in Rolfsen) and then embed $S^1 \times D^2 \to S^1 \times S^2$ in the standard way.