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Let $i:T^{2}\rightarrow S^{1}\times S^{2}$ be an embedding map. If $i(T^{2})$ is inseparable in $S^{1}\times S^{2}$, then $S^{1}\times S^{2}-i(T^{2})\cong (O\cup K)^{c}$. Here $(O\cup K)^{c}$ is a two components link complement in $S^{3}$ such that $O$ is a trivial knot and $K$ can be any unlinked with $O$.

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up vote 5 down vote accepted

I think you mean "non-separating" rather than "inseparable" (or at least I'm not familiar with that terminology), which means as you say that when you cut along the submanifold, you get a connected space. This follows because $i(T^2)$ must be compressible, by the loop theorem since it cannot be $\pi_1$-injective (I'm assuming your map $i$ is smooth or PL). When you perform a compression of $i(T^2)$, you get a non-separating 2-sphere, which much be isotopic to $\{x\}\times S^2$ for some $x\in S^1$. Reversing this compression, you obtain $i(T^2)$ from $S^2$ by adding on a 2-handle to one side of $S^2$. The boundary components of $S^1\times S^2-i(T^2)$ are separated by a 2-sphere which is a push-off of $S^2$ to the side to which the handle is not attached. The complement of $S^2$ is $S^2\times [0,1]$. When you surger $S^2$ to get $T^2$, the complement gets modified by attaching a 1-handle to $S^2\times 0$ (giving a compressible boundary) and removing a 1-handle whose ends are on $S^2\times 1$. When you bore out this 1-handle, you get a knot complement in $S^3$, and thus the torus complement is a split link, with one component unknotted.

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thank you for your response. – Gerson031 Jan 1 '12 at 14:39

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