Dehn surgery for non-compact manifolds

Question

Lickorish theorem states that every closes, orientable, compact 3-manifold can be obtained by surgery on $S^3$. What do we know about surgery for non-compact manifolds? I.e. can we obtain $\mathbb{R}^2 \times S^1$ from surgery on $S^3$, $\mathbb{R}^3$, or at least from surgery on some non-compact manifold?

Answer 1

I'll turn my comments into an answer, since your comment indicates that this is what you want to know.

Surgery preserves compactness, so no, there is no surgery on $S^3$ that produces $\mathbb R^2 \times S^1$.

To prove this, start with a compact manifold $M$ and an embedded solid torus $T \subset M$. Step 1 of surgery is to remove the interior of $T$, so $M - \text{int}(T)$ is a closed subset of $M$, hence compact. Step 2 of surgery is to glue in a different solid torus $T'$, using some homeomorphism $$f : \partial T \to \partial(M - \text{int}(T) = \partial T $$ The resulting surgered manifold is thus the quotient of the disjoint union of $M - \text{int}(T)$ and $T'$ by the relation $x \sim f(x)$, $x \in \partial T$. Since the two pieces of that disjoint union are compact, their union is also compact, and its quotient is compact.