Dehn surgery

Let $M$ be a smooth 3-manifold, and $K\subset M$ a smooth knot. Since $K$ is an embedded submanifold, by the tubular neighborhood theorem there is a closed neighborhood $U$ of $K$ diffeomorphic to the solid torus $D^2\times S^1$. We let $U^{\prime }$ denote the interior of $U$. Now, let $\phi :\partial U\to \partial U$ be an automorphism of the torus, and consider the manifold $M^{\prime }=M\backslash \backslash U^{\prime }{\coprod }_{\phi }U$, which is the disjoint union of $M\backslash \backslash U^{\prime }$ and $U$, with points in the boundary of $U$ identified with their images in the boundary of $M\backslash \backslash U^{\prime }$ under $\phi$.

It’s a bit hard to visualize how this actually results in a different manifold, but it generally does. For example, if $M=S^3$, the 3-sphere, $K$ is the trivial knot, and $\phi$ is the automorphism exchangingmeridians and parallels (i.e., since $U\cong D^2\times S^1$, get an isomorphism $\partial U\cong S^1\times S^1$, and $\phi$ is the map interchanging to the two copies of $S^1$), then one can check that $M^{\prime }\cong S^1\times S^2$ ($S^3\backslash \backslash U$ is also a solid torus, and after our automorphism, we glue the two solid tori, meridians to meridians, parallels to parallels, so the two copies of $D^2$ paste along the edges to make $S^2$).

Every compact 3-manifold can obtained from the $S^3$ by surgery around finitely many knots.