Complement of the Solid Torus in $S^3$ is Again a Solid Torus On pg. 48 of Hatcher's Algebraic Topology, the author writes that the $3$-sphere $S^3$ can be thought of as the union of two solid torus.
First a formal reasoning is given which is

$S^3=\partial D^4 = \partial(D^2\times D^2) = (\partial D^2\times D^2)\cup (D^2\times \partial D^2)$

This is clear. But then the author gives a geometric interpretation which is as follows.
Think of $S^3$ as the one point compactification of $\mathbf R^3$. Let $T=S^1\times D^2$ be the first solid torus. The second solid torus is the closure of the complement of $T$ in $\mathbf R^3$ along with the compatification point at infinity.
The statement in italics is not clear to me. How do we see this as a solid torus?

Here's another way that I like to view this: The solid torus $S^1 \times D^2$ is a donut, and you can fill in the donut hole with a plug ("munchkin" if you're a Dunkin' Donuts fan) homeomorphic to $D^2 \times [-1,1]$. The union of these is isotopic to the standard $D^3 \subset \mathbb{R}^3 \subset S^3$, so the complement (of its interior, technically) is also homeomorphic to $D^3$. The plug $D^2 \times [-1,1]$ intersects this outer $D^3$ along $D^2 \times \{\pm 1\}$. Working in reverse, we see that the complement of the original solid torus is obtained by attaching $D^2 \times [-1,1]$ to the outer $D^3$, together forming a complementary solid torus. 
