The complement of a torus is a torus.

Take $S^3$ to be the three-sphere, that is, $S^3=\lbrace (x_1,x_2,x_3,x_4):x_1^2+x_2^2+x_3^2+x_4^4=1\rbrace$. Using the stereographic projection, $S^3=\mathbb{R}^3\cup \lbrace \infty \rbrace.$ Can someone explain how the complement of the solid torus (centered at the origin) $S^1\times D^2$, where $D^2$ is a 2-disk, is also a torus? I am reading Milnor’s paper “On Manifolds Homeomorphic to the 7-Sphere,” and this is a prerequisite to understand how Milnor glues the surfaces of two tori of the form $S^3 \times D^4$ in $S^6$ to create an exotic $7$-sphere.

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One solid torus is $x_1^2 + x_2^2 \leq \frac{1}{2}.$ The other solid torus is $x_3^2 + x_4^2 \leq \frac{1}{2}.$ The intersection, a torus, is…

The boundary of $x_1^2+x_2^2 < \frac{1}{2}$ (in the $x_1x_2$-plane) is a circle, with equation $x_1^2+x_2^2 = \frac{1}{2}$. Likewise for the boundary of $x_3^2+x_4^2 < \frac{1}{2}$. The boundary of either region is thus the cartesian product of two circles, a torus.