# Why are those two manifolds with boundary diffeomorphic to $D^2\times S^1$?

I have this problem and I don't know why I can't finish it:

Let $S=\{x\in \mathbb{R}^4\mid \vert\vert x \vert\vert=2\}$ the sphere of dimension $3$ and radius $2$. Let $T_+$ (resp. $T_-$) the set of the $x=(x_1,x_2,x_3,x_4)\in S$ such that $x_1^2+x_2^2\geq 2$ (resp. $x_1^2+x_2^2\leq 2$). Prove that $T_+$ and $T_-$ are manifolds with boundary, both diffeomorphic to $D^2\times S^1$ and that \begin{align*} S=T_+\cup T_-&& \text{and} && T_+\cap T_-=\partial T_-= \partial T_+ \end{align*}

I used the function $f\colon S\longrightarrow \mathbb{R}$ sending $f((x_1,x_2,x_3,x_4))=x_1^2+x_2^2-2$, and this has $df_x=(2x_1\ \ 2x_2\ \ 0\ \ 0)$, so it is only zero if $x_1=x_2=0$, and this value doesn't affect $0$ to be a regular value. With this function I have that $T_+$ and $T_-$ are manifolds and that their boundaries are the points where $x_1^2+x_2^2=2$ (this forces $x_3^2+x_4^2=2$, to keep the point in $S$), but I can't see why it is diffeomorphic to $D^2\times S^1$. Did I do something wrong?

Suggestion: Show that stereographic projection from $(0, 0, 0, 2)$ maps the rest of $S^{3}$ diffeomorphically to $\mathbf{R}^{3} \times \{0\}$. The common boundary, $$\{(x_{1}, x_{2}, x_{3}, x_{4}) \in S^{3} : x_{1}^{2} + x_{2}^{2} = 2 = x_{3}^{2} + x_{4}^{2}\},$$ maps to a torus; check that $T_{-}$ maps to the interior, a solid torus. (It's clear $T_{-}$ and $T_{+}$ are diffeomorphic by permuting coordinates.)
• In fact what I see is that the boundary of $T_+$ ($T_-$) is $S^1\times S^1$, not $D^2\times S^1$ – MonsieurGalois Dec 2 '16 at 11:43
• The boundary is $S^{1} \times S^{1}$; the solid tori $T_{\pm}$ are the components of the complement. :) – Andrew D. Hwang Dec 2 '16 at 11:49
• OOOOOOOOOOOOOOOOOOOOOOOOOOH NOW I UNDERSTOOD. I though I needed to prove that $\partial T_{\pm}$ was diffeomorphic to $D^2\times S^1$. – MonsieurGalois Dec 2 '16 at 11:55