Torus diffeomorphic to $S^1\times S^1$. This is an exercise from Guillemin/Pollack's Differential Topology. In a previous exercise, I'm asked to give a complete set of parametrizations of $S^1\times S^1$, which I've succeeded in (I think) by the likes of $$\begin{align}\varphi_1(u,v) &= (u,\sqrt{1-u^2},v,\sqrt{1-v^2}) \\ \varphi_2(u,v) &= (u,-\sqrt{1-u^2},v,\sqrt{1-v^2}) \end{align} \\ \vdots\\ \text{etc.}$$ (defined over $[0,1]^2).$
Then, the torus $\Bbb T_{a,b}$ (my notation) is defined as the set of points in $\Bbb R^3$ that are at distance $b < a$ from the circumference in the plane $\{z = 0\}$ of radius $a$ centered at the origin.
The exercise is to prove that $\Bbb T_{a,b}$ is diffeomorphic to $S^1\times S^1$. I'm very stuck because I can't seem to characterize neighborhoods of points in $\Bbb T_{a,b}$, can someone help? Answers or just hints are welcome (but I might ask follow up questions in the latter case :) ).
 A: $\newcommand{\Num}[1]{\mathbf{#1}}\newcommand{\Reals}{\Num{R}}\newcommand{\Integers}{\Num{Z}}$One approach is to define $\Phi:\Reals^{2} \to T_{a, b}$ by
$$
\Phi(u, v) = \bigl((a + b\cos u) \cos v, (a + b\cos u) \sin v, b\sin u\bigr),
$$
and to observe the following:


*

*If $(u, v)$ and $(u', v')$ are arbitrary points of $\Reals^{2}$, then $\Phi(u, v) = \Phi(u', v')$ if and only if $(u' - u, v' - v) \in (2\pi\Integers) \times (2\pi\Integers)$. Particularly, $\Phi$ factors through the quotient map
$$
\Pi:\Reals^{2} \to (\Reals/2\pi\Integers) \times (\Reals/2\pi\Integers) = S^{1} \times S^{1},
$$
defining a bijection $\overline{\Phi}:S^{1} \times S^{1} \to T_{a, b}$.

*The Jacobian $D\Phi$ has rank two at each point. Consequently, $\Phi$ maps an arbitrary square $(u_{0}, u_{0} + 2\pi) \times (v_{0}, v_{0} + 2\pi)$ diffeomorphically to its image. Since $S^{1} \times S^{1}$ is covered by images under $\Pi$ of such squares, $\overline{\Phi}$ is a diffeomorphism (as a bijection represented locally by diffeomorphisms).
A: Here's a solution to the problem which seems valid.
We will denote the set of points at distance $b$ from the circle of radius $a$ as $Tor(a,b)$. If $(x,y) \in S^1$, then $(ax,ay,0)$ lies on the circle of radius $a$ in the $xy$ plane of $\mathbb{R^3}$, and $u:=(x,y,0)$ is also a unit vector perpendicular to the circle at that point. Now, let $(x′,y′) \in S^1$ belong to the "second" circle. Since arc-tangent (as a map of two arguments, not the ratio) is a smooth map we can use $θ:=arctan(x′,y′)$ in our parametrization. Any point at distance $b$ away from the $a-radius$ $circle$ in $xy$ plane can be thought of $(ax,ay,0)$ plus vector $bu$ rotated in the plane perpendicular to the circle at the $(ax,ay,0)$. This discussion constructs the following (smooth) function from $S^1 \times S^1$ to $Tor(a,b)$: 
$$ϕ(x,y,x′,y′)=(x(a+bx′),y(a+bx′),by′)$$
(note that we used the fact that $cos~θ=x′$,$sin~θ=y′$). 

To invert the function, take any point $(A,B,C) \in Tor(a,b)$. Then $(x_1,x_2)=(\frac{A}{\sqrt{A^2+B^2}},\frac{B}{\sqrt{A^2+B^2}})$ is the corresponding location on the first coordinate circle, and $(x_3,x_4)=(\frac{\sqrt{A^2+B^2}−a}{b},\frac{C}{b})$ is the location on the second coordinate circle (which we've computed directly from the definition of ϕ). Thus, let 
$$ϕ^{−1}(A,B,C)=(\frac{A}{\sqrt{A^2+B^2}},\frac{B}{\sqrt{A^2+B^2}},\frac{\sqrt{A^2+B^2}−a}{b},\frac{C}{b})$$
It's easy to check that $ϕ∘ϕ^{−1}=1$, and $ϕ$ is a restriction of a function smooth over all of $R^4$. $ϕ^{−1}$ is smooth provided we enclose any points on the torus in an open ball of small enough radius not to reach the origin, at which the extension of $ϕ^{−1}$ is not defined. Note that if $b<c$ then the torus will contain points with $A^2+B^ 2=0$, and thus $ϕ^{−1}$ will not be well-defined, destroying the diffeomorphism. 
A: Here is a similar approach that seems easier. The map 
$$(u,v) \rightarrow \frac{1}{\sqrt{2}}(\sin(2\pi u), \cos(2\pi u), \sin(2\pi v), \cos(2\pi v))$$
sends $S^1 \times S^1$ into the unit $3$-sphere. This is a parameterized flat torus. To realize it in 3 space project it using stereographic projection. 
