Diffeomorphism between $T$ (torus) and the cover $S^1 \times S^1$ - Question of the book '' Differential Topology '' of Guillemin and Pollack Here's the question : ''The'' torus is the set of points in $\mathbb{R^3}$ at distance $b$ from the circle of radius a in the xy plane, where $0<b<a$. Prove that these tori are all diffeomorphic to $S^1 \times S^1$. Also draw the cases $b = a$ and $b>a$; why are these not manifolds?
Definitions :


*

*A smooth map $f:X \rightarrow Y$ if subsets if two Euclidean spaces is a diffeomorphism if it is one to on and onto, and if the inverse map $f^{-1}$ is also smooth. 

*A map $f:X\rightarrow \mathbb{R^m}$ defined on an subset $X$ in $\mathbb{R^n}$ is called smooth if it may be locally extended to a smooth map if around ech point $x \in X$ there is an open set $U\subset \mathbb{R^n}$ adn a smooth map $F: U \rightarrow \mathbb{R^m}$ such that $F=f$ on $U\cap X$.

*A diffeomorphism $\phi:U\rightarrow V$ is called a parametrization of an open set $U \subset \mathbb{R^k}$ into an open set $v \subset \mathbb{R^N}$ (Related Subject : manifold)
We already know that the parametrization of the torus: $$\Phi(\phi,\theta):=((b+acos~\phi)cos~\theta),(b+acos~\phi)sin~\theta,asin~\phi),$$ and there exist enough parametrization to cover $S^1 \times S^1 \subset \mathbb{R^4}$ by a previous exercise of the same book.
Here's what has been achieved so far:
$$\Phi(\phi,\theta)=((b+acos~\phi)cos~\theta),(b+acos~\phi)sin~\theta,asin~\phi)$$ $$ = (bcos~\theta \overrightarrow{i}+bsin~\theta \overrightarrow{j}+0 \overrightarrow{k})+(acos~\phi cos~\theta \overrightarrow{i}+acos~\phi sin~\theta \overrightarrow{j}+asin~\phi \overrightarrow{k})$$ $$=(cos~\theta ~(b\overrightarrow{i})+sin~\theta ~(b\overrightarrow{j}))+(cos~\phi ~(a(cos~\theta \overrightarrow{i}+sin~\theta \overrightarrow{j}))+sin~\phi ~(a\overrightarrow{k}))$$
Thus, we get a base change from $C$ to $C'$: 
$$C:=\{\overrightarrow{i},\overrightarrow{j},\overrightarrow{k}\} \longrightarrow C':=\{(b\overrightarrow{i}),(b\overrightarrow{j}),(a(cos~\theta \overrightarrow{i}+sin~\theta \overrightarrow{j})),(a\overrightarrow{k})\}$$
Namely that we have a $\theta$-dependence in the third vector of the base $C'$.
We obtain a new vector $$((cos~\theta,sin~\theta),(cos~\phi,sin~\phi))$$ in the base $C'$. Hence, each point on the torus may be represented by a point on the space $S^1 \times S^1$
Questions : 


*

*Is it enough to prove that the torus and the space $S^1 \times S^1$
are diffeomorphic with $0<b<a$?

*Is it possible, with the joined arguments, to answer the question "Draw the case $ b = a$ and $ b> a $; why are these not manifolds?". I think so, with some analysis on the linear independence and the generator from our new base $ C '$, but I'm not sure. 
For those which have Mathematica, I leave you the codes to verify the images with $ b = a$  and $ b> a $
$$
ParametricPlot3D[{(3 + 3*Cos[p])*Cos[t], (3 + 3*Cos[p])*Sin[t], 
3*Sin[p]}, {p, 0, 2*Pi}, {t, 0, 2*Pi}]$$
$$ParametricPlot3D[{(2 + 3*Cos[p])*Cos[t], (2 + 3*Cos[p])*Sin[t], 
3*Sin[p]}, {p, 0, 2*Pi}, {t, 0, 2*Pi}]$$
Thank you in advance for the help!
 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 $(x_3,x_4)=(\frac{\sqrt{A^2+B^2}−ab}{2},\frac{C}{b})$ is the location on the second coordinate circle (which we've computed directly from the definition of ϕ). Thus, let 
$$ϕ^{−1}(A,B,C)=(\frac{A}{\sqrt{A^2+B^2}},\frac{B}{\sqrt{A^2+B^2}},\frac{C}{b})$$
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. 
