Prove the torus given is a regular surface A subset $ S \subseteq\Bbb R^3 $ is called a regular surface if for every point $p \in S$, there exists a neighborhood $V$ of $p$ in $\Bbb R^3$, an open set $U \subseteq\Bbb R^2$ and a $C^\infty$ function $F\colon U\to\Bbb R^3$ such that

*

*$F:U\to S \cap V $ is a homeomorphism


*The jacobian Matrix $D_uF$ has rank $2$ for all $u \in U$
So this is what I have as a definition for a regular surface and I want to prove
$$T^2 = \{(x,y,z) \in\Bbb R^3 | x = (\cos(u)+2)\cos(v), y = (\cos(u)+2)\sin(v), z =\sin(u), u,v \in\Bbb R)\}$$
I have seen some examples from the book by Docaro, Differential Geometry of Curves and Surfaces. But for this one, I am stuck for showing the first part of the definition holds. Can anyone help with this? I know it is a basic question but I am greatly struggling with this book..
 A: To parametrize a $(R,r)-\textbf{Torus}$ where $r<R$, we revolve a circle of radius $R$ in the $xz$-plane, about the $z$-axis. One can come up with a parametrization by taking: 
$$\Phi(t,s) = \gamma(s)+ r\bigg(\textbf{n}(s) \cos (t) + \textbf{b}(s) \sin(t)\bigg)$$
where $\gamma(s) = \left(0,R \cos \left(\frac{s}{R}\right), R \sin \left(\frac{s}{R}\right)\right)$. The above map simply wraps a tube of radius $r$ about the circle of radius $R$ in the $z = 0$ plane. Simplifying the above we have that:
$$\Phi(t,s) = ((R+r \cos(t)) \cos (s), (R+r \cos(t)) \sin(s), R \sin(t)) = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$
Here you can take $(t,s) \in U=(0,2\pi) \times (0,2\pi)$ so that $\Phi$ is injective. Observe that we've missed various points, namely those in the image of $\Phi(t,\pi/2)$ and $\Phi(0,s)$. 

Now just take another restriction of the plane, i.e $(t,s) \in V=(\epsilon,\epsilon+2\pi) \times (\epsilon,\epsilon+2\pi)$ where $\epsilon$ is small. Hence we have:
$$\{(\Phi|_U,U),(\Phi|_V,V)\}$$
is an atlas on $T^2$ (where you are left to check the other conditions). Use this same method to show that the cylinder is a surface.
