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Subset $ S \subseteq R^3 $ is called regular surface if for every point $p \in S$, there exists a neighborhood V of p in $R^3$, an open set $u \subseteq R^2$ and a $c^\infty$ function $F: u -> R^3$ such that

1) $F:u -> S \cap V $ is a homeomorphism

2) 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

and I want to prove

$$T^2 = \{(x,y,z) \in R^3 | x = (cos(u)+2)cosv, y = (cos(u)+2)sin(v), z = sin(u), u,v \in R)\}$$

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..

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    Hint: the function $F$ is $F(u, v) = (\cos(u)+2)\cos v, (\cos(u)+2)\sin(v), \sin(u)$. – 2017-03-01
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    @JohnHughes: The torus cannot be covered by one chart. All you've given is a parametrization. You need to accompany $F$ with one more chart. – 2017-03-01
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    Right...but if you restrict $u$ and $v$ to various subsets of $\Bbb R^2$ (e.g., any axis-aligned rectangle whose $u$- and $v$-extents are both less than $2\pi$), then the various restrictions are all charts, and the union of their images covers the torus. I was hoping to get OP to recognize that in defining $F$, I'd omitted something critical -- the domain -- and that OP would have to pick it carefully to make sure that on the domain, $F$ was injective. The cat's out of the bag now, though. :) – 2017-03-01
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    I know that. I just didn't want to OP to get confused. The prompt asks her/him to show that for each $p$, there exists an $F$. Since you gave one map without saying much after, I didn't want it to be assume that $F$ covers everything. – 2017-03-01
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    @Meer In the definition of $T^2$ you have $u,v \in \mathbb{R}$. Can you explain why it is not $[0,2\pi)$ or $(-2\pi,2\pi)$ ? – 2017-03-01

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To parametrize a $(R,r)-\textbf{Torus}$ where $r

$$\Phi(t,s) = \gamma(s)+ r\bigg(\textbf{n}(s) \cos (t) + \textbf{b}(s) \sin(t)\bigg)$$

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 $\Phi(0,s)$.

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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.

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    so it does not matter even if it is a set of two homeomorphisms from different open sets? U and V? – 2017-03-01
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    Yeah, it doesn't matter. In general you can't expect to cover a chart with one map. An atlas is given by $(\phi_i,U_i)$ i.e a collection of maps and open sets which are their domains. – 2017-03-01