Finding a smooth atlas of $M=\{ (x,y,z) \in \mathbb{R}^3 \; | \; x^2+y^2-z^2=1 \}$ We know that $M=\{ (x,y,z) \in \mathbb{R}^3 \; | \; x^2+y^2-z^2=1 \}$ is a submanifold of $\mathbb{R}^3$, with dimension $2$. The point is that I have to find an atlas which define the structure of $C^{\infty}$ manifold for $M$. I just discovered these objects, and I have to say that I don't know how to do. As $M$ is a submanifold of dimension $2$, I think I should find some open sets $(U_i)_i$ of $M$, and some open sets $(V_i)_i$ of $\mathbb{R}^2$, and some homeomorphism $\psi : U_i \rightarrow V_i$, such that the transition map are $C^\infty$.
My first idea was to use the theorem of the implicit functions, to have $z$ function of $x,y$ and then I would be able to solve the problem I think. But all I know is that : $(x,y,z) \in M \iff z = +- \sqrt{x^2+y^2-1}$, and from that, I tried several things but it doesn't lead where I wanted.
Someone could help me, and explain me how to do, and what's the idea between thoses objects ?
Thank you !
 A: Consider for instance $U_1=\{(x,y,z)\in M \mid x>0\}$ and $V_1=\{(y,z)\in\mathbb{R}^2\mid y^2-z^2<1\}.$ Then
$\psi_1:U_1\to V_1,$ $\psi_1(x,y,z)=(y,z)$ is a diffeomorphism with inverse $\psi_1^{-1}(y,z)=(\sqrt{1+z^2-y^2},y,z).$ Similarly, let $U_1'=\{(x,y,z)\in M \mid x<0\}.$ Then $\psi_1':U_1'\to V_1$ given by $\psi_1'(x,y,z)=(y,z)$ is a diffeomorphism with inverse $\psi_1'(y,z)=(-\sqrt{1+z^2-y^z},y,z).$ Thus, with $U_1$ and $U_2$ we have covered all points in $M$ with $x>0$ or $x<0.$ To cover the points with $x=0$ you may use the same idea to cover the points with, say, $y>0$ or $y<0.$
A: Hint Since the defining equation has the form $F(x^2 + y^2, z) = C$, $M$ is invariant under rotations about the $z$-axis and in particular it is a surface of revolution. We can take the generating curve to be intersection of $M$ with the half-plane $\{x > 0, y = 0\}$. Substituting gives that this curve is $x^2 - z^2 = 1$, $x > 0$, which we can parameterize (as a subset of the $xz$-plane) as $$\{(\cosh t, \sinh t) : t \in \Bbb R\} .$$ So, we can parameterize $M$ by
$${\bf r} : (t, \theta) \mapsto (\cosh t \cos \theta, \cosh t \sin \theta, \sinh t) ,$$ and we can produce an atlas by computing inverses for appropriate restrictions of ${\bf r}$. A little algebraic manipulation gives that one coordinate chart is $$\phi : (x, y, z) \mapsto \left(\operatorname{artanh} \frac{z}{x^2 + y^2}, \operatorname{arctan} \frac{y}{x}\right) . $$
A: The space is topologically equivalent to a punctured plane, which suggests that we can cover $M$ with a single smooth chart. Since the defining equation has the form $F(x^2 + y^2, z) = C$, $M$ is invariant under rotations about the $z$-axis $Z$, which suggests that we look for a map $M \to \Bbb R^2 - \{ 0 \}$ that is suitably compatible with those rotations (or more precisely, that is equivariant for the two rotation actions.)
In particular, this suggests that if we view $\Bbb R^2$ as the $xy$-plane in $\Bbb R^3$ we map the arc given by the intersection of $M$ with a half-plane bounded by $Z$ to the ray $\Bbb R^2 \cap Z$ of the form
$$(x, y, z) \mapsto \left(f(z) \frac{x}{|x|}, f(z) \frac{y}{|y|}\right)$$
for a suitable bijection $f : \Bbb R$ to $\Bbb R^+$. A natural choice is $f(z) := \exp z$, giving
$$(x, y, z) \mapsto \left(\frac{x}{|x|} \exp z, \frac{y}{|y|} \exp z\right) .$$
A: This embedded surface is a hyperboloid with one sheet $H_1$, which is diffeomorphic (see figure) 


*

*to the unit vertical cylinder $C$, 

*to the doubly punctured unit sphere $S'_2=S_2$ minus its North and South pole,


both through a central projection through the origin (analog of the Mercator mapping from the puntured sphere onto the cylinder). 
Thus any atlas of the cylinder or of the doubly punctured sphere gives, by central projection, an atlas of $H_1$.
Remark : in fact, there is another "natural" diffeomorphism between $C$ and $H$ : it is easily described in cylindrical (!) coordinates :
$$\underbrace{(1,\theta,h)}_{\in C} \to \underbrace{(\sqrt{h^2+1},\theta,h)}_{\in H_1}$$
(please note that the $h$ coordinate is kept the same). 
Another way to achieve the correspondence between these three surfaces is by using the fact that they have a common equation :
$$x^2+y^2-1-m z^2=0 \ \ \text{with} \ \begin{cases}m=1 & \text{for hyperboloid } H_1\\m=0 &\text{for cylinder} \ C\\m=-1 &\text{for punctured sphere } S'_2\\\end{cases}.$$
Otherwise said, they can be considered as level surfaces of 
$$m=\frac{x^2+y^2-1}{z^2}$$
(this is how I have programmed the representation of these surfaces on Fig. 1).

Fig. 1
Remark : the doubly punctured sphere possesses an atlas with two maps using stereographic projection on plane $xOy$ 


*

*one through its South pole (See for example slide 25 of http://www.math.utah.edu/~treiberg/ERD_1-22-2010.pdf), 

*the other one, in a symmetrical way, through its North pole. 

