Find an atlas for $M=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=1+z^2\}$
It is easy for me to check that $M$ is a submanifold of $\mathbb{R}^3$ of dimension $2$ using the following theorem:
Let $F:U \rightarrow \mathbb{R}^m$ be a $C^\infty$ function on an open set $U \subseteq \mathbb{R}^{n+m}$ and let $c \in \mathbb{R}^m$.
Assume that for each $a \in F^{-1}(c)$, the derivative $DF_a:\mathbb{R}^{n+m} \rightarrow \mathbb{R}^m$ is surjective.
Then, $F^{-1}(c)$ has the structure of an $n$-dimensional manifold which is Hausdorff and has a countable basis of open sets.
I know also that the proof of this Theorem is to find an atlas, but in practise I have problems with construction that atlas.
Let $F:\mathbb{R}^3 \rightarrow \mathbb{R}$ s.t. $F(x,y,z)=x^2+y^2-z^2-1$.
We have $DF_a=[2a_1 \quad 2a_2 \quad 2a_3]$, where $a=(a_1,a_2,a_3)$.
Since $(0,0,0) \not \in M$, $DF_a$ has rank $1$, ie, $DF_a$ is surjective.
Then $M=F^{-1}(0)$ is a submanifold of $\mathbb{R}^3$ os dimension $3-1=2$.
Now comes the part I don't understand.
We construct $G:U \rightarrow \mathbb{R}^3$ s.t. $G(x,y,z)=(x^2+y^2-z^2-1,y,z)$.
Then, $$DG_a=\begin{bmatrix} 2a_1 & 2a_2 & -2a_3 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 & 0 & 1 \end{bmatrix}.$$
So the atlas is $\{(U_i,\varphi_i)\}_{1 \leq i \leq 3}$ where $U_i=\{(x,y,z) \in \mathbb{R}^3:x_i \neq 0\}$.
But what is explicitly $\varphi_i:U_i \rightarrow \varphi(U_i)$?
Some help here would be appreciated.
This is a one sheeted hyperboloid. You can find one parametrization rotating the hyperbola $(\cosh u, 0, \sinh u)$ to obtain: $${\bf x}(u,v) = (\cosh u \cos v, \cosh u \sin v, \sinh u), \quad u \in \Bbb R, \quad 0 < v < 2\pi.$$ This leaves out a meridian ($v = 0$). You can cover it taking any $\epsilon > 0$ and making $v \mapsto v+ \epsilon$ above. These two parametrizations will cover the surface.