I have to prove the following:
Consider $M:= \{(x,y,z) \in \Bbb{R}^3 : x^2 + y^2 - z^2 = 1 \}$. Prove that $M$ is a smooth manifold and that the map $\Phi : \Bbb{R} \times (0,2\pi) \hookrightarrow M $ given by $$\Phi(u,v)=(\sqrt{1+u^2} \cos{(v)}, \sqrt{1+u^2} \sin{(v)}, u)$$ is compatible with the inherited atlas of being a submanifold of $\Bbb{R}^3$ .
In order to prove that $M$ is a smooth manifold, let denote $F: \Bbb{R}^3 \hookrightarrow \Bbb{R}$ given by $F(x,y,z) = x^2 + y^2 -z^2 -1 = 0$ . Hence, $F$ is clearly smooth, $M=F^{-1}(0)$ and $\nabla _p F = (2x,2y,-2z)$, so $\nabla _p F = 0 \iff p=0 $ .
If $p=(x,y,z) \in F^{-1}(0) \Rightarrow x^2 + y^2 = z^2 + 1 \geq 1 \Rightarrow x > 0 \lor y>0 \Rightarrow p \neq 0 $ . Therefore, $0$ is a regular value of $F$ and by the Regular Value Theorem $M$ is a smoothe manifold.
For the second part I am having trouble to understund "who" is the inherited atlas in this case.
My teacher defines as follows:
If $N$ is a smooth manifold and $M \subset N$ a smooth submanifold, an inherited chart of $M$ from a given chart $(U, \varphi)$ of $N$ is $(U \cap M, \varphi|_{U\cap M})$. An inherited atlas of $M$ is the union of the inherited charts of an atlas on $N$ .
The usual atlas of $\Bbb{R}^n$ is $\mathcal{A}_n = \{(\Bbb{R}^n, Id_{\Bbb{R}^n})\}$, so the inherited atlas of $M$ would be $\{(M, Id_M)\}$. The problem is that if this is an atlas, $M$ is open and it is not.
Did I misunderstund something? Is there another atlas that I should consider to prove the exercise?
Thank in advance :)
As you write in a comment, a $d$-codimensional smooth submanifold $M$ of an $n$-dimensional smooth manifold $N$ is defined as follows:
Let call such $(U,\varphi)$ an admissible chart for $M$ (this is just an adhoc notation).
The map $r : \tilde{\mathbb R}^{n-d} \to \mathbb R^{n-d}, r(x_1, \ldots, x_n) = (x_{d+1}, \ldots, x_n)$, is a homeomorphism. To each chart $(\varphi,U)$ which is admissible for $M$ we associate the inherited chart $(\tilde \varphi, \tilde U)$ with $$\tilde U = U \cap M,$$
$$\tilde \varphi : U \cap M \xrightarrow{\varphi \mid_{U \cap M}} \varphi(U) \cap \tilde{\mathbb R}^{n-d} \xrightarrow{r \mid_{\varphi(U) \cap \tilde{\mathbb R}^{n-d}}} \tilde V$$ where $\tilde V = r(\varphi(U) \cap \tilde{\mathbb R}^{n-d}) \subset \mathbb R^{n-d}$. Clearly $\tilde \varphi$ is a homeomorphism from the open subset $U \cap M$ of $M$ to the open subset $\tilde V$ of $\mathbb R^{n-d}$, thus it is (topological) chart on $M$.
It is important to understand that the image $\varphi(U \cap M) = \varphi(U) \cap \tilde{\mathbb R}^{n-d}$ is not open in $\mathbb R^n$ (unless $d = 0$); it is the intersection of the open subset $\varphi(U)$ with the $(n-d)$-dimensional linear subspace $\tilde{\mathbb R}^{n-d}$. But this intersection is open in the subspace $\tilde{\mathbb R}^{n-d}$ of $\mathbb R^n$, and therefore it is mapped by $r$ onto an open subset of $\mathbb R^{n-d}$. In other words, the map $\varphi \mid_{U \cap M} : U \cap M \to \varphi(U \cap M)$ cannot be a chart on $M$, we need to compose it with $r$ to get a chart on $M$.
Also observe that this construction only works for charts on $N$ which are admissible for $M$. You cannot take an arbitrary chart on $N$ and expect that it produces a chart on $M$.
By the way, let me mention that it is more usual to set $m = n-d$ and call $M$ an $m$-dimensional smooth submanifold $M$ of $N$.
The collection of all inherited charts on $M$ (based on the admissible charts on $N$) is obviously an atlas on the topological space $M$, but it remains to show that all transition functions are smooth. Once this has been done, you know that the inherited charts provide a smooth atlas on $M$, i.e. that $M$ inherits the structure of a smooth manifold from $N$.
If $M \subset \mathbb R^n$ is a submanifold (as your concrete example $M \subset \mathbb R^3$), then in general $(Id_{\mathbb R^n}, \mathbb R^n)$ will not be an admissible chart for $M$. But your atlas $\mathcal A_n$ generates the standard maximal atlas on $\mathbb R^n$ which consists of all diffeomorphisms (in the sense of multivariable calculus) between open subsets of $\mathbb R^n$. Then you can take all admissible charts for $M$ and perform the above construction. But recall that $\varphi(U \cap M)$ will not be open in $\mathbb R^n$, only relatively open in an $(n-d)$-dimensional subspace.
To solve your problem it is not expedient to try to determine explicitly all charts on $M$. This attempt would be fairly hopeless. This is why I asked for theorems about submanifolds. Also it needs a precise definition of what it means that $\Phi$ is compatible with the inherited atlas.