How to prove that $M := \{(x,y,z) \in \mathbb{R}^3 | x^2+y^2=\cosh(z)\}$ is a 2-dim manifold by using the Implicit function theorem?

Question

Let's say we construct a function: $$F(x,y,z) = x^2+y^2-\cosh(z)$$ Then we could define the manifold such as: $$M := \{(x,y,z) \in \mathbb{R}^3 | F(x,y,z) = 0\}$$

Now, I'm a little lost. Am I right about the following statements? I must hereafter show that:
1) $F$ is differentiable
2) $M$ is a plane in $\mathbb{R}^3$
3) $M =: F^{-1}(\{0\})$

Am I on the right track here? And if so, how could I prove these to statements? Am I missing something?

Answer 1

Let be $F(x,y,z) = x^2 + y^2 - \cosh(z)$. Note that $\nabla F (x) \neq 0$ $\forall (x,y,z) \in F^{-1}({0})$. Indeed, because if $(x,y,z) \in F^{-1}({0})$ and $\nabla F (x,y,z) = 0$, we have that $\frac{\partial F}{\partial z}(x,y,z) = -\sinh{z} = 0 $ and therefore

$$0=\sinh^2{z} =\cosh^2{z} - 1 $$

that is $(x^2 + y^2)^2 = 1$. Thus $x\neq 0$ or $y \neq 0$ and therfore $\nabla F(x,y,z) = (2x,2y,-\sinh{z}) \neq 0$. Follows of Implicit Function Theorem, that if $(x,y,z) \in F^{-1}({0})$ that there are open set $U \subset R^3$ with $(x,y,z) \in U$ such that $F^{-1}({0})\cap U = \{ (u,v) \in V \subset R^2; (u,v, \psi(u,v))\}$, where $\psi : V \to R $ is function $C^{\infty}$ of open set $V$ of $R^2$ on $R$. Thus $F^{-1}({0})$ is a manifold.

Answer 2

Since $F$ is class $C^{k\geq1}$, this is important, it is not enough to differentiate F, it must be class $C^{k\geq1}$. You need to check now that $\frac{\partial F}{\partial x_i}(p)\neq0$, for some i = 1,2,3 and $F(p)=c$, for $c$ a regular value. Then the implicit function theorem, says

$F^{-1}(c)\cap Z$ is the graph of a function defined in an open $\mathbb{R^2}$ and $Z\subset \mathbb{R^3} $ is an open containing $0\in \mathbb{R^3}$ .

In other words, $M=F^{-1}(c)\subset \mathbb{R^2}$ it's a $2$-dimensional hypersurface is class $C^{k\geq1}$.