Checking if $f: \Phi_1 \to \Phi_2 \; ; f_1(\theta, v) \mapsto f_2(\theta, \sinh v)$ is an isometry. Exercise :

Consider the surfaces :
  $$\Phi_1 : f_1(\theta, v) = \left( \cos \theta \cosh v, \sin \theta \cosh v, v\right), \; (\theta, v) \in (0,2 \pi) \times \mathbb R$$
$$\Phi_2 : f_2(\phi, u) = \left( u\cos \phi , u\sin \phi, \phi\right), \; (\theta, v) \in (0,2 \pi) \times \mathbb R$$
  Check if the following mapping is an isometry between them :
  $$f: \Phi_1 \to \Phi_2 \;; f_1(\theta, v) \mapsto f_2(\theta, \sinh v)$$

Thoughts-Question :
To start off, this is a Differential Geometry related question which I am not that experienced, thus if it feels trivial, excuse me.
From my continuous experience, interest and studying of a whole differnt subject (Functional Analysis - Operator Theory), I know very well that a Linear Isometry is essentialy achieved if $\|Av\|_Y = \|v\|_X$ where $A:X \to Y$ is a linear operator. This means that they are distance preserving. It is a global isometry if it also is surjective.
Now, a similar correspondance can be found in Differential Geometry. Specifically, if we have $2$ surfaces, $\Phi_1$ and $\Phi_2$, then the function $f: \Phi_1 \to \Phi_2$ is an isometry if and only if $f:\Phi_1 \to \Phi_2$ is a differentiable mapping which is an inective and surjective local isometry.
Now, I am having a hard time proving the following statements. First of all, I start by constructing my function as stated by the exercise body :
$$f(f_1(\theta,v)) = f_2(\theta, \sinh v)$$
$$\implies$$
$$f(\cos\theta\cosh v, \sin \theta\cosh v, v) = (\sinh v \cos \theta, \sinh v\sin \theta, \theta)$$
So, checking the statements needed, first of all, that $f$ is differentiable. 
Now, how does one show that this $f$ is injective and surjective ?
Also, what about the local isometry ? I know that we can check if it is a local isometry or not, since the fundamental quantities of the fundamental form must oblige the following relations :
$$E_p = E_{f(p)}, \; F_p = F_{f(p)}, \; G_p = G_{f(p)}$$
I am kind of confused on the calculations of the fundamental quantities though. In a solved (but poorly elaborated) example I've seen, one must first calculate the inverse of $f$ and then correlate the argument of $f$ with what it's mapped to. 
I would really appreciate any thorough elaboration which can help me how to handle showing the injectivity, surjectivity but most importantly on how to find the fundamental quantities stated.
 A: Let's try to make the question a bit more precise. Define
$$ S_1 = \{ f_1(\theta, v) \, | \, (\theta, v) \in (0,2\pi) \times \mathbb{R} \} \subseteq \mathbb{R}^3_{x,y,z}, \\
S_2 = \{ f_2(\phi, u) \, | \, (\phi, u) \in (0,2\pi) \times \mathbb{R} \} \subseteq \mathbb{R}^3_{a,b,c}. $$
Then $S_1,S_2$ are both parametric surfaces in $\mathbb{R}^3$. In order to make things less confusing, it is comfortable to think of each $S_i$ as living in a different copy of $\mathbb{R}^3$. To emphasize this point I can give different names to the coordinates of $\mathbb{R}^3$ in which each $S_i$ lives (this what my non-standard notation $\mathbb{R}^3_{x,y,z},\mathbb{R}^3_{a,b,c}$ mean).
The functions $f_i \colon (0,2\pi) \times \mathbb{R} \rightarrow S_1$ given by the formulas above are global parametrizations for the surfaces $S_i$. Let's write
$$ f_1(\theta,v) = (x(\theta,v), y(\theta,v),z(\theta,v)). $$
The function $f_1$ is one-to-one and onto $S_1$ so given a point $p = (x_0,y_0,z_0) \in S_1$, we have a unique point 
$$f^{-1}(p) = f^{-1}(x_0,y_0,z_0) = (\theta(p), v(p)) = (\theta(x_0,y_0,z_0), v(x_0,y_0,z_0))$$
such that
$$ f_1(\theta(p),v(p)) = f_1(\theta(x_0,y_0,z_0),v(x_0,y_0,z_0)) = p $$
and similarly for $f_2$.
Now, let's define a map $F \colon S_1 \rightarrow S_2$ by the formula
$$ F(p) = f_2(\theta(p), \sinh v(p)). $$
This is related to your definition for if we write $p = f_1(\theta,v)$ then
$$ F(f_1(\theta,v)) = f_2(\theta, \sinh v). $$
The local representation of the map $F$ between the surfaces is the map 
$$\tilde{F} = f_2^{-1} \circ F \circ f_1 \colon (0,2\pi) \times \mathbb{R} \rightarrow (0,2\pi) \times \mathbb{R}$$
and is given by
$$ \tilde{F}(\theta,v) = (\theta, \sinh v). $$
You are asked whether $F$ is an isometry between $S_1$ and $S_2$. For $F$ to be an isometry, it needs to satisfy two conditions:


*

*The map $F$ needs to be one-to-one and onto.

*For all $p \in S_1$ and $v,w \in T_p(S_1)$, we should have $\left< v, w \right> = \left< dF|_p(v), dF|_p(w) \right>$. That is, $F$ should infinitesimally preserve the length of tangent vectors. Here, $dF \colon T_p(S_1) \rightarrow T_{F(p)} S_2$ is the differential of the map $F$.


Now, it turns out that instead of checking this directly from the definitions for $F$, you can deduce everything from the representation $\tilde{F}$ of the map $F$. Namely, $F$ will be an isometry if and only if:


*

*The map $\tilde{F}$ needs to be one-to-one and onto.

*The map $\tilde{F}$ needs to preserve the (local representations of the) first fundamental form. For each $(\theta,v) \in (0,2\pi) \times \mathbb{R}$, we should have
$$ E_1(\theta,v) = E_2(\tilde{F}(\theta,v)),  F_1(\theta,v) = F_2(\tilde{F}(\theta,v)),  G_1(\theta,v) = G_2(\tilde{F}(\theta,v)) $$
where $E_i,F_i,G_i$ are the coefficients of the first fundamental form of $S_i$ with respect to the parametrization $f_i$.



How do we calculate the $E_i,F_i,G_i$? By the formulas
$$ E_1(\theta,v) = \left< \frac{\partial f_1}{\partial \theta}, \frac{\partial f_1}{\partial \theta} \right>, \,\, F_1(\theta, v) = \left< \frac{\partial f_1}{\partial \theta}, \frac{\partial f_1}{\partial v} \right>, \,\, G_1(\theta, v) = \left< \frac{\partial f_1}{\partial v}, \frac{\partial f_1}{\partial v} \right> $$
and similarly
$$ E_2(\phi,u) = \left< \frac{\partial f_2}{\partial \phi}, \frac{\partial f_2}{\partial \phi} \right>, \,\, F_2(\phi, u) = \left< \frac{\partial f_2}{\partial \phi}, \frac{\partial f_2}{\partial u} \right>, \,\, G_2(\phi, u) = \left< \frac{\partial f_2}{\partial u}, \frac{\partial f_2}{\partial u} \right>. $$
For example,
$$ E_1(\theta,v) = \left< \frac{\partial f_1}{\partial \theta}, \frac{\partial f_1}{\partial \theta} \right> = \| \left( -\sin \theta \cosh v, \cos \theta \cosh v, 0 \right) \|^2 = \cosh^2(v), \\
E_2(\phi, u) = \left< \frac{\partial f_2}{\partial \phi}, \frac{\partial f_2}{\partial \phi} \right> = \| \left( -\sin \phi u, \cos \phi u, 1 \right) \| = 1 + u^2$$
and we indeed see that 
$$\cosh^2(v) = E_1(\theta,v)) = E_2(\tilde{F}(\theta,v)) = E_2(\theta, \sinh v) = 1 + \sinh^2 v. $$
I'll leave the rest of the calculations for you.
