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.
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:
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:
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.