My question is how to show that the following map is a parametrization:
$$ X(u, v) = x(u) a + y(u) \cos v b - y(u) \sin v c, \qquad (u, v) \in I \times (\theta, \theta + 2 \pi) $$
where $\{a, b, c\}$ is a positively oriented orthonormal basis or the 3-space, $I$ is an open interval of the real line and $x, y: I \longrightarrow \Bbb{R}$ are functions that do not take the zero value for the same $t \in I$.
This is, according to the solution of Exercise (3) of chapter 2 of Curves and Surfaces, 2nd Edition, by Montiel and Ros, a parametrization of a surface of revolution around an axis $R$ in the direction of $a$.
I can see, intuitively, that $X$ is an homeomorphism, but I am failing to write a formal proof. Also, I computed the derivative $$ (dX)_{(u, v)} = \begin{bmatrix} x'(u) & 0 \\ y'(u) \cos v & - y(u) \sin v \\ - y'(u) \sin v & - y(u) \cos v \end{bmatrix} $$ but I am unable to show that it is injective.
Any help will be the most appreciated. Thanks in advance.
EDIT
With the hints by Ted Shifrin I was able to show that the derivative is injective:
Let $(z, w) \in \Bbb{R}^2$. We will show that we cannot have $(dX)_{(u, v)}(z, w) = 0$ unless $(z, w) = 0$. Indeed, if it was the case, we would have \begin{align*} & x'(u)z = 0, \\ & y'(u) \cos v z - y(u) \sin v w = 0, \\ & - y'(u) \sin v z - y(u) \cos v w = 0. \end{align*} If $z = 0$ then $w = 0$. Suppose that $x'(u) = 0$ and $z \neq 0$. Then $y'(u) \neq 0$. If $\cos v = 0$ or $\sin v = 0$ then $z = w = 0$. If not, then $$ w = \frac{y'(u)}{y(u)} \frac{1}{\tan v}z = - \frac{y'(u)}{y(u)} \tan v z \implies z = - \tan^2 v z \implies z = w = 0. $$
I am yet unable to find an inverse function for $X(u, v)$.