I was trying to solve the following problem, but I am not sure about my solution.Problem:
Let $P = \{ (x,y,z) \in \mathbb{R}^3 \mid x = y\}$, and let $f \colon U \subset \mathbb{R}^2 \to \mathbb{R}^3$ be given by
\begin{equation*} f(u,v) = (u + v, u + v, uv) \end{equation*} where $U = \{(u,v) \in \mathbb{R}^2 \mid u > v\}$. Clearly, $f(U) \subset P$. Is $f$ a parametrization of $P$?
Solution: No. This function $f$ does not cover $P$. For example, points of the form $(2x_0, 2x_0, x_0^2)$ where $x_0 > 0$ are never covered by $f$. Is my solution correct? Is it this simple?
This surface lies entirely in a plane $ x=y$ patch, it is simple. The generators have an envelope as a parabola. It is projection of the more familiar patch of parametrization. $$ (x,y,z)=(u + v,\,0, \,u v) $$
where we divide tangent points to vertex meeting point tangent segment proportionately and draw the connecting rays forming the parabolic envelope.
$u,v$ can be interchaged, we get a coincident point for two sets of $ (u,v) $ but still has full cover.
It is worth looking at saddle points of a 3-space surface hyperbolic paraboloid in this projected connection of parametrization $ (u+v,\,u-v, \,uv )$ to appreciate special case covering.