Parametrization of a plane

Question

"this exercice is 2.5 from Do carmo"

Let $P=\{(x,y,z) \in \mathbb{R}^3 ; x=y\}$ (a plane)

and let $X:U\subset \mathbb{R}^2\rightarrow \mathbb{R}^3$ given by $ X(u,v)=(u+v,u+v,u.v)$

where $U=\{(u,v)\in \mathbb{R}^2 ; u>v\}$

the question : is $X$ a parametrization of $P$?

so Do carmo said yes , but for example for the point $(0,0,3)\in P$ we cannot find any $(u,v)\in U$ such that $ X(u,v)=(0,0,3)$ so $X$ can not be a parametrization of $P$, what do you think

Answer 1

Yes it is correct it is not a parametrization indeed $(0,0,a)$ with $a>0$ belongs to the plane but it is not reached by $X(u,v)$ since for $u=-v$ we have $uv=-u^2<0$.

Answer 2

Actually Do Carmo says No (page 480).

The point $(0,0,3)$ is indeed not in the image of $X$ since $u+v=0$ implies $u=-v$ and then $3=-v^2 \leq 0$.

Answer 3

so this the picture from the book may be i have the oldest version