I'm trying to prove that $X:U \subseteq \mathbb{R}^2 \to \mathbb{R}^3$ given by $X(u,v) = (u+v,u+v,uv)$, and $U = \{(u,v) \in \mathbb{R}^2 | u > v \}$ is a parametrization for $P = \{ (x,y,z) \in \mathbb{R}^3 | x = y \}$.
I've already verified that $X$ is differentiable (Since $u+v$ and $uv$ are), $d_{X_q}$ is injective (Since $\{(1,1,v),(1,1,u)\}$ is linearly independent) and that $P$ is a regular surface (since it is the graph of $f:U \to \mathbb{R}, f(x,y) = z$), so if I prove that $X$ is bijective I'll guarantee by a proposition that it is a homeomorphism, then it is a parametrization. (Or, even if I tried to do by definition, I'd also need to prove bijectivity). The problem is I can't do that.
Here's my try: $X$ is injective: If $u_1+v_1 = u_2+v_2$ and $u_1v_1 = u_2v_2$, then $u_1^2+v_1^2 = u_2^2 + v_2^2$, then $|(u_1,v_1)| = |(u_2,v_2)|$, but that's not enough to get $(u_1,v_1) = (u_2,v_2)$.
$X$ is surjective on $P$: Let $(x,x,z) \in P$, then is there $(u,v) \in U$ such that $u+v = x$ and $uv = z$? I can't see why there is this $(u,v)$.
How can I proceed?
Thanks.
From $u_1^2+v_1^2 = u_2^2 + v_2^2$ we get, since $u_1v_1=u_2v_2$,
$u_1^2-2u_1v_1+v_1^2 = u_2^2 -2u_2v_2+ v_2^2$, hence $(u_1-v_1)^2=(u_2-v_2)^2$, therefore $|u_1-v_1|=|u_2-v_2|$. Since $u_i > v_i$ we derive
$u_1-v_1=u_2-v_2$.
Now use again that $u_1+v_1 = u_2+v_2$ to derive $u_1=u_2$ and $v_1=v_2$.
$X$ is not surjective ! We have $(1,1,1) \in P$, but the equations $u+v=1$ and $uv=1$ have no (real) solutions:
from $u+v=1$ we see that $u^2+uv=u$, hence $u^2-u+1=0$. The last equation has no solution in $\mathbb R$.