Exercise :
Let $f(u,v) = (u,v,u^2-v^2), \; (u,v) \in U$ where $U$ is a coherent and open subset of $\mathbb R^2$. Show that $f$ is a parametrization of a $C^\infty$ patch (local surface) of a surface $\Phi$.
Attempt/Question :
I know that in order to show that $f$ is a regular parametrization (I don't know and cannot find the cases just for "parametrization"), one needs to show that the function $f: U \to \Phi$ is injective and onto, while also $f_u \times f_v \neq 0 $ everywhere in $U$.
It is $f_u \times f_v = (-2u,2v,1) \neq \mathbf{0} \; \forall (u,v) \in U$.
I am having trouble showing that $f$ is injective and onto though, any guidance will be appreciated.
Though, the following question arises :
Question : When is $f$ just simply called a parametrization ? I assume, that when $f$ is injective and onto, while also smooth. That takes the $f_u \times f_v \neq \mathbf{0}$ out.
As a starter in Differential Geometry I would really appreciate any thorough explanations.
P.S. : I apologise if any term is mistakenly stated, as I am trying to properly translate the terms and the exercise from Greek.
This is an example of a Monge patch. It's easy to see that $f$ is injective, $f$ is onto $\Phi:=f(U)$ ($\Phi$ was never defined, so I assume that it is $f(U)$; note that it will be a subset of a hyperbolic paraboloid), and $f$ is smooth. As you noted, $f_u\times f_v\neq 0$ on $U$, and so $f$ is regular. So, we've shown everything that we need to.
As for terminology, it differs. I've often seen the definition that a regular map $f:U\rightarrow\mathbb{R}^3$ whose image lies in a surface $M$ is called a parameterization of the region $f(U)$ in $M$, meaning that a patch is an injective parameterization.
EDIT: $f$ is always onto its image, so we just need to show that $f$ is injective. Suppose that $f(x,y)=f(u,v).$ Then, $$(x,y,x^2-y^2)=(u,v,u^2-v^2).$$ Since the coordinates must be equal, we have that \begin{align*} x&=u\\y&=v\\x^2-y^2&=u^2-v^2,\end{align*} and so $x=u, y=v,$ and $(x,y)=(u,v).$