Let $V$ be the open set in the $xy$-plane. Show that the set \begin{align*} \left\lbrace (x, y, z)\in \mathbb{R}^3\, | \, z=0 \text{ y } (x, y)\in V \right\rbrace \end{align*} is a regular surface.
Let be $\textbf{x}\subseteq \mathbb{R}^2 \to \mathbb{R}^3$ be a function given by \begin{align*} \textbf{x}(u, v)=(u, v, 0) \end{align*} Then $\textbf{x}(V)$ is exactly the given surface. The function $\mathbf{x}$ is continuously differentiable because its component functions are polynomials. Also, \begin{align*} \textbf{x}_u(u, v) &= (1, 0, 0) \\ \textbf{x}_v(u, v) &= (0, 1, 0) \end{align*} These two vectors are obviously linearly independent for all $(u, v)\in V$. Condition 1 is clearly satisfied and condition 3 also offers no difficulty. Finally, every point $(x, y, z)$ of the set is in the image by $\textbf{x}$ of the single point $(u, v)=(x, y)\in V$. For this reason $\textbf{x}$ is injective and now it would only be necessary to give an argument to say that $\textbf{x}^{-1}$ is continuous. Thus the given set is a regular surface.
Well, I don't know if my idea is right or how would you solve it?