Okay, i'm studying Differential Geometry and the book defines regular parametrized surface as follows:
Definition: A regular parametrized surface is a map $X:U\subset \mathbb R^2 \to \mathbb R^3$, where $U$ is an open subset of $\mathbb R^2$ such that:
$(a)$ $X$ is differentiable of class $C^{\infty}$;
$(b)$ For all $q = (u, v)\in U$, the differential of $X$ in $q$, $dX_q:\mathbb R^2\to\mathbb R^3$, is injective.
Then the book gives some equivalences for item $(b)$:
$(b.1)$ $dX_q$ is injetive;
$(b.2)$ The jacobian matrix of $X$, $J(u_0,v_0)$, has rank $2$;
$(b.3)$ The vectors $X_u(u_0,v_0)$ and $X_v(u_0,v_0)$ are linearly independent;
$(b.4)$ $X_u(u_0,v_0)\times X_v(u_0,v_0)\neq 0$.
I'm trying to solve this problem:
Let $X(u,v)$ be a regular parametrized surface. Prove that if $F: \mathbb R^3\to\mathbb R^3$ is a diffeomorphism, then $\overline{X}=F\circ X$ is a regular parametrized surface.
So, $(a)$ is satisfied since $X$ and $F$ are differentiable. Now, let $$ X(u,v)=(x(u,v),y(u,v),z(u,v)). $$ Then $\overline{X}(u,v)=F(X(u,v))=F(x(u,v),y(u,v),z(u,v))$.
I tried to show that $\overline{X}_u(u,v)\times \overline{X}_v(u,v)\neq 0$ for all $(u,v)$.
$\overline{X}_u(u,v) = F_x\cdot\dfrac{\partial x}{\partial u}+F_y\cdot\dfrac{\partial y}{\partial u}+F_z\cdot\dfrac{\partial z}{\partial u}$
$\overline{X}_v(u,v) = F_x\cdot\dfrac{\partial x}{\partial v}+F_y\cdot\dfrac{\partial y}{\partial v}+F_z\cdot\dfrac{\partial z}{\partial v}$
Hence $$ \overline{X}_u(u,v)\times \overline{X}_v(u,v) = \require{cancel} \cancelto{0}{\left(F_x\cdot\dfrac{\partial x}{\partial u}\right)\times\left(F_x\cdot\dfrac{\partial x}{\partial v}\right)} + \left(F_y\cdot\dfrac{\partial y}{\partial u}\right)\times\left(F_x\cdot\dfrac{\partial x}{\partial v}\right) + \left(F_z\cdot\dfrac{\partial z}{\partial u}\right)\times\left(F_x\cdot\dfrac{\partial x}{\partial v}\right) + \left(F_x\cdot\dfrac{\partial x}{\partial u}\right)\times\left(F_y\cdot\dfrac{\partial y}{\partial v}\right) + \cancelto{0}{\left(F_y\cdot\dfrac{\partial y}{\partial u}\right)\times\left(F_y\cdot\dfrac{\partial y}{\partial v}\right)} + \left(F_z\cdot\dfrac{\partial z}{\partial u}\right)\times\left(F_y\cdot\dfrac{\partial y}{\partial v}\right) + \left(F_x\cdot\dfrac{\partial x}{\partial u}\right)\times\left(F_z\cdot\dfrac{\partial z}{\partial v}\right) + \left(F_y\cdot\dfrac{\partial y}{\partial u}\right)\times\left(F_z\cdot\dfrac{\partial z}{\partial v}\right) + \cancelto{0}{\left(F_z\cdot\dfrac{\partial z}{\partial u}\right)\times\left(F_z\cdot\dfrac{\partial z}{\partial v}\right)} $$
But how to show that this last expression is $\neq 0$ and i don't even know if what i did is correct.
At first glance, the calculations look correct, but it's not the easiest approach.
I suggest to use (b.3) and the fact that $F$ is a diffeomorphism. Since $X$ is regular, $X_u(u_0,v_0)$ and $X_v(u_0,v_0)$ are linearly independent. Since $F$ is a diffeomorphism, $dF_{(u_0,v_0)}$ is a regular linear map, and hence the vectors $dF_{(u_0,v_0)}\left(X_u(u_0,v_0)\right)$ and $dF_{(u_0,v_0)} \left(X_v(u_0,v_0)\right)$ are also linearly independent.