Proving very basic property of surfaces in $\mathbb{R^3}$ 
Let $X: U \subset \mathbb{R^2} \mapsto\mathbb{R^3}$ be a regular parameterized surface, i.e
a) $X$ is $C^{\infty}$;
b) The differential of $X$ at any $q \in U$, $dX_q:\mathbb{R^2} \mapsto \mathbb{R^3}$ is injective.
Prove that if $F$ is an invertible, $C^{\infty}$, function, then $\overline{X} = F \circ X$ is also a regular parameterized surface.

I don't know what do here because this looks so obvious that a simple "$\overline{X}$ is the composition of two differentiable and injective linear maps so it satisfies a) and b)" would suffice, but unfortunately I don't trust myself enough to consider the exercise done with just that. So, am I correct here? If not, how do I go about proving it? If it's not what I'm thinking of I'm sure there must be another argument that is probably something just as simple.

$\overline{X}$ is clearly differentiable because it's the composition of differentiable maps. Similarly, $d\overline{X}_q = dF_{X(q)}\cdot dX_q$, the composition of injective maps, hence injective, as desired.
