A given surface can be parameterize in many different ways. How to prove that a change in parameters, given a smooth, invertible map between the two parameter domains, does not change a normal vector to the surface? I have the intuition about it, but trying so hard I still can`t prove it strictly.
EDIT: I mean that if we have equivalent parametrizations, that comprize the positive orientation of the surface, then a unit normal vector does not change.
On
In surface theory normal vector is intrinsic depending only on first fundamental form and its dependants $(H^2 = E G -F^2 , L N - M^2)$.
Referring to Weingarten relations and second fundamental form,
$$ H N_1 \times N_2 = (LN - M^2). $$ The latter is invariant by Gauss theorem.( unsubscripted N is of second fundamental form), Normal vector to surface is parametrization-free.
Perhaps you cannot prove it because it's false.
Consider the usual parameterization of the sphere:
$$ (u, v) \mapsto (\cos u \cos v, \sin u \cos v, \sin v) $$ where $- \pi/2 \le v \le \pi/2$.
Now compose it with the map $(u, v) \mapsto (u, -v)$, to get
$$ (u, v) \mapsto (\cos u \cos v, \sin u \cos v, -\sin v) $$
The resulting parameterizations produce opposite normal vectors; the normals point inward in the one, and outward in the other.
To get the result you want, you need the change-of-variables formula to be a smooth invertible map between the domains, and the Jacobian determinant must be positive at each point.