How's the first fundamental form of a surface parameterization with a diffeomorphism? Let $\sigma: U\subset\Bbb{R^2}\to V\subset S$ a parameterization of a surface $S$, and let $g:\tilde{U}\to U$ be a diffeomorphism between open sets of $\Bbb{R^2}$. I need to obtain a formula for the coefficients of the first fundamental form associated to $\sigma\circ g$ in terms of the coefficients of the first fundamental form of $\sigma$.
Since $g$ is a diffeomorphism, then $g^{-1}:U\to\tilde{U}$ exists and is differentiable.
If $g^{-1}(u,v)=(\tilde{u},\tilde{v})$, so $g(\tilde{u},\tilde{v})=(u,v)$, then $\sigma(u,v)=[\sigma\circ(g\circ g^{-1})](u,v)=(\sigma\circ g)(\tilde{u},\tilde{v})$ but I'm stuck at this point. 
Can I say after that $\frac{\partial\sigma}{\partial u}=\frac{\partial(\sigma\circ g)}{\partial\tilde{u}}$?
I think this is wrong and I need to consider $g^{-1}$ but I'm not sure.
 A: It is worthwhile to consider this picture

where we can plainly see the maps involved.
We have $\partial_1=J\sigma e_1$ and $\partial_2=J\sigma e_2$ 
as the generators of the tangent space and
$\sigma\circ g$, which is another parametrization, serves to define another
tangent base 
$$\tilde{\partial_1}=J(\sigma\circ g)e_1\quad,\quad \tilde{\partial_1}=J(\sigma\circ g)e_2$$
But by the Chain Rule $J(\sigma\circ g)=J\sigma\cdot Jg$ then
$$J(\sigma\circ g)e_i=(J\sigma\cdot Jg )e_i$$ 
which evaluated at some point $a$ in the domain of $g$ would give
$$J(\sigma\circ g)|_ae_i=(J\sigma|_{g(a)}\cdot Jg|_a)e_i$$ 
If we choose the coordinate functions $\tilde{v},\tilde{w}$ in the domain of $g$ and $v,w$ in the domain of $\sigma$, then 
$$Jg=
\left[\begin{array}{cc}
\dfrac{\partial v}{\partial\tilde v}&\dfrac{\partial v}{\partial\tilde w}\\
\\
\dfrac{\partial w}{\partial\tilde v}&\dfrac{\partial w}{\partial\tilde w}
\end{array}\right]$$
So $$Jge_1=
\dfrac{\partial v}{\partial\tilde v}e_1+
\dfrac{\partial w}{\partial\tilde v}e_2
\quad ,\quad Jge_2=
\dfrac{\partial v}{\partial\tilde w}e_1+
\dfrac{\partial w}{\partial\tilde w}e_2$$ 
which plugged above
\begin{eqnarray*}
\tilde{\partial_1}&=&
J\sigma\left(\dfrac{\partial v}{\partial\tilde v}e_1+
\dfrac{\partial w}{\partial\tilde v}e_2\right)\\
&=&\dfrac{\partial v}{\partial\tilde v}J\sigma e_1+
\dfrac{\partial w}{\partial\tilde v}J\sigma e_2\\
\\
&=&
\dfrac{\partial v}{\partial\tilde v}\partial_1+
\dfrac{\partial w}{\partial\tilde v}\partial_2
\end{eqnarray*}
and similarly 
$$\tilde{\partial_2}=
\dfrac{\partial v}{\partial\tilde w}\partial_1+
\dfrac{\partial w}{\partial\tilde w}\partial_2$$
So the first fundamental form will be
$$\tilde{g}_{11}=
\left(\dfrac{\partial v}{\partial\tilde v}\partial_1+
\dfrac{\partial w}{\partial\tilde v}\partial_2\right)
\bullet
\left(\dfrac{\partial v}{\partial\tilde v}\partial_1+
\dfrac{\partial w}{\partial\tilde v}\partial_2\right)$$
i.e.
$$\tilde{g}_{11}=
\left(\dfrac{\partial v}{\partial\tilde v}\right)^2g_{11}+
2\dfrac{\partial v}{\partial\tilde v}
\dfrac{\partial w}{\partial\tilde v}g_{12}+
\left(\dfrac{\partial w}{\partial\tilde v}\right)^2g_{22},
$$
similarly with
$$\tilde{g}_{12}=
\dfrac{\partial v}{\partial\tilde v}\dfrac{\partial v}{\partial\tilde w}
g_{11}+
2\dfrac{\partial v}{\partial\tilde v}\dfrac{\partial w}{\partial\tilde w}g_{12}+
\dfrac{\partial w}{\partial\tilde v}\dfrac{\partial w}{\partial\tilde w}
g_{22}.
$$
$$\tilde{g}_{22}=
\left(\dfrac{\partial v}{\partial\tilde w}\right)^2g_{11}+
2\dfrac{\partial v}{\partial\tilde w}\dfrac{\partial w}{\partial\tilde w}g_{12}+
\left(\dfrac{\partial w}{\partial\tilde w}\right)^2g_{22}.
$$
