exterior derivative of 1-form on surface for non-regular mapping? I am studying alone the elementary differential geometry written by Barrett O'Neill. 
This time I'm totally lost. I can't even get any idea from the hint...;(
How is the problem related to Lemma 4.5 ???
I need help...
(I understand how to prove the lemma 4.5. I want the exercise 4.4.6 to be proved...)
image link 1 - the problem 4.4.6, p165
image link 2 - Lemma 4.5, p 161
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Exercise 4.4.6
Let y:E → M be an arbitrary mapping of an open set of $R^2$ into a surface M. If $\phi$ is a 1-form on M, show that the formula $$d\phi(y_u,y_v)=\frac{\partial}{\partial u}(\phi(y_v))-\frac{\partial}{\partial v}(\phi(y_u))$$
is still valid even when y is not regular or one-to-one.
(Hint: In the proof of Lemma 4.5, check that equation (3) is still valid in this case.)
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Lemma 4.5
Let $\phi$ be a 1-form on M. If x and y are patches in M, then $$d_x\phi=d_y\phi$$on the overlap of x(D) and y(E).
(proof)
y=y(u,v) and y=x($\bar u,\bar v)$
$$\frac{\partial}{\partial u}(\phi(y_v))-\frac{\partial}{\partial v}(\phi(y_u))=J\{\frac{\partial}{\partial \bar u}(\phi(x_v))-\frac{\partial}{\partial \bar v}(\phi(x_u))\}$$
where $J=(\frac{\partial \bar u}{\partial u})(\frac{\partial \bar v}{\partial v})-(\frac{\partial \bar u}{\partial v})(\frac{\partial \bar v}{\partial u})$
(this is equation 3 of Lemma 4.5)
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Definition 4.4
Let $\phi$ be a 1-form on a surface M. Then the exterior derivative d$\phi$ of $\phi$ is the 2-form such that for any patch x in M,
$$d\phi(x_u,x_v)=\frac{\partial}{\partial u}(\phi(x_v))-\frac{\partial}{\partial v}(\phi(x_u))$$
 A: Choose a point $(u_0,v_0)$ in $E$. Define $\mathbf{p}=\mathbf{y}(u_0,v_0)\in M$. Since $M$ is a surface, we can choose a patch $\mathbf{x}:D\to\mathbf{R}^3$ in $M$ satisfying $(u_0,v_0)\in D$ and $\mathbf{x}(u_0,v_0)=\mathbf{p}$. Consider the composite $\mathbf{x}^{-1}\mathbf{y}:\mathbf{y}^{-1}(\mathbf{x}(D))\to D$. If $\bar{u}$ and $\bar{v}$ are the Euclidean coordinate functions of $\mathbf{x}^{-1}\mathbf{y}$, then $\mathbf{y}=\mathbf{x}\mathbf{x}^{-1}\mathbf{y}=\mathbf{x}(\bar{u},\bar{v})$. Then deduce by the chain rule that
\begin{aligned}
\mathbf{y}_u=\frac{\partial\bar{u}}{\partial u}\mathbf{x}_u+\frac{\partial\bar{v}}{\partial u}\mathbf{x}_v,\\
\mathbf{y}_v=\frac{\partial\bar{u}}{\partial v}\mathbf{x}_u+\frac{\partial\bar{v}}{\partial v}\mathbf{x}_v,
\end{aligned}
where $\mathbf{x}_u$ and $\mathbf{x}_v$ are henceforth evaluated on $(\bar{u}, \bar{v})$. Then we have
$$(d\phi)(\mathbf{y}_u,\mathbf{y}_v)=J(d\phi)(\mathbf{x}_u,\mathbf{x}_v)$$
where $J$ is the Jacobian $(\partial\bar{u}/\partial u) (\partial\bar{v}/\partial v)-(\partial\bar{u}/\partial v) (\partial\bar{v}/\partial u)$. Thus all we need is the equation
$$\frac{\partial}{\partial u}(\phi(\mathbf{y}_v))-\frac{\partial}{\partial v}(\phi(\mathbf{y}_u))=J\left\{ \frac{\partial}{\partial \bar{u}}(\phi(\mathbf{x}_v))-\frac{\partial}{\partial \bar{v}}(\phi(\mathbf{x}_u)) \right\}.$$
We can verify this equation is still valid in analogous manner from the proof of Lemma 4.5 of the textbook.
