O'Neill's Elementary Differential Geometry contains the following exercise, 3.3 in chapter 4: Prove Corollary 3.4 ("If x and y are overlapping patches in a surface M, there exist differentiable functions $\bar{u}$ and $\bar{v}$ such that $y(u,v)=x(\bar{u}(u,v),\bar{v}(u,v))$ for all (u,v) in the domain of $x^{-1}y$." The proof is supposed to be like that for lemma 3.1, which is:

"If $\alpha:I\rightarrow M$ is a curve and has route in the image x(D) of a single patch x, then there exists unique differentiable functions $a_1$ and $a_2$ such that $\alpha(t)=x(a_1(t),a_2(t))$.

This is what I have for the proof of this lemma:

The coordinate expression is $x^{-1}\alpha$ by definition. It is differentiable since $\alpha$ is a curve. $x^{-1}\alpha$ has Euclidean coordinate functions, call them $a_1$ and $a_2$. They lie in D since $\alpha$ has a route in the image of D. By the definition of the inverse of a function, $\alpha=x(a_1,a_2)$. If $x(a_1,a_2)=x(b_1,b_2)$ then $(a_1,a_2)=(b_1,b_2)$ so $a_1$ and $a_2$ are unique.

Here is what I have for the proof of Corollary 3.4:

The coordinate expression of y is $x^{-1}y$. It is differentiable since y is a mapping. Let $a_1(u,v),a_2(u,v)$ be the Euclidean coordinates of $x^{-1}y$. They lie in D since x and y overlap. The domain of $x^{-1}y$ is the set of points in the domain of y whose images, under y, are also in the image of x. By definition of the inverse of a function, $y=x(a_1(u,v),a_2(u,v))$. If $x(a_1,a_2)=x(b_1,b_2)$ then $(a_1,a_2)=(b_1,b_2)$. So $\bar{u}=a_1$ and $\bar{v}=a_2$.

Is this proof correct?