Diffeomorphism between a regular surface and the plane 
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*Do Carmo states that (example 2, page 74) if $\mathbf x: U\subset\mathbb R^2\rightarrow S$ is a parameterization, then $\mathbf x^{-1}: \mathbf x(U)\rightarrow \mathbb R^2$ is differentiable. Why is this true, please? Thank you!

*Do Carmo continued to say that (same example) for any $p\in\mathbf x(U)$ and any parametrization $\mathbf y: V\subset\mathbb R^2\rightarrow S$ at $p$, we have that $\mathbf x^{-1}\circ\mathbf y: \mathbf y^{-1}(W)\rightarrow \mathbf x^{-1}(W)$, where $W=\mathbf x(U)\cap\mathbf y(V)$ is differentiable. This is understandable since it directly is from the definition of a differentiable function from a regular surface to another regular surface. However, Do Carmo said that this shows that $U$ and $\mathbf x(U)$ are diffeomorphic. Why is this true, please? Thank you!

 A: First part of your question
Basically it's all to do with the inverse function theorem. But more explicitly: 
Let $\mathbf{x}:U \subset \mathbb{R}^2 \rightarrow S$ be a parameterization and $q$ be a point in $U$.
Define $\mathbf{F}:U \times I \rightarrow \mathbb{R}^3$ by 
$$
\mathbf{F}(u,v,t):= \mathbf{x}(u,v) + t\cdot \hat{e}_3, \hspace{1in} t \in I.
$$
Another way to think about $\mathbf{F}$ is as follows: 
If we know that $\mathbf{x}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle$, then 
$$
\mathbf{F}(u,v,t) = \langle x(u,v), y(u,v), z(u,v) +t \rangle.
$$
So, given that $\mathbf{x}(u,v)$ is a parameterization, we know that it must be differentiable, thus $\mathbf{F}(u,v,t)$ is also differentiable.
Furthermore, since $\mathbf{x}(u,v)$ is a parameterization, we can assume without loss of generality, that 
$$
\dfrac{\partial(x,y)}{\partial(u,v)}(q) \not= 0.
$$
Now take the differential of $\mathbf{F}(u,v,t)$ at $q$: 
$$
d\mathbf{F}_q = \left| \begin{array}{ccc}
x_u & x_v & 0 \\
y_u & y_v & 0 \\
z_u & z_v & 1 \end{array} \right|_{q} = \dfrac{\partial(x,y)}{\partial(u,v)}(q).
$$
Thus $d\mathbf{F}_q \not=0$. This satisfies the condition for the inverse function theorem. Hence, there exists an open neighborhood, say $M$, around $\mathbf{F}(q) \in \mathbb{R}^3$ such that $\mathbf{F}^{-1}$ exists and is differentiable. 
But if we set $t=0$ in the expression for $\mathbf{F}(u,v,t)$, then $\mathbf{x}(u,v) = \mathbf{F}(u,v,0)$. Hence $\mathbf{x}^{-1}$ is differentiable around $q$. Since $q$ was arbitrarily chosen, we know that this holds for all points in $U$. Hence $\mathbf{x}^{-1}:\mathbf{x}(U) \rightarrow R^2$ is differentiable.
Second part of your question 
If there exists a diffeomorphism between $U$ and $\mathbf{x}(U)$, then $U$ and $\mathbf{x}(U)$ are diffeomorphic. But we already know that $\mathbf{x}(U)$ is differentiable and invertible by hypothesis; and we just showed that $\mathbf{x}^{-1}$ exists and is differentiable. Thus $\mathbf{x}(u,v)$ is a diffeomorphism, so $U$ and $\mathbf{x}(U)$ are diffeomorphic to each other. 
Remarks 
I like to think about $U$ as a sheet of rubbery paper, and $\mathbf{x}(U)$ as me bending the paper to look like the surface. Then $\mathbf{x}^{-1}$ is just me bending the paper back into its' original shape. Moreover, you want to make sure that your logic does not depend on this special sheet of paper; so you take another rubbery sheet of paper, $V$, and bend that into the shape of the surface as well by $\mathbf{y}(V)$. Now I like to think that if I can bend the original rubbery sheet of paper, $U$, into the other rubbery sheet of paper, $V$, then I have all my bases covered and my logic does not depend on any particular sheet. The Change of Parameters theorem basically tells us that you can bend $U$ into $V$ by $\mathbf{y}^{-1} \circ \mathbf{x}: U \rightarrow V$ or vice-versa if you want to bend $V$ into $U$ by $\mathbf{x}^{-1} \circ \mathbf{y}: V \rightarrow U$. 
