Writing differentiable maps on surfaces

In calculus it is usual to work with smooth functions in terms of `variables', I mean, something as $f(x,y,z)=x^2+\sin(z)+e^y$, in terms of `x', `y' and `z'. However working on surfaces, sometimes (or many), we prefer do not use `variables', specially when we need to compute the derivative of the function. The next example clarifies this issue. 

If $S$ is a surface, define $$f:S\rightarrow{\mathbb R},\ f(p)=|p|^2=\langle p,p\rangle.$$ Here we use `p' instead of the variables. This function measures the square of the distance of the point $p$ to the origin of ${\mathbb R}^3$. We observe that if $p=(x,y,z)$, then $f(x,y,z)=x^2+y^2+z^2$, which is a known differentiable function in ${\mathbb R}^3$, but now $f$ is defined on a surface. If we want to prove that $f$ is differentiable on $S$, first we consider $F:{\mathbb R}^3\rightarrow{\mathbb R}$ the function $F(p)=\langle p,p\rangle$. Since $p\mapsto p$ is the identity, which is differentiable, then $F$ is noting the scalar product of a differentiable map by itself. Then $F$ is differentiable. Finally, $f=F_{|S}$, that is, the restriction on $S$ of a differentiable map of ${\mathbb R}^3$. This proves definitively that $f$ is differentiable.

Other example is the height function. Let $a\in {\mathbb R}^3$ be a unit vector and define $$f:S\rightarrow{\mathbb R},\ f(p)= \langle p,a\rangle.$$ This function measures the square of the distance of the point $p$ to the vector plane $\Pi$ orthogonal to $a$. For this reason, it is named height function. If we write in coordinates and $p=(x,y,z)$, we have $f(x,y,z)=a_1 x+a_2 y+a_3 z$, where $a=(a_1,a_2,a_3)$.  If we want to prove that $f$ is differentiable on $S$ without the use of `x's', define $F:{\mathbb R}^3\rightarrow{\mathbb R}$ the function $F(p)=\langle p,a\rangle$. Since $p\mapsto p$ and $p\mapsto a$ are differentiable maps, then  $F$ is the scalar product of two differentiable vector maps, so $F$ is differentiable. Finally, $f=F_{|S}$, proving that $f$ is differentiable.  

Two possible definition of differentiability between two surfaces

The definition of a differentiable map between two surfaces given in the course is extrinsic. I explain it. Consider  $f:S_1\rightarrow S_2$ a map between two surfaces and $p\in S_1$. Then $f$ is differentiable at $p$ if $i\circ f\circ X:U\rightarrow{\mathbb R}^3$ is differentiable at $q=X^{-1}(q)$, where $i:S_2\rightarrow{\mathbb R}^3$ is the inclusion map (definition I). Here we use strongly that $S_2$ is included in Euclidean space ${\mathbb R}^3$. If one changes the viewpoint, one would request that the definition does not depend if $S_2$ is or is not included in ${\mathbb R}^3$, but only on $S_2$, that is, an intrinsic definition. Then the natural way to do it is by means of parametrizations and the definition would be: $f$ is differentiable at $p$ if $Y^{-1}\circ f\circ X:U\rightarrow W$ is smooth at $q=X^{-1}(p)$, where $X:U\rightarrow S_1$ and $Y_W\rightarrow S_2$ are parametrizations around $p$ and $f(p)$ respectively (definition II). Now it is not important if the surface is included in Euclidean space. 

This allows to extend the above definition to object with similar properties than surfaces, that is, objects with a set of parametrizations between open sets of ${\mathbb R}^n$ and open sets of the object. Then it will appear the concept of manifold of dimension $n$.

Returning, we prove that both definition are equivalents. 

1. (II) $\Rightarrow$ (I). Suppose a such $f$ which is differentiable at $p$ with definition II. When we consider a parametrization $X$ around $p$, then $i\circ f\circ X=(i \circ Y)\circ (Y^{-1} \circ f\circ X)$ and thus, it is the composition of two smooth maps between open sets of Euclidean spaces.
2. (I) $\Rightarrow$ (II). Suppose $f$ which is differentiable at $p$ with definition I, that is, we know $i\circ f\circ X$ is smooth at $q$ for any $X$. Without loss of generality, and fi $Y=(Y_1,Y_2,Y_3)$, we suppose that $$\left|\begin{array}{cc}\frac{\partial Y_1}{\partial u}&\frac{\partial Y_2}{\partial u}\\  \frac{\partial Y_1}{\partial v}& \frac{\partial Y_2}{\partial v}\end{array}\right|\not=0.$$ The Inverse function theorem asserts that the function $$(Y_1,Y_2):W'\rightarrow O', (u,v)\mapsto (Y_1(u,v),Y_2(u,v))$$ is a diffeomorphism between suitable open sets of ${\mathbb R}^2$. Let $\phi=(Y_1,Y_2)^{-1}$. If $(i\circ f\circ X)=(f_1,f_2,f_3)$, then $$Y^{-1} \circ f\circ X (u',v')=\phi^{-1} (f_1(u',v'),f_2(u',v'')),$$ which is differentiable because it is the composition of two differentiable maps.

Finally, we will adopt the definition I because it is more intuitive, although we are loosing `generality'.

Using the theory on differentiability for the properties of differentiability on surfaces

Almost all properties on the differentiability of maps on surfaces are proved by the analogous properties of differentiable maps between open sets of Euclidean spaces. I point out two of them

1. A parametrization of a surface is differentiable. Here we are saying that the parametrization $X_U\subset{\mathbb R}^2\rightarrow V\subset S$ is differentiable, where $V$ is an open set of a surface $S$. In order to clarify the notation, we stand for $Y$ the above map, and $X:U\rightarrow {\mathbb R}^3$ the parametrization. In fact, $Y$ is noting the restriction of $X$ into the codomain. Because $Y$ arrives to a surface, $Y$ is differentiable if $i\circ Y: U\rightarrow{\mathbb R}^3$ is smooth. But this map is just $X$, which it is smooth because is the second property of a parametrization.
2. The inverse of a parametrization is differentiable. Here we mean $X^{-1}:V\rightarrow U\subset{\mathbb R}^2$ is differentiable. Now $X^{-1}$ is a map whose domain is a surface, in fact, the open set $V$ of $S$, which is indeed a surface. By the definition, we have to prove that $X^{-1}\circ Z$ is smooth for some parametrization of $S$. Here we take $Z=X$. Then $X^{-1}\circ X$ is the identity map on the open set $U$, which is trivially smooth.