Considering the following definitions extracted of do Carmo's book Differential Geometry of Curves and Surfaces, prove that a surface of revolution is regular surface.
Def. 1: A parametrized differentiable curve is a differentiable ($C^\infty$) map $\alpha:I\to\mathbb R^3$ of an open interval $I$ of $\mathbb R$ into $\mathbb R^3$. $\alpha$ is said to be regular if $\alpha'(t)\neq 0$ for all $t\in I$.
Def. 2: A subset $S\subset \mathbb{R}^3$ is a regular surface if, for each $p\in S$, there exists a neighborhood $V$ in $\mathbb{R}^3$ and a map $\mathbf{x}: U\to V\cap S$ of an open set $U\subset \mathbb{R}^2$ onto $V\cap S\subset \mathbb R^3$ such that
- $\mathbf{x}$ is differentiable;
- $\mathbf{x}$ is a homeomorphism;
- For each $q\in U$, the differential $\mathrm d\mathbf x_q:\mathbb R^2\to\mathbb R^3$ is one-to-one.
Def. 3: A surface of revolution is a set $S\subset \mathbb R^3$ obtained by rotating a regular connected plane curve $C$ about an axis in the plane which does not meet the curve.
do Carmo's considerations:
Considering the $xz$ plane as the plane of the curve and $z$ axis as the rotation axis, let $$ x=f(v), z=g(v), a<v<b, f(v)>0,$$ be a parametrization for $C$ and denote by $u$ the rotation angle about the $z$ axis. Thus, we obtain a map $$\mathbf x(u,v)=(f(v)\cos u, f(v)\sin u, g(v))$$ from the open set $U=\{(u,v)\in\mathbb R^2; 0<u<2\pi, a<v<b\}$ into $S$.
To show that $\mathbf x$ is a parametrization of $S$ we must check conditions 1, 2 and 3 of Def. 2.
My attempt solution:
Since $U$ is an open set of $\mathbb R^2$ and $\mathbf x(U)=S$, then for all $p\in S$ there exist $q\in U$ such that $\mathbf x(q)=p.$
- $\mathbf x$ is differentiable because is differentiable each coordinate function of $\mathbf x$, for all $q\in U$;
- $\mathrm d\mathbf x_q$ is one-to-one because $$\mathrm d\mathbf x_q=\begin{bmatrix} -f(v)\sin u & f'(v)\cos u\\ f(v)\cos u &f'(v)\sin u \\ 0 & g'(v) \end{bmatrix},$$ so $$\frac{\partial(x,y)}{\partial(u,v)}=-f'(v)f(v)$$ and $$\frac{\partial(y,z)}{\partial(u,v)}=g'(v)\cos u.$$ Therefore $\mathrm d\mathbf x_q$ is one-to-one, because $C$ is regular [$(f'(v))^2+(g'(v))^2\neq 0$] .
- $\mathbf x$ is continuous because the coordinates functions are continuous. To show injectivity, let $(u,v),(z,w)\in U$ be such that $\mathbf x(u,v)=\mathbf x(z,w)$. Then $$\begin{cases}f(v)\cos u=f(w)\cos z\\ f(v)\sin u=f(w)\sin z\\ g(v)=g(w)\end{cases}.$$ So we have $(f(v))^2=(f(w))^2$, that implies $f(v)=f(w)$ since $f(v), f(w)>0$. Therefore $$\begin{cases}\cos u=\cos z\\ \sin u=\sin z\end{cases}\implies u=z.$$
But I can't show that $v=w$. I got stuck here. The continuity of $\mathbf x^{-1}$ is shown by do Carmo.
