Proof that the infinite cylinder is a regular surface. I have to proof that the circular cylinder $M=\{(x,y,z)\in\mathbb{R}^3\mid x^2 + y^2 = r^2\}$ is a regular surface, where $r$ is a constant, $r>0$.
Then I have to see also that $\mathrm x\colon \mathbb{R}^2 \to \mathbb{R}^3, \mathrm x(u,v)= (r \cos u, r\sin u, v)$ is a parameterization of $M$ when we restrict the domain of definition to a suitable open $U$ in $\mathbb{R}^2$ . What part of $M$ covers? Find another parameterization that, along with it, cover to all $M$.
With parameterization I mean:
$(i)\,U$ in $\mathbb{R}^2$ is open and $\mathrm x(U)$ in $M$ is open.
$(ii)\,\mathrm x\colon U \subseteq\mathbb{R}^2 \to \mathbb{R}^3$ is differentiable.
$(iii)\,\mathrm x\colon U\subseteq \mathbb{R}^2 \to x(U)$ in $M$ is an homeomorphism.
$(iv)\,\mathrm x$ is regular.
The hardest part for me is the first one, to see if is a regular surface. And in order to see if is a parameterization the part (ii) and (iv) I know how to do.
Thank you.
 A: Quasi-Definition. Given a function $F:U \subset \mathbb R^n \to \mathbb R^m$ ($U$ an open set of $\mathbb R^n$), we say that $p \in U$ is a critical value  of $F$ if all its partial derivates vanishes at $p$, and $F(p)$ is called a critical value of F. A point of $\mathbb R^m$ wich is not a critical value is called a regular value of $F$.
Proposition. If $f:U\subset \mathbb R^3 \to \mathbb R$ is a differentiable function and $a\in f(U)$ is a regular value of $f$, then $f^{-1}(a)$ is a regular surface in $\mathbb R^3$.
Lets take $f:M \to \mathbb R$ with $f(x,y,z)=x^2+y^2-r^2$. First of all $f$ is differentiable, so we can calculate its critical points. So $$\frac{\partial f}{\partial x}=2x$$ $$\frac{\partial f}{\partial y}=2y$$ $$\frac{\partial f}{\partial z}=0$$ now the critical points of $f$ are of the form $(0,0,z), z\in \mathbb R$.Then $f(0,0,z)=-r^2 \neq 0$. So 0 is a regular value of $f$ and the point $(0,0,z) \notin f^{-1}(0)$. But notice that $f^{-1}(0)=M$ it follows that $m$ is a regular surface.
For the second part you need to prove that $x$ is differentiable and one-to-one, and $\frac{\partial f}{\partial x} \times \frac{\partial f}{\partial y}$ is nowhere zero.
I'm don't know the aswer. But if we define $U=(0,2\pi)\times \mathbb R$ then $x:U \to \mathbb R^3$ covers M except the rect line $x=r$ so we need another paremetrization, let $x':U \to \mathbb R^3$ with $x'(u,v)=(r\sin u,r\cos u, v)$ this one covers M except the rect line $y=r$.So the part that $x$ do not cover, $x'$ cover it, and viceversa.
A: I think you can calculate the Jaccobi matrix of $r$: $D(r)$. And it is easy to see that the rank of Jaccobi $\geq{2}$. That is to say, $r$ is an immersion. Then I think you can use the property of immersion to prove this.
