Vtyjkyuk

First some definitions.

A surface on $R^3$ is a set $Msubset R^3 $with the following properties.

1) For every point on the surface there exist an open nbrhood $V$ for which there exists a $C^infty$, 1-1 and onto function $φ:U rightarrow Vcap M$ so that the inverse of $φ$ is continuous for some $U subset R^2$ open.

2) The $Dφ$ (jacobian) has maximal degree. $φ$ is called local parametrization.

Now the problem. Prove that the cylinder $C:x^2+y^2=1$ is a smooth surface globally parametrized by the $φ(r,θ)=(cos(theta),sin(theta),ln(r))$ . $ φ: R^2 setminus 0 rightarrow C $.(polar coordinates)

Sollution.

1)First im not given any definition for a smooth surface so i believe if the parametrization is smooth then the surface is too. So since $φ$ is smooth hence C is smooth.

2)Now i also dont have a definition for a global parametrization so i believe it means ( and correct me please if im wrong) that if i can find an open set $V$ such that $Vcap M=M$ and an $φ:U rightarrow Vcap M $ that checks the requirements above the $φ$ is indeed a global parametrization.

Now the given parametrization has a jacobian of degree 2. So it checks that. It is continuous and smooth. it is 1-1 because of logr part that guarantees you that it will always give a different point and it is onto because $cos^2θ+sin^2θ=1$. Now i need to find a good $V$ open set that does the work(and does it globally so i though of the whole $R^3$ but it feels like cheating can i do that?) and to find the inverse function to make sure it is continuous. That is where im stuck.
Also Please can you check if my definitions are correct? And can i find them in a proper textbook cause all these come from my teachers notes and didnt find anything on the internet that looked similar when i was looking for cylinder parametrization and a proof for it.

And is this the "real" proof that the cylinder is a surface. (Meaning that there exists the above $φ$ given the above definitions)

You can parametrize the cylinder with 4 charts. One for each hemisphere like so.