First some definitions.
A surface on $R^{3}$ is a set $M\subset 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 $φ:U \rightarrow V\cap M$ 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 $V\cap M=M$ and an $φ:U \rightarrow V\cap 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.