Answer to Question #138613 in Differential Geometry | Topology for anjali g

Question #138613
Show that the circular cylinder S = {(x, y, z) ∈R^3 : y^2+z^2 = 1}can be covered by a single regular surface patch, and hence is a surface.
1
Expert's answer
2020-10-19T16:01:14-0400
"Solution"

A surface patch can be given by "(x,y)\\mapsto (\\frac{x}{r},\\frac{y}{r}, tan (r-\\pi\/2))."

However we must be careful on how we define the open set that this is defined over

Here "\\{x,y|\\sqrt {x^2+y^2} \\in (0,\\pi)\\}."

Also, ."r=+\\sqrt{x^2+y^2}".

Clearly, the map is smooth since "r\\neq 0." Also "r<\\pi."

Hence "tan(r-\\pi\/2)" tan is well defined.

For surjection



Let "(x,y,z)" be a given point on the cylinder. Since tan is continuous and injective on the range "(-\\pi\/2,\\pi\/2)" and unbounded its surjective, and hence we get "r" such that "tan(r-\\pi\/2)=z." So take "a=yr, b=zr." Now "\\sqrt{y^2+z^2}=1\\Rightarrow y,z\\leq 1""\\sqrt{y^2+z^2}=1\\Rightarrow y,z\\leq 1" . Hence "a,b\\leq r<\\pi". Hence the pre-image lies in our given domain.



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