We give the patch as "(x,y)\\mapsto (\\frac{x}{r},\\frac{y}{r}, tan (r-\\pi\/2))." 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 "(a,b,c)" 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)=c." So take "x=ar, y=br." Now "\\sqrt{a^2+b^2}=1\\Rightarrow a,b\\leq 1" . Hence "x,y\\leq r<\\pi". Hence the pre-image lies in our given domain.
b) We append the diagram below.
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