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

Question

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.

Answer 1

$$Solution$$

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.