Surface of a cylinder is a differentiable manifold I'm struggling to understand the concept of a differentiable manifold. Lets look at an example. Let $S\subset\Bbb{R^3},S=\{(x_1,x_2,x_3)\vert \enspace x_1^2+x_2^2=1\}.$ This is a cylindrical surface with a radius of $1$. Let $g:U_1=(0,2\pi)\times\Bbb{R}\to \Bbb{R^3},$ and $h:U_2=(-\pi,\pi)\times \Bbb{R} \to \Bbb{R^3}$, such that $$g(\phi,z) = (\cos\phi, \sin\phi, z)$$
and $$h(\phi,z) = (\cos\phi,\sin\phi, z).$$
Now $S=g(U_1) \cup h(U_2).$ Since $g,h$ are continuously differentiable homeomorphisms, $S$ is a differentiable manifold. Both $g$ and $h$ look like cylindrical coordinate transforms, but why do we need two of them here? What would happen if we exclude $h$?
 A: You're right that the formulas for $g$ and $h$ are identical, but we really need both. For example, if we only used $g$, then because the domain is $(0,2\pi)\times \Bbb{R}$, this parametrization doesn't cover the entire cylinder $S$; i.e $S\setminus g(U_1)$ is non-empty (geometrically, it is a vertical line). The concept of a manifold requires us to show that around EVERY point of $S$, we can find an appropriate parametrization (very roughly speaking, if you take any "small" region of $S$, then it must "look" very nice).
But like I said, if we use $g$ alone, we fail to describe what happens at all those points on the line. This is why we need $h$ to guarantee that even at those points which we originally missed, everything is still nicely behaved.
Of course, when we use $h$, we also fail to cover the whole of $S$, we again miss out a vertical line. The thing is this is a different vertical line; so although both $g$ and $h$ miss out stuff, together their images cover $S$.
Otherwise, think of the graph of $f(x)=|x|$. This is "nice and smooth" everywhere except at the origin, where it is pointy and "badly behaved".
