I somewhere found that we can take $U$ an annulus instead of a disc
where
$U=\{ (u,v): 0 < u^2+v^2 < π \}$. Can anyone please explain me that how a cylinder can be covered with a single surface patch from $U$?
[Math] Circular cylinder $S=\{ (x,y,z) : x^2+y^2=1 \}$ can be covered with a single surface patch.
differential-geometrysurfaces
Related Solutions
This is only a topology problem. Since that $M\subset \mathbb{R}^{3}$, $M$ have a relative topology induced by the topology on $\mathbb{R}^{3}$, $i.e.$, the open sets (or open neighborhood) $V$ on $M$ are open subsets on $\mathbb{R}^{3}$ intersection $M$ ($V=U\cap \mathbb{R}^{3}$). If $\chi: D\rightarrow M$ is a proper patch then this is a continuous function one-to-one with continuos inverse (topologically speaking), then $\chi$ is a homeomorphism between $D$ and $M$. Then $\chi(E)$ is open if $E\subset D$ is open. Another way is, W.L.G. $\chi(E)\subset \chi(D)\cap\rho(F)$ with $\rho:F\rightarrow M$ other proper patch. By your corolary $(\rho^{-1}\circ\chi)(E)$ is open since $\rho^{-1}\circ\chi$ is a difeomorphism on $\mathbb{R}^{3}$ and then $\rho\circ(\rho^{-1}\circ\chi)(E)$ is open on $M$, but $\rho\circ(\rho^{-1}\circ\chi)(E)=\chi(E)$ since $\rho$ and $\chi$ are one-to-one.
Your question isn't clear, but I assume you mean that for a cylinder you can cut a piece of paper into a single shape and fold (roll) it into a cylinder, but that you cannot do that for a sphere. [For the cylinder, the flat shape is a rectangle with two disks touching opposite sides of length equal to the circumference of each disk.]
Here's is the case for a cone:
Here's the case for a cube:
The reason for this is that the Gaussian curvature of the cylinder is zero everywhere, whereas for a sphere it is zero nowhere.
A surface of zero Gaussian curvature (such as a cone, cube, etc.) can be "cut and flattened out onto a plane" (and the converse). A surface with non-zero Gaussian curvature (such as a sphere, or ellipsoid) cannot.
Best Answer
Any point in the annulus $U$ is uniquely of the form $(t \cos \theta, t \sin \theta)$ for some real $t \in (0,\sqrt{\pi}), \theta \in [0,2\pi).$ Map this point to the point of the cylinder $(x,y,z)=(\cos \theta, \sin \theta, \cot t^2).$ This is clearly a subset of the cylinder as it satisfies $x^2+y^2=1.$ Also, because $\theta$ ranges in $[0,2\pi),$ for any fixed $z$ the entire slice of the cylinder at that $z$ level gets covered. Finally, because the cotangent of $t^2$ for $t \in (0,\sqrt{\pi})$ takes on every real value, every level $z$ indeed gets hit, showing the result of mapping the annulus as above covers the whole cylinder.