Why is the cylinder surface on $\Bbb R^3$ orientable?

Question

Why is the cylinder surface on $\Bbb R^3$ orientable? Please can someone explain me clearly?

Answer 1

If the cylinder, $C$, is given by the equation $x^2+y^2=1$ in $\mathbb{R}^3$, we can give an atlas of $C$ by $$\begin{array}{ccc}U_1=C\setminus\{x=-1\}&&U_2=C\setminus\{x=1\}\\\varphi_1:U_1\to (-\pi,\pi)\times\mathbb{R}&&\varphi_2:U_2\to (0,2\pi)\times\mathbb{R}\\(x,y,z)\mapsto(\theta,z)&& (x,y,z)\mapsto(\theta,z)\end{array}$$

where $\theta$ is the polar angle of the point $(x,y)\in\mathbb{R}^2$. The transition functions are then given by:

$$\psi_{12}=\varphi_1\circ\varphi_2^{-1}:(0,\pi)\cup(\pi,2\pi)\times\mathbb{R}\to(-\pi,0)\cup(0,\pi)\times\mathbb{R}\\[.1in](\theta,z)\mapsto\begin{cases}(\theta,z)&0<\theta<\pi\\[.1in](\theta-2\pi,z)&\pi<\theta<2\pi\end{cases}\\[.4in]\psi_{21}=\varphi_2\circ\varphi_1^{-1}:(-\pi,0)\cup(0,\pi)\times\mathbb{R}\to(0,\pi)\cup(\pi,2\pi)\times\mathbb{R}\\[.1in](\theta,z)\mapsto\begin{cases}(\theta,z)&0<\theta<\pi\\[.1in](\theta+2\pi,z)&-\pi<\theta<0\end{cases}$$

At this point, it would be very helpful to draw some pictures of $U_1$, $U_2$, and the open subsets of $\mathbb{R}^2$ to which they are homeomorphic. This will help you visualize the transition functions.

Now, we want to find the Jacobian of the transition functions. If we write $\psi_{12}=(f_1,f_2)$, then the Jacobian is defined as follows:

$$\operatorname{Jac}(\psi_{12})=\begin{pmatrix}\frac{\partial f_1}{\partial\theta}&\frac{\partial f_1}{\partial z}\\\frac{\partial f_2}{\partial\theta}&\frac{\partial f_2}{\partial z}\end{pmatrix}$$

Looking at how we defined $\psi_{12}$, we see that

$$f_1(\theta,z)=\begin{cases}\theta&0<\theta<\pi\\\theta-2\pi&\pi<\theta<2\pi\end{cases}$$

and $f_2(\theta,z)=z$. Taking partial derivatives of these functions gives the Jacobian

$$\operatorname{Jac}(\psi_{12})=\begin{pmatrix}\frac{\partial f_1}{\partial\theta}&\frac{\partial f_1}{\partial z}\\\frac{\partial f_2}{\partial\theta}&\frac{\partial f_2}{\partial z}\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$

You can also show this is the case for $\psi_{21}$.