I understand this proof to my question. But what if I have the following definition of "orientable manifold"?
Let $M$ be a manifold of class $C^k$, where $1\leq k \leq +\infty$. An atlas $\mathcal{A}$ over $M$ is called coherent if the coordinate changes preserve the orientation of $\mathbb R^m$, that is, if $\mathrm{x}_\alpha,\mathrm{x}_\beta\in\mathcal A$, then $\mathrm{x}_\beta\circ\mathrm{x}_\alpha^{-1}$ has a positive Jacobian for every point of $\mathrm{x}_\alpha(U_\alpha\cap U_\beta)$. We will say that M is an orientable manifold if M admits a coherent atlas.
I know that every surface can be locally parameterized by $\varphi(u,v)=(f(v)\cos(u), f(v)\sin(u),g(v))$, where $0<u<2\pi$ and $a<v<b$, but how can I use the definition of an orientable manifold to prove that every surface of revolution is orientable? This is where I get a bit stuck...
Of course my idea is the following: Let $x_1(u,v)=(f(v)\cos(u), f(v)\sin(u),g(v))$ defined in $U_1=\{(u,v):0<u<2\pi,a<v<b\}$ and $x_2(u,v)=(f(v)\cos(u), f(v)\sin(u),g(v))$ defined in $U_1=\{(u,v):-\pi<u<\pi,a<v<b\}$ these two parameterizations satisfy that $x_1(U_1)\cup x_2(U_2)=M$. So taking the cards $x_1^{-1}$ and $x_2^{-1}$ the following should be fulfilled, this is where I don't know how to attack it:
- I am getting very complicated in calculating the inverse
- I am getting complicated with the issue of order since I initially gave a parameterization but to do the accounts I need cards, therefore I am getting complicated with that.
- For a moment I thought that the inverse could be $x_1^{-1}(x,y,z)=(\sqrt{x^2+y^2},\arctan(y/x))$, but for example $x_1^{-1}\circ x_1$ is not the identity, at least I see it