Parametrizations of a surface of revolution

Consider $S$ the surface of revolution obtained by rotating the planar curve $\alpha(t)=(f(t),0,g(t))$, $t\in I$ which it is contained in the halfplane $y=0, x\geq 0$. We know that $\alpha$ is regular and we have two possible types of curves, i) $\alpha$ is an embedding or ii) $\alpha$ is a simple closed curve. The surface $S$ is $X(I\times{\mathbb R})$, where $$X(t,\theta)=(f(t)\cos\theta,f(t)\sin\theta,g(t)).$$ In order to prove that $S$ is a surface, it appears the problem about how many of parametrizations are needed and how to prove that they are homeomorphisms.

First consider that $\alpha$ is an embedding. We compute the inverse function of $X$ so we will discover what is the right domain of $X$. By the injectivity, it is necessary that $\theta$ moves in an interval of length $2\pi$ at most. Thus a possibility is $X:U_1:=I\times (0,2\pi)\rightarrow X(U_1)$. The set $X(U_1)$ is an open set of $S$ because $$X(U_1)=S-\alpha(I)=S-S\cap(\{y=0,x\geq 0\}).$$ Here we use that $f(t)>0$. Because we need to cover the curve $\alpha(I)$, then the other parametrization is $Y:U_2:=I\times (-\pi,\pi)\rightarrow X(U_2)$. For $Y$ we have to remove `the other side' of $S$, that is $\Phi_\pi(\alpha(I))$, where $\Phi_\theta$ is the rotation of angle $\theta$. In other words, $Y(U_2)=S-(S\cap(\{y=0,x\leq 0\})$.

We compute the inverse of $X$ (or $Y$). We have to write $t$ and $\theta$ in terms of $x,y,z$ from the next equations $$\left\{\begin{array}{l}x=f(t)\cos\theta\\ y=f(t)\sin\theta\\ z=g(t)\end{array}\right.$$ From $x^2+y^2=f(t)^2$, we obtain $f(t)=\sqrt{x^2+y^2}$, and using that $\alpha$ is an embedding, then $t=\alpha^{-1}(\sqrt{x^2+y^2},0,z)$. For $\theta$ we have two possibilities:

1. If we want to write `something' with the arc tangent, then it would be $\theta=\mbox{arc tan}(y/x)$. But in such a case, $\theta$ is defined in $(-\pi/2,\pi/2)$. Because the initial interval is $(0,2\pi)$, we have to change the domain of $X$. We write now $X:U_1:=I\times(-\pi/2,\pi/2)\rightarrow X(U_1)$. The only difference is that we have to prove that $X(U_1)$ is an open set of $S$. But by the picture, $X(U_1)=S-(S\cap \{x\leq 0\})$. Other parametrization is with $I\times (\pi/2,3\pi/2)$ where it is possible to define the inverse of the tangent. And what about the points with $\theta=\pi/2$ or $\theta=3\pi/2$? Here, the $x$-coordinate of the point $(x,y,z)\in S$ vanishes. Now we consider that inverse of the cotangent and taking $\theta={\mbox arc cot}(x/y)$. We need two more domains, namely, $I\times (0,\pi)$ and $I\times (\pi,2\pi)$. Finally, we observe that, using the inverse functions of the tangent of the cotangent function, we need 4 parametrizations: all them write 'very similar', but  the domain goes changing.
2.  If we do not want to use trigonometric functions, we can do the following. If one looks the picture of a surface of revolution, it is clear that the parametrization $X$ can be defined in $I\times (0,2\pi)$, because the angle $\theta$ is well defined. The problem appeared in how to catch the variable $\theta$. The trigonometric functions have added a bit of confusion. Other way is the following.                                                                                                                        Consider $\beta:(0,2\pi)\rightarrow {\mathbb S}^1-\{(0,0\})$ a parametrization of the circle minus one point. The key is the following: the map $\beta$ is a homeomorphism! It is clear that $\beta$ is one-to-one and continuous. One could do the following argument. It is well known that  ${\mathbb S}^1$ minus one point is homeomorphic to the real line ${\mathbb R}$, which it is homeomorphic to the interval $(0,2\pi)$. But the problem is if $\beta$ is, indeed, a homeomorphism. The only trouble is about the inverse. But $\beta$ is an open map because the image of an interval of $(0,2\pi)$   is an open set of ${\mathbb S}^1$.             Once proved that $\beta$ is a homeomorphism, from $x=f(t)\cos\theta$ and $y=f(t)\sin\theta$ we conclude $$\theta=\beta^{-1}\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right).$$ Then $$X^{-1}(x,y,z)=\left(\alpha^{-1}(\sqrt{x^2+y^2}),\beta^{-1}\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right)\right).$$ Thus we need 2 parametrizations.

If $\alpha$ is a simple closed curve, for covering the curve $\alpha$ by embeddings, we need two parametrizations, namely, $\alpha:(0,T)\rightarrow \alpha(0,T)$ and $\alpha:(T/2,3T/2)\rightarrow \alpha(T/2,3T/2)$. Then for the surface, we need 4 parametrizations.

There is another elegant argument. We observe that when one has proved that $X$ is a parametrization, then we think $Y$ as a `rotation' of $X$. Recall that if $S$ is a parametrization and $\psi$ is a diffeomorphism of ${\mathbb R}^3$, then $\psi(S)$ is a surface. But the parametrizations of $\psi(S)$ are of type  $\psi\circ X$, where $X$ is a parametrization of $S$. With this idea in mind, consider now $\Phi_\theta$ the rotation about the $z$-axis of angle $\theta$. By the definition of $S$, we have $S=\Phi_\theta(S)$. Suppose that we have proved that $X:U_1:=I\times (-\pi/2,\pi/2)\rightarrow {\mathbb R}^3$ is a parametrization of $S$: it is only of a part of $S$, exactly, $V_1=X(U_1)$. Now we take $\theta\in{\mathbb R}$. Because $\Phi_\theta$ is a homeomorphism, it is clear that $\Phi_\theta\circ X$ satisfies the properties of a parametrization, where now the coordinate open is $\Phi_\theta(V_1)$. We point out that $\Phi_\theta(V_1)$ is an open set of $S$. Therefore, if we are going taking many $\theta$, we cover all the surface $S$ by coordinate opens of type $\Phi_\theta(U_1)$. In fact, it is enough only one rotation!, namely, $\Phi_\pi$. With this argument, the effort lies only in the first parametrization $X$: the other ones are obtained `by rotating' $X$.

Finally, we conclude that it is clear that the parametrizations $X$ hold to prove that $S$ is a surface of revolution, for example, taking very small domains, or rotations of $X$. However, the proof is a bit tedious if one wants to write precise arguments.