Take $S=\{(x,y,z)\in \mathbb{R}^3\ |\ x^2+y^2=1\}$ and define $\sigma:\mathbb{R}^2\rightarrow S,\ \sigma(u,v)=(\cos u, \sin u, v).$
If $U= \{ (u,v)\in \mathbb{R}^2\ |\ 0<u<2\pi\}$, then $\sigma_{|U}:U\rightarrow \sigma(U)$ is a homeomorphism. Why is that?
More generally I don't know how to prove that trigonometric parametrizations such as this have continuous inverses (on the right domain and codomain). Can someone help?
Define $\phi : \mathbb R \to S^1, \phi(u) = e^{iu} = (\cos(u),\sin(u))$. Here $S^1 =\{(x,y) \in \mathbb R^2 \mid x^2 + y^2 = 1 \} \subset \mathbb R^2$ is the unit circle. It is well-known that $\phi(u) = \phi(u')$ iff $u - u' = 2k\pi$ for some $k \in \mathbb Z$ and $\phi([0,2\pi)) = S^1$. Thus $\phi \mid_{[0,2\pi)} : [0,2\pi) \to S^1$ is a continuous bijection (but of course no homeomorphism since the domain is non-compact, but the range is compact).
We claim that $\phi \mid_{(0,2\pi)} : (0,2\pi) \to \phi((0,2\pi)) = S^1 \setminus \{(1,0)\}$ is a homeomorphism. To show this, it suffices to show that $\phi \mid_{(0,2\pi)}$ is an open map. So let $V \subset (0,2\pi)$ be open. Then $C = [0,2\pi] \setminus V$ is compact, hence $\phi(C) \subset S^1$ is compact and $W = S^1 \setminus \phi(C)$ is open in $S^1$. But $\phi(C) = \phi((0,2\pi) \setminus V)) \cup \phi(\{0,2\pi\}) = \phi((0,2\pi) \setminus V)) \cup \{(1,0)\}$, thus $$W = (S^1 \setminus \{(1,0)\}) \setminus \phi((0,2\pi) \setminus V)) = \phi((0,2\pi)) \setminus \phi((0,2\pi) \setminus V)) \\ = \phi(0,2\pi) \setminus ((0,2\pi) \setminus V)) = \phi(V). $$ We have $\sigma \mid_U = \phi \mid_{(0,2\pi)} \times id_{\mathbb R}$ which shows that it is a homeomorphism.
