On the proof that $x^2+y^2-z^2=1$ is a topological manifold

Question

Let $S=\{(x,y,z)\in\mathbb{R}^3\ |\ x^2+y^2-z^2=1\}$ and, for every $\theta \in \mathbb{R}$: $$L_\theta = \{(x,y,z)\in \mathbb{R}^3\ |\ (x-z)\cos\theta=(1-y)\sin\theta,\ (x+z)\sin\theta=(1+y)\cos\theta\}$$ One can easily prove that $L_\theta=\{(\sin2\theta,-\cos2\theta,0)+t(\cos2\theta,\sin2\theta,1)\ |\ t,\theta\in \mathbb{R}\}$ for all $\theta \in \mathbb{R}$ and that $\bigcup_{\theta \in \mathbb{R}} L_\theta=S.$

This proves that the map $\sigma:\mathbb{R}^2\rightarrow S, \ (\theta,t)\mapsto (\sin2\theta+t\cos 2\theta, -\cos 2\theta+t\sin2\theta, t)$ is well defined. I'm convinced of the fact that one can find a family $\{U_i\}_{i \in I}$ of open sets of $\mathbb{R}^2$ such that the restriction of $\sigma$ to $U_i$ (which I'll indicate with $\sigma_i:U_i\rightarrow \sigma(U_i)$) is a homeomorphism for all $i \in I$ and such that $\bigcup_{i\in I}\sigma(U_i)$, but I'm having trouble coming up with such an atlas. Does this lead to some viable solution? How would one define the charts?

Answer 1

If you don't care about constructing explicit chart maps, you could say that $S = F^{-1}(0)$ is the zero set of the polynomial equation $F(x,y,z) = x^2 + y^2 -z^2 -1$. Since $F$ is smooth, and the Jacobian $\begin{pmatrix} 2x & 2y & -2z\end{pmatrix}$ has full rank on $S$, you can locally write it as the graph of a function using the Implicit Function Theorem. That gives you a local homeomorphism to $\mathbb{R}^2$.

Answer 2

$x^2+y^2=1+z^2\subset \Bbb{R}^3$ is isomorphic to the cylinder $X^2+Y^2=1\subset \Bbb{R}^3$ through

$$(x,y,z) \to (\frac{x}{\sqrt{1+z^2}},\frac{y}{\sqrt{1+z^2}},z),\qquad (X,Y,Z)\to (X\sqrt{1+Z^2},Y\sqrt{1+Z^2},Z)$$

You need two charts to cover the cylinder $$\phi_1(u,v) = (\cos u,\sin u,v),\qquad \phi_2(u,v) = (-\cos u,-\sin u,v),\qquad (u,v)\in (-\pi,\pi)\times \Bbb{R}$$ $\phi_i^{-1}\phi_j$ is clearly smooth because $\phi_1(u,v)=\phi_2(u\pm\pi,v)$