Manifold diffeomorphic to $\mathbb{S}^1\times\mathbb{R}$.

Question

Let $H=\{(x, y, z)\in\mathbb{R}^3 | \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}−\dfrac{z^2}{c^2}=1\}$. Prove that $H$ is a manifold diffeomorphic to $\mathbb{S}^1\times\mathbb{R}$.

Already I tried to be spread but not how to begin the diffeomorphism

Answer 1

Consider the mapping $f : \Bbb S^1 \times \Bbb R \to H$ given by $$f(x,y,z) = (ax\sqrt{1 + z^2}, by\sqrt{1 + z^2}, cz)$$ It has an inverse $g : H \to \Bbb S^1 \times \Bbb R$ given by $$g(x,y,z) = \left(\frac{cx}{a\sqrt{c^2 + z^2}}, \frac{cy}{b\sqrt{c^2 + z^2}}, \frac{z}{c}\right)$$ Indeed,

\begin{align}g(f(x,y,z)) &= g(ax\sqrt{1 + z^2}, by\sqrt{1 + z^2}, cz)\\ &= \left(\frac{cax\sqrt{1 + z^2}}{a\sqrt{c^2 + (cz)^2}}, \frac{cby\sqrt{1 + z^2}}{b\sqrt{c^2 + (cz)^2}}, \frac{cz}{c}\right)\\ &= \left(\frac{cx\sqrt{1 + z^2}}{c\sqrt{1 + z^2}}, \frac{cy\sqrt{1 + z^2}}{c\sqrt{1 + z^2}}, z\right)\\ &= (x,y,z) \end{align}

and

\begin{align} f(g(x,y,z)) &= f\left(\frac{cx}{a\sqrt{c^2 + z^2}}, \frac{cy}{b\sqrt{1 + z^2}}, \frac{z}{c}\right)\\ &= f\left(\frac{cx}{\sqrt{c^2 + z^2}}\sqrt{1 + \frac{z^2}{c^2}}, \frac{cy}{\sqrt{c^2 + z^2}}\sqrt{1 + \frac{z^2}{c^2}}, z\right)\\ &= f\left(\frac{cx}{\sqrt{c^2 + z^2}}\frac{\sqrt{c^2 + z^2}}{c}, \frac{cy}{\sqrt{c^2 + y^2}} \frac{\sqrt{c^2 + y^2}}{c}, z\right)\\ &= (x,y,z) \end{align}

Both $f$ and $g$ are smooth since their components are smooth. So $f$ is a diffeomorphism.