Two hyperbolas in $\mathbb R^2$ are diffeomorphic, but the corresponding hyperbolic surfaces in $\mathbb R^3$ are not. Is there a deeper reason? Consider the two hyperbolas $x^2-y^2=1$ and $x^2-y^2=-1$. Looking at these pictures, they are clearly diffeomorphic via $(x,y)\mapsto (y,x)$, which is also easily verified through a calculation.
 
However, consider these two hyperbolic surfaces with apparently the same structure: $\rho^2-z^2=1$ and $\rho^2-z^2=-1$ in cylindrical coordinates ($\rho^2=x^2+y^2$). One is connected, while the other is not, so they aren't homeomorphic, not to mention diffeomorphic.
 
My question:
Is there some deeper reason why this is so? Of course I can construct paths connecting any two points on the first surface, and I can show that the second surface is non-trivially covered by the upper and lower half space, which don't intersect. So proving that these aren't diffeomorphic is not hard. What I'm interested in is a deeper reason for the apparent difference between the 2d and 3d case.
 A: $\newcommand{\Reals}{\mathbf{R}}$Let $m$ and $n$ be positive integers, $(x, y)$ a general point of $\Reals^{m} \times \Reals^{n}$, and consider the "hyperboloid"
$$
H_{\pm}^{m,n} = \{(x, y) : |x|^{2} - |y|^{2} = \pm 1\}.
$$
Since swapping $x$ and $y$ changes the sign of the condition defining the hyperboloid, we have $H_{\pm}^{m,n} \simeq H_{\mp}^{n,m}$.
For $k \geq 1$, let $S^{k-1}$ denote the set of unit vectors in $\Reals^{k}$ and let $B^{k}$ denote the open unit ball in $\Reals^{k}$. As shown below, $H_{+}^{m,n}$ is diffeomorphic to $S^{m-1} \times B^{n}$. (To check consistency: Either hyperbola in the plane is $H_{\pm}^{1,1} \simeq S^{0} \times B^{1}$, a two-point space cross an interval; the two-sheeted hyperboloid in space is $H_{+}^{1,2} \simeq S^{0} \times B^{2}$; and the one-sheeted hyperboloid is $H_{-}^{1,2} \simeq H_{+}^{2,1} \simeq S^{1} \times B^{1}$.)
Consequently, $H_{+}^{m,n} \simeq H_{-}^{m,n}$ if and only if $m = n$. Particularly, the hyperboloids $H_{\pm}^{n-1,1}$ are diffeomorphic if and only if $n - 1 = 1$, i.e., $n = 1$.


To see that $H_{+}^{m,n} \simeq S^{m-1} \times B^{n}$, consider the quadratic function
$$
f(x, y) = |x|^{2} - |y|^{2}.
$$
For later use, note that $f$ is homogeneous of degree $2$, in the sense that for all $(x, y)$, we have $f(e^{t}x, e^{t}y) = e^{2t}f(x, y)$ for all real $t$.
The zero set of $f$ is the cone (union of lines through the origin) comprising all $(x, y)$ such that $|x| = |y|$. The intersection of this cone with the sphere $S^{m+n-1}(\sqrt{2})$ of radius $\sqrt{2}$ in $\Reals^{m+n}$, i.e., the locus $\{(x, y) : |x|^{2} + |y|^{2} = 2\}$, is the product of unit spheres
$$
S^{m-1} \times S^{n-1} = \{(x, y) : |x| = |y| = 1\}.
$$
Since $f$ is homogeneous of degree $2$, if $(x, y) \neq (0, 0)$ is arbitrary then the ray through the origin and $(x, y)$ hits the hyperboloid $H_{+}^{m,n}$ precisely once if $|x|^{2} - |y|^{2} > 0$ and does not hit $H_{+}^{m,n}$ otherwise. Projection along rays from the origin (radial lines in the diagram) defines a diffeomorphism from $H_{+}^{m,n}$ to $S^{m-1} \times B^{n}$.
