Transversality of a function to a sphere I'm working through problem 6-9 in Lee Smooth Manifolds and I'm stuck.
Let $F: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ defined by $(x,y) \mapsto (e^{y}\cos(x), e^{y}\sin(x), e^{-y})$. For which positive values of $r$ is $F$ transverse to the sphere $S_{r}(0) \subset \mathbb{R}^3$. For which positive values $r$ is $F^{-1}(S_{r}(0)$ an embedded submanifold of $\mathbb{R}^2$.
I started by computing the differential $dF$ given by the Jacobian
$\left(\begin{array}{cc}
-e^y \sin(x) & e^y\cos(x) \\
e^y\cos(x) & e^y \sin(x) \\
0 & -e^{-y} \end{array} \right)$.
It seems like this matrix always has full column rank. So can we conclude that this is true for all values of $r$? It's also difficult to tell what the level sets of $F$ are. 
 A: Apologies in advance if this doesn't answer your question in the way you're asking, but I've always found it most intuitive to think about whether two manifolds are transversal by thinking about them geometrically, at least in $\mathbb{R}^3$. If you look at a parametric plot of the map $F$, it appears pretty clear that there is a particular radius $r$ for which the 2-sphere $S_r(0)$ is tangent to the parametric map, which will be the value of $r$ for which the manifolds are not transversal.
Perhaps elaborating a bit more, at any point of intersection, we'll have a tangent plane to the sphere and similarly a tangent plane to the parametrized image. As long as these are not the same plane, their sum will yield $\mathbb{R}^3$, and so we only have to worry about the value of $r$ for which the intersection of the sphere and the parametrized image is exactly the tangent points.
Adding a bit more: the equation $x^2 + y^2 - 1/z^2 = 0$ defines the manifold given by the image of the parametric map. This has a gradient of $(2x,2y,2/z^3)$ at each point. Meanwhile, the gradient of the constraint for the sphere is $(2x,2y,2z)$. For the tangent planes of the two manifolds to be equal, these gradients must be equal. We see that this clearly happens at $z = 1$, at which point the $x,y$ trace out a unit circle and the sphere and parametric image are tangent at every point on this circle, i.e. $(\cos\theta, \sin\theta, 1)$. This means the radius of the sphere must be $r = \sqrt{2} = \sqrt{x^2 + y^2 + z^2} =\sqrt{\cos^2\theta + \sin^2\theta + 1}$. Thus, we've found the value of $r$ for which the two manifolds are not transversal.
