Problem: I'd like to parametrize the manifold given by $\{(x,y,z)\in{\mathbb R}^{3}\,|\, x^2 + y^2 - z^2 = a\}$ for $a < 0$. The two mappings we'd use are $f(x,y) = (x,y,\sqrt{x^2 + y^2 - a})$ and a similar map for the "bottom part."
It was noted that I should prove that these are diffeomorphisms by showing that $d(f)_{(x,y)}:R^{2}\to T(M_{1})_{f(x,y)}$ with $M_{1}$ being the "top part" is nonsingular as a linear transformation and then apply the inverse function theorem. My problem is showing this map is non-singular: what exactly is the derivative here? I thought that it might be something like
$\left(\begin{array}{ccc} 1 & 0 & \frac{-x}{\sqrt{x^2 + y^2 - a}}\\ 0 & 1 &\frac{-y}{\sqrt{x^2 + y^2 - a}}\end{array}\right)$
but this isn't square and so cannot (I think!) be singular or non-singular by definition. Is there something I'm missing here?
Let's go partway through the problem. Let $f(p') = p$, where $p' \in \mathbb R^2$ and $p \in M$. The Jacobian is then
$$\underline f(a) = a \cdot \nabla' f(p')$$
You can then find $\underline f(e_1)$ and $\underline f(e_2)$. These two vectors define the tangent space. We know the tangent space must be a plane, so the cross product of these vectors (both of which are in $\mathbb R^3$) gives the normal to the plane that is the tangent space at the point $p$.
Now then, to ensure that this is the case, $\underline f(a) \neq 0$ for any $a \neq 0$. Why is this so? Because then you could take $b,c$ such that $a = b-c$ and then $\underline f(b) = \underline f(c)$. This would make the Jacobian, as a linear operator, no longer bijective. You can't "invert" it in the usual matrix algebra sense anyway, as it isn't square, but the Jacobian of $f$ and the Jacobian of $f^{-1}$ have the property that, when used consecutively, they can project vectors in $\mathbb R^3$ onto the tangent space.