What's the normal space of the manifold $z = x^2 + y^2$?
Let's say I have a continous function $g: M \rightarrow S^n$ that sends every point on the manifold to the unit normal vector. I want to know what the image of $g$ is, and for every $v \in Im(g)$, I want to describe the set $g^{-1}(v)$. By picturing the manifold, I can guess that $Im(g)$ will be the lower half of the sphere. I want to prove this formally:
We denote $f(x,y,z) = x^2 + y^2 - z$.
$D_f(x,y,z) = (2x,2y,-1)$.
The normal vector is a normalized vector in the image of $D_f$, but I'm not sure how to continue from here.
Any help would be appreciated!
Hint/Direction: Nothing to do with differential geometry, but elementary set theory here. You have the mapping $g : \mathbb R^2 \to \mathbb S^2$, where
$$g(x,y) = \left(\frac{2x}{\sqrt{4x^2+4y^2+1}}, \frac{2y}{\sqrt{4x^2+4y^2+1}},\frac{-1}{\sqrt{4x^2+4y^2+1}} \right).$$
Now you want to show that the image is the lower hemisphere, which is
$$\mathbb S^2_- =\{ (a, b, c) \in \mathbb S^2 : c<0\}.$$
So for all $(a, b, c) \in \mathbb S^2 _-$, you want find $(x, y)$ so that
$$\frac{2x}{d} = a,\ \ \ \frac{2y}{d} =b, \ \ \ \frac{-1}{d} = c,\ \ \text{where } d= \frac{1}{\sqrt{4x^2+4y^2+1}}.$$
The computation don't seem to be difficult here.