I have confusions about how exactly the process of showing homeo-/diffeomorphism differs between embedded manifolds and ones that are just on their own. Here we have a hemisphere in $\mathbb{R}^3$ and the unit ball in $\mathbb{R}^2$.
So, the open upper hemisphere, $S = \{x \in \mathbb{R}^3 \mid \|x\|_2 = 1 \text{ and } x_3 > 0 \}$, and the open unit ball $B^2(0,1)$.
Obviously there is a diffeomorphism between the two: you flatten the sphere and that's about it, but I'm getting stuck on details about the dimensions and the implicit function theorem.
The only way of finding a diffeomorphism I know is by the inverse function theorem, for which I need an invertible Jacobian and $C^1$ function, but let's say I map the sphere as such: $f(x,y,z) = (x,y)$. The Jacobian is not square and not invertible.
Using the implicit function theorem, I couldn't really figure out how to use it, here. What exactly to parameterise, all that. $x^2 + y^2 + z^2 - 1 = 0$. No getting a useful Jacobian, nor a map to $\mathbb{R}^2$. There's probably some way to work it through here, but I've not been able to figure it out.
I was also looking at $f(x,y) = (x,y,\sqrt{1-x^2-y^2})$, which seems very promising. Maintains the zeros of the implicit function, at least. Well, I'd appreciate some help concerning specifically these ideas. Thanks.
Identify $(x,y,0)\cong (x,y).$ It is easy to check that the projection $\pi(x,y,z)=(x,y,0)$ maps the upper hemisphere bijectively onto the disk. In fact, the inverse is $\pi^{-1}(x,y,0)=(x,y,\sqrt{1-x^2-y^2})$ so they are homeomorphic. However, this map does not provide a diffeomorphism (why? Hint: what happens at the boundary?)
But the stereographic projection $\sigma:\mathbb R^2\setminus \{(0,0,1)\}\to \mathbb R^2\times \{0\}\cong \mathbb R^2$ from the north pole $\sigma(x,y,z) = \frac{(x,y,0)}{1-z}$ has inverse $\sigma^{-1}(x,y,0)=\frac{(2x,2y,\sqrt{x^2+y^2}-1)}{\sqrt{x^2+y^2}+1},$ and you can check that $\sigma^{-1}$ maps the closed disk diffeomorphically onto the lower hemisphere.