Conformally mapping $D^{n}$ to the upper-half plane $H^{n}$

Question

Defining the upper half-plane $H^{n} = \left\{ \ x = (x_{1}, \ldots, x_{n}) \in \mathbb{R}^{n} \ | \ x_{n} \geq 0 \ \right\}$.

In complex analysis there's this notion of conformally mapping the unit disk $D^{2}$ to the upper half plane $H^{2}$.

But is this true in general that $D^{n}$ (including its insides) can be conformally mapped to $H^{n}$? (this would have to be a Mobius transformation, via Liouville's theorem)

I feel like the stereographic projection should be able to do the trick somehow, but I'm not certain if this is indeed a conformal mapping. On top of this, the stereographic projection only maps the boundary of $\partial D^{n} = S^{n-1}$ to the boundary of $\partial H^{n}$...

EDIT: Just saw on wikipedia that the stereographic projection is indeed conformal. This resolves the first part of my question...what's left unclear to me is how do I map points on the inside of the disk to the parts of the upper half plane not on its boundary?

Answer 1

Wikipedia gives the isometry between the Poincaré half plane model and the Poincaré disk model. Although the formulas given are for the two-dimensional case, they extend in the obvious way to the $n$-dimensional case, which are also isometric.

A point $(x,y)$ in the disk model maps to $$\left( \frac{2x }{x^2+ (1-y)^2} \ , \ \frac{1-x^2-y^2}{x^2+ (1-y)^2} \right) \,$$ in the halfplane model.

Since the boundary of the half-plane or disk is not included in these models, the formula is for the interior.