Write $\mathbb{R}^{n,1} \approx \mathbb{R}^{n} \times \mathbb{R}$. Consider the matrix $I_{n,1} = \mathrm{diag}(1, 1, \ldots, 1, -1)$.
Consider the following Lie groups:
$SO(n) = \left\{ \ A \in M_{n \times n}(\mathbb{R})\ |\ A^{T}A = A A^{T} = I,\ \mathrm{det}(A) = 1\ \right\}$
$SO^{+}(n,1) = \left\{\ A \in M_{(n+1)\times(n+1)}(\mathbb{R})\ |\ A^{T} I_{n,1} A = I_{n,1},\ \mathrm{det}(A) = 1 \ \mathrm{and}\ A_{(n+1)(n+1)}>0 \ \right\}$
$\mathbb{H} = \left\{\ x \in \mathbb{R}^{n,1}\ |\ B(x,x) = -1, \ x_{n+1} > 0\ \right\}$
With the bilinear form $B(x,y) = - x_{n+1} y_{n+1} + \sum_{j=1}^{n} x_{j} y_{j}$.
I'm trying to show that $SO^{+}(n,1) / SO(n) \approx \mathbb{H}$. The way I think I should do this is using the first isomorphism theorem (I don't know how else to do it!).
So I believe I need to find a mapping $f : \mathbb{H} \to SO^{+}(n,1)$ which is surjective and such that $\mathrm{ker}(f) = SO(n)$. If I could find such a mapping then I'd be done - but what is this mapping?
My thinking is to define $f$ as: $$ f(\mathbb{x},x_{n+1}) = \left[ \begin{matrix} \alpha(\mathbf{x}) & \mathbf{x} \\ \mathbf{x}^{T} & x_{n+1}^{2} \end{matrix} \right] $$
But I don't know how to define the function $\alpha$ (which I think should be a surjective mapping onto $SO(n)$).
Can someone help me out here? Or is there a simpler way to show that $SO^{+}(n,1) / SO(n) \approx \mathbb{H}$?