proof that $PSL(2,\mathbb{R})$ is $SO^+(2,1)$

Sorry if this is too elementary for this forum:

We have two models of 2 dimensional hyperbolic space:

The upper sheet of a hyperboloid in $\mathbb{R}^3$ with the metric induced by the Minkowski metric on $R^3$. Define $O^+(2,1)$ to be the subgroup of $GL(3, \mathbb{R})$ that preserves the Minkowski metric and takes the upper sheet of the hyperboloid to itself. Then $O^+(2, 1)$ acts by isometries on this model. $SO^+(2,1)$ is then the intersection of $O^+(2, 1)$ with $SL(3, \mathbb{R})$.

A second model is the upper half plane. Then $SL(2,\mathbb{R})$ acts by isometries on this model when each element of $SL(2,\mathbb{R})$ is viewed as a Möbius transformation. $PSL(2,\mathbb{R})$ is $SL(2,\mathbb{R})/\langle \pm I \rangle$ (i.e., $SL(2,\mathbb{R})$ mod the two element subgroup consisting of the identity and minus the identity).

How do I prove that these two groups are isomorphic as groups? Lie groups?

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Related: math.stackexchange.com/q/491455 (a map from SO(1,3) to PSL(2,C) etc) – Grigory M Dec 18 '13 at 20:40

$\mathrm{SL}_2(\mathbb{R})$ acts on its Lie algebra $\mathfrak{sl}_2$ by conjugation, which yields an injective morphism $\mathrm{PSL}_2(\mathbb{R}) \rightarrow \mathrm{Aut}(\mathfrak{sl}_2)$. Moreover, the elements in the image of this morphism preserve the Killing form, which is a non-degenerate quadratic form, and is not anisotropic (I'm not sure if "isotropic" is the good word here), and so it is of signature (2,1) or (1,2), and both types yield the same groups of isometries.
So we have an injective (and obviously smooth) morphism $\mathrm{PSL}_2(\mathbb{R}) \rightarrow \mathrm{O}(2,1)$, and since $\mathrm{SL}_2(\mathbb{R})$ is connected, this morphism actually goes to the connected component of the identity in $\mathrm{O}(2,1)$, that is $\mathrm{SO}^+(2,1)$. By injectivity and comparing dimensions, it is a diffeomorphism.