 Solution

Use the real three-dimensional space of traceless matricesThus $-\det X$ is a quadratic form of signature $(2,1)$. The adjoint action $X\mapsto AX{A}^{-1}$ of $SL(2,\mathbb{R})$ preserves trace and determinant. Connectedness puts its image in $S{O}_0(2,1)$.

A kernel element commutes with every traceless matrix and with the identity, hence with every real matrix; it is therefore scalar. Determinant one leaves $\pm I$. The derivative is injective and both groups have dimension three, so the image is open and equals the identity component. Consequently