$\ker(\Phi:\Sl(2,\C)\rar G)=\pm1$ and $\im(\Phi)$ is the group of Lorentz transformations on $(E,g)$ (cf. exam).
1. If $\Phi(X)=1$, then for all $A\in E$: $XAX^*=A$, which implies that $XX^*=\s_0$, i.e. $X\in\SU(2)$ and hence it commutes with all $A\in E$. This in turn shows that it's a multiple of $\s_0$. Since $X\in\Sl(2,\C)$, it follows that $X=\pm\s_0$.
2. $\Phi(X)$ is a Lorentz transformation: By polar decomposition $X=UR$ for some $U\in\UU(2)$ and some positive definite $R$; since $1=\det(U)\det(R)$ and $\det R > 0$ we conclude that $\det U=1$, i.e. $U\in\SU(2)$. By exam: $\Phi(X)$ must be a Lorentz transformation - the orientation is defined by $\s_0,\s_3,\s_1,\s_2$.
3. For any Lorentz transformation $L$ on $(E,g)$ there is some $X\in\Sl(2,\C)$ such that $\Phi(X)=L$. That's the hard part! By exam it suffices to show that any rotation about $\s_0$ is of the form $A\mapsto XAX^*$ for some $X\in\SU(2)$. Denoting by $F$ the subspace $\s_0^\perp$, we get a homomorphism $\SU(2)\rar\SO(3)$ mapping $X\in\SU(2)$ to the euclidean isometry $F\rar F$, $A\mapsto XAX^*$. But this is just the double-covering map $\SU(2)\rar\SO(3)$ (cf. e.g. wikipedia).