Consider the Lie algebras $\mathfrak {so} (3)$ and $\mathfrak {su} (2)$ and let $f : \mathfrak {so} (3) \longrightarrow \mathfrak {su} (2)$ be the Lie algebra homomorphism given by $$f \left (\begin{pmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{pmatrix} \right ) = \frac {1} {2} \begin{pmatrix} iz & y - ix \\ -y - ix & -iz \end{pmatrix}$$ $x,y,z \in \mathbb R.$ Show that there does not exist any smooth Lie group homomorphism $F : SO (3) \longrightarrow SU(2)$ such that $F(\exp (X)) = \exp (f(X)),$ for all $X \in \mathfrak {so} (3).$
If we can show that then by Lie's theorem it yields an alternative way of showing that the matrix Lie group $SO(3)$ is not simply connected. But I am having hard time showing the non-existence of such a smooth $F.$ Could anyone help me out here?
Thanks for your time.