 Solution

For fixed $X\in L(G)$, set $b(t)=\varphi (e^{tX})$. The homomorphism property gives $b(s+t)=b(s)b(t)$ and $b(0)=I$. Its initial derivative is $L(\varphi )(X)=D{\varphi }_I(X)$. The one-parameter subgroup uniqueness theorem givesFor the Adjoint representation, ${\mathrm{Ad}}_{g}Y=gY{g}^{-1}$. ThusThe differential $L(\varphi )$ is a real linear map of Lie algebras; the bracket identity follows by differentiating the compatibility of conjugation with $\varphi$.