A differential $d\rho$ of a Lie homomorphism $\rho:G\to H$ commutes with the Lie bracket. Some clarification. Let $G,H$ be Lie groups, $\rho:G\to H$ be a Lie group homomorphism, and $\psi_g:G\to G$ be an automorphism given by $\psi_g(a)=gag^{-1}$. Consider $X,Y\in T_eG$. Then we define a Lie bracket $[X,Y]:=ad(X)(Y)$ where $ad$ is the differential at the identity of the map
$$Ad:G\to Aut(T_eG)$$
where $Ad(g)$ is equal to $d(\psi_g)_e$.
I want to show that $(d\rho)_e([X,Y])=[(d\rho)_e(X),(d\rho)_e(Y)]$ i.e.
$$(d\rho)_e(ad(X)(Y))=ad((d\rho)_e(X))((d\rho)_e(Y)).$$
Somehow, it follows from the fact that $\rho$ and $d\rho$ respect the adjoint representation, but I don't see it directly.
Any comment/clarification/hints will be greatly appreciated.

The fact $f^*[X,Y]=[f^*X,f^*Y]$ is true for every differentiable map defined on a manifold.
Pushforward of Lie Bracket
