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.
 A: As Tsemo Aristide points out, in terms of left-invariant vector field it follows from a very general rule for Lie brackets of $f$-related fields.
If you want to establish this directly from the $\text{ad}$ definition, you can do so in a few steps. I'll use tildes to denote objects in $H$.
First, we note that conjugation maps commute with $\rho$, i.e. $\rho(\psi_ga)=\widetilde{\psi}_{\rho(g)}\rho(a)$, or, alternately
$$
\rho\circ\psi=\widetilde{\psi}\circ(\rho\times\rho)
$$
This follows directly from the fact that $\rho$ is a group homomorphism. Thinking of both sides of the above equation as maps $G\times G\to H$ and differentiating with respect to the second argument, we have an equality of maps $G\times T_eG\to T_eH$ $$
d_e\rho\circ\text{Ad}(g)=\widetilde{Ad}(\rho(g))\circ d_e\rho
$$
We can equivalently interpret these as maps $G\to\text{Hom}(T_eG,T_eH)$ where $\text{Hom}$ here denotes the space of linear maps. Differentiating this expression again, we have
$$
d_e\rho\circ\text{ad}(v)=\widetilde{\text{ad}}(d_e\rho(v))\circ d_e\rho
$$
Were we use the fact that, under the canonical identification $T_A\text{Hom}(T_eG,T_eH)\cong\text{Hom}(T_eG,T_eH)$ the differential of left/right multiplication by a fixed matrix is left/right multiplication by the same matrix. Adding a dummy vuriable $u\in T_eG$, we arrive at the desired equality.
$$
d_e\rho(\text{ad}(v)(u))=\widetilde{\text{ad}}(d_e\rho(v))(d_e\rho(u))
$$
A: The fact $f^*[X,Y]=[f^*X,f^*Y]$ is true for every differentiable map defined on a manifold.
Pushforward of Lie Bracket
