Tangent map of homomorphism of Lie groups preserves commutator

I am trying to prove the following proposition:

Let $\phi:G\rightarrow H$ be a homomorphism of Lie groups, and $X,Y\in T_eG$. Then, $$T_e\phi([X,Y]_G) = [T_e\phi X,T_e\phi Y]_H$$

So we want to prove $T_e\phi\circ ad^G(X)(Y) = ad^{H}(T_e\phi X,T_e\phi Y)$.
It is clear that $\phi\circ C^G_x = C^H_{\phi(x)} \circ \phi$. Applying $T_e$ to both sides and applying the chain rule yields $$T_e\phi\circ \text{Ad}^G(x) = \text{Ad}^H(\phi(x)) \circ T_e\phi$$ I now want to apply $T_e$ again and apply the chain rule to get the wanted result. But in order to do that I have to write both sides as functions of $G$. The LHS is already in that form. Applying $T_e$ to the $T_e\phi\circ \text{Ad}^G$ yields $T_e\phi\circ \text{ad}^G$. I don't know how to proceed with the RHS: how to write it as a composition in order to apply the tangent operation and the chain rule.

$$Ad_H(\phi(x))\circ T_e\phi = (R_{T_e\phi}\circ Ad_H\circ\phi)(x)$$ Hence $T_e\phi\circ Ad_G = R_{T_e\phi}\circ Ad_H\circ\phi$. The LHS was taken care of in the question.
Note: $R_{T_e\phi}$ is right composition with the linear map $T_e\phi$ (which is itself linear).
Taking the tangent map of $R_{T_e\phi}\circ Ad_H\circ\phi$ at the identity and applying the chain rule: $$R_{T_e\phi}\circ ad_H\circ T_e\phi$$ Hence $T_e\phi\circ ad_G=R_{T_e\phi}\circ ad_H\circ T_e\phi$. Applying both sides to $X$ and then to $Y$ we get $$T_e\phi[X,Y]_G=[T_e\phi X,T_e\phi Y]_H$$
You need to use that $ad : \mathfrak{gl}(\mathfrak{g}) \to \mathfrak{gl}(\mathfrak{g})$ is the derivative of the map $Ad : GL(\mathfrak{g}) \to GL (\mathfrak{g})$. This will involve chasing around lots of diagrams involving induced maps from $\phi$.
• To start with, what map does $\phi$ induce from $GL(\mathfrak{g}) \to GL(\mathfrak{h})$? – Jonathan Mar 6 '18 at 23:28