I want to show that if $\phi:G\to H$ is a Lie group homomorphism, then $d\phi:\mathfrak{g}\to\mathfrak{h}$ is a Lie algebra homomorphism. The proof given in the classroom is the following: let $x \in G$. Since $\phi$ is a Lie group homomorphism, we have that
$$\phi(xyx^{-1}) = \phi(x) \phi(y) \phi(x)^{-1} \tag{$1$}$$
for all $y \in G$. Differentiating $(\ast)$ with respect to $y$ at $y = 1$ in the direction of $Y \in \mathfrak{g}$ gives us
$$d\phi(\mathrm{Ad}(x) Y) = \mathrm{Ad}(\phi(x)) d\phi(Y). \tag{$2$}$$
Differentiating $(2)$ with respect to $x$ at $x = 1$ in the direction of $X \in \mathfrak{g}$, we obtain
$$d\phi(\mathrm{ad}(X) Y) = \mathrm{ad}(d\phi(X)) d\phi(Y) \tag{3}$$
and so:
$$d\phi([X, Y]_{\mathfrak{g}}) = [d\phi(X), d\phi(Y)]_{\mathfrak{h}}. \tag{4}$$
$(1)\implies(2)$, because, if $C_z:G\to G$ is $C_z(g)=zgz^{-1}$, (i use same notation for $C_z:H\to H$ if $z\in H$), (1) is: $\phi \circ C_x=C_\phi(x)\circ\phi$, differentiating wrt $e\in G$, $e=1_G$, and evaluated in $Y\in\mathfrak{g}$, i have (2). $(3)\implies (4)$ is obvious. But i can't prove that $(2)\implies (3)$. Can someone show me the calculation for $(2)\implies(3)$ please?
Any help is appreciated and sorry for the bad English.
Best Answer
Let $t\in\mathbb R$ and $X\in\mathfrak g$. Then\begin{align}\mathrm d\phi\left(\operatorname{Ad}\left(e^{tX}\right)Y\right)&=\operatorname{Ad}\left(\phi\left(e^{tX}\right)\right)\mathrm d\phi(Y)\text{ by (2)}\\&=\operatorname{Ad}\left(e^{t\mathrm d\phi(X)}\right)\mathrm d\phi(Y).\end{align}Now, differentiating with respect to $t$ at $t=0$ will give you $(3)$.