[proofplan] We first prove that the differential of $\varphi$ intertwines the adjoint actions of $G$ and $H$. This follows by differentiating the conjugation identity $\varphi(gxg^{-1})=\varphi(g)\varphi(x)\varphi(g)^{-1}$ at the identity in the variable $x$. Then we insert the curve $g(t)=\exp_G(tX)$ and use naturality of the exponential to rewrite $\varphi(g(t))$ as $\exp_H(t\,d\varphi(X))$. Differentiating the resulting identity at $t=0$ and using that the [infinitesimal adjoint action is the Lie bracket](/theorems/8822) gives the desired bracket identity. [/proofplan]

[step:Intertwine the adjoint actions by differentiating conjugation] For each $g\in G$, let \begin{align*} C_g:G\to G \end{align*} denote the conjugation map $C_g(x)=gxg^{-1}$. For each $h\in H$, let \begin{align*} C_h^H:H\to H \end{align*} denote the conjugation map $C_h^H(y)=hyh^{-1}$. By definition of the adjoint representation, recalled in [citetheorem:8821], \begin{align*} \operatorname{Ad}_g=d(C_g)_{e_G}:\mathfrak g\to\mathfrak g \end{align*} and \begin{align*} \operatorname{Ad}_h=d(C_h^H)_{e_H}:\mathfrak h\to\mathfrak h. \end{align*} Since $\varphi$ is a [group homomorphism](/page/Group%20Homomorphism), for every $g,x\in G$, \begin{align*} \varphi(C_g(x))=\varphi(gxg^{-1})=\varphi(g)\varphi(x)\varphi(g)^{-1}=C_{\varphi(g)}^H(\varphi(x)). \end{align*} Thus, as smooth maps $G\to H$, \begin{align*} \varphi\circ C_g=C_{\varphi(g)}^H\circ\varphi. \end{align*} Taking differentials at $e_G$ and using the chain rule gives, for every $X\in\mathfrak g$, \begin{align*} d\varphi(\operatorname{Ad}_g X)=\operatorname{Ad}_{\varphi(g)}(d\varphi(X)). \end{align*} [/step]

[step:Specialize the adjoint identity to exponential curves] Fix $X,Y\in\mathfrak g$. Define a smooth curve \begin{align*} \gamma_X:\mathbb R\to G \end{align*} by $\gamma_X(t)=\exp_G(tX)$. Substituting $g=\gamma_X(t)$ into the adjoint-intertwining identity from the previous step gives \begin{align*} d\varphi(\operatorname{Ad}_{\exp_G(tX)}Y)=\operatorname{Ad}_{\varphi(\exp_G(tX))}(d\varphi(Y)) \end{align*} for every $t\in\mathbb R$. By exponential naturality [citetheorem:8802], applied to the Lie group homomorphism $\varphi:G\to H$ and the element $X\in\mathfrak g$, \begin{align*} \varphi(\exp_G(tX))=\exp_H(t\,d\varphi(X)). \end{align*} Therefore \begin{align*} d\varphi(\operatorname{Ad}_{\exp_G(tX)}Y)=\operatorname{Ad}_{\exp_H(t\,d\varphi(X))}(d\varphi(Y)) \end{align*} for every $t\in\mathbb R$. [/step]

[step:Differentiate at the identity parameter to recover the bracket]Because $d\varphi:\mathfrak g\to\mathfrak h$ is linear, differentiating the identity \begin{align*} d\varphi(\operatorname{Ad}_{\exp_G(tX)}Y)=\operatorname{Ad}_{\exp_H(t\,d\varphi(X))}(d\varphi(Y)) \end{align*} at $t=0$ gives \begin{align*} d\varphi\left(\frac{d}{dt}\bigg|_{t=0}\operatorname{Ad}_{\exp_G(tX)}Y\right)=\frac{d}{dt}\bigg|_{t=0}\operatorname{Ad}_{\exp_H(t\,d\varphi(X))}(d\varphi(Y)). \end{align*} By the identification of the infinitesimal adjoint action with the bracket [citetheorem:8822], applied in $G$, \begin{align*} \frac{d}{dt}\bigg|_{t=0}\operatorname{Ad}_{\exp_G(tX)}Y=[X,Y]. \end{align*} Applying the same result in $H$ gives \begin{align*} \frac{d}{dt}\bigg|_{t=0}\operatorname{Ad}_{\exp_H(t\,d\varphi(X))}(d\varphi(Y))=[d\varphi(X),d\varphi(Y)]. \end{align*} Substitution yields \begin{align*} d\varphi([X,Y])=[d\varphi(X),d\varphi(Y)]. \end{align*}[/step]

[guided]We now extract the Lie bracket by differentiating the adjoint identity along an exponential curve. The identity obtained above says that, for every real number $t$, \begin{align*} d\varphi(\operatorname{Ad}_{\exp_G(tX)}Y)=\operatorname{Ad}_{\exp_H(t\,d\varphi(X))}(d\varphi(Y)). \end{align*} Here the right-hand side uses exponential naturality [citetheorem:8802]: the theorem applies because $\varphi:G\to H$ is a Lie group homomorphism, $X\in\mathfrak g$, and $t\in\mathbb R$, so \begin{align*} \varphi(\exp_G(tX))=\exp_H(t\,d\varphi(X)). \end{align*} The purpose of this one-parameter identity is that the derivative of the adjoint action at the identity is exactly the infinitesimal adjoint action, which is the Lie bracket. Since $d\varphi:\mathfrak g\to\mathfrak h$ is a [linear map](/page/Linear%20Map), differentiation commutes with applying $d\varphi$ to the curve \begin{align*} t\mapsto \operatorname{Ad}_{\exp_G(tX)}Y. \end{align*} Differentiating both sides at $t=0$ gives \begin{align*} d\varphi\left(\frac{d}{dt}\bigg|_{t=0}\operatorname{Ad}_{\exp_G(tX)}Y\right)=\frac{d}{dt}\bigg|_{t=0}\operatorname{Ad}_{\exp_H(t\,d\varphi(X))}(d\varphi(Y)). \end{align*} We now identify both derivatives. By the theorem that the infinitesimal adjoint action is the bracket [citetheorem:8822], applied to the Lie group $G$, the curve $t\mapsto \operatorname{Ad}_{\exp_G(tX)}Y$ has derivative \begin{align*} \frac{d}{dt}\bigg|_{t=0}\operatorname{Ad}_{\exp_G(tX)}Y=[X,Y]. \end{align*} The same theorem applies to the Lie group $H$, with $d\varphi(X),d\varphi(Y)\in\mathfrak h$, and gives \begin{align*} \frac{d}{dt}\bigg|_{t=0}\operatorname{Ad}_{\exp_H(t\,d\varphi(X))}(d\varphi(Y))=[d\varphi(X),d\varphi(Y)]. \end{align*} Substituting these two derivative identities into the differentiated adjoint identity yields \begin{align*} d\varphi([X,Y])=[d\varphi(X),d\varphi(Y)]. \end{align*} This is precisely the statement that the differential at the identity, $d\varphi=d\varphi_{e_G}$, preserves Lie brackets.[/guided]

[step:Conclude that the differential is a Lie algebra homomorphism] The differential $d\varphi:\mathfrak g\to\mathfrak h$ is linear because it is the differential of a smooth map between manifolds at a point, viewed between tangent vector spaces. The previous step proves that it preserves the Lie bracket for arbitrary $X,Y\in\mathfrak g$. Hence $d\varphi$ is a [Lie algebra](/page/Lie%20Algebra) homomorphism. [/step]