[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*}
[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]
[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]
