[proofplan]
We first check that the differential of a Lie [group homomorphism](/page/Group%20Homomorphism) is taken between the correct tangent spaces by proving that homomorphisms send identity elements to identity elements. The identity law follows by differentiating the identity map at the identity. The composition law is exactly the chain rule at $e_G$, using $\varphi(e_G)=e_H$. Finally, bracket preservation shows that these maps are morphisms in the category of Lie algebras, not merely linear maps between tangent spaces.
[/proofplan]
[step:Show that homomorphisms send identity elements to identity elements]
Let $\varphi:G\to H$ be a Lie group homomorphism. Since $\varphi$ preserves multiplication,
\begin{align*}
\varphi(e_G)=\varphi(e_Ge_G)=\varphi(e_G)\varphi(e_G).
\end{align*}
Multiplying on the left by $\varphi(e_G)^{-1}$ in $H$ gives
\begin{align*}
e_H=\varphi(e_G).
\end{align*}
Therefore the differential at the identity is a [linear map](/page/Linear%20Map)
\begin{align*}
d\varphi_{e_G}:T_{e_G}G\to T_{\varphi(e_G)}H=T_{e_H}H.
\end{align*}
[guided]
We need first to justify the target of the differential. A smooth map $\varphi:G\to H$ has differential at $e_G$ of the form
\begin{align*}
d\varphi_{e_G}:T_{e_G}G\to T_{\varphi(e_G)}H.
\end{align*}
For this to become a map between Lie algebras, the target must be $T_{e_H}H$, so we must prove $\varphi(e_G)=e_H$.
Because $\varphi$ is a group homomorphism, it preserves multiplication. Applying this to the equality $e_Ge_G=e_G$ gives
\begin{align*}
\varphi(e_G)=\varphi(e_Ge_G)=\varphi(e_G)\varphi(e_G).
\end{align*}
The element $\varphi(e_G)$ is an element of the group $H$, hence it has an inverse $\varphi(e_G)^{-1}$. Multiplying the displayed equality on the left by $\varphi(e_G)^{-1}$ gives
\begin{align*}
e_H=\varphi(e_G).
\end{align*}
Thus
\begin{align*}
T_{\varphi(e_G)}H=T_{e_H}H,
\end{align*}
and the differential at the identity is indeed a linear map
\begin{align*}
d\varphi_{e_G}:T_{e_G}G\to T_{e_H}H.
\end{align*}
This is the map denoted $\operatorname{Lie}(\varphi)$.
[/guided]
[/step]
[step:Identify the differential of the identity map with the identity on the Lie algebra]
Let
\begin{align*}
\operatorname{id}_G:G\to G
\end{align*}
be the identity map. For every tangent vector $X\in T_{e_G}G$, the differential of the identity map satisfies
\begin{align*}
d(\operatorname{id}_G)_{e_G}(X)=X.
\end{align*}
Hence
\begin{align*}
\operatorname{Lie}(\operatorname{id}_G)
=d(\operatorname{id}_G)_{e_G}
=\operatorname{id}_{T_{e_G}G}
=\operatorname{id}_{\operatorname{Lie}(G)}.
\end{align*}
[/step]
[step:Apply the chain rule at the identity to prove compatibility with composition]
Let
\begin{align*}
\varphi:G\to H
\end{align*}
and
\begin{align*}
\psi:H\to K
\end{align*}
be Lie group homomorphisms. By the previous identity calculation,
\begin{align*}
\varphi(e_G)=e_H.
\end{align*}
The chain rule for smooth maps, applied to the composition $\psi\circ\varphi:G\to K$ at the point $e_G$, gives
\begin{align*}
d(\psi\circ\varphi)_{e_G}=d\psi_{\varphi(e_G)}\circ d\varphi_{e_G}.
\end{align*}
Substituting $\varphi(e_G)=e_H$ yields
\begin{align*}
d(\psi\circ\varphi)_{e_G}=d\psi_{e_H}\circ d\varphi_{e_G}.
\end{align*}
By the definition of $\operatorname{Lie}$ on morphisms,
\begin{align*}
\operatorname{Lie}(\psi\circ\varphi)=\operatorname{Lie}(\psi)\circ\operatorname{Lie}(\varphi).
\end{align*}
[/step]
[step:Use bracket preservation to place the identities in the category of Lie algebras]
For every Lie group homomorphism $\varphi:G\to H$, [citetheorem:8803] gives
\begin{align*}
d\varphi_{e_G}([X,Y])=[d\varphi_{e_G}(X),d\varphi_{e_G}(Y)]
\end{align*}
for all $X,Y\in\operatorname{Lie}(G)$. Thus $\operatorname{Lie}(\varphi)$ is a [Lie algebra](/page/Lie%20Algebra) homomorphism. The two identities proved above are therefore identities of Lie algebra homomorphisms, so the assignments $G\mapsto\operatorname{Lie}(G)$ and $\varphi\mapsto\operatorname{Lie}(\varphi)$ define a functor from Lie groups to real Lie algebras.
[/step]
