I'd like to show that if $F : G \to G$ is a Lie group homomorphism which is also a diffeomorphism, one has that $d_eF: \mathfrak{g} \to \mathfrak{g}$ is a Lie-Algebra isomorphism, in other words a bijective linear map $\mathfrak{g} \to \mathfrak{g}$ such that
$$
d_eF([v,w]) =[d_eF(w),d_eF(w)].\tag1
$$
Let me know if the following proof has a flaw :
Firstly, we need to check that $d_eF$ really sends $\mathfrak{g}\cong T_eG$ to itself. This simply follows because $F(e)=e$, indeed $F$ is a group homomorphism. The fact that $d_e$ is an isomorphism simply follows because $F$ is a diffeomorphism and differentials of diffeomorphisms are vector space isomorphisms. Finally $(1)$ simply follows from the fact that $F$ is a diffeomorphism, indeed arbitrary diffeomorphisms from a manifold to itself preserve the bracket of vector fields.
Somehow something seems slightly off about this proof. Appreciate if someone could give me feedback.
Best Answer
As Mariano's comment states, you need this operation to be on vector fields to define the Lie bracket, but in fact more is true. If $\phi:G\rightarrow H$ is $\textit{any}$ Lie group homomorphism, then there is an induced Lie algebra homomorphism which we denote by $\phi_*:\mathfrak{g}\rightarrow \mathfrak{h}$.
To see this, first note that for any $X\in\mathfrak{g}$, i.e. a left invariant vector field, that we can push $X$ forward by: $$ \begin{align} (\phi_*X)_h=D_eL_h\circ D_e\phi(X_e) \end{align} $$ In other words, $\phi_*X$ is the smooth map $H\rightarrow \mathfrak{h}$ given by the composition: $$ \begin{align} H\xrightarrow{e}\{e\}\xrightarrow{X}T_eG\xrightarrow{D_e\phi}T_eH\xrightarrow{\omega}\mathfrak{h} \end{align} $$ where $e$ is the trivial homomorphism sending every element of $H$ to $e$, and $\omega$ is the isomorphism $T_eH\leftrightarrow \mathfrak{h}$, given by: $$ \begin{align} \omega(v)_h=D_eL_h(v) \end{align} $$ for all $h\in H$, and $v\in T_eH$. Essentially, using the group structure of $G$ and $H$, and the homomorphism properties of $\phi$, we have the bypassed the general need for a smooth inverse to define the pushforward of a vector field, though this only holds for left invariant vector fields.
Using this construction, it is clear that $\phi_*:\mathfrak{g}\rightarrow \mathfrak{h}$ it is clear that $\phi_*$ is a linear map, so to show it is a Lie algebra homomorphism, i.e. $$ \begin{align} \phi_*[X_1,X_2]=[\phi_*X_1,\phi_*X_2] \end{align} $$ it suffices to show that $\phi_*X$ is $\phi$ related to $X$, i.e. $$ \begin{align*} (\phi_*X)_{\phi(g)}=D_g\phi(X_g) \end{align*} $$ which I invite you to prove yourself, as it is not difficult from the definition of $\phi_*$.