Let $G$ be a connected Lie group with Lie algebra $\mathfrak g$. Aut(G) is the group of automorphisms of G. It is known that the automorphism group Aut($\mathfrak g$) is a Lie group and a closed subgroup of GL$(\mathfrak g)$ The tangent map (at $e$) of any automorphism of $G$ is an automorphism of $\mathfrak g$. I want to show that the induced map
$\DeclareMathOperator{\Aut}{\operatorname{Aut}} D : \Aut(G) \to \Aut(\mathfrak g)$
is an injective group homomorphism.
I have to check two things: that $D$ is a group homomorphism and that it is injective.
Proving that it is a group homomorphism:
$D : \Aut(G) \to \Aut(\mathfrak g): f \mapsto T_ef$
I have that for $f,g \in \Aut(G)$,
$D(f \circ g)=T_e(f \circ g)=T_{g(e)}f\circ T_eg=T_ef\circ T_eg=D(f)\circ D(g)$
Now for the injectivity part
I have to check that the kernel is trivial
Following the corrections in the comments I think the kernel is defined like this :
$ \operatorname{Ker}(D)=\{f\in \Aut(G): Df=id_{T_eG}\}=\{f\in \Aut(G): T_ef=id_{T_eG}\}$
where $id_{T_eG}$ is the identity over $T_eG$ which is the identiy element of the group $Aut(\mathfrak g)$
I am stuck at this point. How do I conclude from here that $f=id_G$?
I know $T_e id_G=id_{T_eG}$ . So I have that $Df=T_ef=id_{T_eG}=T_e id_G$ from which I am not sure how to justify that $T_ef=T_e id_G$ implies $f=id_G$. Besides don't I need to use the connectedness of the Lie group?
Hint: Note that essentially by definition, for any $x\in\mathfrak g$ and $f\in \operatorname{Aut}(G)$, we have $\exp(Dfx)=f\exp(x)$.
Then use the fact that $G$ is connected to conclude that if $Df=\operatorname{id}_{\mathfrak g}$, then $f$ is the identity.
Hint 2: Suppose $Df$ is the identity and consider the set $\{g\in G\mid f(g)=g\}$.