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 have already proven that the following induced map
$\DeclareMathOperator{\Aut}{\operatorname{Aut}} D : \Aut(G) \to \Aut(\mathfrak g)$
is an injective group homomorphism
Now suppose $(f_n)_{n=1}^{\infty}$ is a sequence in $Aut(G)$, such that $(T_e(f_n))_{n=1}^{\infty}$ converges in $Aut(\mathfrak g)$ say to $\psi$. Show that $ f := lim_{n\to \infty} f_n$ exists as an continuous smooth automorphism of the abstract group $G$
How do I do this ? The only thing I can think about is to use the exponential map to relate the differential to the function, but I am not sure how to do this if that is useful
I also think that the $f_n$ must be smooth right? Otherwise the $(T_e(f_n))$ would not make sense, would they?