Differential structure of the group of automorphism of a Lie group

[padawan]

I am working on this

I am having trouble with b and c:

b) Suppose $(f_n)_{n=1}^{\infty}$ is a sequence in $Aut(G)$, such that $(T_e(f_n))_{n=1}^{\infty} \to \psi$ converges in $Aut(\mathfrak g)$

I want to show that $f := \lim_{n\to \infty} f_n$ exists as an continuous automorphism of the abstract group $G$

First of all the $f_n$ are smooth, because they are in Aut(G), so they are Lie group isomorphisms, and therefore smooth by definition

Then by a known property of the exp:

$f_n(\exp(X))=\exp(T_ef_n X), \forall X \in T_eG$ taking the limit yields:

$\lim_{n\to \infty} f_n(\exp(X))=\exp(\psi(X))\in G, \forall X \in T_eG$

$\lim_{n\to \infty} f_n(g)=\exp(\psi(X))=:f(g)\in G, \forall g$ in a nbhd of the identity

I know that the image of the exponential map generates the connected component of the identity $G_e$,and by connectedness this coincides with G so:
$G=G_e=<(T_eG)>=<\exp(\psi(T_eG))>=<\exp(T_eG)>$

This means that convergence is actually valid in the whole group, because if I have convergence in the generating set, I must have convergence in the generated set.

I hope this is correct,if not please tell me. Still I have to prove that it is bijective and a group homomorphism. I am not sure how to do this

b) How do I argue from here that $f$ is continuous? First I thought it was automatic, but then I recalled from analysis that pointwise convergence of a sequence of functions does not imply continuity. So I am clueless about how to proceed here as well