The book I'm reading leaves as exercise to prove that every lie algebra homomorphism of a simply connected Lie group, such that $\exp$ is surjective, induces a group homomorphism such that its differential at the origin coincides with the homomorphism. It suggests to use the fact that $\lim_{m\to\infty}(\exp(tX/m)\exp(tY/m))^m = \exp(t(X+Y))$. The book only deals with linear Lie groups by the way. If $f$ is the lie algebra homomorphism, my idea was to define:
$$\phi(\exp(tX)) = \exp(t(f(X))$$
which is well defined for small $t\in\mathbb{R}$, that is, there is a neighbourhood of the identity where $\phi$ this map is well defined (since $\exp$ is injective there). It is also ok to show it is a local homomorphism. The idea is then to show it is well defined for all $t$, and so to all $G$, since $\exp$ is surjective. But I'm having trouble showing it is well defined. I'm aware I'm supposed to use the fact the group is simply connected, but I'm not sure how. I thought about using that since $\exp(t_1X)=\exp(t_2Y)$, I can construct a closed path from that point to the identity and shrink it to the identity, but I'm not sure how that helps.
Edit: Another Idea I just had is to define $\phi$ in a neighbourhood of the Identity just like I did, and then extend it to $\tilde{\phi}$ in $G$ by defining:
$$\tilde{\phi}(x) = (\phi(\exp(X/m)))^m$$
Where $m\in\mathbb{N}$ is big enough such that the expression above is well defined, and $\exp(X) = x$. I believe this would work also if I could show it is well defined, that is, $\exp(X)=\exp(Y)\implies \forall m, \phi(\exp(X/m))^m=\phi(\exp(Y/m))^m$, but having the same problem here.
I'm aware there is a rather long proof for the case where $\exp$ is not necessarily surjective, but I'm wondering if there is an easier one for this case.