Suppose $G,H$ are Lie groups and $F:G\to H$ is a Lie group homomorphism. Suppose $\mathfrak f: \mathfrak g\to\mathfrak h$ is the induced Lie algebra homomorphism, show that we can identify $\DeclareMathOperator{\ker}{ker}\ker\mathfrak f$ with $\DeclareMathOperator{\Lie}{Lie}\Lie(\ker F)$ in the sense that $$\ker\mathfrak f=\{X\in \mathfrak g\mid X_e\in d(\iota)_e(T_e(\ker F))\}\cong \Lie(\ker F).$$ in which $\iota : \ker F\to G$ is the inclusion map.
I'm aware that $\mathfrak f(X)=0\iff \mathfrak f(X)_e = d(F)_e(X_e)=0$. If $X\in \Lie(\ker F)$ then it is equal to saying $\exists v\in T_e(\ker F)$ such that $X_e=d(\iota)_e(v)$ and $$d(F)_e(X_e)=d(F)_e(d(\iota)_e(v))=d(F\circ\iota)_e(v)=d(F|_{\ker F})_e(v)$$ where $F|_{\ker F}\equiv e\in G$ is a constant map, hence its differential vanishes and $d(F|_{\ker F})_e(v)=0$. And therefore we have proved $\Lie(\ker F)\subset \ker\mathfrak f$ under our identification.
What about the other direction? Given only $d(F)_e(X_e)=0$ there seems to be no easy way to extract more information from it. Any help? Thanks in advance.
You can use that the exponential maps commute with the map on the Lie algebra / Lie groups.
Suppose $X \in Lie(G)$, with $f(X) = 0$. If you look at the one parameter subgroup of $G$ generated by $X \in ker f$ you will see that it lies in $Ker F$. The computation is: $F(exp(tX)) = exp(tf(X)) = exp(0) = Id$. So this means that $X \in Lie(Ker(f))$.