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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.

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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 $X \in Lie(Ker(f))$.

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  • $\begingroup$ Thank you. Would you please explain what do you mean by the one parameter subgroup of $G$ generated by $X\in \ker \mathfrak f$? $\endgroup$ – Vim Apr 24 '17 at 5:58
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    $\begingroup$ @Vim If you have a vector $X$ in the Lie algebra, you can use it to "generate"a one dimensional subgroup of $G$ using the exponential map. You can get it by extending $X$ to a left invariant vector field on $G$, and then "solving" the differential equation (this defines the exponential map). Alternatively, if you work with matrix groups (why not), this subgroup of $G$ equal to the matrix exponentials $e^{tX} = 1 + tX + t^2 X^2/ 2! + \ldots$ for $t \in \mathbb{R}$. $\endgroup$ – Lorenzo Najt Apr 24 '17 at 6:01

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