Exactness of Lie algebra exact sequence

Question

If $G\rightarrow H\rightarrow K$ is an exact sequence of Lie groups, then I want to show that the induced sequence $\mathfrak{g}\rightarrow\mathfrak{h}\rightarrow\mathfrak{k}$ in Lie algebras is exact. How can we do this? Perhaps using the exponential map, which gives a commutative diagram between these sequences. Is this the right way to proceed?

Answer 1

If we denote $f_1:G\rightarrow H$ and $f_2:H\rightarrow K$ the maps in the exact sequence then, since $f_2\circ f_1$ is constant, its differential in the identity is null, it follows that :

$$T_{1_H}f_2\circ T_{1_G}f_1=0 $$

So that $Im(T_{1_G}f_1)\subseteq Ker(T_{1_H}f_2)$.

Assume that $X\in Ker(T_{1_H}f_2)$, write $\gamma(t):=exp(tX)$ it is a one-parameter subgroup of $H$. Since $X \in Ker(T_{1_H}f_2)$ we have that $f_2\circ \gamma$ is constant, but $f_2(\gamma(0))=1_K$ so that $\gamma$ is a one-parameter subgroup of $Ker(f_2)$. Because the sequence is exact, it is a one parameter subgroup of $Im(f_1)$ and hence its derivative $X$ in $t=0$ must belong to $Im(T_{1_G}f_1)$ whence the reverse inclusion.

I used here the exponential... maybe it is easier to deal with this using the first inclusion and some dimension argument...