A question about some exact sequence of Lie algebras

Question

M. Schottenloher in his book "A mathematical introduction to conformal field theorey" in Remark 4.3 says that: For every central extension of Lie algebras $0\to \mathfrak{a}\to \mathfrak{h}\overset{\pi}{\to} \mathfrak{g}\to 0$, there is a linear map $\beta: \mathfrak{g}\to \mathfrak{h}$ with $\pi \circ \beta = id_{\mathfrak{g}}$ ($\beta$ is in general not a Lie algebra homomorphism). How can I define $\beta$?

Answer 1

The map $\beta$ is called a section. And that exists if the short exact sequence splits. In your case then it will be a trivial extension. The explicit mapping will depend on the algebra itself but usually it should be clear once the exact sequence is defined.(It's the right inverse of the projection map $\pi$).

Note/Edit 1: A right inverse $\beta$ (as morphism in $\textbf{Set}$) always exists (but not unique unless $\mathfrak{a}=0$) as $\pi$ is surjective. But it also has to be (in this case) a morphism in $\textbf{LieAlg}$ to be a section.

For an example one can look at the (non-split) short exact sequence in $\textbf{Ab}$: $$0 \to 2\mathbf{Z} \to \mathbf{Z} \to \mathbf{Z/2Z} \to 0$$

Note/Edit 2 (to include Torsten helpful comments): The question asks for linear section so in fact we are looking at $\textbf{Vect}$. But every short exact sequence in $\textbf{Vect}$ splits, which can be proven by linear algebra. See here.