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Let $\mathfrak{b}$ and $\mathfrak{g}$ be finite dimensional Lie algebras and let $(\tilde{\mathfrak{g}},j,\phi)$ be a Lie algebra extension of $\mathfrak{g}$ by $\mathfrak{b}$, so we have a short exact sequence of Lie algebras$$ 0\to\mathfrak{b}\underbrace{\to}_{j}\tilde{\mathfrak{g}}\underbrace{\to}_{\phi}\mathfrak{g}\to0 $$ as in https://www.staff.science.uu.nl/~ban00101/lecnotes/repq.pdf, page 15. Then, the author chooses a linear map $\xi:\mathfrak{g}\to\tilde{\mathfrak{g}}$ such that $\phi\circ\xi=Id_{\mathfrak{g}}$. If $\xi$ is a Lie algebra homomorphism, it means that the central extension splits but it is just assumed to be linear. Why does this map exist? I read on here that every short exact sequence of vector spaces splits, is there an intuitive reason why that is true?

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  • $\begingroup$ It suffices to define $\xi$ on a basis of $\frak{g}$. $\endgroup$ – Arnaud D. Oct 25 '19 at 12:46
  • $\begingroup$ I have to define it using $\phi$ I suppose? $\endgroup$ – Lucas Smits Oct 25 '19 at 12:48
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    $\begingroup$ If I understand correctly the question is unrelated to Lie algebras... $\endgroup$ – YCor Oct 25 '19 at 20:03
  • $\begingroup$ I added an answer for vector spaces and $R$-modules. $\endgroup$ – Dietrich Burde Oct 26 '19 at 8:19
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Extensions of vector spaces split, but not extensions of Lie algebras in general. So "this homomorphism $\xi$" need not exist. Consider an example, i.e., a non-split extension of the $3$-dimensional Heisenberg Lie algebra:

Non-split extension of Lie algebras?

For vector spaces, or more generally for $R$-modules see here:

Why any short exact sequence of vector spaces may be seen as a direct sum?

A question about split short exact sequence of modules

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