Let $G$ be a Liegroup and $H$ a Lie subgroup of $G$. Then we find a Liegroup homomorphism $i \colon H \to G$ and the induced map $i_* \colon \mathfrak{h} \to \mathfrak{g}$ between the corresponding Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ is a Lie-algebra homomorphism.
Is it always possible to find a map $$\pi \colon \mathfrak{g} \to \mathfrak{h}$$ such that $\pi \circ i_* = \operatorname{id}_{\mathfrak{h}}$ and such that $\pi$ is a Lie algebra homomorphism?
If not, are there some concrete assumptions (on the Lie groups or Lie algebras) under which such a map $\pi$ exists?
Such a Lie algebra homomorphism is called a section. It need not exist in general. It means that the short exact sequence $$ 0\rightarrow \mathfrak{a} \rightarrow \mathfrak{g} \xrightarrow{\pi} \mathfrak{h} \rightarrow 1 $$ splits. This is equivalent to saying that $\mathfrak{g}$ is a semidirect product of $\mathfrak{h}$ and $\mathfrak{a}$, i.e., $\mathfrak{g}\cong \mathfrak{h}\ltimes \mathfrak{a}$. Assuming that $\mathfrak{a}$ is abelian, the assumption to assure that the sequence splits is, that the cohomology group $H^2(\mathfrak{h},\mathfrak{a})$ is trivial.