 Solution

For the Lie algebra two-cocycle ${\alpha }_c$ of part (i), the central extension of a Lie algebra has a basis $x,y,z$ and Lie bracketsPut $y^{\prime }=y+cz$. This is still a basis together with $x,z$, andConsequently $⟨x,y^{\prime }⟩$ is a Lie subalgebra isomorphic to the affine Lie algebra of the line. It commutes with $kz$ and has zero intersection with $kz$, so both summands are ideals of a Lie algebra andThis also holds for the stricter equivalence of central extensions of a Lie algebra. Indeed, choose the linear functional $t$ with $t(x)=0$, $t(y)=c$. Since $[x,y]=y$ in $\mathfrak{s}$, the change of section in part (a) gives ${\alpha }_c^{\prime }(x,y)=c-t(y)=0$. Thus every Lie algebra two-cocycle here is a Lie algebra two-coboundary, so $H^2(\mathfrak{s},k)=0$ in Lie algebra cohomology. The explicit map from the split central extension of a Lie algebra sends $x$ to $x$, $y$ to $y+cz$, and $z$ to $z$; it fixes the central line and induces the identity on the quotient Lie algebra.