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Choose a basis $x,y$ of the two-dimensional abelian Lie algebra $\mathfrak{a}$, and lifts with the same names. Every alternating bilinear form is determined by $c=\alpha (x,y)$, and every such form is a Lie algebra two-cocycle because every Lie bracket in $\mathfrak{a}$ is zero. Thus every central extension of a Lie algebra has a basis $x,y,z$ withIf $c=0$, all Lie brackets vanish, so this is the direct sum of Lie algebras $\mathfrak{a}\oplus kz\cong k^3$. If $c\ne 0$, replace the central basis vector by $z^{\prime }=cz$. Then $[x,y]=z^{\prime }$, which is the Heisenberg Lie algebra ${\mathfrak{h}}_3(k)$. This change of basis shows that every nonzero $c$ gives the same Lie algebra isomorphism class.

These two Lie algebras cannot be isomorphic: their derived algebras have dimensions zero and one, respectively, and a Lie algebra isomorphism preserves the derived algebra. Moreover, the Heisenberg Lie algebra cannot be a direct sum of Lie algebras $kz\oplus \mathfrak{a}$, since that direct sum of Lie algebras would have zero Lie bracket. ThereforeHere “up to isomorphism” means the Lie algebra isomorphism class of the middle Lie algebra, allowing a change of the identifications at the ends of the short exact sequence. If instead equivalence of central extensions of a Lie algebra must induce the identity on both $kz$ and $\mathfrak{a}$, the parameter $c$ cannot change: the formula in part (a) has $t([x,y])=0$. That stricter notion has one class for each $c\in k$, rather than two. This distinction explains the convention required by the question.

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.