= Solution
{c}
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 brackets>
$$
[x,y]=y+cz,\qquad[x,z]=[y,z]=0.
$$
Put $y'=y+cz$. This is still a <basis> together with $x,z$, and
$$
[x,y']=[x,y]+c[x,z]=y+cz=y'.
$$
Consequently $\langle x,y'\rangle$ 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> and
$$
\boxed{\widetilde{\mathfrak s}\cong\mathfrak s\oplus kz:\quad\text{exactly one class, which is trivial}.}
$$
This 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 $\alpha'_c(x,y)=c-t(y)=0$. 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>.
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