= Equivalence of central extensions of a Lie algebra
Two <central extensions of a Lie algebra> $0\to\mathfrak z\to E\to\mathfrak g\to0$ and $0\to\mathfrak z\to E'\to\mathfrak g\to0$ are equivalent if a <Lie algebra isomorphism> $E\to E'$ commutes with both maps in the <short exact sequences>, inducing the identity on the specified $\mathfrak z$ and $\mathfrak g$. This is finer than asking only that $E$ and $E'$ be isomorphic as <Lie algebras>.
For $\mathfrak z=kz$, choose linear sections and identify the two <vector spaces> with $\mathfrak g\oplus kz$. A <linear map> compatible with both ends necessarily has the form $(x,uz)\mapsto(x,(u+t(x))z)$ for a <linear functional> $t$. If the two <Lie algebra two-cocycles> are $\alpha,\alpha'$, preservation of the <Lie bracket> is exactly $\alpha'(x,y)=\alpha(x,y)+t([x,y])$. Hence equivalence means that their difference is a <Lie algebra two-coboundary>. If changes of the identifications at the ends are allowed, an <automorphism of a Lie algebra> at either end can identify further classes. For a two-dimensional <abelian Lie algebra> and a one-dimensional central line, the finer classes are parametrized by $k$, whereas the middle <Lie algebras> have only the zero and nonzero classes, represented by the <abelian Lie algebra> $k^3$ and the <Heisenberg Lie algebra>.
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