Second cohomology classifies abelian extensions

Given an extension, lift a finite basis of $\mathfrak{g}$ and extend linearly to a section $s:\mathfrak{g}\to \mathfrak{e}$. The formula $x\cdot m=[s(x),m]$ is independent of the lift because $M$ is abelian, and it is the fixed action. Define ${\omega }_s(x,y)=[s(x),s(y)]-s([x,y])\in M$. It is alternating. Expanding Jacobi for the three lifts and collecting their $M$-components gives exactly $d{\omega }_s=0$ with [L1]'s signs.

Conversely, for a $2$-cocycle $\omega$, put on $M\oplus \mathfrak{g}$ the bracket $[(m,x),(n,y)]=(xn-ym+\omega (x,y),[x,y])$. Alternation is immediate. The $\mathfrak{g}$-component of Jacobi vanishes by Jacobi in $\mathfrak{g}$; the $M$-component is precisely $d\omega$, so it vanishes. Thus this is an extension inducing the fixed action. Replacing $\omega$ by $\omega +dt$ gives the equivalent extension through $(m,x)\mapsto (m-t(x),x)$.

A second section is $s^{\mathrm{\prime }}=s+t$ for a linear map $t:\mathfrak{g}\to M$. Since $M$ is abelian, direct expansion gives ${\omega }_{s^{\mathrm{\prime }}}={\omega }_s+dt$. Hence the class $[{\omega }_s]$ is independent of the finite-basis section and is preserved by extension equivalences.

Starting from an extension, the map $(m,x)\mapsto m+s(x)$ identifies the construction in step 1.2 with the original extension; starting from a cocycle recovers it from the canonical section. These operations are inverse on equivalence classes. Finally, $[\omega ]=0$ exactly when a section change makes $\omega$ zero, and then the section is a Lie homomorphism; by [L3] this is exactly a split extension. If $\mathfrak{g}=0$ or $M=0$, the constructions reduce to the unique zero class and the evident split extension.