The Heisenberg algebra from a two-cocycle

Since both the bracket of $\mathfrak{a}$ and its action on $kz$ are zero, every term in the Chevalley–Eilenberg differential of a $2$-cochain vanishes. Equivalently, the possible target $C^3(\mathfrak{a},kz)$ is already zero because ${\bigwedge }^3\mathfrak{a}=0$. Hence $d\omega =0$.

The same triviality makes the differential $C^1\to C^2$ zero, so the nonzero form $\omega$ is not a coboundary. Indeed $C^2(\mathfrak{a},kz)$ is one-dimensional, generated by $x^{\ast }\wedge y^{\ast }\otimes z$.

On $kz\oplus \mathfrak{a}$ define $[(a,u),(b,v)]=(\omega (u,v),0)$; this is the usual cocycle-extension formula here because the action and the bracket of $\mathfrak{a}$ are both zero. It is bilinear and alternating, and Jacobi holds because every bracket lies in $kz\oplus 0$, which brackets to zero. Its kernel $kz\oplus 0$ is abelian, its quotient is $\mathfrak{a}$, and the induced action is trivial, so it is the abelian extension represented by $\omega$. Its only nonzero basis bracket is $[(0,x),(0,y)]=(z,0)$. Renaming the three basis vectors gives exactly the displayed Heisenberg relations. This checks the zero and one-dimensional boundary spaces explicitly and uses no choice.