The gl(1|1) bracket data define a Lie superalgebra

The bracket table is homogeneous: even–even brackets lie in the even part, even–odd brackets in the odd part, and odd–odd brackets in the even part. Its odd–odd entries are symmetric and its even–odd reverse entries are their negatives, so it satisfies graded skew symmetry in [F2].

For homogeneous $x,y,z$, put $J(x,y,z)=(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }}[x,[y,z]]+(-1{)}^{\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}[y,[z,x]]+(-1{)}^{\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}[z,[x,y]]$. Graded skew symmetry gives $J(x,y,z)=(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }}([x,[y,z]]-[[x,y],z]-(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }}[y,[x,z]])$ and $J(y,x,z)=-(-1{)}^{\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }y\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}J(x,y,z)$; cyclic rotation preserves $J$ by its formula. Hence the derivation Jacobi identity is equivalent to $J=0$, and it suffices to check representative basis triples up to permutation. For representatives $(h_i,h_j,x)$ with $x=e$ or $f$, the commuting diagonal actions of $h_1,h_2$ have weights $(1,-1)$ and $(-1,1)$, so $[h_i,[h_j,x]]=[h_j,[h_i,x]]$. All-even brackets vanish, so all-even triples also satisfy Jacobi.

Put $c=h_1+h_2$. By [F1], $[h_i,c]=[e,c]=[f,c]=0$. For a triple $(h_i,e,f)$, the Jacobi identity reduces to $[h_i,c]=[[h_i,e],f]+[e,[h_i,f]]$; the two terms on the right are opposite multiples of $c$, so their sum is zero. For $(h_i,e,e)$ or $(h_i,f,f)$, every term vanishes because $[e,e]=[f,f]=0$. This checks all basis triples with one even and two odd entries.

For three odd basis entries, each inner bracket is either zero or $c$, and every outer bracket of $c$ with $e$ or $f$ is zero by step 2.1. Hence every nested-bracket term vanishes. Together with the representative even and mixed checks from step 1.2 and step 2.1, permutation reduction and multilinearity give signed Jacobi for all homogeneous triples.

For any odd $x$, put $A=[x,[x,x]]$. Applying the signed Jacobi identity from [F2] to $(x,x,x)$ gives $A=[[x,x],x]-A$, while graded skew symmetry and the even parity of $[x,x]$ give $[[x,x],x]=-A$. Hence $3A=0$, and $A=0$ because $\mathbb{C}$ has characteristic zero. This proves the additional odd self-Jacobi clause, so the bracket satisfies the full complex Lie-superalgebra axioms.

The subspaces $\mathfrak{h},{\mathfrak{n}}^{+},{\mathfrak{n}}^{-}$ are graded subalgebras: $\mathfrak{h}$ is abelian, and $[e,e]=[f,f]=0$. Their direct sum is $E$. The table gives $[h_i,e]\in \mathbb{C}e$, so $\mathfrak{h}\oplus {\mathfrak{n}}^{+}$ is a subalgebra as well. These are precisely the conditions in [F3].