The two-dimensional gl(1|1) action is well-defined

The actions of $h_1,h_2$ preserve the parity of $v,w$, and the actions of $e,f$ reverse it. Thus the candidate action has the parity required in [F4].

The operators $h_1,h_2$ are diagonal in the basis $v,w$, so they commute, matching $[h_1,h_2]=0$. On $v$, both the bracket actions $[h_i,e]v$ and the operator commutators $(h_{i}e-e{h}_i)v$ vanish for $i=1,2$, since $[h_i,e]$ is a multiple of $e$, $ev=0$, and each $h_{i}v$ is a scalar multiple of $v$. On $w$, the commutators $h_{1}e-e{h}_1$ and $h_{2}e-e{h}_2$ give $(a-(a-1))(a+b)v=(a+b)v=[h_1,e]w$ and $(b-(b+1))(a+b)v=-(a+b)v=[h_2,e]w$. Likewise, $h_{1}f-f{h}_1$ and $h_{2}f-f{h}_2$ on $v$ give $((a-1)-a)w=-w=[h_1,f]v$ and $((b+1)-b)w=w=[h_2,f]v$, and both bracket actions and commutators vanish on $w$. Hence all even–odd bracket relations in [F1] hold.

Since $e^2=f^2=0$ on both basis vectors, the odd self-brackets act as zero. The anticommutator $ef+fe$ sends $v$ to $(a+b)v$ and $w$ to $(a+b)w$; this equals the action of $h_1+h_2$ on both vectors. Therefore the odd–odd relations in [F1] hold.

By bilinearity, the checked basis-pair identities imply the module identity in [F4] for every homogeneous pair. For $h=c_1{h}_1+c_2{h}_2$, its action on $v$ is $(c_{1}a+c_{2}b)v={\lambda }_{a,b}(h)v$ by [F1] and [F3]. The vector $v$ is nonzero and even, is killed by $e$, and $e$ spans ${\mathfrak{n}}^{+}$ by [F2]; hence [F5] makes it a highest-weight vector. Since $fv=w$, it generates both basis vectors.