Complete coderivations form a filtered differential graded Lie algebra

Fix a degree $d$. For a uniformly Cauchy sequence $T_m$ in $E^d$, each $T_m(c)$ is Cauchy in $C_B$, because $T_m-T_j\in F^{p}E$ eventually for every $p$. For homogeneous $c$ of degree $j$, define $T(c)={\lim }_m{T}_m(c)$ in $(C_B{)}^{j+d}$; extend over the finite homogeneous decomposition of each $c$. Limits in the fixed output degree preserve addition, $B$-linearity, and each closed $F^q{C}_B$. In particular $T$ has degree $d$. Thus $T$ preserves the filtration and is continuous. If $T_m-T_j\in F^{p}E$ for all sufficiently large $j$, passing to the limit on every $F^q{C}_B$ gives $T_m-T\in F^{p}E$. More directly, a compatible family of cosets in $E^d\mathrm{/}(F^{p}E{)}^d$ determines compatible evaluation cosets on each $c$ in every $C_B\mathrm{/}{F}^n{C}_B$, without choosing simultaneous representatives. Their unique limit defines $T$; evaluating on $F^q{C}_B$ modulo $F^{p+q}$ proves that its class is the prescribed one in $E^d\mathrm{/}(F^{p}E{)}^d$. This proves degreewise inverse-limit completeness as well; if $T\in {\bigcap }_p{F}^{p}E$, every $T(c)$ belongs to ${\bigcap }_p{F}^p{C}_B=0$, proving separation.

If $P\in F^{p}E$ and $Q\in F^{q}E$, then $PQ(F^r{C}_B)\subseteq P(F^{q+r}{C}_B)\subseteq F^{p+q+r}{C}_B$, proving multiplicativity. The coderivation equation, the counit equation, and evaluation at $1_C$ are closed conditions: reduce each equation modulo $F^n$ in its explicitly completed target, where sufficiently close maps agree. Consequently $\mathfrak{l}$ and every $F^p\mathfrak{l}$ are closed and complete in each fixed degree for their induced filtration.

Put $p=\mathrm{\mid }P\mathrm{\mid }$, $q=\mathrm{\mid }Q\mathrm{\mid }$. For coderivations, the graded tensor composition rule gives $\mathrm{\Delta }PQ=(PQ\otimes 1+P\otimes Q+(-1{)}^{pq}Q\otimes P+1\otimes PQ)\mathrm{\Delta }$. Subtracting $(-1{)}^{pq}\mathrm{\Delta }QP$ cancels the two mixed terms and gives $\mathrm{\Delta }[P,Q]=([P,Q]\otimes 1+1\otimes [P,Q])\mathrm{\Delta }$. Also $[P,Q](1_C)=0$ and $ϵ[P,Q]=0$. Thus coderivations are closed under the commutator, and multiplicativity proves the additive filtration bound.

The commutator is graded skew because $[P,Q]=-(-1{)}^{pq}[Q,P]$. For $\mathrm{\mid }R\mathrm{\mid }=r$, direct expansion of $[P,[Q,R]]$ is $PQR-(-1{)}^{qr}PRQ-(-1{)}^{p(q+r)}QRP+(-1{)}^{p(q+r)+qr}RQP$. The expansion of $[[P,Q],R]$ is $PQR-(-1{)}^{pq}QPR-(-1{)}^{(p+q)r}RPQ+(-1{)}^{(p+q)r+pq}RQP$. The expansion of $(-1{)}^{pq}[Q,[P,R]]$ is $(-1{)}^{pq}QPR-(-1{)}^{pq+pr}QRP-(-1{)}^{qr}PRQ+(-1{)}^{qr+pr}RPQ$. The $QPR$ and $RPQ$ terms cancel in the last two expressions; the other four agree with the first expression. This proves the graded Jacobi identity in derivation form, with all signs explicit.

For homogeneous $T$ of degree $t$, expanding $[D,[D,T]]$ gives $D^{2}T-T{D}^2$, since the two $DTD$ terms have coefficients $-(-1{)}^t$ and $+(-1{)}^t$. Hence $d^{2}T=0$. Substituting $P=D$ into the Jacobi identity proved in step 4.1 gives $d[P,Q]=[dP,Q]+(-1{)}^{\mathrm{\mid }P\mathrm{\mid }}[P,dQ]$. Because $D$ preserves each coefficient filtration, composition with $D$ and hence $d$ preserve every $F^{p}E$ and every $F^p\mathfrak{l}$.

The preceding steps prove all graded Lie, differential, degreewise completeness, and filtration axioms on $F^1\mathfrak{l}$. In each quotient by $F^n$, any bracket with at least $n$ positive-order entries vanishes by the additive bound. Thus the positive filtration is pronilpotent as well as complete, as claimed.