Filtered L∞-morphisms preserve completed polynomial gauge equivalence

For any dg scalar algebra $A$ put , with filtration the kernels of the projections. Each projection is surjective: lift the finitely many coefficients of any algebraic tensor to $P$, giving a compatible element of the inverse limit. The quotient by its kernel is therefore exactly the displayed algebraic tensor product. The limit is complete and separated, degreewise. Tensoring over the field preserves exact sequences, so each graded filtration layer is its corresponding algebraic tensor product with $A$. Here every input lies in $P$, so the additive filtration bounds also show independence of the chosen representatives modulo $F^{N}W$. The operations in the statement are well defined on every quotient, and the input filtration indices add under each operation; they thus extend continuously to the limit.

To verify the signs and identities on algebraic tensors, move each scalar $a_i$ past every later $W$ factor $x_j$, then apply the $W$ operation and multiply the scalars in their original order. This gives precisely the exponent ${\sum }_{i<j}\mathrm{\mid }{a}_i\mathrm{\mid }\mathrm{\mid }{x}_j\mathrm{\mid }$. This movement respects substitution: the product of its signs in two substituted operations is the sign of the composite permutation, including the sign of moving a degree-one inner operation across earlier inputs. Consequently the part of each shifted Jacobi identity involving only the extended $q_n$ is the original $W$ identity multiplied by the common ordered scalar product. Its sum is zero. Write $D(x\otimes a)=(-1{)}^{\mathrm{\mid }x\mathrm{\mid }}x\otimes d_{A}a$ for the scalar differential. For an extended operation of degree $d$, let $E={\sum }_{i<j}\mathrm{\mid }{a}_i\mathrm{\mid }\mathrm{\mid }{x}_j\mathrm{\mid }$. The coefficient of the scalar product with $d_A{a}_i$ in the output differential has sign exponent $E+{\sum }_j\mathrm{\mid }{x}_j\mathrm{\mid }+d+{\sum }_{j<i}\mathrm{\mid }{a}_j\mathrm{\mid }$. The same coefficient obtained by applying the tensor differential to the $i$th input first has exponent $E+{\sum }_j\mathrm{\mid }{x}_j\mathrm{\mid }+{\sum }_{j<i}\mathrm{\mid }{a}_j\mathrm{\mid }$: the extra scalar crossing signs are ${\sum }_{j>i}\mathrm{\mid }{x}_j\mathrm{\mid }$, and the tensor differential contributes ${\sum }_{j<i}(\mathrm{\mid }{x}_j\mathrm{\mid }+\mathrm{\mid }{a}_j\mathrm{\mid })+\mathrm{\mid }{x}_i\mathrm{\mid }$. Thus the two expressions differ by $(-1{)}^d$. For $d=1$ this proves that $D$ anticommutes with each extended $q_n$. Also $D^2=0$. Thus the full coderivation with unary term $q_1^A$ squares to zero. This checks the scalar-differential terms as well as the original Jacobi terms.

The same permutation and substitution verification transfers $FQ=RF$ to scalar tensors. Here each $f_n$ has degree zero, so the scalar differential commutes with $f_n$; its Leibniz expansion supplies precisely all the terms in which a scalar differential appears on an input. Thus the stated $f_n^A$ form a filtered morphism. These are finite identities on each algebraic tensor and each fixed arity. They descend to every quotient and hold on the inverse limit by continuity. Any dg algebra map $\rho :A\to B$ preserves scalar products, scalar differentials and degrees, so $\mathrm{id}\widehat{\otimes }\rho$ commutes with all these operations. Applying [F1] to the extended algebras proves their curvature and composition identities; in particular ${\rho }_{\ast }(f_{\ast }^A(z))=f_{\ast }^B({\rho }_{\ast }(z))$.

For $A=\mathrm{\Omega }$, every total-degree-zero element in the positive ideal has a unique expression $z=w(t)-a(t)dt$ with the degrees in the statement. All these functions are polynomials at each quotient and their degrees may grow with $N$. Differentiation, multiplication and evaluation at any $\lambda \in k$ are defined quotientwise and respect the transition maps. The scalar differential of $w$ contributes $w^{\mathrm{\prime }}(t)dt$. The input $-a(t)dt$ appearing once in $q_{n+1}^{\mathrm{\Omega }}$ contributes $-q_{n+1}(w(t{)}^n,a(t))dt$: place this input last, so it crosses no later $W$ factor. Because $w$ and $-adt$ both have total degree zero, its $n+1$ possible positions contribute equally, changing $1\mathrm{/}(n+1)!$ to $1\mathrm{/}n!$. Two such inputs vanish because $(dt{)}^2=0$, and the scalar differential of $a(t)dt$ vanishes for the same reason. Hence $c_{q^{\mathrm{\Omega }}}(z)=c_q(w(t))+(w^{\mathrm{\prime }}(t)-q_1^{w(t)}(a(t)))dt$. In particular the equations in the statement are exactly the Maurer–Cartan equation for $z$.

Applying the identical one-marked-input computation to $f_{\ast }^{\mathrm{\Omega }}$ gives $f_{\ast }^{\mathrm{\Omega }}(z)=u(t)-b(t)dt$, where $u=f_{\ast }w$ and $b={\sum }_{n\ge 0}{f}_{n+1}(w^n,a)\mathrm{/}n!$. Each summand of the latter series lies in filtration $n+1$; the corresponding statement for $u$ is [F1]. Thus only finitely many terms survive in each quotient, and these terms are polynomials. By curvature transformation over $\mathrm{\Omega }$ from step 3.1, $z$ Maurer–Cartan implies $u-bdt$ Maurer–Cartan. The curvature decomposition of step 3.2 applied on $Z$ now proves $c_r(u)=0$ and $u^{\mathrm{\prime }}=r_1^u(b)$.

Evaluation ${\mathrm{ev}}_{\lambda }:\mathrm{\Omega }\to k$, $t\mapsto \lambda$, $dt\mapsto 0$, is a dg algebra map. Naturality in step 3.1 yields ${\mathrm{ev}}_{\lambda }{f}_{\ast }^{\mathrm{\Omega }}(z)=f_{\ast }(w(\lambda ))$. In particular the endpoint values are precisely those in the statement. If the input path is specified as the unique flow supplied by [F3], it satisfies the equations in step 3.2; its image satisfies the corresponding target flow equations and is the target flow for $b$ and initial value $f_{\ast }(w(0))$ by [F3]. This proves the assertion for every such input parameter as well.

Every generating endpoint pair for the equivalence relation in [F2] maps, by the path and endpoint calculations, to another such pair. Finite chains of generating pairs and their reverses therefore map to chains of the same kind; equality maps to equality. This proves preservation of the generated equivalence relation without claiming that a concatenation itself is polynomial. Composition and identities over $\mathrm{\Omega }$ follow from step 3.1 and [F1]. Explicitly, for successive $g$ the output parameter is ${\sum }_{j\ge 0}{g}_{j+1}(u^j,b)\mathrm{/}j!={\sum }_{n\ge 0}{h}_{n+1}(w^n,a)\mathrm{/}n!$: in the composite partition formula exactly one $f$ block contains $a$. Thus the point and parameter components both compose, as do the induced quotient maps. Every equality holds modulo each $F^N$ and hence in the separated inverse limit; no globally polynomial bound or analytic integration was invoked.