Basepoint variables act chain homotopically on a knot grid

The sum defining $H_{\mathrm{\ell }}$ is finite because there are finitely many states and at most two torus rectangles for each ordered state pair. It is ${\mathbb{F}}_2[U_1,\dots ,U_n]$-linear by its coefficient formula. For each counted rectangle $r$, [F4] gives $M(\mathbf{x})-M(\mathbf{y})=1-2n_{\mathbb{O}}(r)$; the coefficient has Maslov degree $-2n_{\mathbb{O}}(r)$ by [F2]. Therefore each term of $H_{\mathrm{\ell }}(\mathbf{x})$ has Maslov degree $M(\mathbf{x})-1$. Also [F4] gives $A(\mathbf{y})-n_{\mathbb{O}}(r)=A(\mathbf{x})-n_{\mathbb{X}}(r)$, and $X_{\mathrm{\ell }}\in \mathrm{Int}(r)$ implies $n_{\mathbb{X}}(r)\ge 1$. Thus $H_{\mathrm{\ell }}$ does not raise the Alexander filtration.

Expand ${\mathrm{\partial }}^{-}{H}_{\mathrm{\ell }}+H_{\mathrm{\ell }}{\mathrm{\partial }}^{-}$ and group its terms by their two-rectangle domain $\psi$. If $\psi$ is non-annular, [F3] pairs its two decompositions. In each decomposition, when $X_{\mathrm{\ell }}$ has multiplicity zero neither ordered term is selected; when it has multiplicity one exactly one of the two rectangles contains it and exactly one ordering is selected; when it has multiplicity two both rectangles contain it and the two possible placements of $H_{\mathrm{\ell }}$ cancel in characteristic two. The paired decomposition has the same domain, $O$-weight, and multiplicity at $X_{\mathrm{\ell }}$, so every non-annular contribution cancels.

A returning two-rectangle domain is a thin horizontal or vertical annulus by [F3]. The only thin annuli containing $X_{\mathrm{\ell }}$ are the horizontal one in its row and the vertical one in its column. In the horizontal annulus the unique $O$-marking is $O_i$, so its coefficient is $U_i$; in the vertical annulus the unique $O$-marking is $O_j$, so its coefficient is $U_j$. Each annulus has one decomposition and contributes its coefficient times the input generator. Hence the returning contribution is $(U_i+U_j)\mathbf{x}$. Together with step 1.2 this proves the chain homotopy identity.

The marking route in [F1] orders the variables cyclically, and each adjacent pair in that cycle is consecutive as defined in the statement. Any two indices are joined by a finite path of adjacent pairs. By [F5], each multiplication map is a chain map for the differential; adding the homotopy maps from step 2.1 along the path gives a filtered chain homotopy between multiplication by $U_i$ and multiplication by $U_j$.