Setting a to zero in a closed KR factorization gives the layer-by-layer Koszul complex

Specialize $a=0$. By [L1] each local factorization has $d^2=a\cdot (\text{linear form})\cdot \mathrm{i}\mathrm{d}$ and setting $a=0$ makes $d^2=0$ and kills the $a$-linear entries. By [L2] the arc row $(a,x_1-x_2)$ becomes the $\mathbb{Z}\mathrm{/}2$-graded complex with even part $S$, odd part $S\{-1,1\}$, even-to-odd map $0$ and odd-to-even map $x_1-x_2$; the wide-edge tensor of rows $(a,\beta )$ and $(0,\gamma )$ becomes the tensor product of the two rows $(0,\beta )$ and $(0,\gamma )$.

Identify each local $a=0$ complex with the folding of the Koszul complex of its relations. The arc of labels $x_1,x_2$ gives, with $f=x_1-x_2$, the folding of $K(f;S)$: $K_1=S\{-1,1\}\stackrel{f}{\to }{K}_0=S$ has even part $S$ and odd part $S\{-1,1\}$ with the same maps. The wide edge gives the folding of $K(\beta ,\gamma ;S)$: place the Koszul symbols for $\beta$ and $\gamma$ in the shifted bidegrees $(-1,1)$ and $(-1,3)$, so that $K_0=S$, $K_1=S\{-1,1\}\oplus S\{-1,3\}$, $K_2=S\{-2,4\}$ and every differential has bidegree $(1,1)$; the tensor product of the two rows $(0,\beta ),(0,\gamma )$ has even part $S\oplus S\{-2,4\}$, odd part $S\{-1,1\}\oplus S\{-1,3\}$ and differentials matching those of $K(\beta ,\gamma )$ up to a global sign on the odd part.

Tensor over the layers. The $a=0$ specialization of a tensor product of factorizations is the tensor product of the $a=0$ specializations, because the product differential is the sum over the factors of the local differentials tensored with the other factors; the Koszul signs on the product are the Koszul signs of the total complex. By 1.2 the result is the tensor product of the foldings of the local Koszul complexes, which is the folding of the tensor product of those Koszul complexes.

Apply Koszul concatenation. Take the local relations in the layer order, at layer $i$ first the two wide-edge relations ${\beta }_i,{\gamma }_i$ and then the differences $x_{i,j}-x_{i-1,j}$ for $j\mathrm{\notin }\{s,s+1\}$, and put the closure differences last. By [L5] the tensor product of the local Koszul complexes over the shared polynomial ring is isomorphic, with Koszul signs, to the Koszul complex of the concatenated sequence, which is $K(D)$. Combining with step 2.1 gives an isomorphism of $\mathbb{Z}\mathrm{/}2$-graded $S$-complexes from $C(D){\mathrm{\mid }}_{a=0}$ to the folding of $K(D)$, up to the global sign on the odd part that a change of basis absorbs.

Read off the shifts and the closure hypothesis. In the arc row the odd generator $1\otimes 1$ of $S\{-1,1\}$ has bidegree $(-1,1)$, and in the wide edge the two odd generators have bidegrees $(-1,1)$ and $(-1,3)$, exactly the shifts displayed in [L2] for the middle terms; the differential $S\{-1,1\}\to S$ has bidegree $(1,1)$ because $x_1-x_2$ has bidegree $(0,2)$, and likewise ${\gamma }_i:S\{-1,3\}\to S$ has bidegree $(1,1)$ because $x_1{x}_2-x_3{x}_4$ has bidegree $(0,4)$. For the closed graph the potential vanishes by [L3] and there are no boundary points, so no boundary variable remains and the sequence has exactly $(r+1)m$ elements; the isomorphism of step 3.1 is therefore an identification of the $a=0$ specialization of $C(D)$ with the folding of $K(D)$, carrying the displayed shifts.