Markings do not change the Khovanov-Rozansky complex

Removing a mark: the top configuration of Figure 10. The Koszul matrix of $C({\mathrm{\Gamma }}_1)$ has rows $(a,x_1+x_5-x_3-x_4)$, $(0,x_1{x}_5-x_3{x}_4)$ and $(a,x_2-x_5)$. Apply the row operation $[13{]}_1$ to the first and third rows: they become $(a,x_1+x_5-x_3-x_4+x_2-x_5)=(a,x_1+x_2-x_3-x_4)$ and $(a-a,x_2-x_5)=(0,x_2-x_5)$, while the quadratic row is unchanged; the operation is an isomorphism of factorizations by [F2]. The bottom row $(0,x_2-x_5)$ has coefficient $-1$ on $x_5$, a unit; $x_5$ may still occur in the quadratic row, to which the ensuing substitution must also be applied; it is internal because $w=a(x_1+x_2-x_3-x_4)$ does not involve $x_5$. By the variable-exclusion clause of [F2] the row may be deleted and $x_5$ replaced by $x_2$ in every remaining row, leaving rows $(a,x_1+x_2-x_3-x_4)$ and $(0,x_1{x}_2-x_3{x}_4)$, which is the Koszul matrix of $C({\mathrm{\Gamma }}_2)$. Hence $C({\mathrm{\Gamma }}_1)\cong C({\mathrm{\Gamma }}_2)$ in ${\mathrm{h}\mathrm{m}\mathrm{f}}_w$, with the same potential by [F1]; the other local pairs of Figure 10 are the symmetric cases with the roles of the rows exchanged.

Compatibility with the crossing differential. The first complex of formula (10), written in Koszul form, has common first and third rows $(a,x_1+x_5-x_3-x_4)$ and $(a,x_2-x_5)$, with second row $(0,(x_5-x_3)(x_4-x_5))$ in the source and $(0,x_5-x_3)$ in the target, with differential $\mathrm{I}\mathrm{d}\otimes \psi (x_4-x_5)\otimes \mathrm{I}\mathrm{d}$. Applying the row operation $[13{]}_1$ to both matrices simultaneously gives an isomorphic complex whose matrices have first row $(a,x_1+x_2-x_3-x_4)$, second row unchanged and third row $(0,x_2-x_5)$; the differential is the identity on this third row. Substituting the internal variable $x=x_2-x_5$, both matrices have identical bottom rows $(0,x)$ on which the differential acts by the identity, so the variable-exclusion clause of [F2] deletes that row and sets $x_5=x_2$, reducing the ground ring to $R=\mathbb{Q}[a,x_1,x_2,x_3,x_4]$ and leaving the complex $(a,x_1+x_2-x_3-x_4)\otimes (0,(x_2-x_3)(x_4-x_2))\stackrel{\psi (x_4-x_2)}{\to }(a,x_1+x_2-x_3-x_4)\otimes (0,x_2-x_3)$, which is precisely the second complex of formula (10). The two complexes are therefore chain homotopy equivalent. The reverse crossing map ${\psi }^{\mathrm{\prime }}(x_4-x_5)$ is likewise the identity on the first and third exterior factors, so the same row change and substitution give ${\psi }^{\mathrm{\prime }}(x_4-x_2)$. This proves compatibility for both crossing signs. The second pair of Figure 11 is obtained by exchanging the exterior edge labels; that relabelling carries each of these row operations, substitutions and maps to the corresponding formulas, proving the equivalence for $i=2$.

Conclusion. Every change of marking decomposes into the local moves of Figure 10, each of which changes $C$ by an isomorphism or a chain homotopy equivalence as in step 1.1, and step 2.1 shows that these local equivalences can be chosen compatibly with the crossing differentials ${\chi }_1$, so the two-term complexes of Figure 11 are chain homotopy equivalent. Composing the local equivalences along any finite sequence of marking changes gives a chain homotopy equivalence $C(D^{\mathrm{\prime }})\simeq C(D)$ in $K({\mathrm{h}\mathrm{m}\mathrm{f}}_w)$ with the same potential, all three gradings being preserved because every operation is a homogeneous change of basis or a substitution by a linear form of bidegree $(0,2)$; passing to cohomology gives an isomorphism $H(D^{\mathrm{\prime }})\cong H(D)$ with no shift. No Axiom of Choice is used.