Invariance under the braid-like Reidemeister IIa move

The splitting criterion. The four resolution corners form a complex with $C({\mathrm{\Gamma }}_{00})$ in degree $-1$, $C({\mathrm{\Gamma }}_{10})\oplus C({\mathrm{\Gamma }}_{01})$ in degree $0$ and $C({\mathrm{\Gamma }}_{11})$ in degree $1$, with the shifts supplied by the crossing cones. Suppose $f_1$ identifies $C({\mathrm{\Gamma }}_{00})$ with a summand $I$ of $C({\mathrm{\Gamma }}_{10})=I\oplus M$, and $f_2{\mathrm{\mid }}_M:M\to C({\mathrm{\Gamma }}_{11})$ is invertible. Elementary changes of coordinates first cancel the block $C({\mathrm{\Gamma }}_{00})\to I$ and then the block $M\to C({\mathrm{\Gamma }}_{11})$. For an invertible block $\phi$, subtracting its other row and column entries using ${\phi }^{-1}$ makes the differential block diagonal; the surviving block is the Schur complement. The equation $d^2=0$ makes adjacent components to the canceled pair zero in these coordinates, so its identity pair is contractible. Here the sole final term is $C({\mathrm{\Gamma }}_{01})$ in degree $0$, and it has zero differential; off-diagonal cube maps introduce no further term. Thus the two invertible blocks suffice to prove $C(D_1)\simeq C({\mathrm{\Gamma }}_{01})=C(D_2)$.

Koszul form of the diagram and its first reduction. In the standard Koszul bases the three factorizations have four rows: $C({\mathrm{\Gamma }}_{00})$ has rows $(a,x_1+x_2-x_5-x_6)$, $(0,x_2-x_6)$, $(a,x_5+x_6-x_3-x_4)$, $(0,(x_6-x_4)(x_3-x_6))$; $C({\mathrm{\Gamma }}_{10})$ has the same rows with second row $(0,(x_2-x_6)(x_5-x_2))$; $C({\mathrm{\Gamma }}_{11})$ has the same rows as $C({\mathrm{\Gamma }}_{10})$ with last row $(0,x_6-x_4)$; the maps are $f_1=\mathrm{I}\mathrm{d}\otimes {\psi }^{\mathrm{\prime }}(x_5-x_2)\otimes \mathrm{I}\mathrm{d}\otimes \mathrm{I}\mathrm{d}$ and $f_2=\mathrm{I}\mathrm{d}\otimes \mathrm{I}\mathrm{d}\otimes \mathrm{I}\mathrm{d}\otimes \psi (x_3-x_6)$. Apply the row operation $[13{]}_1$ to all three matrices simultaneously: the first rows become $(a,x_1+x_2-x_3-x_4)$ and the third rows $(0,x_5+x_6-x_3-x_4)$ in all three; the common third rows can be deleted and the internal variable $x_5$ excluded by $x_5\mapsto x_3+x_4-x_6$. The diagram becomes the tensor product of the row $(a,x_1+x_2-x_3-x_4)$ with the diagram of two-row matrices $[x_2-x_6,(x_6-x_4)(x_3-x_6)]\stackrel{g_1}{\to }[(x_2-x_6)(x_3+x_4-x_6-x_2),(x_6-x_4)(x_3-x_6)]\stackrel{g_2}{\to }[(x_2-x_6)(x_3+x_4-x_6-x_2),x_6-x_4]$ over $R^{\mathrm{\prime }}[x_6]$, $R^{\mathrm{\prime }}=\mathbb{Q}[a,x_1,x_2,x_3,x_4]$, where $g_1={\psi }^{\mathrm{\prime }}(x_3+x_4-x_6-x_2)\otimes \mathrm{I}\mathrm{d}$ and $g_2=\mathrm{I}\mathrm{d}\otimes \psi (x_3-x_6)$; each step is an isomorphism or a chain homotopy equivalence by [F2].

Quadratic polynomial division. Put $v=x_6$, $Q(v)=(v-x_4)(x_3-v)$ and $S=R^{\mathrm{\prime }}[v]$. Its leading coefficient is the unit $-1$, so polynomial division gives an $R^{\mathrm{\prime }}$-module decomposition $S=QS\oplus (R^{\mathrm{\prime }}1\oplus R^{\mathrm{\prime }}v)$. In the zero-first-entry Koszul row $(0,Q)$, multiplication by $Q$ maps the odd copy of $S$ isomorphically to $QS$ in the even copy. Cancel these pairs over $R^{\mathrm{\prime }}$; the surviving even copy is $S\mathrm{/}(Q)$, free over $R^{\mathrm{\prime }}$ on $1,v$. In tensoring this row with another zero-first-entry row $(0,f)$, the same cancellation gives the surviving differential induced by $f$ on $S\mathrm{/}(Q)$: the quotient map commutes with multiplication by $f$, and the invertible $Q$ block removes all its complementary components by the elementary Schur-complement calculation of step 1.1. Apply this simultaneously to the first two factorizations and their coefficient maps. The third bottom row $(0,v-x_4)$ reduces by the linear exclusion lemma [F3] to evaluation $v=x_4$. The reduced diagram thus has first differential multiplication by $x_2-v$ on $R^{\mathrm{\prime }}1\oplus R^{\mathrm{\prime }}v$, second differential multiplication by $(x_2-x_4)(x_3-x_2)$ on that same rank-two module, and third differential multiplication by this last element on $R^{\mathrm{\prime }}$. Its vertical maps are $1$ and $x_3+x_4-v-x_2$ on the respective components of $g_1$, and evaluation $v=x_4$ on both components of $g_2$. The quadratic reduction is polynomial division, rather than an application of the linear exclusion clause.

Splitting the reduced diagram. In the reduced diagram the first factorization has a contractible summand $R^{\mathrm{\prime }}\stackrel{1}{\to }{R}^{\mathrm{\prime }}(x_2-x_6)$, whose removal leaves the rank-one row $R^{\mathrm{\prime }}(x_6+x_2-x_3-x_4)\stackrel{(x_2-x_4)(x_2-x_3)}{\to }{R}^{\mathrm{\prime }}$; the middle factorization is the direct sum of the two factorizations $R^{\mathrm{\prime }}\stackrel{(x_2-x_4)(x_3-x_2)}{\to }{R}^{\mathrm{\prime }}$ and $R^{\mathrm{\prime }}(x_6+x_2-x_3-x_4)\stackrel{(x_2-x_4)(x_3-x_2)}{\to }{R}^{\mathrm{\prime }}(x_6+x_2-x_3-x_4)$, isomorphic to the first up to a grading shift; and the third factorization is the rank-one row over $R^{\mathrm{\prime }}$ with differential $(x_2-x_4)(x_3-x_2)$. The map $g_1$ takes the reduced first factorization isomorphically onto the second summand of the middle factorization, and $g_2$ restricts to an isomorphism from the first summand of the middle factorization onto the third factorization; thus $f_1$ is an isomorphism onto a direct summand and $f_2$ restricts to an isomorphism from a complement, as required by step 1.1, so $C(D_1)$ is isomorphic in $K({\mathrm{h}\mathrm{m}\mathrm{f}}_w)$ to the direct sum of two contractible complexes and $C({\mathrm{\Gamma }}_{01})\cong C(D_2)$. Every map in the computation is homogeneous of bidegree $(0,0)$ between the shifted rows, so no grading shift occurs.

Conclusion and mirror image. Steps 1.2-3.1 verify the splitting criterion of step 1.1 for the pair of Figure 15, giving $C(D_1)\cong C(D_2)$ in $K({\mathrm{h}\mathrm{m}\mathrm{f}}_w)$ with no shift of the trigrading, hence an isomorphism of trigraded cohomology for the two braid-like IIa diagrams. The mirror-image move is the same computation with the signs of all potentials reversed and the roles of the two sides exchanged, which leaves the conclusion unchanged; both diagrams have potential $w$ in Figure 15 and potential $-w$ after all orientations are reversed, and no step of the argument uses the Axiom of Choice.