The wide-edge morphisms chi-zero and chi-one

The map ${\chi }_0$ commutes with the differentials. Multiplying the displayed matrices over the commutative ring $R$ gives $Q_0{U}_0^0=\left(\begin{array}{cc}a(x_4-x_2)& x_3{x}_4-x_1{x}_2\\ 0& x_1+x_2-x_3-x_4\end{array}\right)=U_0^1{P}_0$ and $Q_1{U}_0^1=\left(\begin{array}{cc}(x_4-x_2)(x_1-x_4)& (x_4-x_2)(x_2-x_3)\\ -a& a\end{array}\right)=U_0^0{P}_1,$ as is checked entry by entry using $x_i{x}_j=x_j{x}_i$; hence ${\chi }_0$ intertwines the two differentials and is a morphism of factorizations.

The map ${\chi }_1$ commutes with the differentials. Likewise $P_0{U}_1^0=\left(\begin{array}{cc}a& (x_3-x_2)(x_4-x_2)\\ a& (x_1-x_4)(x_4-x_2)\end{array}\right)=U_1^1{Q}_0$ and $P_1{U}_1^1=\left(\begin{array}{cc}{x}_1+x_2-x_3-x_4& x_1{x}_2-x_3{x}_4\\ 0& a(x_4-x_2)\end{array}\right)=U_1^0{Q}_1,$ so ${\chi }_1$ is a morphism of factorizations.

Bidegrees. Every entry of the four matrices is homogeneous, and an entry of bidegree $(p,q)$ in the $i$-th row and $j$-th column represents the map from the $j$-th summand of the source to the $i$-th summand of the target of total bidegree $(p,q)+(\text{target shift}-\text{source shift})$. Reading the shift tables: in $U_0^0$ the scalar $x_4-x_2$ has bidegree $(0,2)$ and maps the unshifted summand $R$ to the unshifted summand $R$, while $1$ has bidegree $(0,0)$ and maps $R\{-2,2\}$ to $R\{-2,4\}$, whose shift difference is $(0,2)$; in $U_0^1$ the entries $x_4,-x_2$ have bidegree $(0,2)$ and the entries $-1,1$ map between the summands $R\{-1,1\}$, $R\{-1,3\}$ whose shift difference is $(0,2)$. So ${\chi }_0$ has bidegree $(0,2)$. For ${\chi }_1$ the entries of $U_1^0$ are $1$ and $x_4-x_2$, which map the summands $R$, $R\{-2,4\}$ to $R$, $R\{-2,2\}$ with shift difference $(0,-2)$ in the second column, compensated by the coefficient of bidegree $(0,2)$, and $U_1^1$ maps $R\{-1,1\},R\{-1,3\}$ to $R\{-1,1\}$ with the entries $1,x_2,1,x_4$ contributing the compensating bidegrees; so ${\chi }_1$ has bidegree $(0,0)$.

The Koszul forms. By [F1] and [F2] the matrix of $C({\mathrm{\Gamma }}_0)$ has rows $(a,x_1-x_4)$ and $(a,x_2-x_3)$; the operation $[12{]}_1$ replaces them by $(a,x_1+x_2-x_3-x_4)$ and $(0,x_2-x_3)$. The matrix of $C({\mathrm{\Gamma }}_1)$ has rows $(a,z)$ and $(0,q)$ with $z=x_1+x_2-x_3-x_4$ and $q=x_1{x}_2-x_3{x}_4$; the operation $[21{]}_{-x_2}$ on the ordered pair $(0,q),(a,z)$ replaces them by $(0,q-x_{2}z)$ and $(a,z)$. Expanding $q-x_{2}z=x_1{x}_2-x_3{x}_4-x_2(x_1+x_2-x_3-x_4)=(x_2-x_3)(x_4-x_2)$ shows that the two standard forms are $C({\mathrm{\Gamma }}_0)\cong (a,x_1+x_2-x_3-x_4)\otimes (0,x_2-x_3),C({\mathrm{\Gamma }}_1)\cong (a,x_1+x_2-x_3-x_4)\otimes (0,(x_2-x_3)(x_4-x_2)),$ with the same first row and with the second factors related by multiplication by $x_4-x_2$.

The flip morphisms and the composites. Put $y=x_4-x_2$ and let $\psi (y)\text{\hspace{0.17em}⁣}:(0,yz)\to (0,z)$ be the morphism whose first-term component is the identity and whose middle component is multiplication by $y$; it commutes with the differentials because $y\cdot 0=0\cdot 1$ and $1\cdot yz=z\cdot y$. Let ${\psi }^{\mathrm{\prime }}(y)\text{\hspace{0.17em}⁣}:(0,z)\to (0,yz)$ be the morphism whose first-term component is multiplication by $y$ and whose middle component is the identity; then $1\cdot 0=0\cdot y$ and $y\cdot z=yz\cdot 1$. Multiplying the matrices of the change of basis in step 2.1 against the standard product bases identifies the conjugates of ${\chi }_0$ and ${\chi }_1$ with $\mathrm{I}\mathrm{d}\otimes {\psi }^{\mathrm{\prime }}(y)$ and $\mathrm{I}\mathrm{d}\otimes \psi (y)$ respectively, as in Lemma 2 of the source. The composites satisfy $\psi (y){\psi }^{\mathrm{\prime }}(y)=y\cdot \mathrm{i}\mathrm{d}$ and ${\psi }^{\mathrm{\prime }}(y)\psi (y)=y\cdot \mathrm{i}\mathrm{d}$ on both components, hence ${\chi }_1{\chi }_0=\mathrm{I}\mathrm{d}\otimes (y\cdot \mathrm{i}\mathrm{d})$ and ${\chi }_0{\chi }_1=\mathrm{I}\mathrm{d}\otimes (y\cdot \mathrm{i}\mathrm{d})$ on the Koszul standard forms. They are nonzero even modulo homotopy: specialize $a=0$, $x_1=x_4$, $x_3=x_2$. Both factorization differentials then vanish, while $y=x_4-x_2$ remains nonzero in $\mathbb{Q}[x_2,x_4]$. A null-homotopy would specialize to $y\mathrm{i}\mathrm{d}=dH+Hd=0$, a contradiction. Thus both composites are nonzero endomorphisms of bidegree $(0,2)$ and act as multiplication by $x_4-x_2\in R$; in particular they are not the identity and the two maps are not inverse to each other.