Koszul row operations and variable exclusion preserve homotopy type

Row operations. Model the Koszul factorization on the exterior algebra of a free module with basis $e_i$, with differential $d={\sum }_i{a}_i(e_i\wedge -)+{\sum }_i{b}_i{\iota }_i$, where ${\iota }_i$ is contraction by the dual basis. The exterior and contraction operators anticommute for distinct indices and satisfy ${\iota }_i(e_i\wedge -)+(e_i\wedge -){\iota }_i=1$, giving $d^2={\sum }_i{a}_i{b}_i$. The basis change $e_i\mapsto e_i+\lambda e_j$, $e_j\mapsto e_j$ induces an invertible exterior-algebra map. Expressing $d$ in that basis gives exactly $(a_i,b_i+\lambda b_j),(a_j-\lambda a_i,b_j)$. This intertwines the differentials; in the graded case it preserves the grading precisely under the degree compatibility stated above.

Polynomial remainders. Replace $y$ by $y+\mu$ to reduce $(0,y-\mu )$ to $(0,y)$. For every other row write $a_i=a_i^{\mathrm{\prime }}+y{A}_i$ and $b_i=b_i^{\mathrm{\prime }}+y{B}_i$, where $a_i^{\mathrm{\prime }},b_i^{\mathrm{\prime }}\in R^{\mathrm{\prime }}$ are the values at $y=0$ and $A_i,B_i\in R^{\mathrm{\prime }}[y]$ are polynomial quotients; no linearity in $y$ is assumed. Since the excluded row has product zero and $w\in R^{\mathrm{\prime }}$, subtraction of the value at zero yields $y{\sum }_i(A_i{b}_i^{\mathrm{\prime }}+a_i^{\mathrm{\prime }}{B}_i+y{A}_i{B}_i)=0$. Multiplication by $y$ is injective in $R^{\mathrm{\prime }}[y]$, so ${\sum }_i(A_i{b}_i^{\mathrm{\prime }}+a_i^{\mathrm{\prime }}{B}_i+y{A}_i{B}_i)=0$.

Exclusion of the row. Pair $(a_i,b_i)$ with the distinguished row using step 1.1 with parameter $-B_i$. After doing this for every $i$, the rows become $(a_i,b_i^{\mathrm{\prime }})$ and $({\sum }_i{a}_i{B}_i,y)$. Swap the entries of the distinguished row, with the corresponding parity shift, and use the dual row operation $(a_i,b_i^{\mathrm{\prime }}),(y,A)\mapsto (a_i-A_{i}y,b_i^{\mathrm{\prime }}),(y,A+A_i{b}_i^{\mathrm{\prime }})$. This dual operation is another exterior-basis change (or the previous operation after exchanging wedge and contraction), and preserves the product. The ordinary rows now equal $(a_i^{\mathrm{\prime }},b_i^{\mathrm{\prime }})$, while the distinguished row equals $(y,{\sum }_i(a_i{B}_i+A_i{b}_i^{\mathrm{\prime }}))=(y,0)$ by step 1.2. Undoing its entry swap cancels the parity shift and leaves $(0,y)$. Thus the original factorization is isomorphic to $(a^{\mathrm{\prime }},b^{\mathrm{\prime }})\otimes (0,y)$ over $R^{\mathrm{\prime }}[y]$.

Standard form of graph factorizations. By [F2] the matrix of $C(\mathrm{\Gamma })$ has the $m_1+m_2$ linear rows $(a,z)$ contributed by the arcs and the first row of each wide edge, and the $m_2$ quadratic rows $(0,q)$ contributed by the second row of each wide edge; permute rows so the linear rows come first. Applying the operation of clause (1) to the first row and each further linear row $p$ with $\lambda =1$ replaces the pair $(a,z_1),(a,z_p)$ by $(a,z_1+z_p),(0,z_p)$, so afterwards the first row is $(a,{\sum }_{\text{linear rows}}z)$ and every other linear row is $(0,z)$ with its original $z$. The internal labels occur twice with opposite signs and cancel in the sum, while each boundary label occurs once, so the first row is $(a,{\sum }_p{ϵ}_p{x}_p)$; the quadratic rows are untouched.

The splitting. Over $R^{\mathrm{\prime }}$ the row $(0,y)$ presents the complex $0\to R^{\mathrm{\prime }}[y]\stackrel{y}{\to }{R}^{\mathrm{\prime }}[y]\to 0$, which is the direct sum of the two-term complexes $0\to R^{\mathrm{\prime }}{y}^j\stackrel{y}{\to }{R}^{\mathrm{\prime }}{y}^{j+1}\to 0$ for $j\ge 0$ and the rank-one complex $0\to R^{\mathrm{\prime }}\to 0$; for $j\ge 0$ the map $y\text{\hspace{0.17em}⁣}:R^{\mathrm{\prime }}{y}^j\to R^{\mathrm{\prime }}{y}^{j+1}$ is an isomorphism, so those summands are contractible and contribute nothing to the homotopy type. In a tensor product with $(a^{\mathrm{\prime }},b^{\mathrm{\prime }})$ the contractible summands remain contractible, hence $(a,b{)}^{\mathrm{\prime }}\simeq (a^{\mathrm{\prime }},b^{\mathrm{\prime }})\otimes R^{\mathrm{\prime }}\simeq (a^{\mathrm{\prime }},b^{\mathrm{\prime }})$ in ${\mathrm{h}\mathrm{m}\mathrm{f}}_w(R^{\mathrm{\prime }})$, which proves clause (2); the equivalence forgets the variable $y$ and is not an isomorphism of $R$-factorizations because $C(\mathrm{\Gamma })$ has infinite rank over $R^{\mathrm{\prime }}$.

Closed graphs. If $\mathrm{\Gamma }$ is closed, step 2.2 makes the first row $(a,0)$ and every other row has first entry zero. The latter rows include both the remaining linear entries and every quadratic entry; write their tensor product as $K$. The first row is $\mathbb{Q}[a]\stackrel{a}{\to }\mathbb{Q}[a]\{-1,1\}\stackrel{0}{\to }\mathbb{Q}[a]$. Its cohomology is the odd-parity copy of $\mathbb{Q}\{-1,1\}$. More explicitly, as a complex over $\mathbb{Q}$ it is the direct sum of contractible pairs $\mathbb{Q}{a}^j\stackrel{a}{\to }\mathbb{Q}{a}^{j+1}\{-1,1\}$ and the remaining constant in odd parity. Since $K$ has no $a$ in its entries, tensoring this splitting with $K$ leaves its specialization at $a=0$, with the first row's shift and odd parity retained. This is exactly the Koszul complex on all remaining linear and quadratic entries, with its cyclic grading folded into parity; multiplication by $a$ is zero on the resulting cohomology.