\section{Associative families and completion of the proof}
\label{sec:binary-families}

Under the inductive hypothesis of Corollary~\ref{cor:zero-supported-coverage}, the cocycles supported on zero ray products span the positive-degree Hochschild cohomology $\HH^2(A_0)_{>0}$. In this section we organize all such cocycles into a single polynomial family of associative multiplications, verify that every specialization with parameters in $\{0,1\}$ has a normed multiplicative basis, and apply Proposition~\ref{prop:family-criterion} to recover the original algebra $A$ and close the induction of Proposition~\ref{prop:table-weight-lifting}.

\subsection{Indicator cocycles on zero products}
\label{subsec:binary-families}

The stem product $r\circledast s$ specifies the only possible nonzero value for a deformation term on a pair of rays $(r,s)$ with $rs=0$. Imposing the Hochschild cocycle condition on all such pairs simultaneously turns out to require only pairwise equalities between their scalar coefficients. These equalities partition the pairs with $rs=0$ and $r\circledast s\ne0$ into equivalence classes whose indicator sums form a basis of the supported cocycle space. Their cohomology classes need not be independent or nonzero; we select the subfamily used in the proof after constructing the full family.

We first record this phenomenon as a general observation about two compatible associative multiplication tables on the same basis.

\begin{lemma}
\label{lem:two-associative-tables}
Let $\mathsf T$ be a finite set with an absorbing element $0$, and let $m$ and $m_{\circledast}$ be two associative multiplications on $\mathsf T$ that share a common finite set of orthogonal idempotents summing to the identity of both linearizations. Suppose that $m_{\circledast}(a,b)=m(a,b)$ whenever $m(a,b)\ne0$. Set
\[
  Z=\{(a,b)\in(\mathsf T\setminus\{0\})^2:m(a,b)=0\text{ and }m_{\circledast}(a,b)\ne0\},
\]
and for each $z=(a,b)\in Z$ let $\varphi_z$ be the bilinear cochain on the free abelian group $\Z[\mathsf T\setminus\{0\}]$ that sends $(a,b)$ to $m_{\circledast}(a,b)$ and vanishes on all other basis pairs. Assume that all nested insertions vanish:
\[
  \varphi_z(\varphi_w(a,b),c)=0=\varphi_z(a,\varphi_w(b,c))
  \qquad\text{for all }z,w\in Z\text{ and }a,b,c\in \mathsf T\setminus\{0\}.
\]
Then there is an equivalence relation $\sim$ on $Z$ such that, over any commutative ring $R$, the Hochschild two-cocycles of $(R[\mathsf T\setminus\{0\}],m)$ of the form $\sum_{z\in Z}\lambda_z\varphi_z$ ($\lambda_z\in R$) are precisely the $R$-linear combinations of the class sums
\[
  \mu_E=\sum_{z\in E}\varphi_z\qquad(E\in Z/{\sim}).
\]
Moreover, the deformed multiplication $m+\sum_{E\in Z/{\sim}}T_E\mu_E$ is associative over the polynomial ring $R[T_E:E\in Z/{\sim}]$, and at every specialization with $T_E\in\{0,1\}$ the set $\mathsf T\setminus\{0\}$ is a multiplicative basis.
\end{lemma}
\begin{proof}
Set $\varphi=\sum_{z\in Z}\varphi_z$, so that $m_{\circledast}=m+\varphi$. Because both $m$ and $m_{\circledast}$ are associative and all quadratic terms $\varphi(\varphi(a,b),c)$ and $\varphi(a,\varphi(b,c))$ vanish by hypothesis, the Hochschild coboundary $d\varphi$ with respect to $m$ vanishes identically over $\Z$.

Now consider a general $R$-linear combination $\mu=\sum_{z\in Z}\lambda_z\varphi_z$ and evaluate its coboundary on a triple of nonzero basis elements $a,b,c\in \mathsf T\setminus\{0\}$ (writing $xy$ for $m(x,y)$):
\[
  (d\mu)(a,b,c)=a\mu(b,c)-\mu(ab,c)+\mu(a,bc)-\mu(a,b)c.
\]
If $abc\ne0$, then $ab\ne0$, $bc\ne0$, $(ab)c\ne0$, and $a(bc)\ne0$, so all four terms vanish because $\mu$ is supported on $Z$. When $abc=0$, the surviving terms depend on whether $ab$ and $bc$ are zero:
\[
  \begin{array}{c|c|c}
  ab & bc & (d\mu)(a,b,c) \\\hline
  0 & 0 & a\mu(b,c)-\mu(a,b)c \\
  0 & \ne0 & \mu(a,bc)-\mu(a,b)c \\
  \ne0 & 0 & a\mu(b,c)-\mu(ab,c) \\
  \ne0 & \ne0 & \mu(a,bc)-\mu(ab,c).
  \end{array}
\]
In every row, $(d\mu)(a,b,c)$ is the difference of at most one positive term and at most one negative term, each of which is a single coefficient $\lambda_z$ times a nonzero element of $\mathsf T$ (since $a\varphi_{(b,c)}(b,c)$ lies in $\mathsf T$). Specializing to $\lambda_z=1$ for all $z\in Z$, the identity $d\varphi=0$ over $\Z$ shows that on each triple $(a,b,c)$ either both terms vanish or both terms are $\lambda_z$ and $\lambda_{z'}$ times the exact same basis element of $\mathsf T\setminus\{0\}$. Consequently, $(d\mu)(a,b,c)=0$ imposes either no condition or a simple equality $\lambda_z=\lambda_{z'}$. Generating an equivalence relation $\sim$ on $Z$ by these equalities proves that the cocycles of the form $\sum_{z\in Z}\lambda_z\varphi_z$ have $R$-basis $\{\mu_E:E\in Z/{\sim}\}$.

Finally, the associator of $m+\sum_E T_E\mu_E$ has zero linear part in the variables $T_E$ because each $\mu_E$ is a cocycle, and zero quadratic part because all nested insertions vanish. When every $T_E$ is specialized to $0$ or $1$, the deformed product of $a,b\in \mathsf T\setminus\{0\}$ equals $m(a,b)$ if $(a,b)\notin Z$, and equals either $0$ or $m_{\circledast}(a,b)$ if $(a,b)\in Z$; in every case it belongs to $\mathsf T$.
\end{proof}

We now specialize Lemma~\ref{lem:two-associative-tables} to the ray category $\mathcal P$ of $A_0=k[\mathcal P]$. Let $Z$ be the set of composable pairs $z=(r,s)$ of nonidentity rays of $\mathcal P$ satisfying $rs=0$ and $r\circledast s\ne0$, and for each $z=(r,s)\in Z$ let $\varphi_z$ denote the bilinear cochain that sends $(r,s)$ to $r\circledast s$ and vanishes on all other pairs of rays (including identities).

\begin{proposition}
\label{prop:indicator-cocycles}
Let $Z$ and $\varphi_z$ be as above for $A_0=k[\mathcal P]$ in Setting~\ref{set:projective-companion}. Over any commutative coefficient ring $R$:
\begin{enumerate}[label={\upshape(\arabic*)}]
\item Every nontrivial cocycle condition on a linear combination $\sum_{z\in Z}\lambda_z\varphi_z$ ($\lambda_z\in R$) takes the form $\lambda_z=\lambda_{z'}$.
\item If $\sim$ denotes the equivalence relation on $Z$ generated by these equalities, then the Hochschild two-cocycles of the form $\sum_{z\in Z}\lambda_z\varphi_z$ have $R$-basis
\begin{equation}
  \mu_E=\sum_{z\in E}\varphi_z\qquad(E\in Z/{\sim}).
  \label{eq:indicator-cocycles}
\end{equation}
\item For any two cochains $\mu,\nu$ in the span of $\{\varphi_z:z\in Z\}$ and any rays $a,b,c$,
\begin{equation}
  \mu(\nu(a,b),c)=0,\qquad \mu(a,\nu(b,c))=0.
  \label{eq:mixed-insertions-zero}
\end{equation}
\end{enumerate}
\end{proposition}
\begin{proof}
We first verify~(3). By Corollary~\ref{cor:active-zero-pairs}, any nonzero value $u=r\circledast s$ of $\nu(a,b)$ spans the endpoint socle of its corner bimodule and is therefore annihilated by both endpoint radicals. For any composable ray $c$, if $uc\ne0$, then $(u,c)\notin Z$ and hence $\mu(u,c)=0$; if instead $uc=0$, then $uq_y^nc=0$ for all $n\geq1$ because $uq_y=0$, so $u\circledast c=0$ by Proposition~\ref{prop:intrinsic-stem-product} and again $\mu(u,c)=0$. The identity $\mu(a,\nu(b,c))=0$ follows dually.

Now view the ordinary ray product $m(r,s)=rs$ and the stem product $m_{\circledast}(r,s)=r\circledast s$ as binary operations on the finite set $\mathsf T$ of rays of $\mathcal P$ (including $0$). By Proposition~\ref{prop:intrinsic-stem-product}, $m_{\circledast}$ is associative and agrees with $m$ whenever $m(r,s)\ne0$, so $m_{\circledast}=m+\sum_{z\in Z}\varphi_z$. Since all nested insertions vanish by~(3), Lemma~\ref{lem:two-associative-tables} immediately yields~(1) and~(2).
\end{proof}

We call the cocycles $\mu_E$ of~\eqref{eq:indicator-cocycles} the \emph{indicator cocycles} on $A_0$. Mixtures of ordinary and stem ray products, together with interpolating families between the standard algebra and its modified forms, appear in \cite[Sections~9.2 and~9.5--9.7]{BGRS}; here the indicator cocycles provide polynomial coordinates prior to passing to Hochschild cohomology, and Lemma~\ref{lem:two-associative-tables} guarantees associativity over any commutative coefficient ring. The next proposition establishes the associativity and basis properties of this full polynomial family, from which Corollary~\ref{cor:zero-supported-coverage} will select a subfamily spanning $\HH^2(A_0)_{>0}$.

\begin{proposition}
\label{prop:zero-product-family}
Let $A_0=k[\mathcal P]$ satisfy Setting~\ref{set:projective-companion}, let $E$ range over the equivalence classes in $Z/{\sim}$ defining the indicator cocycles $\mu_E$ in~\eqref{eq:indicator-cocycles}, and let $(T_E)_{E\in Z/{\sim}}$ be independent polynomial variables. On the free $k[T_E]$-module having the nonzero rays of $\mathcal P$ as basis, define the multiplication
\begin{equation}
  r\cdot s=
  \begin{cases}
  rs, & \text{if }rs\ne0\text{ in }\mathcal P,\\
  T_E(r\circledast s), & \text{if }(r,s)\in E\in Z/{\sim},\\
  0, & \text{otherwise},
  \end{cases}
  \label{eq:affine-ray-family}
\end{equation}
where $r\circledast s$ is the stem product of Definition~\ref{def:stem-product}. Then:
\begin{enumerate}[label={\upshape(\arabic*)}]
\item The multiplication~\eqref{eq:affine-ray-family} is associative over $k[T_E]$.
\item Every scalar fibre (obtained by specializing the variables $T_E$ to elements of $k$) is a positive filtered deformation of $A_0$.
\item Every scalar fibre has ray category $\mathcal P$, with the nonzero rays of $\mathcal P$ as representatives, and every binary fibre (with $T_E\in\{0,1\}$ for all $E$) has a normed multiplicative basis consisting of the nonzero rays of $\mathcal P$.
\end{enumerate}
\end{proposition}
\begin{proof}
Assertion~(1) follows directly from Lemma~\ref{lem:two-associative-tables} and Proposition~\ref{prop:indicator-cocycles}. For~(2), the vertex idempotents are unchanged, and for every pair $(r,s)\in E$ with middle vertex $y$, Corollary~\ref{cor:active-zero-pairs} shows that $r\circledast s=rq_ys$, so the indicator term $\varphi_{(r,s)}$ has weight
\[
  \chi_E=\ell(r\circledast s)-\ell(r)-\ell(s)=\ell(q_y),
\]
which has positive categorical degree $\deg_\JJ(q_y)>0$ (and is constant across each equivalence class $E$ because the defining equations $\lambda_z=\lambda_{z'}$ arise from homogeneous triples). Thus every scalar fibre is a positive filtered deformation of $A_0$.

For~(3), Corollary~\ref{cor:active-zero-pairs} shows that $(r,s)\in Z$ requires both $r$ and $s$ to be generating rays between distinct vertices. Consequently, all products involving an endomorphism ray $q_x^j$ or a proper ray remain unchanged in every scalar fibre. In particular, the local rings $R_x$ and their left and right actions on every corner bimodule coincide with those of $A_0$, so the basis elements represent the exact same $(R_x^\times\times R_y^\times)$-orbits and generate the same endpoint subbimodules $U_r$. Furthermore, for any composable rays $r\colon x\to y$ and $s\colon y\to z$ and any $n\in\rad R_y$, the product $rn$ is a linear combination of proper rays, so $(rn)\cdot s=(rn)s$; hence the middle-radical insertion subspace $M(r,s)=R_xr(\rad R_y)sR_z$ of Definition~\ref{def:ray-product} is identical in $A_0$ and in the fibre algebra. Because each deformation term $T_E(r\circledast s)=T_E(rq_ys)$ lies inside $M(r,s)$, the deformed product $r\cdot s$ belongs to $M(r,s)$ if and only if $rs=0$ in $A_0$ (and when $rs\ne0$, $r\cdot s=rs$). Thus every scalar fibre has ray category $\mathcal P$.

Finally, at any binary specialization $T_E\in\{0,1\}$, all products of basis elements are either basis elements or zero. Because the endpoint subbimodule chains are unchanged, every corner of a two-sided ideal is an endpoint subbimodule and is therefore spanned by a subset of the basis rays. Intersecting with powers of the radical shows that the basis is normed.
\end{proof}

The weight calculation in part~(2) also describes the deformation
geometrically: on each pair $r:x\to y$, $s:y\to z$ in $E$, the new
product is obtained by inserting the local generator $q_y$ between
$r$ and $s$. Thus the weight $\chi_E$ of an indicator cocycle is a
local generator weight $\ell(q_y)$, constant over all pairs in $E$.

\begin{example}
\label{ex:zero-product-fibre}
In the nine-dimensional algebra $B$ of Example~\ref{ex:zero-ray-span}, replacing the single zero product $\alpha\beta=0$ by $\alpha\mathbin{\cdot_t}\beta=tu$ (where $u=\alpha q\beta=[\alpha]\circledast [\beta]$) while keeping all other basis products fixed yields the one-parameter family
\[
  B(t)=kQ/(q^2,\ \alpha\beta-t\alpha q\beta).
\]
The same nine paths form a basis of $B(t)$, and at $t=0$ and $t=1$ they form a normed multiplicative basis; assigning degree~$1$ to $q,\alpha,\beta$ makes $B(t)$ a positive filtered deformation of $B$. Here the linear change of basis $\alpha'=\alpha-t\alpha q$ satisfies $\alpha'\beta=0$ and $\alpha'q=\alpha q$, giving an isomorphism $B(t)\cong B$ for every $t\in k$; equivalently, the indicator cocycle $\varphi_{(\alpha,\beta)}=d\delta$ is the coboundary of the one-cochain $\delta(\alpha)=\alpha q$. Thus, in contrast to Example~\ref{ex:nonstandard-pair}, an individual indicator cocycle $\mu_E$ may be a coboundary, and we use Corollary~\ref{cor:zero-supported-coverage} to select indicator cocycles whose cohomology classes span $\HH^2(A_0)_{>0}$.
\end{example}