Right orthogonality and finite inverse words do not prove generation

The specified compositions are associative: a composable string containing a nonidentity $\delta :E\to D$ has no further nonzero nonidentity factor, and identities act as units. The differential is zero and obeys the Leibniz rule. Thus these data define a strictly unital DG category. The matrix $q$ with sole entry $q_{D,E}=\delta$ is strictly upper triangular and satisfies $q^2=0$, so $U=(D\oplus E,q)$ is a finite twisted object. The inclusion $\epsilon :D\to U$ is closed because $q$ vanishes on the $D$-summand. The normalized bar over the full subcategory generated by $D$ has no nonidentity letters, since $\mathrm{End}(D)=\mathbb{C}{1}_D$; its evaluation map is therefore $Y_{\epsilon }:Y_D\to Y_U$.

At $E$, the right representable $Y_U(E)=\mathrm{Hom}(E,U)$ has basis $a:E\to E\hookrightarrow U$ in degree zero and $\epsilon \delta$ in degree one, with differential $da=\epsilon \delta$. Hence it is acyclic. Also $\mathrm{Hom}(E,D)[1]=\mathbb{C}\delta [1]$ is concentrated in degree zero with zero differential, so its class is nonzero.

A degree-zero map $r:U\to D$ is multiplication by some $\lambda \in \mathbb{C}$ on the $D$-summand and is zero on the $E$-summand, since ${\mathrm{Hom}}^0(E,D)=0$. The twisted Hom differential is $dr=-rq=-\lambda \delta$. Thus $dr=0$ forces $\lambda =0$, whereas $r\epsilon =1_D$ forces $\lambda =1$; no closed $r$ can satisfy the co-unit equation. Moreover ${\mathrm{Hom}}^{-1}(D,D)=0$, so there is no $s$ with $1_D-r\epsilon =ds$ that repairs this failure.

The bar map $Y_{\epsilon }:Y_D\to Y_U$ is an objectwise chain map. At $D$ it is the identity $C\to C$, because $\mathrm{Hom}(D,U)=\mathbb{C}\epsilon$; its cone is contractible. At $E$ its source is $C\delta$ in degree one and it sends $\delta$ to $\epsilon \delta$. The cone contains the degree-zero cycle $(a,-\delta [1])$, since the two differential contributions are both $\epsilon \delta$ and cancel. Projection $\pi :U\to E$ is closed, and the induced cone map sends this cycle to $1_E$. The E-evaluation cone has no other cohomology: its only differential pairs the source copy of $\delta [1]$ with $\epsilon \delta$, leaving the class of $a-\delta [1]$. Consequently the cone map is a quasi-isomorphism at E and D, hence a quasi-isomorphism of these finite right modules. The cone is represented by $E$.

In the quotient that makes $D$ contractible, adjoin ${ϵ}_D\in {\mathrm{Hom}}^{-1}(D,D)$ with $d{ϵ}_D=1_D$. The element $v=a-\epsilon {ϵ}_D\delta$ has degree zero and $dv=\epsilon \delta -\epsilon (d{ϵ}_D)\delta =0.$ Since $\pi \epsilon =0$ and $\pi a=1_E$, one has $\pi v=1_E$. The inverse-letter word therefore retains the identity of $E$, even though the ordinary old-$U$ test is acyclic. The calculation proves that finite quotient words and right-side acyclicity alone do not supply the missing full co-unit nullhomotopy. No nontrivial quotient word can occur in $\mathrm{Hom}(E,E)$: every word through $D$ would require a nonzero final morphism $D\to E$, which does not exist. Thus $1_E$ remains nonboundary in the quotient. This calculation asserts neither existence nor nonexistence of an additional geometric primitive.

Form a finite upper history complex $Q_U^H=Q_U^0\oplus W$, where $Q_U^0=\mathrm{Hom}(E,U)=\mathbb{C}{a}^0\to \mathbb{C}(\epsilon \delta {)}^1$ is the ordinary acyclic complex from step 2.1, $d_{0}a=\epsilon \delta$, and the added localization-word space $W=\mathbb{C}{v}^0$ has $d_W=0$, with zero off-diagonal block $B=0$. The contraction $h_U(\epsilon \delta )=a,h_U(a)=0$ retracts $Q_U^0$ to $T_U=0$, so $T_U^0=0$. Take $Q_P^H=\mathbb{C}{c}^{-1}\stackrel{d_P}{\to }\mathbb{C}{\beta }^0$, $d_{P}c=\beta ,d_P\beta =0$, with $h_P\beta =c,h_{P}c=0$; direct substitution gives $d_P{h}_P+h_P{d}_P=1$. Define the degree-one map $J:Q_P^H\to Q_U^H$ by $Jc=v,J\beta =0$. It obeys $d_{U}J+J{d}_P=0$ because $d_{U}v=0=J\beta$. Hence the full differential $D(a,b)=(d_{U}a+Jb,d_{P}b)$ squares to zero. Then $F=(0,\beta )$ is a closed full-block arrow, and $D(0,c)=(v,\beta )$ gives $F-D(0,c)=(-v,0)$. Its lower contraction gives the upper residual $w=-{\pi }_{W}J{h}_P\beta =-v\mathrm{.}$ The class of $v$ is nonzero in $H^0(W)$ because $W^{-1}=0$ and $v$ is a nonzero basis vector. Thus the actual transfer formulas admit a finite chain-level model with acyclic ordinary upper test and $T_U^0=0$, but nonzero history residual and no $z\in W^{-1}$ with $d_{W}z=w$. This construction is an algebraic countermodel to the implication, not a geometric history count. It leaves every separate geometric primitive-count obligation unresolved.