A vanishing Euler class does not imply contractibility of a Soergel complex

Given: $n\ge 2$, $R=\mathbb{Q}[x_1,\dots ,x_n]$, and the type-A Soergel category with unit bimodule $R$.

Categorified braid-line model with an explicit realization boundary: $R$ is the unit object of ${\mathrm{S}\mathrm{B}\mathrm{i}\mathrm{m}}_n$, and bounded cochain complexes are considered in $K^b({\mathrm{S}\mathrm{B}\mathrm{i}\mathrm{m}}_n)$.

Split Grothendieck rings of the type-A Soergel categories: the split Grothendieck group identifies isomorphic objects and has $[X\oplus Y]=[X]+[Y]$; the unit bimodule $R$ is a nonzero object.

Euler classes of Rouquier complexes are homotopy invariant and multiplicative: the Euler class of a bounded complex is $\chi (C)={\sum }_m(-1{)}^m[C^m]$.

Complexes, homotopies and contractibility in an additive category: a contracting homotopy satisfies $1_{C^m}=d^{m-1}{h}^m+h^{m+1}{d}^m$ in every cochain degree $m$.

Take $C^0=C^1=R$ as unshifted graded $(R,R)$-bimodules, $d^0=0$, and all other terms and differentials zero. Compute cohomology in the abelian category of graded $(R,R)$-bimodules (equivalently, graded modules over $R^e=R{\otimes }_{\mathbb{Q}}{R}^{\mathrm{o}\mathrm{p}}$), so $H^m(C)=\ker (d^m)\mathrm{/}\mathrm{im}(d^{m-1})$.