Equal Euler classes do not by themselves prove homotopy-equivalent complexes

Given: A simple reflection $s_i$, the bimodule $B_i=R{\otimes }_{R^{s_i}}R(1)$ with generators $u=1\otimes 1$ of degree $-1$ and $w_0=1\otimes {\delta }_i$, ${\delta }_i={\alpha }_i\mathrm{/}2$, the standard graph bimodule $R_{s_i}$ of Standard graph bimodules, support filtrations and characters, and the two-term complexes $Z_i=[B_i\stackrel{0}{\to }R(1)]$ and $F_i=[B_i\stackrel{{\epsilon }_i}{\to }R(1)]$ of The positive and negative Rouquier generator complexes.

Equal classes. $\chi (F_i)=[B_i]-[R(1)]$ in $K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}({\mathrm{S}\mathrm{B}\mathrm{i}\mathrm{m}}_n)$; the alternating class is the alternating sum of the term classes, so the zero-differential complex with the same two terms in the same cohomological degrees has the same class $\chi (Z_i)=[B_i]-[R(1)]$, and under the identification $\mathrm{\Phi }$ of the split Grothendieck ring with the Hecke algebra this class corresponds to $H_i-v$. (Euler classes of Rouquier complexes are homotopy invariant and multiplicative, Decategorification of a Rouquier complex is the Hecke braid generator)

Cohomology of the generator complex. $H^0(F_i)=\ker {\epsilon }_i\cong R_{s_i}(-1)$ and $H^1(F_i)=\mathrm{coker}{\epsilon }_i=0$, all other cohomology of $F_i$ being zero. (Rouquier generator complexes have canonical derived graph models)

Cohomology of the zero-differential complex. For the bounded complex $Z_i$ with zero differential, $H^0(Z_i)=B_i$, $H^1(Z_i)=R(1)$ and all other cohomology is zero. (Cohomology object of a cochain complex)

Left $R$-ranks. $B_i$ is finite free of rank two as a left $R$-module, while $R_{s_i}$ and its shifts are free of rank one as left $R$-modules; an isomorphism of graded bimodules restricts to an isomorphism of left $R$-modules, and isomorphic free modules have equal rank. (The Soergel bimodule $B_i$ of a simple reflection, Standard graph bimodules, support filtrations and characters)

Invariance of cohomology. Homotopy equivalent complexes have isomorphic cohomology objects, because homology factors through the homotopy category. (Homology factors uniquely through the homotopy category, Complexes, homotopies and contractibility in an additive category)