The minimal Rouquier differential supplies the injectivity input of the weak-Lefschetz substitute

The rank-one retraction. In the right basis $(c_{\mathrm{i}\mathrm{d}},c_s)$ of $B_s^R$ we have ${\epsilon }_s(c_{\mathrm{i}\mathrm{d}}a+c_{s}b)=a+b{\alpha }_s$ for $a,b\in S$. The section $\sigma (r)=c_{\mathrm{i}\mathrm{d}}\cdot r$ satisfies ${\epsilon }_s\sigma (r)=r$, so ${\epsilon }_s$ is split surjective and $B_s^R=\sigma (S(1))\oplus \ker ({\epsilon }_s)$ as right $S$-modules; the subcomplex $\sigma (S(1))\stackrel{\cong }{\to }S(1)$ of $F_s$ is contractible, and $\ker ({\epsilon }_s)$ is a direct summand of the complex $F_s$ concentrated in degree $0$. The kernel is $\{c_{\mathrm{i}\mathrm{d}}(-{\alpha }_{s}b)+c_{s}b\}=\{(c_s-c_{\mathrm{i}\mathrm{d}}{\alpha }_s)b\}$, so it is free of rank one as a right $S$-module with generator $c_s-c_{\mathrm{i}\mathrm{d}}{\alpha }_s=({\alpha }_s\otimes 1-1\otimes {\alpha }_s)\mathrm{/}2$ of internal degree $1$; hence $\ker ({\epsilon }_s)\cong S(-1)$ as a graded right $S$-module and $F_s\simeq S(-1)$ in the homotopy category of complexes of graded right $S$-modules.

The injectivity criterion. Let $C$ be as in (2) and let $c\in \overline{C^0}$ be homogeneous of internal degree $D$ with $\overline{d^0}(c)=0$. Then $c$ is a cocycle, so it represents a class $[c]\in H^0(\bar{C})$; since $H^{-1}(\bar{C})=0$ and $C^{-1}=0$, the zeroth cohomology is the cokernel of no incoming map and $[c]$ vanishes if and only if $c$ lies in the image of $\overline{d^{-1}}$, which is zero because $C^{-1}=0$. Hence $[c]\mathrm{\ne }0$ forces $D$ to be the single internal degree $d$ in which $H^0(\bar{C})$ is concentrated; therefore a nonzero kernel element can occur only in degree $d$, and the map is injective in every other degree, in particular in degrees $\le d-1$. If $C^{\mathrm{\prime }}$ is homotopy equivalent to $C$ with $C^{\mathrm{\prime }-1}=0$ and is obtained by cancelling contractible two-term summands whose source lies in positive cohomological degree, then $C^{\mathrm{\prime }0}=C^0$ for the cancellation in degrees $\ge 1$, the reductions $\overline{C^{\mathrm{\prime }}}$ and $\bar{C}$ are homotopy equivalent by [F3], so $H^{-1}(\overline{C^{\mathrm{\prime }}})=0$ and $H^0(\overline{C^{\mathrm{\prime }}})$ is concentrated in degree $d$, and the same argument applies to $C^{\mathrm{\prime }}$. This proves (2).

The retraction for a reduced word. Write $\underset{‾}{x}=s_1\cdots s_m$. For $m=0$ the complex is $S$. For $m>0$, tensor the right-$S$-linear homotopy equivalence $F_{s_1}\simeq S(-1)$ of step 1.1 with the remaining bimodule complex $G=F_{s_2}\cdots F_{s_m}$. This is legitimate because the maps and homotopies on the first factor are right linear; signed totalization preserves the homotopy identities. As a complex of right modules, $S(-1){\otimes }_{S}G\cong G(-1)$. Repeating on the first remaining factor gives $F_{s_1}\cdots F_{s_m}\simeq S(-m)$. No left action is assigned to a right-module retraction and no right-linear map is tensored as a map on the second factor. Forgetting the left action of the bimodule homotopy equivalence from $F_x$ to this word complex proves (1). Applying the additive reduction functor preserves these homotopy identities, so the reduced cohomology is a single copy of $\mathbb{R}$ in internal degree $m$.

Application to $F_x{F}_s$. Let $xs>x$ and assume $\mathrm{S}(\le x)$. By [F2] the concatenated word is reduced for $xs$, so $F_x{F}_s\cong F_{xs}$ in $K^b$, and by step 2.1 applied to $xs$ the complex $F_{xs}$ is homotopy equivalent over right $S$-modules to $S(-\mathrm{\ell }(xs))=S(-\mathrm{\ell }(x)-1)$ concentrated in cohomological degree $0$. Hence $C:=F_x{F}_s$ satisfies the hypotheses of (2) with $d=\mathrm{\ell }(x)+1$: it is bounded below, $C^{-1}=0$, and its reduction has $H^{-1}=0$ with $H^0$ one copy of $\mathbb{R}$ in internal degree $\mathrm{\ell }(x)+1$. By step 1.2 the reduced first differential $\overline{d^0}$ is injective in internal degrees $\le \mathrm{\ell }(x)$, which is (3); note that $F_x{F}_s$ has the terms $C^0=B_x^R{B}_s^R$ and $C^1={}^1{F}_x{B}_s\oplus B_x^R(1)$ by [F2], and that no minimality of $F_x{F}_s$ was used, only the homotopy equivalence with $F_{xs}$.

Stability under cancellation and the descent caveat. If the zero-parameter hard-Lefschetz proof cancels a contractible summand of $F_x{F}_s$ whose source lies in cohomological degree $\ge 1$ (the shifted descent piece), the surviving complex $C^{\mathrm{\prime }}$ has $C^{\mathrm{\prime }-1}=0$, is homotopy equivalent to $C$ and has $\overline{C^{\mathrm{\prime }}}$ homotopy equivalent to $\bar{C}$, so step 1.2 applies and the reduced first differential of $C^{\mathrm{\prime }}$ is still injective in internal degrees $\le \mathrm{\ell }(x)$; this is the "splitting off contractible summands with zero degree-zero source" clause of (2). On the other hand the form identity $⟨b,L_{\zeta }{b}^{\mathrm{\prime }}⟩=⟨\phi (b),\phi (b^{\mathrm{\prime }}){⟩}_{\gamma }$ and the commutation $\phi (L_{\zeta }b)=L\phi (b)$ do not follow from the right-module retraction: they are identities between maps of bimodules involving the induced forms and the deleted-word factorization, and they must be verified in the hard-Lefschetz theorem using [F4]; for $\zeta =0$ on a descent summand the Lefschetz form of the shifted piece is identically zero, so the theorem restricts the weak-Lefschetz substitute to the kernel of $d_3$ instead of the whole target. This completes the proof of (3) and (4).