Wall-crossing under the Soergel functor is tensoring with the coinvariant bimodule

Set $V_{\nu }=\mathrm{Hom}(P_{\nu },-)$. It is exact and induces the singular quotient equivalence with finite $C^s$-modules by the same finite-presentation argument $P_{\nu }{\otimes }_{C^s}-⊣V_{\nu }$ used in [F2]. Adjunction and the chosen identification in [F1] give $V_{\nu }{T}_{\mathrm{o}\mathrm{n}}M\cong \mathrm{Hom}(T_{\mathrm{o}\mathrm{u}\mathrm{t}}{P}_{\nu },M)\cong V_{\lambda }M$. Precomposition by an element of $\mathrm{End}(P_{\nu })$ becomes precomposition by its translated endomorphism of $P_{\lambda }$, which is the inclusion $C^s\hookrightarrow C$ in [F1]. Hence this is an isomorphism of $C^s$-module functors $V_{\nu }{T}_{\mathrm{o}\mathrm{n}}\cong {\mathrm{Res}}_{C^s}^C{V}_{\lambda }$.

Both translations preserve the kernels of these quotient functors. For $T_{\mathrm{o}\mathrm{n}}$ this follows from step 1.1. For the return functor, $T_{\mathrm{o}\mathrm{n}}{P}_{\lambda }$ is projective-injective by [F3]. In the singular block its indecomposable summands are copies of $P_{\nu }$: the unique simple Verma is the antidominant one (the same integral Verma embedding argument as in the regular block), every projective has a Verma flag, so every simple submodule of a projective has this label; an indecomposable injective is its injective hull, by the finite-socle argument in the Struktursatz proof [F2]. Thus $T_{\mathrm{o}\mathrm{n}}{P}_{\lambda }=P_{\nu }^{\oplus m}$ for some finite $m$. If $V_{\nu }N=0$, adjunction gives $V_{\lambda }{T}_{\mathrm{o}\mathrm{u}\mathrm{t}}N=\mathrm{Hom}(T_{\mathrm{o}\mathrm{n}}{P}_{\lambda },N)=0$. The exact adjunction $T_{\mathrm{o}\mathrm{u}\mathrm{t}}⊣T_{\mathrm{o}\mathrm{n}}$ therefore descends to the two Serre quotients: the unit, counit and their triangle identities descend.

Under the quotient equivalences, the onto-wall functor is restriction by step 1.1. Its descended left adjoint is uniquely isomorphic to extension of scalars $C{\otimes }_{C^s}-$: the elementary adjunction sends a map $C{\otimes }_{C^s}N\to M$ to $n\mapsto f(1\otimes n)$, with inverse $c\otimes n\mapsto cg(n)$. Thus $V_{\lambda }{T}_{\mathrm{o}\mathrm{u}\mathrm{t}}\cong C{\otimes }_{C^s}{V}_{\nu }$ as module functors, with the $C$-action established by adjunction rather than just a vector-space dimension count. Composing with step 1.1 proves $V{\theta }_s\cong C{\otimes }_{C^s}V$, equivalently tensoring by the $(C,C)$-bimodule $C{\otimes }_{C^s}C$.

Let $J$ be the ideal of $S^s$ generated by $S_{+}^W$. Then $I=SJ=J\oplus \alpha J$ under [F4], so $C=(S^s\mathrm{/}J)\oplus \alpha (S^s\mathrm{/}J)$ and averaging over $s$ identifies $S^s\mathrm{/}J=C^s$. Thus $C$ is free over $C^s$ with generators in degrees $0,2$. The natural tensor map $S{\otimes }_{S^s}C\to C{\otimes }_{C^s}C$ is an isomorphism: its left $I$ acts as zero since the generators $S_{+}^W$ balance to zero on the right; factoring the left quotient and the image of $S^s$ therefore gives inverse maps on pure tensors. Hence $B_s{\otimes }_{S}C\cong (C{\otimes }_{C^s}C)(1)$ as graded bimodules. As left $C$-modules this is $C(1)\oplus C(-1)$, not $C(1)\oplus C(3)$, because a degree-two generator contributes the external shift $(-2)$. Its dimension is $2\mathrm{\mid }W\mathrm{\mid }$, whereas $B_s{\otimes }_S\mathbb{C}$ has the two centered degrees $-1,1$. This proves the polynomial comparison and the dimension distinction without imposing any grading on category $\mathcal{O}$.