The positive lift b_w is not a monoid homomorphism

The rank-one data: since a braid pair requires two distinct letters, the braid-pair set of $(\{s\},m)$ is empty by [F1], so ${\equiv }^{+}$ is the diagonal and $A^{+}=S^{\ast }$ is the free monoid on the single generator $s$, with elements $[s^k]$ for $k\ge 0$ and $[s^i]=[s^j]$ exactly when $i=j$, because $L([s^k])=k$ by [F3]. The relator set of $W$ is $\{s^2\}$, so $W=⟨s\mid s^2=1⟩=\{1,s\}$ with $\mathrm{\ell }(1)=0$ and $\mathrm{\ell }(s)=1$ (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action).

The positive lifts are $b_1=[\epsilon ]$ (the empty word is a reduced expression of $1$ because $\mathrm{\ell }(1)=0$) and $b_s=[s]={\sigma }_s$ (the one-letter word is a reduced expression of $s$ because $\mathrm{\ell }(s)=1$), by the definition of $b$ and step 1.1.

But $b_s{b}_s=[s][s]=[ss]$ by the product rule of [F1], while $b_{s\cdot s}=b_{s^2}=b_1=[\epsilon ]$ since $s^2=1$ in $W$; the two are different elements of $A^{+}$, because $L([ss])=2\mathrm{\ne }0=L([\epsilon ])$ by [F3]. Hence $b_s{b}_s\mathrm{\ne }{b}_{s\cdot s}$, so the assignment $b$ is not a monoid homomorphism: it fails to preserve the product $s\cdot s=1$.

The same defect appears at the group level: $(\gamma \circ b)(s{)}^2=\gamma ([ss])={\sigma }_s^2$ while $(\gamma \circ b)(s^2)=\gamma ([\epsilon ])=1_A$, and ${\sigma }_s^2\mathrm{\ne }{1}_A$ because $\mathrm{deg}({\sigma }_s^2)=2\mathrm{\ne }0=\mathrm{deg}(1_A)$ by [F3] and [F4], a group homomorphism preserving the identity. Thus $\gamma \circ b$ is not a group homomorphism $W\to A$.

The failure is not an artefact of the rank-one computation: for every finite Coxeter matrix with $S\mathrm{\ne }\mathrm{\varnothing }$ and every $s\in S$, $[ss]$ is the class of the word $ss$ in $A^{+}(S,m)$, and its length satisfies $L([ss])=2\mathrm{\ne }0=L([\epsilon ])$ by [F3]; combined with $s^2=1$ in $W$ and $\mathrm{\ell }(s)=1$ (Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action) this gives $b_s{b}_s\mathrm{\ne }{b}_{s^2}$ for the ambient monoid, and no parabolic reduction is needed. The general theorem records that $b_u{b}_v=b_{uv}$ holds exactly on the pairs with $\mathrm{\ell }(uv)=\mathrm{\ell }(u)+\mathrm{\ell }(v)$ by [F5], so the refuted statement is obtained by dropping that length hypothesis.

Scope and choice: this counterexample is a finite computation in the rank-one system plus the stated supplier clauses; it constructs no topological model, claims nothing about embeddings of $A^{+}$ into $A$, and uses no choice.