The generic rank-one DAHA polynomial representation is faithful

Write each ${\mathsf{T}}_i=a_i{\sigma }_i+b_i$ as in [F1], with $b_i(z)\in K(z)$. The rational function $a_i(z)$ is nonzero: its numerator $t{u}_i(z)-t^{-1}$ is nonzero because $u_i$ is a nonconstant monomial and $t\mathrm{\ne }0$. In the expansion of an alternating word $T_{i_1}\cdots T_{i_r}{\pi }^{ϵ}$, choosing the substitution term $a_{i_j}{\sigma }_{i_j}$ from every ${\mathsf{T}}_{i_j}$ gives the affine substitution $g=s_{i_1}\cdots s_{i_r}{\pi }^{ϵ}$ with a nonzero rational coefficient ${\gamma }_g(z)$, a product of transforms of the nonzero $a_{i_j}$. Every other term omits at least one reflection and therefore has affine Coxeter length strictly less than $r$. The semidirect-product normal form in [F4] makes the top substitutions for distinct alternating words and ${\pi }^{ϵ}$ distinct.

Suppose a finite linear combination of the spanning words in [F2] acts as zero, and choose the largest affine length $r$ among words with a nonzero coefficient. For any fixed top substitution $g$ of length $r$, step 1.1 shows that its coefficient in the crossed-product expansion is exactly ${\gamma }_g(z){\sum }_m{c}_m{z}^m$: shorter words and all lower terms have smaller length. By [F3], the coefficient of each substitution must vanish. Since ${\gamma }_g\mathrm{\ne }0$, the Laurent polynomial ${\sum }_m{c}_m{z}^m$ is zero, so all its coefficients vanish. Repeating for the finitely many top substitutions of length $r$ and then descending in length shows that every coefficient in the original relation is zero.

The spanning family in [F2] has linearly independent images by step 2.1. Every element of the representation kernel is a finite linear combination of that spanning family, so it is zero. Thus the representation is injective and hence faithful.