A zero Temperley–Lieb cell form can have a nonzero simple module

There is exactly one rank-two link state with zero defects, namely the arc joining sites $1$ and $2$. Call it $x$. Thus $W_{2,0}(k,0)=kx$. The four cyclic boundary points have three pairings; the pairing with alternating endpoints is crossing, while the other two are the identity matching and $d_1$ with its top cap and bottom cup. Hence those two diagrams are the full basis of ${\mathrm{T}\mathrm{L}}_2(k,0)$.

The identity acts as the identity on $x$. Stacking $x$ above $d_1$ closes the top cap of $d_1$ with the arc of $x$, producing one loop and leaving the same link state, so $x\cdot d_1=\delta x=0$. Since $1$ and $d_1$ span ${\mathrm{T}\mathrm{L}}_2(k,0)$, every algebra element acts on the one-dimensional space $kx$ by a scalar. The module is nonzero and simple.

Reflecting $x$ and gluing it to itself creates one loop, so ${\varphi }_{2,0}(x,x)=\delta =0$. Since $x$ is the sole basis vector, bilinearity makes the cell form identically zero and its cell-form quotient $W_{2,0}\mathrm{/}\mathrm{rad}({\varphi }_{2,0})$ is zero. But $W_{2,0}=kx$ is itself a nonzero simple module by step 2.1. This refutes the statement.