An invertible Yang–Baxter solution that is not a universal R-matrix

Multiplication and unit on $H=\mathbb{Q}$ are the usual ones; $\mathrm{\Delta }$, $\epsilon$, and $S$ are the identity scalar maps with the tensor-unit identification. They preserve multiplication and unit, both coalgebra composites send $q$ to $q(1\otimes 1\otimes 1)$, both counit contractions return $q$, and both antipode contractions give $q$. Thus these maps make $H$ a Hopf algebra.

The tensor $R$ has inverse $\frac{1}{2}(1\otimes 1)$. Each leg tensor is $2(1\otimes 1\otimes 1)$, so both sides of QYBE are $8(1\otimes 1\otimes 1)$. Therefore $R$ is an invertible QYBE solution.

The first hexagon has left side $(\mathrm{\Delta }\otimes \mathrm{id})(R)=2(1\otimes 1\otimes 1)$, whereas $R_{13}{R}_{23}=4(1\otimes 1\otimes 1)$. These tensors differ over $\mathbb{Q}$, so the first hexagon fails and $R$ is not quasitriangular.

By [F3], every quasitriangular element must satisfy QYBE, and step 1.2 gives an invertible QYBE solution for which step 1.3 fails a defining hexagon. Hence QYBE is necessary but not sufficient for quasitriangularity.