I'm reading a proof that in a quasi-triangular Hopf algebra $H$, $(S\otimes 1)Q$ is $\operatorname{Ad}$-invariant. Here $Q=\tau(R)R$, where $R$ is the invertible element in $H\otimes H$ satisfying all the quasi-triangular properties.
It goes, if $R=R'^1\otimes R'^2$ is the second copy of $R$ being transposed in the definition of $Q$, $$ \begin{align*} \operatorname{Ad}_h(S\otimes 1)(Q)) &= h_1(SQ^1)Sh_2\otimes h_3Q^2Sh_4\\ &= h_1(SR^1)(SR'^2)Sh_2\otimes h_3R'^1R^2Sh_4\\ &= h_1(SR^1)Sh_3(SR'^2)\otimes R'^1h_2R^2Sh_4\\ &= h_1Sh_2(SR^1)(SR'^2)\otimes R'^1R^2h_3Sh_4\\ &= \epsilon(h)(S\otimes 1)(Q) \end{align*} $$
For the last equality, shouldn't $h_1Sh_2=\epsilon(a)$ and $h_3Sh_4=\epsilon(b)$, where $\Delta h=a\otimes b$, which would give a coefficient of $\epsilon(a)\epsilon(b)=\epsilon(ab)$ in front? Not $\epsilon(h)$?
Yes, you're right, but $\varepsilon(ab)=\varepsilon(h)$ - so the proof you quoted is also right. For: $$\varepsilon(ab)=\varepsilon(a)\varepsilon(b)=\varepsilon(\varepsilon(a)b)=\varepsilon(h).$$ The 2nd equality used that $\varepsilon$ is $k$-linear.