If $H$ is a Hopf algebra, do we have $H^{cop}$ is a Hopf algebra? Let $H=(H, m, u, \Delta, \epsilon, S)$ be a Hopf algebra, see for example the lecture notes, where $m$ is the multiplication, $u$ is the unit, $\Delta$ is the comultiplication, $\epsilon$ is the counit, $S$ is the antipode. Let $H^{cop} = (H, m, u, \Delta^{op}, \epsilon, S)$, where $m,u,\epsilon,S$ are the same maps as the maps in $H$, and $\Delta^{op}=\tau \circ \Delta$, $\tau$ is the flip map. Is $H^{cop}$ a Hopf algebra? I am asking this question because in the book: Hopf algebras and their actions on rings, pages 213, 214, it is said that $t_{M,N}$ is a braiding if $H^{cop}$ is a Hopf algebra. Why the condition is not "$H$ is a Hopf algebra" but "$H^{cop}$ is a Hopf algebra"? Are "$H$ is a Hopf algebra " and "$H^{cop}$ is a Hopf algebra" equivalent? Thank you very much. ![enter image description here]( ![enter image description here](

For a Hopf algebra $H$, $H^{cop}$ may or may **not** be a Hopf algebra. **If it is** , then the maps $t_{M,N}$ above define a braiding.