My book gives this Lagrangian:
$$ L = -|\partial \phi|^2 -V(\phi) -\bar \psi_L \not \partial \psi_L -\bar \psi_R \not \partial \psi_R -g(\phi \bar \psi_L \psi_R + \phi^* \bar \psi_R \psi_L) $$
It's supposed to have $U(1)$ symmetry, with $\phi$ having charge $+1$, $\psi_L$ having $+1/2$ and $\psi_R$ having $-1/2$, which is the reason for those cubic terms.
However, the book also says that
$$ \psi_{L/R} = \frac{1}{2}(1\pm\gamma_5)\psi $$
so I deduce that
$$ \bar \psi_L \psi_R = \frac{1}{4} \bar \psi (1+\gamma_5)(1-\gamma_5)\psi = \frac{1}{4}\bar \psi (1-\gamma_5^2)\psi = 0 $$
because $\gamma_5^\dagger = \gamma_5$ and $\gamma_5^2 = 1$.
It's can't be zero otherwise the interaction terms given by the book are wrong. What happened?
EDIT:
$$ \bar \psi_L = \psi_L^\dagger \gamma_0 = \frac{1}{2}\psi^\dagger (1+\gamma_5)\gamma_0 = \frac{1}{2}\psi^\dagger \gamma_0 \gamma^0 (1+\gamma_5)\gamma_0 = \frac{1}{2}\bar \psi (1+\gamma_5) $$
using $\gamma_0 \gamma^0 = 1$ and $\gamma^0 \gamma_5 \gamma_0=\gamma_5$.