$E$-field between dielectric and plate in parallel plate capacitor and Feynman's claim

Feynman says about $E_{0}$ below Note that the field $E_{0}$ between the metal plate and the surface of the dielectric is higher than the field $E$ ; it corresponds to $\sigma _{free}$ alone. $\sigma _{pol}$ is the surface charge density induced by polarizing the dielectric. Why does $E_{0}$ correspond to $\sigma _{free}$ alone ? It strikingly makes no sense from the schematic. If we apply Gauss's law, we can clearly see that both $\sigma_{free}$ and $\sigma _{pol}$ contribute to $E_{0}$ , not just $\sigma_{free}$ .