Internal discrete fibrations

Question

The nLab page for discrete fibrations says that an internal functor is an internal discrete fibration iff the square given there is 'Cartesian' -- does this mean that it is cartesian as a morphism in the arrow category, or simply that it is a pullback?

Answer 1

They mean the same thing. If a category $\mathcal{E}$ has pullbacks, then a square in $\mathcal{E}$ is cartesian as a morphism in $\mathcal{E}^{\to}$ (with respect to $\mathsf{cod} : \mathcal{E}^{\to} \to \mathcal{E}$) if and only if it is a pullback square in $\mathcal{E}$.