Normality is equivalent to positive pressing down

If $F$ is normal and every fibre of a regressive $f:S\to \kappa$ is small, all their complements belong to $F$. Their diagonal belongs to $F$ and must meet $S$. At an intersection point $\alpha$, its value $f(\alpha )<\alpha$ forces it into the complement of its own fibre, a contradiction.

Conversely let $A_{\xi }\in F$ and suppose their diagonal $D$ is not in $F$. Then $S=\kappa \setminus D$ is positive and excludes zero. For $\alpha \in S$, take the least $\xi <\alpha$ with $\alpha \mathrm{\notin }{A}_{\xi }$. This defines a regressive map. A positive fibre would be disjoint from the corresponding filter member $A_{\xi }$, impossible. Hence $D\in F$.

For $(A_i{)}_{i<\mu }$ in $F$, $\mu <\kappa$, pad by $\kappa$ at all remaining indices. Its diagonal, intersected with $[\mu ,\kappa )$, is in $F$ and is contained in ${\bigcap }_{i<\mu }{A}_i$. Upward closure proves completeness. For $\mu =0$ the intersection is $\kappa$.