[proofplan] We argue by contradiction. If every fibre of the regressive map is $U$-small, then the complement of each fibre inside $S$ is $U$-large. Normality closes $U$ under diagonal intersections, so there is a $U$-large set of points $\xi$ that avoid the $\alpha$-th fibre for every $\alpha < \xi$. Choosing a nonzero such point contradicts regressiveness, because $f(\xi)$ itself is one of the ordinals below $\xi$. [/proofplan] [step:Assume every fibre is $U$-small and form its $U$-large complement] Suppose, toward a contradiction, that for every $\alpha < \kappa$, \begin{align*} f^{-1}(\{\alpha\}) \notin U. \end{align*} For each $\alpha < \kappa$, define \begin{align*} A_\alpha := S \setminus f^{-1}(\{\alpha\}). \end{align*} Since $U$ is an ultrafilter on $\kappa$ and $f^{-1}(\{\alpha\}) \notin U$, its complement $\kappa \setminus f^{-1}(\{\alpha\})$ belongs to $U$. Since $S \in U$, the finite-intersection property of filters gives \begin{align*} A_\alpha = S \cap \bigl(\kappa \setminus f^{-1}(\{\alpha\})\bigr) \in U. \end{align*} Thus $(A_\alpha)_{\alpha < \kappa}$ is a $\kappa$-sequence of members of $U$. [guided] We begin by negating the conclusion. If the theorem were false, then no single ordinal value would appear on a $U$-large set of points. Formally, for every $\alpha < \kappa$, \begin{align*} f^{-1}(\{\alpha\}) \notin U. \end{align*} To use normality, we want a family of sets that are themselves $U$-large. For each $\alpha < \kappa$, define the complement of the $\alpha$-th fibre inside $S$: \begin{align*} A_\alpha := S \setminus f^{-1}(\{\alpha\}). \end{align*} Because $U$ is a normal measure, it is in particular an ultrafilter on $\kappa$. Thus if a set is not in $U$, then its complement is in $U$. Applying that to $f^{-1}(\{\alpha\})$, we get $\kappa \setminus f^{-1}(\{\alpha\}) \in U$. Since $S \in U$ as well, closure under finite intersections yields \begin{align*} A_\alpha = S \cap \bigl(\kappa \setminus f^{-1}(\{\alpha\})\bigr) \in U. \end{align*} So each fibre-complement is $U$-large. That is exactly the input needed for the diagonal-intersection form of normality. [/guided] [/step] [step:Take a diagonal intersection of the complements] Define \begin{align*} D := \Delta_{\alpha < \kappa} A_\alpha = \{\xi < \kappa : \xi \in A_\alpha \text{ for every } \alpha < \xi\}. \end{align*} Since $U$ is normal, it is closed under diagonal intersections, and therefore $D \in U$. Because $U$ is nonprincipal, $\kappa \setminus \{0\} \in U$. Hence \begin{align*} D \cap (\kappa \setminus \{0\}) \in U. \end{align*} In particular, this set is nonempty, so choose $\xi \in D \cap (\kappa \setminus \{0\})$. [guided] Normality is used here in its diagonal-intersection form. The family $(A_\alpha)_{\alpha < \kappa}$ has already been shown to lie in $U$, so normality says that the set of points lying in every earlier member of the family is still $U$-large: \begin{align*} D := \Delta_{\alpha < \kappa} A_\alpha = \{\xi < \kappa : \xi \in A_\alpha \text{ for every } \alpha < \xi\} \in U. \end{align*} We also need a nonzero point, because regressiveness is required only at nonzero elements of $S$. Since a normal measure is nonprincipal, the singleton $\{0\}$ is not in $U$, so its complement $\kappa \setminus \{0\}$ is in $U$. The intersection of two sets in a filter is again in the filter, hence \begin{align*} D \cap (\kappa \setminus \{0\}) \in U. \end{align*} A $U$-large set cannot be empty, so we may choose a point $\xi$ in this intersection. By construction, $\xi \neq 0$ and $\xi$ belongs to every $A_\alpha$ with $\alpha < \xi$. [/guided] [/step] [step:Read off the value of the regressive map at the chosen point] Since $\xi \in D$ and $\xi \neq 0$, we have $\xi \in A_0$, because $0 < \xi$. By the definition of $A_0$, this implies $\xi \in S$, so $f(\xi)$ is defined. Because $f:S \to \kappa$ is regressive and $\xi$ is a nonzero element of $S$, \begin{align*} f(\xi) < \xi. \end{align*} Set $\beta := f(\xi)$. Then $\beta < \xi$, so the definition of $D$ yields $\xi \in A_\beta$. But $A_\beta = S \setminus f^{-1}(\{\beta\})$, so $\xi \notin f^{-1}(\{\beta\})$. This says $f(\xi) \neq \beta$, contradicting the definition of $\beta$. Therefore the assumption that every fibre is outside $U$ is false. Hence there exists $\alpha < \kappa$ such that \begin{align*} f^{-1}(\{\alpha\}) \in U. \end{align*} [guided] We now exploit the choice of $\xi$. Since $\xi \in D$, the defining property of the diagonal intersection says that $\xi$ lies in every $A_\alpha$ with $\alpha < \xi$. Because we also chose $\xi \neq 0$, the index $\alpha = 0$ is allowed, so $\xi \in A_0$. But $A_0 = S \setminus f^{-1}(\{0\})$, and being in $A_0$ already forces $\xi \in S$ because $A_0 \subseteq S$. Thus $f(\xi)$ is defined, and regressiveness applies: \begin{align*} f(\xi) < \xi. \end{align*} Now set \begin{align*} \beta := f(\xi). \end{align*} The inequality $f(\xi) < \xi$ means exactly that $\beta < \xi$, so the diagonal-intersection property applies again, now with the index $\beta$. Since $\beta < \xi$, we get $\xi \in A_\beta$. Unpacking the definition of $A_\beta$ gives \begin{align*} \xi \notin f^{-1}(\{\beta\}). \end{align*} But $\beta$ was defined to be $f(\xi)$, so $\xi \in f^{-1}(\{\beta\})$ by definition. This contradiction shows that our starting assumption was impossible. Hence some fibre $f^{-1}(\{\alpha\})$ must belong to $U$, which is exactly the conclusion of the theorem. [/guided] [/step]