[proofplan] We argue by contradiction. If every fibre of the regressive map is nonstationary, then each fibre is avoided by some club subset of $\kappa$. Taking the diagonal intersection of these clubs gives a new club $C$, and stationarity of $S$ provides a nonzero $\beta \in S \cap C$. Regressiveness gives $f(\beta) < \beta$, so the definition of the diagonal intersection forces $\beta$ into the very club chosen to avoid its fibre, a contradiction. [/proofplan] [step:Assume all fibres are nonstationary and choose avoiding clubs] Suppose, toward a contradiction, that for every $\xi < \kappa$ the fibre \begin{align*} S_\xi := f^{-1}(\{\xi\}) = \{\alpha \in S : f(\alpha) = \xi\} \end{align*} is nonstationary in $\kappa$. By the definition of nonstationarity, for each $\xi < \kappa$ there exists a club set $C_\xi \subseteq \kappa$ such that \begin{align*} C_\xi \cap S_\xi = \varnothing. \end{align*} Here "club" means closed and unbounded in the ordinal $\kappa$. [/step] [step:Form the diagonal intersection and verify that it is club] Define the diagonal intersection \begin{align*} C := \Delta_{\xi < \kappa} C_\xi := \{\beta < \kappa : \text{for every } \xi < \beta,\ \beta \in C_\xi\}. \end{align*} We claim that $C$ is club in $\kappa$. First, $C$ is closed. Let $\lambda < \kappa$ be a limit point of $C$, meaning that $C \cap \lambda$ is unbounded in $\lambda$. To prove $\lambda \in C$, fix $\xi < \lambda$. Since $C \cap \lambda$ is unbounded in $\lambda$, choose $\beta \in C$ with $\xi < \beta < \lambda$. Because $\beta \in C$ and $\xi < \beta$, we have $\beta \in C_\xi$. Since this can be done for arbitrarily large $\beta < \lambda$, and since $C_\xi$ is closed in $\kappa$, it follows that $\lambda \in C_\xi$. As $\xi < \lambda$ was arbitrary, $\lambda \in C$. Second, $C$ is unbounded. Fix $\alpha < \kappa$. We construct an increasing sequence $(\alpha_n)_{n \in \mathbb{N}}$ of ordinals below $\kappa$. Choose $\alpha_1 < \kappa$ with $\alpha < \alpha_1$. Suppose $\alpha_n$ has been chosen. Since $\alpha_n + 1 < \kappa$ and $\kappa$ is regular, the intersection \begin{align*} D_n := \bigcap_{\xi \leq \alpha_n} C_\xi \end{align*} is club in $\kappa$: it is an intersection of fewer than $\kappa$ many club subsets of $\kappa$. Hence choose $\alpha_{n+1} \in D_n$ with $\alpha_n < \alpha_{n+1}$. Define \begin{align*} \delta := \sup_{n \in \mathbb{N}} \alpha_n. \end{align*} Because $\kappa$ is regular uncountable, the supremum of this [countable set](/page/Countable%20Set) of ordinals below $\kappa$ is still below $\kappa$, so $\delta < \kappa$. Also $\delta > \alpha$. We verify that $\delta \in C$. Let $\xi < \delta$. Since $\delta = \sup_{n \in \mathbb{N}} \alpha_n$, there exists $n \in \mathbb{N}$ such that $\xi \leq \alpha_n$. For every $m \geq n+1$, the construction gives $\alpha_m \in C_\xi$. Thus a tail of the increasing sequence $(\alpha_m)$ lies in the [closed set](/page/Closed%20Set) $C_\xi$, and its supremum is $\delta$. Therefore $\delta \in C_\xi$. Since $\xi < \delta$ was arbitrary, $\delta \in C$. Hence $C$ is unbounded in $\kappa$. Thus $C$ is club in $\kappa$. [guided] The diagonal intersection is designed to guarantee the following: if $\beta \in C$ and some index $\xi$ lies below $\beta$, then $\beta$ belongs to the particular club $C_\xi$. This is exactly what will interact with regressiveness later. Define \begin{align*} C := \Delta_{\xi < \kappa} C_\xi := \{\beta < \kappa : \text{for every } \xi < \beta,\ \beta \in C_\xi\}. \end{align*} We prove directly that $C$ is club in $\kappa$. First, $C$ is closed. Let $\lambda < \kappa$ be a limit point of $C$, so $C \cap \lambda$ is unbounded in $\lambda$. To prove $\lambda \in C$, we must show that $\lambda \in C_\xi$ for every $\xi < \lambda$. Fix such a $\xi$. Since $C \cap \lambda$ is unbounded in $\lambda$, there is some $\beta \in C$ satisfying $\xi < \beta < \lambda$. By the definition of $C$, the inequality $\xi < \beta$ implies $\beta \in C_\xi$. In fact, the same argument gives points of $C_\xi$ arbitrarily close below $\lambda$. Since $C_\xi$ is closed in $\kappa$, the limit point $\lambda$ belongs to $C_\xi$. Because this holds for every $\xi < \lambda$, we get $\lambda \in C$. Second, $C$ is unbounded. Fix $\alpha < \kappa$. We want to find an element of $C$ above $\alpha$. Construct an increasing sequence $(\alpha_n)_{n \in \mathbb{N}}$ below $\kappa$ as follows. Choose $\alpha_1 < \kappa$ with $\alpha < \alpha_1$. Given $\alpha_n$, define \begin{align*} D_n := \bigcap_{\xi \leq \alpha_n} C_\xi. \end{align*} The set of indices $\{\xi : \xi \leq \alpha_n\}$ has cardinality less than $\kappa$, and $\kappa$ is regular. Therefore the intersection of these fewer than $\kappa$ many club sets is again club in $\kappa$. In particular $D_n$ is unbounded, so choose $\alpha_{n+1} \in D_n$ with $\alpha_n < \alpha_{n+1}$. Now define \begin{align*} \delta := \sup_{n \in \mathbb{N}} \alpha_n. \end{align*} The regular uncountability of $\kappa$ is used here: a countable increasing sequence of ordinals below $\kappa$ has supremum still below $\kappa$. Hence $\delta < \kappa$, and by construction $\delta > \alpha$. We show $\delta \in C$. Let $\xi < \delta$. Since the sequence $(\alpha_n)$ is cofinal in $\delta$, choose $n \in \mathbb{N}$ such that $\xi \leq \alpha_n$. For every $m \geq n+1$, the construction gives $\alpha_m \in D_{m-1}$, and since $\xi \leq \alpha_n \leq \alpha_{m-1}$, we have $\alpha_m \in C_\xi$. Thus all sufficiently far terms of the sequence lie in $C_\xi$. Because $C_\xi$ is closed and $\delta$ is the supremum of that tail, $\delta \in C_\xi$. Since this holds for every $\xi < \delta$, we have $\delta \in C$. Therefore, above every $\alpha < \kappa$ there is some $\delta \in C$, so $C$ is unbounded. Together with closedness, this proves that $C$ is club in $\kappa$. [/guided] [/step] [step:Choose a nonzero point in the stationary set and the diagonal club] Since $S$ is stationary in $\kappa$ and $C$ is club in $\kappa$, the intersection $S \cap C$ is nonempty. Moreover, it contains a nonzero ordinal. Indeed, the set $\kappa \setminus \{0\}$ contains the club interval $[1,\kappa) := \{\gamma < \kappa : 1 \leq \gamma\}$, so $S \cap C \cap [1,\kappa)$ is nonempty. Choose \begin{align*} \beta \in S \cap C \cap [1,\kappa). \end{align*} Then $\beta \in S$, $\beta \in C$, and $\beta \neq 0$. [/step] [step:Use regressiveness to force membership in the forbidden club] Define \begin{align*} \xi := f(\beta). \end{align*} Since $f: S \to \kappa$ and $\beta \in S$, we have $\xi < \kappa$. Since $\beta \neq 0$ and $f$ is regressive, we also have \begin{align*} \xi = f(\beta) < \beta. \end{align*} Because $\beta \in C$ and $\xi < \beta$, the definition of the diagonal intersection gives \begin{align*} \beta \in C_\xi. \end{align*} On the other hand, $\xi = f(\beta)$ and $\beta \in S$, so \begin{align*} \beta \in S_\xi = f^{-1}(\{\xi\}). \end{align*} Thus $\beta \in C_\xi \cap S_\xi$, contradicting the choice of $C_\xi$ as a club disjoint from $S_\xi$. [/step] [step:Conclude that some fibre is stationary] The contradiction shows that the assumption that every fibre $S_\xi$ is nonstationary is false. Therefore there exists $\xi < \kappa$ such that \begin{align*} f^{-1}(\{\xi\}) = \{\alpha \in S : f(\alpha) = \xi\} \end{align*} is stationary in $\kappa$. [/step]