Proof $(1)$ is trivial. For $(2)$ , suppose that f is almost one-to-one on $A\in U$ , and let for each $\alpha <\kappa \ g(\alpha )=\sup f^{-1}[\alpha +1]\cap A$ . Then for each $\xi \in A\ g(f(\xi ))=\sup f^{-1}[f(\xi )+1]\cap A\geq \xi $ , hence $[g\circ f]_U\geq [\text {id}]_U$ . For the other direction, let g be a monotone function such that $[g\circ f]_U\geq [\text {id}]_U$ . Then there is a set $A\in U$ such that for each $\alpha \in A$ , $g\circ f(\alpha )\geq \alpha $ . Hence, if $\beta \in f^{-1}[\alpha ]$ , then $g(\alpha )\geq g(f(\beta ))\geq \beta $ , hence $f^{-1}[\alpha ]\subseteq g(\alpha )+1$ .

Fact 2.13. $\sum _U{\langle } U_\alpha {\rangle }_{\alpha <\kappa }\equiv _{RK} U\text {-}\lim {\langle } U_\alpha {\rangle }_{\alpha <\kappa }$ iff ${\langle } U_\alpha {\rangle }_{ \alpha <\kappa }$ is discrete mod U.

Proof See [Reference Kanamori24, Corollary 5.15].

Proof By induction on $\alpha $ . Clearly, for $\alpha =0$ , $k(f_0)(\kappa )=0$ and if $k(f_\alpha )(\kappa )=\alpha $ , then by elementarity $k(f_{\alpha +1})(\kappa )=\alpha +1$ . For limit $\eta $ , if $\mathrm {cf}(\eta )<\kappa $ , then the functions used in the definition of $f_\eta $ are ${\langle } f_{\eta _i}{\rangle }_{i<\mathrm {cf}(\eta )}$ are pointwise mapped by k; that is, $k({\langle } f_{\eta _i}\mid i<\mathrm {cf}(\eta ){\rangle })={\langle } k(f_{\eta _i})\mid i<\mathrm {cf}(\eta ){\rangle }$ . It follows by elementarity and the definition of $f_\eta $ that $k(f_\eta )(\kappa )=\sup _{i<\mathrm {cf}(\eta )}k(f_{\eta _i})(\kappa )$ . Hence, by the induction hypothesis, $k(f_\eta )(\kappa )=\sup _{i<\mathrm {cf}(\eta )}\eta _i=\eta $ . If $\mathrm {cf}(\eta )=\kappa $ , then the sequence ${\langle } f_{\eta _i}\mid i<\kappa {\rangle }$ is stretched by k to $k({\langle } f_{\eta _i}\mid i<\kappa {\rangle })={\langle } f^{\prime }_{\eta _i}\mid i<k(\kappa ){\rangle }$ but for every $i<\kappa $ , as $k(i)=i$ , we have $f^{\prime }_{\eta _i}=k(f_{\eta _i})$ . Again by the definition of $f_\eta $ , elementarity, and the induction hypothesis, we conclude that

$$ \begin{align*}k(f_\eta)(\kappa)=\sup_{i<\kappa}f^{\prime}_{\eta_i}(\kappa)=\sup_{i<\kappa}k(f_{\eta_i})(\kappa)=\sup_{i<\kappa}\eta_i=\eta.\\[-37pt]\end{align*} $$

Proof Suppose otherwise, and let $C_X$ for every $X\subseteq \kappa $ be the club guaranteed by item $(2)$ of $\diamondsuit ^*(S)$ . Then $C_X\in U$ . Also, for each $\alpha \in S$ , let ${\langle } I^\alpha _i{\rangle }_{i<\mathrm { cf}(\alpha )}$ be a partition of $\mathcal {A}_\alpha $ such that $|I^\alpha _i|<\alpha $ . Now for each $X\subseteq \kappa $ , consider the function $f_X:C_{X}\cap S\rightarrow \kappa $ defined by $f_X(\alpha )=i<\mathrm {cf}(\alpha )$ for the unique i such that $X\cap \alpha \in I^\alpha _i$ . Since $cf^{M_U}([id]_U)\leq \operatorname {\mathrm {crit}}(j_U)$ , there is a function $\pi :\kappa \rightarrow On$ such that $i<\mathrm {cf}(\alpha )\leq \pi (\alpha )$ and $[\pi ]_U=\operatorname {\mathrm {crit}}(j_U)$ . It follows that there is $A_X\subseteq C_{X}\cap S$ , $A_X\in U$ and $\gamma _X<\kappa $ such that for every $\alpha \in A_X$ , $f_X(\alpha )=\gamma _X$ . There are $2^\kappa $ -many subsets with the same $\gamma _X=\gamma ^*$ . Now apply Galvin’s property to those $2^\kappa $ -many sets in order find $\kappa $ -many distinct subsets of $\kappa $ , ${\langle } X_\xi {\rangle }_{\xi <\kappa }$ for which $A^*:=\bigcap _{\xi <\kappa }A_{X_\xi }\in U$ . Now for each $\alpha \in A^*\cap S$ , $|I^\alpha _{\gamma ^*}|<\alpha $ . Since $\kappa $ is regular, we may apply Födor’s lemma to find a stationary set $S'\subseteq A^*\cap S$ and $\theta <\kappa $ such that $|I^\alpha _{\gamma ^*}|=\theta $ for each $\alpha \in S'$ . Consider ${\langle } X_i{\rangle }_{i<\theta ^{+}}$ and for each $i\neq j<\theta ^{+}$ , let $\beta _{i,j}<\kappa $ be high enough so that $X_i\cap \beta _{i,j}\neq X_j\cap \beta _{i,j}$ . Take any $\alpha \in S'\setminus \sup _{i\neq j<\theta ^{+}}\beta _{i,j}$ . To reach a contradiction, note that on one hand, since $\alpha \in S'$ , $|I^{\alpha }_{\gamma ^*}|=\theta $ . On the other hand, for every $i\neq j<\theta ^{+}$ , $X_i\cap \alpha \in I^{\alpha }_{\gamma ^*}$ and the sets $X_i\cap \alpha $ are all distinct.

Proof The witnessing $\diamondsuit ^*_{\mathrm{thin}}(U)$ -sequence is just the sequence ${\langle } \mathcal {A}_\alpha {\rangle }_{\alpha <\kappa }$ representing A in $M_U$ . Clearly, condition $(1)$ is equivalent to the fact that for every $S\subseteq \kappa $ , $\{\alpha <\kappa \mid S\cap \alpha \in \mathcal {A}_\alpha \}\in U$ . By Proposition 2.3, condition $(2)$ is equivalent to the function $\alpha \mapsto |\mathcal {A}_\alpha |$ not being almost one-to-one mod U.

Proof Assume toward contradiction that W has the Galvin property. Enumerate $\mathcal {A}_\alpha =\{A_{\alpha ,i}\mid i<|\mathcal {A}_\alpha |\}$ . For every set X, there is $B_X\in W$ such that for every for every $\alpha \in B_X$ , $X\cap \alpha \in \mathcal {A}_\alpha $ . By our assumption, there are $\kappa $ -many distinct sets $\{X_i\mid i<\kappa \}$ such that $B:=\bigcap _{i<\kappa }B_{X_i}\in W$ . Note that the key property of B is that for every $i<\kappa $ and for all $\alpha \in B$ , $X_i\cap \alpha \in \mathcal {A}_\alpha $ . Since the function $\alpha \mapsto |\mathcal {A}_\alpha |$ is not almost one-to-one mod W, there is $\theta <\kappa $ and an unbounded subset $B'\subseteq B$ such that for every $\alpha \in B'$ , $|\mathcal {A}_\alpha |=\theta $ . Consider $\{ X_i\mid i<\theta ^+\}$ . For every $i\neq j<\theta ^+$ , find $\alpha _{i,j}<\kappa $ such that $X_i\cap \alpha _{i,j}\neq X_j\cap \alpha _{i,j}$ and take $\alpha ^*=\sup _{i,j<\theta ^+}\alpha _{i,j}$ . By regularity of $\kappa $ , $\alpha ^*<\kappa $ . Since $B'$ is unbounded there exists some $\beta ^*\in B'$ with $\beta ^*>\alpha ^*$ . It follows that for every $i<\theta ^+$ , $X_i\cap \beta ^*\in \mathcal {A}_{\beta ^*}$ , and also for every $i\neq j$ , since $\alpha _{i,j}<\beta ^*$ , $X_i\cap \beta ^*\neq X_j\cap \beta ^*$ . It follows that $i\mapsto X_i\cap \beta ^*$ is a one-to-one function from $\theta ^+$ into $\mathcal {A}_{\beta ^*}$ . This contradicts the fact that $\beta ^*\in B'$ and thus $|\mathcal {A}_{\beta ^*}|=\theta $ .

Proof It remains to show that $\alpha \mapsto |\mathcal {A}_\alpha |$ is not one-to-one on a set in U. If $A\in U$ , then A is stationary since $\text {Club}_\kappa \subseteq U$ . By Födor applied to the function $\alpha \mapsto |\mathcal {A}_\alpha |$ restricted to A, there is an unbounded subset $S'\subseteq A$ and $\theta <\kappa $ such that for every $\alpha \in S'$ , $|\mathcal {A}_\alpha |=\theta $ . In particular, $\alpha \mapsto |\mathcal {A}_\alpha |$ is not almost one-to-one on A.

Proof Assume that U is a non p-point Dodd sound ultrafilter. Let ${\langle }\mathcal {A}_\alpha {\rangle }_{\alpha <\kappa }$ be the sequence obtained by Theorem 3.7. Note that for every $S\subseteq \kappa $ , the sequence ${\langle } S\cap \alpha {\rangle }_{\alpha <\kappa }$ satisfies condition $(a)$ of Theorem 3.7. By the theorem, we conclude that for U-almost every $\alpha $ , $S\cap \alpha \in \mathcal {A}_\alpha $ . It follows that $j_U(S)\cap [id]_U\in [\alpha \mapsto \mathcal {A}_\alpha ]$ and $\{j_U(S)\cap [id]_U\mid S\subseteq \kappa \}\subseteq [\alpha \mapsto \mathcal {A}_\alpha ]_U$ . Similarly, from the implication $(b)$ to $(a)$ we deduce that $[\alpha \mapsto \mathcal {A}_\alpha ]_U\subseteq \{j_U(S)\cap [\text {id}]_U\mid S\subseteq \kappa \}$ . By Dodd soundness, the function $j^{[\text {id}]_U}\colon P(\kappa )\rightarrow \{j_U (S)\cap [\text {id}]_U\mid S\subseteq \kappa \}$ defined by $j^{[\text {id}]_U}(S)=j(S)\cap [\text {id}]_U$ belongs to $M_U$ . Thus $M_U\models |[\alpha \mapsto \mathcal {A}_\alpha ]_U|=2^\kappa $ . Finally, $\alpha \mapsto |\mathcal {A}_\alpha |$ cannot be an almost one-to-one function mod U: otherwise, the class of any unbounded function $\kappa \leq [\pi ]_U$ would also be an almost one-to-one mod U. To see this, suppose that $[\tau ]_U=\kappa $ , then $[\alpha \mapsto 2^{\tau (\alpha )}]_U=2^\kappa $ and by our assumption, this is represented by an almost one-to-one function mod U Footnote ⁹ . Let $X\in U$ be the set witnessing that $\alpha \mapsto 2^{\tau (\alpha )}$ is almost one-to-one mod U. Also we let $Y\in U$ be such that for every $\alpha \in Y$ , $\tau (\alpha )\leq \pi (\alpha )$ . We claim that $\pi \restriction X\cap Y$ is almost one-to-one as for any $\gamma <\kappa $ ,

$$ \begin{align*}\{\alpha<\kappa\mid \pi(\alpha)&<\gamma\}\cap X\cap Y\subseteq\{\alpha<\kappa\mid \tau(\alpha)<\gamma\}\cap X\cap Y\subseteq\\&\subseteq\{\alpha<\kappa\mid 2^{\tau(\alpha)}\leq 2^\gamma\}\cap X\cap Y.\end{align*} $$

The right most set is bounded by the choice of X. We conclude that U is a p-point contradiction.

Proof One direction follows from the previous lemma. Let us prove the other, note that $\alpha \mapsto |\mathcal {A}_\alpha |$ cannot be bounded on a set in U, just otherwise, suppose that $\theta <\kappa $ is such that $B^*:=\{\alpha <\kappa \mid |\mathcal {A}_\alpha |\leq \theta \}\in U$ . Take any $\theta ^+$ -many sets $\{ X_i\mid i<\theta ^+\}$ such that there is $\gamma <\kappa $ such that for all $i\neq j<\theta ^+$ , $X_i\cap \gamma \neq X_j\cap \gamma $ . For each $i<\theta ^+$ , Denote by $B_i:=\{\alpha <\kappa \mid X_i\cap \alpha \in \mathcal {A}_\alpha \}\in U$ . By $\kappa $ -completeness and fineness, there is $\gamma ^*\in B^*\cap (\bigcap _{i<\theta ^+}B_i)\setminus \gamma $ . It follows that $|\mathcal {A}_{\gamma ^*}|=\theta $ but also for each $i<\theta ^+$ , $X_i\cap \gamma ^*\in \mathcal {A}_{\gamma ^*}$ are all distinct sets. Contradiction. We conclude that $\alpha \mapsto |\mathcal {A}_\alpha |$ is an unbounded function mod U which is also not almost one-to-one according to $(1)$ . Hence, U is not a p-point.

Proof The argument of Lemma 3.5 gives this stronger result.

Proof Fix a partition of $\kappa $ into sets $S_\xi \in W_\xi $ . For each $\xi < \kappa $ , let $\langle \mathcal A^\xi _\alpha \rangle _{\alpha < \kappa }$ witness that $\diamondsuit ^*_{\mathrm{thin}}(W_\xi )$ . Then let $\mathcal A_\alpha = \mathcal A^\xi _\alpha $ , where $\xi < \kappa $ is unique such that $\alpha \in S_\xi $ . Fixing $A\subseteq \kappa $ , we would like to show that $B:=\{\alpha <\kappa \mid A\cap \alpha \in \mathcal {A}_{\alpha }\}\in U\text {-}\lim {\langle } W_\xi {\rangle }_{\xi <\kappa }$ . For any $\xi <\kappa $ , then $B_\xi :=\{\alpha \in S_\xi \mid A\cap \alpha \in \mathcal {A}^\xi _\alpha \}\in W_\xi $ . Since for each $\alpha \in S_\xi $ , $\mathcal {A}_\alpha =\mathcal {A}^\xi _\alpha $ , we conclude that $B_\xi \subseteq B$ and therefore $B\in W_\xi $ . It follows that $B\in U\text {-}\lim {\langle } W_\xi {\rangle }_{\xi <\kappa }$ . It remains to show that $c(\alpha )=|\mathcal {A}_\alpha |$ is not almost one-to-one on any set $B\in W$ . Suppose otherwise, and let $B\in W$ witness that c is almost one-to-one. Pick any $\xi <\kappa $ such that $B\in W_\xi $ to reach a contradiction note that $B\cap S_\xi \in W_\xi $ , and the function c is almost one-to-one on this set. However, for every $\alpha \in B\cap S_\xi $ , $\mathcal {A}^\xi _\alpha =\mathcal {A}_\alpha =c(\alpha )$ and so $\alpha \mapsto |\mathcal {A}^\xi _\alpha |$ is almost one-to-one on $B\cap S_\xi $ , contradicting $\diamondsuit ^*_{\mathrm{thin}}(W_\xi )$ .

Proof Denote $\mathcal {A}=\{j(A)\cap \lambda \mid A\subseteq \kappa \}\in M$ . Enumerate $V_\kappa $ in V, $f\colon \kappa \rightarrow V_\kappa $ such that for every $x\in V_\kappa $ , $f^{-1}[x]$ is unbounded in $\kappa $ . Since $\mathcal {A}\in (V_{j(\kappa )})^M$ , there is $j(\kappa )>\xi \geq \lambda $ such that $j(f)(\xi )=\mathcal {A}$ . By similar arguments, we can ensure that there are some functions $g, h: \kappa \rightarrow \kappa $ such that for the same $\xi $ we will also have $\kappa =j(g)(\xi )$ and $\lambda =j(h)(\xi )$ . Let $D=D(j,\xi )$ , $j_D\colon V\rightarrow M_D$ be the ultrapower, and let $k_D\colon M_D\rightarrow M$ be the factor map $k_D([\phi ]_D)=j(\phi )(\xi )$ . Note that

$$ \begin{align*}\lambda=k_D([h]_D), \ \kappa=k_D([g]_D), \ \mathcal{A}=k_D([f]_D)\end{align*} $$

and therefore $\kappa ,\lambda ,\mathcal {A}\in Im(k_D)$ . It follows that $\operatorname {\mathrm {crit}}(k_D)>\kappa $ and $[g]_D=\kappa $ . Since

$$ \begin{align*}k_D([h]_D)=\lambda\leq\xi=k_D([id]_D),\end{align*} $$

the elementarity of $k_D$ implies that $[h]_D\leq [id]_D$ . Recall that for any function $\phi \colon \kappa \rightarrow \kappa $ , $j(\phi )(\kappa )<\lambda $ thus by elementarity of $k_D$ ,

$$ \begin{align*}(*)\text{ for any function }\phi\colon \kappa\rightarrow \kappa, \ j_D(\phi)(\kappa)<[h]_D.\end{align*} $$

By our initial assumption, $\lambda>j(\alpha \mapsto 2^\alpha )(\kappa )=2^\kappa $ and since $M\models |\mathcal {A}|=2^\kappa $ ,

$$ \begin{align*}M_D\models |[f]_D|=2^{[g]_D}<[h]_D.\end{align*} $$

Denote by $B_\alpha =f(\alpha )$ , note that and fix a set $X^*\in D$ such that if $\alpha \in X^*$ then $|B_\alpha |=2^{g(\alpha )}<h(\alpha )$ . Pick any $2^\kappa $ distinct subsets of $\kappa $ , ${\langle } A_\alpha {\rangle }_{ \alpha <2^\kappa }$ , then $j(A_\alpha )\cap \lambda \in \mathcal {A}$ and by elementarity $j_D(A_\alpha )\cap \lambda '\in B$ . It follows that

$$ \begin{align*}X_\alpha:=\{\xi<\kappa\mid A_\alpha\cap h(\xi)\in B_\xi\}\in D.\end{align*} $$

We claim that ${\langle } X_\alpha {\rangle }_{ \alpha <2^\kappa }$ witness that $\neg \mathrm{Gal}(U,\kappa ,2^\kappa )$ . Otherwise, there is $I\in [2^\kappa ]^\kappa $ such that $X_I:=\cap _{i\in I}X_i\in D$ . Let us argue that there must be $\theta <\kappa $ such that

$$ \begin{align*}\sup\{h(\xi):\xi\in X_I\cap X^*,\ 2^{g(\xi)}<\theta\}=\kappa.\end{align*} $$

To see this, assume otherwise, then for each $\theta <\kappa $ , we can define

$$ \begin{align*}\rho(\theta)=\sup\{h(\xi)\mid \xi\in X_I\cap X^*, \ 2^{g(\xi)}\leq 2^\theta\}\end{align*} $$

then $\rho \colon \kappa \rightarrow \kappa $ is well-defined. Since $2^{j_D(g)([id]_D)}=2^{[g]_D}=2^\kappa $ , we conclude that $j_D(\rho )(\kappa )\geq j_D(h)([\text {id}]_D)=[h]_D$ , contradicting $(*)$ . We proceed as before, find $\beta \in X_I\cap X^*$ such that the restriction of $\theta $ -many of the sets in I to $h(\beta )$ are distinct. This produces a contradiction.

Proof By the definition of $\xi $ in the proof of Theorem 4.1, we can choose $\xi =[\text {id}]_U$ and the theorem ensures that $U=D(j_U,[\text {id}]_U)$ is non-Galvin.

Proof Fix any ${\langle } X_\alpha {\rangle }_{\alpha <2^\kappa }$ sequence of distinct subsets of $\kappa $ . $[\alpha \mapsto A_\alpha ]_U=A$ and $[f]_U=\lambda =j_U(f)([id]_U)$ . By our assumption,

$$ \begin{align*}B_\alpha=\{\xi<\kappa\mid X_\alpha\cap f(\xi)\in A_\alpha\}\in U.\end{align*} $$

We claim that ${\langle } B_\alpha {\rangle }_{ \alpha <2^\kappa }$ witness that $\neg \mathrm{Gal}(U,\kappa ,2^\kappa )$ . Otherwise, there is $I\in [2^\kappa ]^\kappa $ such that $B_I:=\cap _{i\in I}B_i\in U$ . Consider the map $\xi \mapsto |A_\xi |$ , note that $|A_\xi |\leq \pi (\xi )$ where $j_U(\pi )([\text {id}]_U)=|A|$ , and therefore there must be $\theta <\kappa $ such that

$$ \begin{align*}\sup\{f(\xi):\xi\in B_I,\ \pi(\xi)<\theta\}=\kappa.\end{align*} $$

Just assume otherwise, then for each $\theta <\kappa $ , we can define

$$ \begin{align*}g(\theta)=\sup\{f(\xi)\mid \xi\in B_I, \ \pi(\xi)\leq\theta\}\end{align*} $$

then $g\colon \kappa \rightarrow \kappa $ is well-defined. Since $j_U(\pi )([\text {id}]_D)=|A|$ we conclude that $j_U(g)(|A|)\geq j_U(f)([\text {id}]_D)=\lambda $ , contradicting condition $(2)$ . Now the continuation is as before, we find $\beta \in B_I$ such that $f(\beta )$ is high enough so that the restriction of $\theta ^{+}$ -many of the sets in I to $f(\beta )$ are distinct. This produces a contradiction.

Proof Let $k:M_U\rightarrow M_W$ be an elementary embedding such that $j_W=k\circ j_U$ and $A,\lambda $ witnessing $\diamondsuit ^-_{\mathrm{thin}}(U)$ . For every $S\subseteq \kappa $ , we have

$$ \begin{align*}j_W(S)\cap k(\lambda)=k(j_U(S)\cap\lambda)\in k(A).\end{align*} $$

Hence $\{j_W(S)\cap k(\lambda )\mid S\subseteq \kappa \}\subseteq k(A)\in M_W$ . By elementarity, $|k(A)|^{M_W}=k(|A|^{M_U})$ . Suppose toward contradiction that there is a function $g:\kappa \rightarrow \kappa $ such that $j_W(g)(k(|A|^{M_U}))\geq k(\lambda )$ , then $k(j_U(g)(|A|))\geq k(\lambda )$ and by elementarity if k, $j_U(g)(|A|)\geq \lambda $ , contradiction.

Proof Fix a partition of $\kappa $ into sets $S_\xi \in W_\xi $ . For each $\xi < \kappa $ , let $\langle \mathcal A^\xi _\alpha \rangle _{\alpha < \kappa }$ and $f_\xi $ witness that $\diamondsuit ^-_{\mathrm{thin}}(W_\xi )$ . Then let $\mathcal A_\alpha = \mathcal A^\xi _\alpha $ , where $\xi < \kappa $ is unique such that $\alpha \in S_\xi $ and $f(\alpha )=f_\xi (\alpha )$ . Let $A\subseteq \kappa $ , we would like to show that $B:=\{\alpha <\kappa \mid A\cap f(\alpha )\in \mathcal {A}_{\alpha }\}\in U\text {-}\lim {\langle } W_\xi {\rangle }_{\xi <\kappa }$ . Take any $\xi <\kappa $ , then $B_\xi :=\{\alpha \in S_\xi \mid A\cap f_\xi (\alpha )\in \mathcal {A}^\xi _\alpha \}\in W_\xi $ . Since for each $\alpha \in S_\xi $ and $f(\alpha )=f_\xi (\alpha )$ , $\mathcal {A}_\alpha =\mathcal {A}^\xi _\alpha $ , we conclude that $B_\xi \subseteq B$ and therefore $B\in W_\xi $ . It follows that $B\in U\text {-}\lim {\langle } W_\xi {\rangle }_{\xi <\kappa }$ . It remains to show that $c(\alpha )=|\mathcal {A}_\alpha |$ is in a lower sky than f. Suppose otherwise and let $g:\kappa \rightarrow \kappa $ such that for some $B\in W$ , $\alpha \in B\rightarrow g(c(\alpha ))\geq f(\alpha )$ . Pick any $\xi <\kappa $ such that $B\in W_\xi $ to reach a contradiction note that $B\cap S_\xi \in W_\xi $ , and for every $\alpha \in B\cap S_\xi $ , $g(|A^\xi |_\alpha )=g(c(\alpha ))\geq f(\alpha )=f_\xi (\alpha )$ . However, the sky $\alpha \mapsto |\mathcal {A}^\xi _\alpha |$ is below the sky of $f_\xi $ , contradicting the choice of $f_\xi $ .

Proof Let $W^*=[\xi \mapsto W_\xi ]_U$ . By our assumption,

$$ \begin{align*}M_U\models W^*\text{ is an ultrafilter over }j_U(\kappa)\text{ and } \diamondsuit^-_{\mathrm{thin}}(W^*).\end{align*} $$

Let $j_{W^*}:M_{U}\rightarrow M_{W^*}$ be the ultrapower of $M_U$ by $W^*$ . It follows that there is $A\in M_{W^*}$ and $\lambda <j_{W^*}(j_U(\kappa ))$ such that $\{j_{W^*}(S)\cap \lambda \mid S\in P(j_U(\kappa ))^{M_U}\}\subseteq A$ and there is no function $f:j_U(\kappa )\rightarrow j_U(\kappa )\in M_U$ such that $j_{W^*}(f)(|A|^{M_{W^*}})\geq \lambda $ . Note that $M_{W^*}=M_{\sum _U{\langle } W_\xi {\rangle }_{\xi <\lambda }}$ and $j_{\sum _U{\langle } W_\xi {\rangle }_{\xi <\lambda }}=j_{W^*}\circ j_U$ . We claim that A and $\lambda $ witness that $\diamondsuit ^-_{\mathrm{thin}}(\sum _U{\langle } W_\xi {\rangle }_{\xi <\lambda })$ . Indeed, for any $X\subseteq \kappa $ , $j_U(X)\in P(j_U(\kappa ))^{M_U}$ and therefore $j_{W^*}(j_U(X))\cap \lambda \in A$ . Similarly, for any function $f:\kappa \rightarrow \kappa $ , $j_U(f):j_U(\kappa )\rightarrow j_U(\kappa )\in M_U$ and therefore $j_{W^*}(j_U(f))(|A|^{M_{W^*}})<\lambda $ .

Proof We may assume $\sup (A) = \sup i"\kappa ^+$ and $A\cap i(\kappa ) = \kappa $ . Let U be the ultrafilter derived from j using A. Let $\bar A = [\text {id}]_U$ , and let $k_U : M_U\to M$ be the unique elementary embedding with $k_U\circ j_U = j$ and $k_U(\bar A) = A$ . Note that $\kappa $ and $i(\kappa )$ are in the range of $k_U$ since these ordinals are definable in M using A as a parameter: $\kappa $ is the least ordinal not in A, and $i(\kappa ) = |\sup (A)|^M$ . Therefore $k_U(\kappa ) = \kappa $ . Let $\rho $ be such that $k_U(\rho ) = i(\kappa )$ .

Let E be the extender of length $\rho $ derived from $j_U$ . Let $k_E : M_E\to M_U$ denote the unique factor embedding with $k_E \circ j_E = j_U$ and $k_E\restriction \rho = \text {id}.$

We will verify (1), (2), (3), and (4). Of course, (1) is true essentially by the definition of $k_E$ .

For (2), note that $\operatorname {\mathrm {crit}}(j_U) = \kappa $ since $k_U\circ j_U = j$ and $k_U(\kappa ) = \kappa $ . The fact that $\operatorname {\mathrm {crit}}(k_E) \geq \rho $ follows from the definition of $k_E$ . To see $\operatorname {\mathrm {crit}}(k_E) = \rho $ and $k_E(\rho ) = j_U(\kappa )$ , we will show thatFootnote ¹²

$$\begin{align*}\mathrm{Hull}^{M_U}(j_U"V\cup \rho)\cap j_U(\kappa) = \rho.\end{align*}$$

This will establish that $\operatorname {\mathrm {crit}}(k_E) =\rho $ and $k_E(\rho ) = j_U(\kappa )$ since $k_E$ is the inverse of the transitive collapse of $\text {Hull}^{M_U}(j_U"V\cup \rho )$ . To prove this equality, it suffices to show the inclusion $\text {Hull}^{M_U}(j_U"V\cup \rho )\cap j_U(\kappa )\subseteq \rho $ .

Since $k\circ i=j$ and since $\operatorname {\mathrm {crit}}(k)=i(\kappa )$ , we have $\text {Hull}^M(j"V\cup i(\kappa ))\subseteq k"N$ , and therefore $\text {Hull}^M(j"V\cup i(\kappa ))\cap j(\kappa ) \subseteq k"N\cap j(\kappa ) = i(\kappa )$ . Since

$$ \begin{align*}k_U"[\mathrm{Hull}^{M_U}(j_U"V\cup \rho)]\subseteq \mathrm{Hull}^M(j"V\cup i(\kappa)),\end{align*} $$

we have

$$\begin{align*}\mathrm{Hull}^{M_U}(j_U"V\cup \rho) \subseteq k_U^{-1}[\mathrm{Hull}^M(j"V\cup i(\kappa))].\end{align*}$$

In particular,

$$ \begin{align*}\mathrm{Hull}^{M_U}(j_U"V\cup \rho)\cap j_U(\kappa) &\subseteq k_U^{-1}[\mathrm{Hull}^M(j"V\cup i(\kappa))\cap j(\kappa)] \\ &= k_U^{-1}(i(\kappa)) = \rho.\end{align*} $$

Since $k_E(\rho ) = j_U(\kappa )> \rho $ and $k_E \circ j_E = j_U$ , it follows $\rho = j_E(\kappa )$ . Note also that $\rho < j_U(\kappa )$ , and so the fact that $k_E(\rho ) = j_U(\kappa )$ implies $k_E(\rho ) \neq \rho $ and hence $\operatorname {\mathrm {crit}}(k_E) = \rho $ .

For (3), the inner model $M_U$ is closed under $\kappa $ -sequences since it is the ultrapower of V by a $\kappa $ -complete ultrafilter. The inner model $M_E$ is closed under $\kappa $ -sequences by Lemma 5.3, since $\mathrm {cf}(\rho )> \kappa $ and ${}^\kappa \rho \subseteq M_E$ . To see that $\mathrm {cf}(\rho )> \kappa $ , note that $\mathrm {cf}(i(\kappa ))> \kappa $ since N satisfies that $i(\kappa )$ is regular and N is closed under $\kappa $ -sequences. Therefore M satisfies that $\mathrm {cf}(i(\kappa ))> \kappa $ . By the elementarity of $k_U$ , and since $k_U(\rho ) = i(\kappa )$ , $M_U$ satisfies $\mathrm {cf}(\rho )> \kappa $ . Here, we use that $k_U(\kappa ) = \kappa $ .

Finally, we verify (4). By elementarity of $k_U$ , since $|A| < i(\kappa )$ , we have $|\bar A| < \rho $ . So we just have to show that $j_E"\kappa ^+\subseteq \bar A.$ Suppose $\alpha \in j_E"\kappa ^+.$ We claim that $k_U(\alpha )\in \text {ran}(i)$ . Let ${\preceq }$ be a well order of $\kappa $ of order type $j_E^{-1}(\alpha )$ . Then $j_E({\preceq })$ has order type $\alpha $ . Note that

$$\begin{align*}k_U(j_E({\preceq})) = k_U(j_U({\preceq})\cap \rho) = j({\preceq})\cap i(\kappa) = i({\preceq}).\end{align*}$$

Thus $k_U(\alpha )$ , which is the order type of $k_U(j_E({\preceq }))$ is equal to the order type of $i({\preceq })$ , which is in the range of i. It follows that $(k_U\circ j_E)"\kappa ^+\subseteq i"\kappa ^+\subseteq A$ . Since $k_U(\bar {A}) = A$ , we conclude that $j_E"\kappa ^+\subseteq k_U^{-1}[A]\subseteq \bar A$ .

Proof We use the notation of 5.1. As before, we can fix an ordinal $\nu <j(\kappa )$ such that for some sequence $\vec {A} = \langle A_\alpha \rangle _{\alpha < \kappa }$ such that $A = j(\vec {A})_\nu $ and for some sequence $\vec {\kappa } = \langle \kappa _\alpha \rangle _{\alpha < \kappa }$ , $i(\kappa ) = j(\vec {\kappa })_\nu .$ Let $U=D(j,\nu )$ be the ultrafilter on $\kappa $ derived from j using $\nu $ . Since $\operatorname {\mathrm {crit}}(j)=\kappa $ , U is a $\kappa $ -complete ultrafilter over $\kappa $ . We will show $\neg \mathrm {Gal}({U,\kappa ,\kappa ^+})$ .

Let $\langle f_\xi \rangle _{\xi < \kappa ^+}$ denote the sequence of canonical functions on $\kappa $ (see Definition 2.16). For $\xi < \kappa ^+$ , define

$$\begin{align*}B_\xi = \{\alpha < \kappa : f_\xi(\kappa_\alpha)\in A_\alpha\}.\end{align*}$$

Note that $B_\xi \in U$ since

$$\begin{align*}j(B_\xi) = \{\alpha < j(\kappa) : j(f_\xi)(j(\vec \kappa)_\alpha) \in j(\vec A)_\alpha\}\end{align*}$$

and

$$\begin{align*}j(f_\xi)(j(\vec \kappa)_\nu) = j(f_\xi)(i(\kappa)) = k(i(f_\xi))(i(\kappa)) = i(\xi)\in A=j(\vec{A})_\nu.\end{align*}$$

The point here is that in N, $\vec g = i(\vec f)$ is the sequence of canonical functions on $i(\kappa )$ , and since $\operatorname {\mathrm {crit}}(k) = i(\kappa )$ , by Proposition 2.17, for any $\eta < i(\kappa ^+)$ , $k(g_\eta )(i(\kappa )) = \eta $ . The fact that $k(i(f_\xi ))(i(\kappa )) = i(\xi )$ follows from this observation when $\eta = i(\xi )$ (and thus $i(f_\xi ) = g_{i(\xi )}$ ).

Suppose $\sigma \subseteq \kappa ^+$ and $\bigcap _{\xi \in \sigma } B_\xi \in U$ . We must show that $|\sigma | < \kappa $ . Since $|A|^M < i(\kappa )$ , it suffices to show that $i(\sigma )\subseteq A$ : then $\text {ot}(i(\sigma )) < \text {ot}(A) < i(\kappa )$ , and hence $N\vDash \text {ot}(i(\sigma )) < i(\kappa )$ , which by elementarity implies $\text {ot}(\sigma ) < \kappa $ .

The proof that $i(\sigma )\subseteq A$ is similar to the calculation in the previous paragraph: Since $\bigcap _{\xi \in \sigma } B_\xi \in U$ , for all $\eta \in j(\sigma )$ , $j(\vec f)_\eta (i(\kappa )) \in A$ . Fix $\xi \in i(\sigma )$ , and we will prove that $\xi \in A$ . We have $k(\xi )\in j(\sigma )$ , so $j(\vec f)_{k(\xi )}(i(\kappa ))\in A$ . But $j(\vec f)_{k(\xi )} = k(g_\xi )$ , hence $k(g_\xi )(i(\kappa )) = \xi $ . It follows that $\xi \in A$ .

Proof Let U be a normal ultrafilter on $\kappa $ . Since $|P^{M_U}(P_\kappa (\kappa ^+))| = 2^\kappa $ , there is a transitive model M with

$$ \begin{align*}P^{M_U}(P_\kappa(\kappa^+))\subseteq M, \ |M|=2^\kappa.\end{align*} $$

By Hayut’s result [Reference Hayut19, Corollary 6], there is a transitive model N, an elementary embedding $j_0\colon M\rightarrow N$ , with $\operatorname {\mathrm {crit}}(j_0)=\kappa $ along with some $s\in N$ , $s\subseteq j_0(\kappa )^+$ such that $j_0"\kappa ^+\subseteq s$ with $|s|^N<j_0(\kappa )$ . Define $\mathcal {W}$ the $\kappa $ -complete ultrafilter on $P_\kappa (\kappa ^+)$ derived from $j_0$ and s. Note that $\mathcal {W}$ is fine since $j_0"\kappa ^+\subseteq s$ and it measures all the subsets of $P_\kappa (\kappa ^+)$ in $M_U$ . Let $j_{\mathcal {W}}\colon M_U\rightarrow M_{\mathcal {W}}$ be the ultrapower of $M_U$ by $\mathcal {W}$ defined in V, and $j\colon V\to M_{\mathcal {W}}$ be the embedding $j = j_{\mathcal {W}}\circ j_U$ . Let $\lambda = j_{\mathcal {W}}(\kappa )<j(\kappa ),$ and let E be the extender of length $\lambda $ derived from j.

Proof For any $X\subseteq \kappa $ , we have that

$$ \begin{align*}j(X)\cap \lambda=j_{\mathcal{W}}(j_U(X))\cap j_{\mathcal{W}}(\kappa)=j_{\mathcal{W}}(j_U(X)\cap\kappa)=j_{\mathcal{W}}(X).\end{align*} $$

Thus for all $\alpha <\lambda $ , $\alpha \in j(X)$ iff $\alpha \in j_{\mathcal {W}}(X)$ .

Finally, let $i\colon V\to N_E$ be the ultrapower of V by E and $A = [\text {id}]_{\mathcal {W}}\in M_{\mathcal {W}}$ . We claim that $i,j,A$ witness that $\kappa $ is a non-Galvin cardinal. Indeed, $i(\kappa )\geq \lambda $ . To see that $i(\kappa )\leq \lambda $ , we compute the ultrapower $i'$ of $M_U$ by E, and since $M_U$ is closed under $\kappa $ -sequences, it follows that $i(\kappa )=i'(\kappa )$ . By the previous claim, $j_{\mathcal {W}}$ also factors through $i'$ and thus $j_{\mathcal {W}}(\kappa )=k'(i'(\kappa ))\geq i'(\kappa )=i(\kappa )$ , as wanted.

By the usual argument about the derived extender, the factor map $k\colon N_E\rightarrow M_{\mathcal {W}}$ has critical point $i(\kappa )$ (see, for example, [Reference Jech21, Lemma 20.29(ii)]). Also, $M_{\mathcal {W}}\models |A|<j_{\mathcal {W}}(\kappa )=i(\kappa )$ and since $\mathcal {W}$ is fine, $j_{\mathcal {W}}"\kappa ^+\subseteq A$ .

Proof Note that $i(U)\in N_E$ is a normal measure on $i(\kappa )$ , let $X\in i(U)$ be any set, $k(X)\in j(U)=j_{\mathcal {W}}(j_U(U))$ . Note that $j_{\mathcal {W}}(j_U(U))$ is generated by $j_{\mathcal {W}}" j_U(U)$ by Theorem 6 and Corollary 8 of [Reference Blass11, Section 3]. Therefore, there is a set $Y\in j_U(U)$ such that $j_{\mathcal {W}}(Y)\subseteq k(X)$ . Since U is normal, there is a set $A\in U$ such that $j_U(A)\subseteq ^* Y$ and $j(A)\subseteq ^* k(X)$ , which in turn implies that $i(A)\subseteq ^* X$ . Now we note that $i(A)\in R$ , where R is the $N_E$ -ultrafilter (external) derived from k and $j_{\mathcal {W}}(\kappa )$ . We conclude that $i(U)\subseteq R$ and thus that $i(U)=R$ (as two $N_E$ -ultrafilters). So k factors through $j_{i(U)}$ and $k'\colon M_{i(U)}\rightarrow M_{\mathcal {W}}$ has critical point $>j_{\mathcal {W}}(\kappa )$ (since $k'(j_{\mathcal {W}}(\kappa ))=k'([\text {id}]_{i(U)})=k(id)(j_{\mathcal {W}}(\kappa ))=j_{\mathcal {W}}(\kappa )$ ). To conclude the claim, let $\alpha <\kappa ^+$ and $f\colon \kappa \rightarrow \kappa $ be the canonical function such that $j_U(f)(\kappa )=\alpha $ , then

$$ \begin{align*}j_{\mathcal{W}}(\alpha)=j_{\mathcal{W}}(j_U(f)(\kappa))=j(f)(j_{\mathcal{W}}(\kappa))=k(i(f))(j_{\mathcal{W}}(\kappa)).\end{align*} $$

By elementarity, $i(f)\colon i(\kappa )\rightarrow i(\kappa )$ is the canonical function for $i(\alpha )$ . Since $j_{i(U)}$ is the ultrapower by a normal ultrafilter over $j_{\mathcal {W}}(\kappa )$ , we conclude that

$$ \begin{align*}k(i(f))(j_{\mathcal{W}}(f))=k'(j_{i(U)}(i(f)))(j_{\mathcal{W}}(\kappa))=k'(i(\alpha))=i(\alpha)\end{align*} $$

as desired.

Proof Derive the extender E from $\lambda ,$ i.e., $E={\langle } E_a\mid a\in [\lambda ]^{<\omega }{\rangle }$ , where $E_a$ is an ultrafilter over $[\kappa ]^{|a|}$ defined by