Proof. Let $\mu <\lambda $ be a cardinal. Let G be $\mathbb {P}$ -generic over V. Let $f:\mu \longrightarrow \mathrm {On}$ enumerate a set of ordinals in $V[G]$ of size $\mu $ . Let $f=\dot {f}^G$ for some $\mathbb {P}$ -name $\dot {f}$ . Let $\phi $ be the sentence $\forall \kappa <\lambda \forall C\in [\mathrm {ssup}(\mathrm {rge}(f))]^\kappa (\mathrm {rge}(f) \not \subseteq C)$ . Suppose, towards a contradiction, $p\Vdash \phi $ . Let $N\in {\mathcal U}$ with $\dot {f}$ , p, $\mathbb {P}\in N$ , $\mu \subseteq N$ and ${\overline {\overline {N}}}=\kappa <\lambda $ .

For each $\alpha <\mu $ let $D_\alpha =\{ r\in \mathbb {P}| \exists \beta r\Vdash \dot {f}(\alpha )=\beta \text { or }r \text { is incompatible with }p\}$ . Then for each $\alpha <\mu $ we have that $D_\alpha $ is dense in $\mathbb {P}$ and $D_{\alpha }$ is definable from $\dot {f}$ , $p,$ and $\mathbb {P}$ and hence is an element of N.

For each $\alpha <\kappa $ let $C_\alpha = \{ \beta \in \mathrm {On}| \exists r\in \mathbb {P}\cap N r\Vdash \dot {f}(\alpha )=\beta \}$ , so $C_\alpha \subseteq N$ .

By the hypothesis on $\mathbb {P}$ let $p^*\le p$ be $(N,\mathbb {P})$ -generic. Then, by the definition of $(N,\mathbb {P})$ -genericity we have for each $\alpha <\mu $ that $N\cap D_\alpha $ is predense below $p^*$ : if $q\le p^*$ there is some $r\in N$ and some $\beta \in \mathrm {On}$ such that $r\Vdash \dot {f}(\alpha )=\beta $ and there is some $s\in \mathbb {P}$ with $s\le q$ , r. Hence for all $\alpha <\mu $ and for all $q\le p^*$ we have that there is some $s\le q$ such that $s\Vdash \dot {f}(\alpha )\in C_\alpha $ (since r forces this) and thus $p^*\Vdash \dot {f}(\alpha )\in C_\alpha $ .

Set $C=\bigcup _{\alpha <\mu }C_\alpha $ . It is clear that $C\in V$ and from the previous paragraph we have that $p^*\Vdash \mathrm {rge}(\;\dot {f})\subseteq C$ . Moreover, as $C_\alpha \subseteq N$ for each $\alpha <\mu $ we have that ${\overline {\overline {C}}}\le \mu .\kappa = \kappa <\lambda $ . This gives the sought after contradiction. $\blacktriangle _{2.3}$

Proof. As $M\in N$ we have $M\cap \mathrm {On} \in N$ . Let $\gamma =\mathrm {otp}(M\cap \lambda )$ . As $H_\lambda \models \gamma =\mathrm {otp}(M\cap \lambda )$ and $N\prec H_\lambda $ one also has $N\models \gamma =\mathrm {otp}(M\cap \lambda )$ . Since $\gamma <\delta ^N$ we have $\mathrm {otp}((M\cap \lambda )^N)=\mathrm {otp}(M\cap \lambda )$ , and hence $(M\cap \lambda )^N =\{ \alpha \in M\cap \lambda | \alpha \in N\} =M\cap \lambda $ . $\blacktriangle _{2.7}$

Proof. By Lemma 2.7 we have $M\cap \lambda \subseteq N$ . Let $\langle x_\alpha | \alpha <\lambda \rangle $ enumerate $(\lambda )^{<\kappa ^{+}}$ . For all $\alpha <\lambda $ we have that $\alpha \in M$ if and only if $x_\alpha \in ((\lambda )^{<\kappa ^{+}})^M$ , and similarly for N. Thus $((\lambda )^{<\kappa ^{+}})^M\subseteq ((\lambda )^{<\kappa ^{+}})^N$ . $\blacktriangle _{2.9}$

Proof. Immediate. $\blacktriangle _{2.11}$

Proof. The proof is by induction on j for each fixed i. If $j=i+1$ the result is immediate from the definition of ${\mathcal {F}}_{ii+1}$ . (Recall that if ${\mathcal {F}}_{ii+1}=\{ \mathrm {id},h_i\}$ is an almost exact amalgamation pair then $\mathrm {ssup}(\mathrm {rge}(h_i))=\theta _i\cup h_i``\theta _i < \theta _{i+1}$ .) If $j=j'+1$ for some $j'>i$ then let $f=g\cdot h$ for some $h\in {\mathcal {F}}_{ij'}$ and $g\in {\mathcal {F}}_{j'j}$ . Again by the definition of ${\mathcal {F}}_{j'j}$ we have that g is not cofinal and hence neither is f. Finally, if j is a limit ordinal then choose any $j'\in (i,j)$ ; then $f=g\cdot h$ for some $h\in {\mathcal {F}}_{ij'}$ and $g\in {\mathcal {F}}_{j'j}$ . By the inductive hypothesis h is not cofinal, hence neither is f. $\blacktriangle _{3.2}$

Proof. By the definition of an almost exact sequence $\mathrm {otp}(N \cap \lambda )$ should be a successor ordinal, contrary to the fact that $(N,\in )\prec (H_{\lambda },\in ).$ $\blacktriangle _{3.3}$

Proof. By induction on j for each i. If $j=i+1$ the lemma is immediate from the two possible cases for the structure of ${\mathcal {F}}_{ii+1}$ . If $j=i'+1$ for some $i'>i$ , note that one can factor f, g as $f=f_1\cdot f_0$ and $g=g_1\cdot g_0$ with $f_0$ , $g_0\in {\mathcal {F}}_{ii'}$ and $f_1$ , $g_1\in {\mathcal {F}}_{i'i'+1}$ . If ${\mathcal {F}}_{i'i'+1}$ is a singleton then this instance of the lemma is immediate from the lemma for i and $i'$ . Otherwise, ${\mathcal {F}}_{i'i'+1}$ is an amalgamation pair with splitting point $\sigma _{i'}$ and either $f_1=g_1$ , in which case the lemma is again immediate by the induction hypothesis, or $f_1\ne g_1$ and hence $\mathrm {ssup}(f_0``\xi )=\mathrm {ssup}(g_0``\tau )\le \sigma _{i'}$ and so the lemma is once again true. If $j\le \zeta $ is a limit ordinal then there is some $k\in [i,j)$ , $f' \in {\mathcal {F}}_{ik}$ , $g' \in {\mathcal {F}}_{ik}$ and $h \in {\mathcal {F}}_{kj}$ such that $f=h\cdot f'$ and $g=h\cdot g'$ . If $\mathrm {ssup}(f``\xi )=\mathrm {ssup}(g``\tau )$ then $\mathrm {ssup}(f'``\xi )=\mathrm {ssup}(g'``\tau )$ . By the inductive hypothesis the lemma is true for k, $f'$ and $g'$ . Hence, applying h on both sides, it is true of j, f and g. If $j=\zeta $ then $\mathrm {ssup}(F\cdot f``\xi )=\mathrm {ssup}(F\cdot g``\tau )$ if and only if $\mathrm {ssup}(f``\xi )=\mathrm {ssup}(g``\tau )$ , and so the lemma for $F\cdot f$ and $F\cdot g$ is true by the lemma for f and g. $\hphantom {w}$ $\blacktriangle _{3.4}$

Proof. For $M\in {\mathcal N}^p$ we have $\pi ^M\upharpoonright \mathrm {On} = F^p\cdot f^M$ . Note that $\pi ^M\upharpoonright \delta ^M=\mathrm {id}$ , $\pi ^M(\delta ^M)=\kappa ^{+}$ and that $\forall i< i^M \theta _i<\delta ^M$ . Observe that it is now immediate that for $i\in \{ 1,2,3,4,5\}$ we have that $\bullet _i$ holds if and only if $\bullet ^{\prime }_i$ does, where the $\bullet ^{\prime }_i$ are given by

• $\bullet ^{\prime }_1$ $\forall i\le j <i^M {\mathcal {F}}^p_{ij}\subseteq \overline {M},$

• $\bullet ^{\prime }_2$ $\forall i <i^M {\mathcal {F}}^p_{ii^M}\subseteq \overline {M},$

• $\bullet ^{\prime }_3$ $\forall N\in {\mathcal N}^p \forall \gamma <\delta ^N \forall g\in (\delta ^N)^{\gamma } (i^N\le i^M \Longrightarrow ,$

  • $(g\in \overline {N} \Longleftrightarrow g\in \overline {M})),$

• $\bullet ^{\prime }_4$ $\forall N\in {\mathcal N}^p \forall \gamma <\delta ^N \forall g\in (\theta _{i^N})^{\gamma } (i^N=i^M \Longrightarrow $

  • $(g\in \overline {N} \Longleftrightarrow g\in \overline {M}))$ ,

• $\bullet ^{\prime }_5$ $\forall N\in {\mathcal N}^p \forall \gamma <\delta ^N \forall g\in (\theta _{i^N})^{\gamma } (i^N<i^M \Longrightarrow $

  • $(g\in \overline {N} \Longrightarrow g\in \overline {M})).$

  The conclusion now follows as ${\circ }_{1}$ is the conjunction of $\bullet ^{\prime }_1$ and $\bullet ^{\prime }_2$ and ${\circ }_{2}$ is the conjunction on $\bullet ^{\prime }_3$ , $\bullet ^{\prime }_4,$ and $\bullet ^{\prime }_5$ . $\blacktriangle _{3.6}$

Proof. By $\bullet _1$ for p applied to $\theta _{i_N'}$ and N one has $\mathrm {otp}(N'\cap \lambda )<\delta ^N$ . By hypothesis $N'\in N$ , so, by Lemma 1.7, $\mathrm {rge}(F\cdot f^{N'}) = N'\cap \lambda \subseteq N\cap \lambda $ $= \mathrm {rge}(F\cdot f^{N})$ . As F is order preserving $\mathrm {rge}(\;f^{N'})\subseteq \mathrm {rge}(\;f^N)$ . By the last property of ${\mathcal {F}}_{i_{N'}\zeta }$ given in Definition 3.5, we may factor $f^{N'}$ as $h\cdot f$ where $h\in {\mathcal {F}}_{i_N\zeta }$ and $f\in {\mathcal {F}}_{i_{N'}i_N}$ . Then $h\upharpoonright \mathrm {ssup}(\mathrm {rge}(\;f)) = f^N\upharpoonright \mathrm {ssup}(\mathrm {rge}(\;f))$ , by Lemma 3.4, so $f^{N'} = f^N\cdot f$ . $\blacktriangle _{3.7}$

Proof. Let $x\in H_\chi $ and $p\in \mathbb {P}$ . Suppose N is a very good model with x, $p\in N$ , $\{ g\in (\theta ^p_{i^M})^{\gamma }| M\in {\mathcal N}^p \&\ \gamma <\delta ^M \&\ F^p\cdot f^M\cdot g\in M \} \subseteq N$ , and $\mathrm {cf}(N\cap \lambda )=\kappa $ . (It is easy to find such an N by the downward Lowenheim-Skolem theorem.)

As $\theta ^p_{\zeta ^p}\in N\cap \kappa ^{+}$ one has that $\theta ^p_{\zeta ^p}\subseteq N$ . Since $F^p\in N$ we have that $F^p``\theta ^p_{\zeta ^p}\subseteq N$ .

Let $\theta ^*=\mathrm {otp}(N\cap \lambda )$ , let $F^*$ be the inverse of the transitive collapse of $N\cap \lambda $ and let $f^*={F^*}^{-1}\cdot F^p$ . Note that the definition of $f^*$ makes sense by the observation of the previous paragraph. Now define $p^*=(S^*,{\mathcal {F}}^*,{\mathcal N}^*)$ by $S^*=\langle \langle \theta ^p_{i} | i\le \zeta ^p \rangle \frown \theta ^*, \langle {\mathcal {F}}^p_{ij} | i\le j\le \zeta ^p \rangle \frown \langle \{ f^*\cdot f | f\in {\mathcal {F}}^p_{i\zeta ^p} \} | i\le \zeta ^p\rangle \rangle $ , ${\mathcal {F}}^*=\{ (F^*\cdot f, \theta ^{p^*}_i) | i\le \zeta ^p+1 \&\ f\in {\mathcal {F}}^*_{i\zeta ^p+1} \}$ and ${\mathcal N}^*={\mathcal N}\cup \{ N\}$ .

Proof. The witnessing pair for N is $(\zeta ^{p^*},\mathrm {id})$ . The only thing to check is that N satisfies $\bullet _1$ – $\bullet _5$ .

$\bullet _1$ : It is immediate that ${\mathcal {F}}^*_{ij}\subseteq N$ for all $i\le j\le \zeta ^p$ as ${\mathcal {F}}^*_{ij}={\mathcal {F}}^p_{ij}$ for all $i\le j\le \zeta ^p$ and ${\mathcal {F}}^p_{ij}\subseteq N$ since $p\in N$ , $\zeta ^p$ and each ${\mathcal {F}}^p_{ij}$ has size ${<}\kappa $ and ${{\hspace {-0.0001pt}}}^{<\kappa }N \subseteq N$ .

$\bullet _2$ : It is also immediate that $F^*\cdot f^*\cdot g =F^p\cdot g\in N$ for all $g\in {\mathcal {F}}_{i\zeta ^p}$ as for each $g\in {\mathcal {F}}_{i\zeta ^p}$ one has $g\in N$ , as mentioned in the proof of $\bullet _1$ , and $F^p\in N$ (since ${\mathcal {F}}^p\in N$ has size ${<}\kappa $ and $^{<\kappa }N\subseteq N$ , and $F^p=F^p\cdot \mathrm {id}\in {\mathcal {F}}^p$ ).

$\bullet _3$ – $\bullet _5$ : Lemma 2.9 gives $\bullet _3$ , $\bullet _4$ is vacuously true, and $\bullet _5$ is true by the construction of N. $\blacktriangle _{4.2}$

Proof. Let $D\in N$ be a dense subset of $\mathbb {P}$ .

Let $q\le p^*$ , and let $m^*\le \zeta ^{q}$ be such that $\theta ^q_{m^*} = \theta ^*$ .

By the definition of small SMS we have two cases: $m^*=0$ or $m^*$ is a successor ordinal. For if $m^*$ were to be a limit ordinal we would have that $\mathrm {cf}(\theta ^*)<\kappa $ , contradicting the fact that $\mathrm {cf}(N\cap \lambda )=\kappa $ .

In the former case we take $q\upharpoonright N= \mathbb {1}$ . Clearly in this case $q\upharpoonright N\in \mathbb {P}$ .

Now suppose (until the end of Claim (4.4)) that we are in the other case, so that $m^* = m +1$ for some $m< \zeta ^{q}$ . In this case $\theta ^q_{m+1}=\theta ^*$ . Note, as remarked in Lemma 3.3, that we must have that ${\mathcal {F}}^q_{mm+1}$ is the singleton $\{ f^q_{m}\}$ .

Define $q\upharpoonright N$ as follows. Let $S^{q\upharpoonright N}= \langle \langle \theta ^q_i| i\le m\rangle ,\langle {\mathcal {F}}^q_{ij}| i\le j\le m\rangle \rangle $ . Let $F^{q\upharpoonright N}= F^*\cdot f^q_{m}$ and ${\mathcal {F}}^{q\upharpoonright N}= \{ (F^*\cdot f^q_m \cdot f,\theta ^q_i)| i\le m \&\ f\in {\mathcal {F}}^q_{im}\}$ . Let ${\mathcal N}^{q\upharpoonright N}={\mathcal N}^q \cap N$ . Set $q\upharpoonright N=(S^{q\upharpoonright N},{\mathcal {F}}^{q\upharpoonright N},{\mathcal N}^{q\upharpoonright N})$ .

Proof. We start by showing that the ‘ ${\mathcal {F}}$ ’-component of $q\upharpoonright N$ is exactly the intersection of the ‘ ${\mathcal {F}}$ ’-component of q with N.

Proof. In one direction, if $i\le m$ and $f\in {\mathcal {F}}^q_{im}$ one has that

$$\begin{align*}F^{q\upharpoonright N} \cdot f = F^* \cdot f^q_m \cdot f = F^q \cdot f^{p^*q} \cdot f^q_m \cdot f,\end{align*}$$

where $(m+1,f^{p^*q})$ witnesses that $N\in {\mathcal N}^q$ and the second equality holds since $f^{p^*q}$ witnesses that $q \le p^*$ and so by the definition of the $\le $ -relation $F^* = F^q\cdot f^{p^*q}$ . Consequently $F^{q\upharpoonright N} \cdot f$ is of the form $F^q\cdot g$ for some $g\in {\mathcal {F}}^q_{i\zeta ^q}$ and $F^{q\upharpoonright N} \cdot f\in N$ (by $\bullet _2$ for N and q). Thus ${\mathcal {F}}^{q\upharpoonright N}\subseteq {\mathcal {F}}^q\cap N$ .

In the other direction, suppose\ $i\le \zeta ^q$ , $g\in {\mathcal {F}}^q_{i\zeta ^q}$ and $(F^q\cdot g,\theta _i)\in N$ . Then $i\le m$ , since $\theta _i\in N$ , and so $F^q\cdot g``\theta _i\subseteq N$ . Write g as $g= l\cdot f^q_{m}\cdot f$ where $f\in {\mathcal {F}}^q_{im}$ and $l\in {\mathcal {F}}^q_{m+1\zeta ^q}$ . Then $F^q\cdot l\upharpoonright (f^q_m \cdot f``\theta ^q_i) = F^q\cdot f_N\upharpoonright (f^q_m \cdot f``\theta ^q_i)$ , where $(m+1,f_N)$ witnesses that $N\in {\mathcal N}^q$ (and hence $f_N$ is the map $f^{p^*q}$ ), by Lemma 1.10. Thus $F^q\cdot g = F^q\cdot f^{p^*q}\cdot f^q_m \cdot f = F^{*}\cdot f^q_m\cdot f = F^{q\upharpoonright N}\cdot f$ . Hence ${\mathcal {F}}^q\cap N\subseteq {\mathcal {F}}^{q\upharpoonright N}$ . $\blacktriangle _{4.5}$

Proof. Let $m=\zeta ^{q\upharpoonright N}$ . Let $k:m+1\longrightarrow \zeta ^s+1$ witness that $s\le q\upharpoonright N$ . Extend k so that $\mathrm {dom}(k) =\zeta ^q+1$ by setting $k(m+1+i)=\zeta ^s+1+i$ for $i\in \mathrm {otp}((m, \zeta ^q])$ .

Set $\zeta ^r=\zeta ^s+ \mathrm {otp}((m, \zeta ^q])$ and $\theta ^r_i=\theta ^s_i$ for $i\le \zeta ^s$ and $\theta ^r_{\zeta ^s+1+i}=\theta ^q_{m+1+i}$ for $i\in \mathrm {otp}((m, \zeta ^q])$ . The sequence $\langle \theta ^r_i| i<\zeta ^r\rangle $ is increasing because $\theta _{\zeta ^s}\in N$ and hence $\theta _{\zeta ^s}<\delta ^N<\mathrm {otp}(N\cap \lambda )=\theta ^q_{m+1}$ .

Let $F^q\cdot f_N$ be the inverse of the transitive collapse of $\mathrm {otp}(N\cap \lambda )$ , where $(m+1,f_N)\in {\mathcal {F}}^q_{m+1\zeta ^q}$ witnesses that $N\in {\mathcal N}^q$ .

Again, as $s\in N$ one has that $F^s\in N$ and hence $F^s``\theta _{\zeta ^s}\subseteq N$ . Thus one can make the following definition.

For $i\le j\le \zeta ^s$ let ${\mathcal {F}}^r_{ij}={\mathcal {F}}^s_{ij}$ . Let ${\mathcal {F}}^r_{\zeta ^s\zeta ^s+1} = \{ f_N^{-1}\cdot ({F^q})^{-1}\cdot F^s\}$ . Hence, in the notation for small SMSs from Definition (3.1), $f^r_{\zeta ^s} = f_N^{-1}\cdot ({F^q})^{-1}\cdot F^s$ . For $1\le i\le j\le \mathrm {otp}((m, \zeta ^q])$ let ${\mathcal {F}}^r_{\zeta ^s+i\zeta ^s+j}={\mathcal {F}}^q_{ij}$ . Finally, for $i\le j\le \zeta ^r$ with $i\le \zeta ^s$ and $j>\zeta ^s$ set ${\mathcal {F}}^r_{ij} = \{ f \cdot f^r_{\zeta ^s} \cdot g | g\in {\mathcal {F}}^r_{i\zeta ^s} \&\ f\in {\mathcal {F}}^r_{\zeta ^s+1j} \}$ .

Then $S^r= \langle \langle \theta ^r_i| i\le \zeta ^r\rangle , \langle {\mathcal {F}}^r_{ij}| i\le j\le \zeta ^r\rangle \rangle $ is a small SMS.

Next, set $F^r=F^q$ and ${\mathcal {F}}^r= \{ (F^r\cdot f, \theta _i)| i\le \zeta ^r\&\ f\in {\mathcal {F}}_{i\zeta ^r}\}$ .

Lastly, set ${\mathcal N}^r={\mathcal N}^s\cup {\mathcal N}^q$ .

In order to finish the proof we must show that $\bullet _1$ – $\bullet _5$ hold for r. If $M\in {\mathcal N}^s$ then $\bullet _1$ – $\bullet _5$ hold for r since they do for s. So suppose $M\in {\mathcal N}^q\setminus {\mathcal N}^s$ .

We shall write $i^M$ for $(i^M)^r$ , $i^N$ for $(i^N)^r$ , and so on, in the remainder of the proof.

$\bullet _1$ . Let $f\in {\mathcal {F}}^r_{ij}$ with $i\le j<i^M$ .

Case 1(i). $i\le j< i^N \le i^M$ . Then $f\in {\mathcal {F}}^s_{ij}$ . As $s\in N$ we have ${\mathcal {F}}^s_{ij}\in N$ and so $f\in N$ (because both s and ${\mathcal {F}}^s_{ij}$ are of size ${<}\kappa $ ). As $s\in N$ we have $\theta ^s_j<\delta ^N$ and so $f\in ((\kappa ^{+})^{<\kappa ^{+}})^N$ . As $i^N\le i^M$ we have $((\kappa ^{+})^{<\kappa ^{+}})^N\subseteq ((\kappa ^{+})^{<\kappa ^{+}})^M$ by $\bullet _3$ for q. Thus $f\in M$ . $\vphantom {i}$ $\blacktriangle _{\text {(Case 1(i))}}$

Case1(ii). $i\le j = i^N<i^M$ . In this case we can factor f as $f=f_m\cdot g$ with $g\in {\mathcal {F}}^r_{im}={\mathcal {F}}^s_{im}$ and $f_m$ is the (unique) map in ${\mathcal {F}}^r_{mm+1}$ . By definition $f_m = (F^r\cdot f^N)^{-1}\cdot F_s$ , so $F^r\cdot f^N\cdot f_m=F^s$ . Since $s\in N$ and s, ${\mathcal {F}}^s$ are of size ${<}\kappa $ , we have that $F^s\in N$ . As $\theta _{\zeta ^s}<\delta ^N$ , and by the case hypothesis that $i^N<i^M$ , we can apply $\bullet _5$ for q (to N and M), obtaining that $f_m\in M$ . However, we also have that $g\in M$ by Case 1(i). Hence $f=f_m\cdot g\in M$ . $\vphantom {i}$ $\blacktriangle _{\text {(Case 1(ii))}}$

Case 1(iii). $i<i^N<j<i^M$ . In this case we can factor f as $f=h\cdot g$ with $g\in {\mathcal {F}}^r_{ii^N}$ and $h\in {\mathcal {F}}^r_{i^Nj}$ . Then $h\in M$ by $\bullet _1$ for q and $g\in M$ by Case 1(ii). Hence $f\in M$ . $\blacktriangle _{\text {(Case 1(iii))}}$

Case 1(iv). $i^N\le i\le j<i^M$ . In this case ${\mathcal {F}}^r_{ij}={\mathcal {F}}^q_{ij}$ and so $f\in M$ by $\bullet _1$ for q.

$\blacktriangle _{\text {(Case 1(iv)), (Verification of } \bullet _1)}$

$\bullet _2$ . Let $i<i^M$ and $f\in {\mathcal {F}}^r_{ii^M}$ .

Case2(i). $i^N\le i$ . Then $\bullet _2$ for f and M holds by $\bullet _2$ for q. $\blacktriangle _{\text {(Case 2(i))}}$

Case 2(ii). $i<i^N<i^M\!{.}$ Factor f as $f=h\cdot f_m\cdot g$ where $h\in {\mathcal {F}}^r_{i^Ni^M}$ and $g\in {\mathcal {F}}^r_{im}$ . Then $F^r\cdot f^M\cdot h\in M$ by $\bullet _2$ for q. Also, $f_m\cdot g\in M$ by $\bullet _1$ for r. So $F^r\cdot f^M\cdot h\cdot f_m\cdot g=F^r\cdot f^M\cdot f\in M.$ $\blacktriangle _{\text {(Case 2(ii))}}$

Case 2(iii). $i<i^N=i^M$ . Factor f as $f=f_m\cdot g$ where $g\in {\mathcal {F}}^r_{im}$ . As in Case 2(ii), $g\in M$ by $\bullet _1$ for r. As in the proof of Case 1(ii), $F^r\cdot f^N\cdot f_m = F^s\in N$ . So $F^r\cdot f^M\cdot f_m\in M$ by $\bullet _4$ for q. $\blacktriangle _{\text {(Case 2(iii)), (Verification of } \bullet _2)}$

Now let $K\in {\mathcal N}^r$ . If $K\in {\mathcal N}^q$ then each of $\bullet _1$ – $\bullet _5$ holds for r by the corresponding property for q. So suppose $K\in {\mathcal N}^s\setminus {\mathcal N}^q$ .

$\bullet _3$ . By Lemma 2.9 we have that $((\lambda )^{<\kappa ^{+}})^K\subseteq N$ . By $\bullet _3$ for q we have $((\kappa ^{+})^{<\kappa ^{+}})^N\subseteq ((\kappa ^{+})^{<\kappa ^{+}})^M$ . Thus $((\kappa ^{+})^{<\kappa ^{+}})^K\subseteq ((\kappa ^{+})^{<\kappa ^{+}})^M$ . $\blacktriangle _{\text {(Verification of } \bullet _3)}$

$\bullet _4$ . As $i^K<i^M$ there is nothing to check here. $\blacktriangle _{\text {(Verification of } \bullet _4)}$

$\bullet _5$ . Suppose $\gamma <\delta ^K$ , $g\in (\theta _{i^K})^\gamma $ and $F^r\cdot f^K\cdot g\in K$ . By Lemma 2.9, since s, ${\mathcal N}^s$ and K are all elements of N, we have $((\lambda )^{<\kappa ^{+}})^K\subseteq N$ , and so $F^r\cdot f^K\cdot g\in N$ .

We also have that $f^K=f^N\cdot f_m\cdot (\;f^K)^s$ , so $F^r\cdot f^K = F^r\cdot f^N\cdot (\;f_m\cdot (\;f^K)^s)$ . Thus $F^r\cdot f^K\in N$ by $\bullet _2$ for r applied to N and $f_m\cdot (\;f^K)^s$ .

Hence g, $ = (F^r\cdot f^N)^{-1} \cdot (F^r\cdot f^N\cdot g)$ , $\in N$ . But then $g\in ((\kappa ^{+})^{<\kappa ^{+}})^N$ and so $g\in M$ by $\bullet _3$ for q. $\blacktriangle _{\text {(Verification of } \bullet _5)}$

Thus $r=(S^r,{\mathcal {F}}^r,{\mathcal N}^r)\in \mathbb {P}$ . $\blacktriangle _{4.6}$

Proof. Let $\chi $ be a cardinal such that $(2^{{\overline {\overline {\mathbb {P}}}}})^{+} \le \chi $ , $x\in H_\chi $ and $\mu <\lambda $ . Suppose that D is a dense set in $\mathbb {P}$ and x, p, D, $\mathbb {P}\in N$ where $N\prec H_{\chi }$ , $\mu \le \delta ^N=N\cap \lambda \in \lambda $ and $\mathrm {cf}(\delta ^N)\ge \kappa ^{+}$ , ${\overline {\overline {N}}}<\lambda $ , and $N^{\kappa }\subseteq N$ .

Proof. As $\theta ^q_{\zeta ^q}<\kappa ^{+}$ we have that $F^q``\theta ^q_{\zeta ^q}$ is of size at most $\kappa $ , so since $N^\kappa \subseteq N$ we have that $Y=N\cap F^q``\theta ^q_{\zeta ^q} \in N$ . Also, as $N\cap \lambda \in \lambda $ we have that Y is an initial segment of $F^q``\theta ^q_{\zeta ^q}$ . Also, again as $N^\kappa \subseteq N$ , each $X_i^q$ for $i\in Z^q$ is an element of N.

Let $\phi (x)$ be the formula which is the conjunction of the following subformulae: $x\in \mathbb {P}$ ; $S^x=S^q$ ; $x\le p$ ; Y is an initial segment of $F^x``\theta ^x_{\zeta ^x}$ ; $Z^x=Z^q$ ; and $\langle X_i^x| i\in Z^x\rangle =\langle X_i^q| i\in Z^q\rangle $ .

Since q witnesses $H_{\chi }\models \exists x \phi (x)$ there is some $s\in N$ such that $N\models \phi (s)$ .

It is easy now to amalgamate the morass parts of s and q - our comments about Y have ensured that we have a head-tail-tail amalgamation which we can use in the morass. We give the details in the next-but-one paragraph.

(We believe the term ‘head-tail-tail’ amalgamation was first coined in [Reference Baumgartner and Shelah4]. The general idea is easily grasped: it is amalgamation of two objects, in a situation where is there some notion of ordering on the elements of their union, where the two objects share with a common “initial segment” and where all of the remainder of one of the objects “comes before” all of the remainder of the other.)

In more detail, define r by setting $\zeta ^r=\zeta ^q+1$ , $S^r\upharpoonright \zeta ^r = S^q = S^s$ , $\theta ^r_{\zeta ^r}=\mathrm {otp}(\mathrm {rge}(F^s)\cup \mathrm {rge}(F^q))+1$ , ${\mathcal {F}}^r_{\zeta ^q\zeta ^r}=\{ \mathrm {id}, h^r_{\zeta ^q}\}$ , where the splitting point of ${\mathcal {F}}^r_{\zeta ^q\zeta ^r}$ , $\sigma ^r_{\zeta ^q}$ , is $\mathrm {otp}(Y)$ , ${\mathcal {F}}^r_{i\zeta ^r}=\{ f\cdot g| f\in {\mathcal {F}}^r_{\zeta ^q\zeta ^r}\&\ g\in {\mathcal {F}}^r_{i\zeta ^q}\}$ for $i<\zeta ^q$ , $F^r\upharpoonright \mathrm {otp}(\mathrm {rge}(F^s)\cup \mathrm {rge}(F^q))$ is the inverse of the transitive collapse of $\mathrm {rge}(F^s)\cup \mathrm {rge}(F^q)$ , $F^r (\mathrm {otp}(\mathrm {rge}(F^s)\cup \mathrm {rge}(F^q)))=\mathrm {ssup}(\mathrm {rge}(F^s)\cup \mathrm {rge}(F^q))$ , and ${\mathcal N}^r={\mathcal N}^s\cup {\mathcal N}^q$ .

Proof. The property ${\circ }_{1}$ holds for r immediately from ${\circ }_{1}$ for q and s. We will be finished if we show that ${\circ }_{2}$ also holds for r.

Suppose that $M\in {\mathcal N}^q$ and $K\in {\mathcal N}^s$ . (In all other cases ${\circ }_{2}$ holds by virtue of holding for q or s.)

Since $Z^s=Z^q$ there are $M'\in {\mathcal N}^s$ and $K'\in {\mathcal N}^q$ such that $i^M=i^{M'}$ and $i^K=i^{K'}$ and, as s, $q\in \mathbb {P}$ we have that ${\circ }_{2}$ holds for each of the two pairs K, $M'$ and $K'$ , M. Thus, if $i^K\le i^M$ we have that $X^q_{i^K}=X^s_{i^K}$ implies that ${\circ }_{2}$ holds for K, M, while if $i^K\ge i^M$ then $X^q_{i^M}=X^s_{i^M}$ implies that ${\circ }_{2}$ holds for K, M. $\blacktriangle _{5.4}$

Proof. As $2^{\kappa }\le \lambda $ the forcing has the $\lambda ^{+}$ chain condition. $\blacktriangle _{5.5}$

Proof. Suppose that ${\mathcal A}$ is an antichain in $\mathbb {P}$ and ${\overline {\overline {\mathcal {A}}}}=\lambda $ . Let $\chi $ be large enough so that ${\mathcal A}$ , $\mathbb {P}\in H_\chi $ (e.g., $\chi = (2^{\lambda })^{+}$ ). Let $N\prec H_{\chi }$ be such that $\delta ^N=N\cap \lambda \in \lambda $ and $\mathrm {cf}(\delta ^N)\ge \kappa ^{+}$ , ${\overline {\overline {N}}}<\lambda $ , and $N^{\kappa }\subseteq N$ and ${\mathcal A}$ , $\mathbb {P}\in N$ . Choose $q\in {\mathcal A}\setminus N$ . Since every condition in $\mathbb {P}$ is $(N, \mathbb {P})$ -generic, q certainly is. Thus there is some $s\in {\mathcal A}\cap N$ compatible with q. This contradicts the assumption that ${\mathcal A}$ is an antichain. Thus we have shown that $\mathbb {P}$ contains no antichains of size $\lambda $ . $\blacktriangle _{5.6}$

Proof. Let $\mu $ be a cardinal with $\mu <\kappa $ . Let $\langle p_\alpha | \alpha <\mu \rangle $ be a descending sequence of conditions from $\mathbb {P}$ . For $\alpha <\beta <\mu $ let $k_{\alpha \beta }$ witness that $p_\beta \le p_\alpha $ . Note that if $\alpha <\beta <\gamma <\mu $ then $k_{\alpha \gamma }=k_{\beta \gamma }\cdot k_{\alpha \beta }$ .

We start by constructing a small SMS segment S into which each $S^{p_\alpha }$ is embedded.

Define two relations $\sim $ and $<$ on the set $\{ (\alpha ,i)| \alpha <\mu \&\ i\le \zeta ^{p_\alpha }\}$ , by $(\alpha ,i)\sim (\beta ,j)$ if $\alpha \le \beta $ and $k_{\alpha \beta }(i)=j$ or $\beta \le \alpha $ and $k_{\beta \alpha }(j)=i$ , and, similarly, $(\alpha ,i) < (\beta ,j)$ and $\alpha \le \beta $ and $k_{\alpha \beta }(i)<j$ or $\beta \le \alpha $ and $k_{\beta \alpha }(j)<i$ . Clearly $\sim $ is an equivalence relation and $<$ is a linear ordering which respects $\sim $ , inducing a linear ordering on the $\sim $ -equivalence classes ${\mathcal E}=\{ [(\alpha ,i)]_\sim | \alpha <\mu \&\ i\le \zeta ^{p_\alpha }\}$ .

For $x=[(\alpha ,i)]_\sim \in {\mathcal E}$ let $\theta _x = \theta ^{p_\alpha }_i$ . Note, by the definition of $\sim $ and of $\le _{\mathbb {P}}$ , that $\theta _x$ is independent of the choice of the pair $(\alpha ,i)$ . Note, also, that $x = [(\alpha ,i)]_{\sim } < y = [(\beta ,j)]_{\sim }$ if and only if $\theta _x<\theta _y<\kappa ^{+}$ . Consequently, $<$ well-orders ${\mathcal E}$ .

For x, $y\in {\mathcal E}$ define ${\mathcal {F}}_{xy}=\bigcup \{ f\in {\mathcal {F}}^{p_\alpha }_{ij}| \alpha <\mu \&\ x=[(\alpha ,i)]_\sim \&\ y=[(\alpha ,j)]_\sim \}$ . Let ${\mathcal {F}}_x= \{ F^{p_\alpha }\cdot f| x=[(\alpha ,i)]_\sim \&\ f\in {\mathcal {F}}^{p_\alpha }_{i\zeta ^p_{\alpha }} \}$ .

Let $\{ x_{\xi }| \xi <\zeta '\}$ be the $<$ -increasing ordering of ${\mathcal E}$ . We must have that $\zeta '<\kappa $ since $\overline {\overline {\{ (\alpha ,i)| \alpha <\mu \&\ i\le \zeta ^{p_\alpha }\}}} < \kappa $ .

Let $\zeta = \bigcup \zeta '$ . That is, if there is a $<$ -maximal element of ${\mathcal E}$ , and so $\zeta '$ is a successor, set $\zeta $ to be the predecessor of $\zeta '$ ; otherwise, let $\zeta =\zeta '$ .

For $\alpha <\mu $ define $k_\alpha :\zeta ^{p_{\alpha }}+1\longrightarrow \zeta +1$ by $k_\alpha (i)= $ that $\xi $ such that $x_\xi = [(\alpha ,i)]_{\sim }$ for $i<\zeta ^{p_{\alpha }}$ .

For $\xi <\zeta '$ let $\theta _\xi =\theta _{x_\xi }$ . For $\nu \le \xi <\zeta '$ let ${\mathcal {F}}_{\nu \xi }={\mathcal {F}}_{x_\nu x_\xi }$ . If $\zeta '=\zeta +1$ and $x_\zeta =[(\alpha ,i)]$ for $\alpha <\mu $ and $i=\zeta ^{p_\alpha }$ , let $F= F^{p_{\alpha }}$ .

If $\zeta =\zeta '$ let $\theta _\zeta = \mathrm {otp}(\bigcup \{ \mathrm {rge}(F^{p_\alpha })| \alpha <\mu \})$ , let F be the inverse of the transitive collapse of $\bigcup \{ \mathrm {rge}(F^{p_\alpha })| \alpha <\mu \}$ and for $\xi <\zeta $ let ${\mathcal {F}}_{\xi \zeta } = \bigcup \{ F^{-1}\cdot F^{p_{\alpha }}\cdot f| \alpha <\mu \&\ k_\alpha (i)=\xi \&\ f\in {\mathcal {F}}^{p_\alpha }_{i\zeta ^{p_\alpha }} \}$ .

We shall show that the structure $\langle \langle \theta _\xi | \xi \le \zeta \rangle , \langle {\mathcal {F}}_{\nu \xi }| \nu \le \xi \le \zeta \rangle \rangle $ has almost all of the properties of a small SMS and then show how to gently interleave limit levels of small cofinality where necessary.

We start with the structure of the sets ${\mathcal {F}}_{\xi \xi +1}$ for $\xi <\zeta $ .

Given $\xi <\zeta $ let $\alpha <\mu $ be such that for some $i<\zeta ^{p_{\alpha }}$ one has $x_\xi =[(\alpha ,i)]_\sim $ and $x_{\xi +1}=[(\alpha ,i+1)]_\sim $ . Then, by the definition of $\mathbb {P}$ , ${\mathcal {F}}^{p_\alpha }_{ii+1}$ is either a singleton (consisting of an order-preserving map) or an amalgamation pair, and for all $\beta \in (\alpha ,\mu )$ we have $k_{\alpha \beta }(i+1)=k_{\alpha \beta }(i) + 1$ and ${\mathcal {F}}^{p_\beta }_{k_{\alpha \beta }(i)k_{\alpha \beta }(i)+1}= {\mathcal {F}}^{p_\alpha }_{ii+1}$ . Hence we must then have that ${\mathcal {F}}_{\xi \xi +1}= {\mathcal {F}}^{p_\alpha }_{ii+1}$ and is a singleton (consisting of an order-preserving map) or an amalgamation pair. (Note that, by the definition of small SMS, if ${\mathcal {F}}_{\xi \xi +1}={\mathcal {F}}^{p_\alpha }_{ii+1} =\{ f_i\}$ is a singleton then $f_i$ is not cofinal.)

Now we show the ${\mathcal {F}}_{\nu \xi }$ have the required factorization property.

Suppose $\nu \le \tau \le \xi <\zeta '$ and $f\in {\mathcal {F}}_{\nu \tau }$ and $g\in {\mathcal {F}}_{\tau \xi }$ . As $\mu $ is a limit ordinal there is some $\alpha <\mu $ with some $i\le j \le l\le \zeta ^{p_\alpha }$ such that $k_\alpha (i)=\nu $ , $k_\alpha (j)=\tau $ and $k_\alpha (l)=\xi $ (i.e., $x_\nu =[(\alpha ,i)]_\sim $ , $x_\tau =[(\alpha ,j)]_\sim $ and $x_\xi =[(\alpha ,l)]_\sim $ ), $f\in {\mathcal {F}}^{p_{\alpha }}_{ij}$ and $g\in {\mathcal {F}}^{p_{\alpha }}_{jl}$ . Thus $g\cdot f\in {\mathcal {F}}^{p_{\alpha }}_{il}$ , and hence $g\cdot f\in {\mathcal {F}}_{\nu \xi }$ .

Similarly, suppose $\nu \le \tau \le \xi <\zeta $ and $h\in {\mathcal {F}}_{\nu \xi }$ . There is some $\alpha <\mu $ and $i\le j \le l\le \zeta ^{p_{\alpha }}$ such that $k_\alpha (i)=\nu $ , $k_\alpha (\;j)=\tau $ , $k_\alpha (l)=\xi $ and $h\in {\mathcal {F}}^{p_{\alpha }}_{il}$ . But then, as $S^{p_{\alpha }}$ is a small SMS, there are $g\in {\mathcal {F}}^{p_{\alpha }}_{jl}$ , $\subseteq {\mathcal {F}}_{\tau \xi }$ , and $f\in {\mathcal {F}}^{p_{\alpha }}_{ij}$ , $\subseteq {\mathcal {F}}_{\nu \tau }$ , such that $g\cdot f= h$ .

Now suppose that $\zeta =\zeta '$ and $\nu \le \tau <\zeta $ . If $f\in {\mathcal {F}}_{\nu \tau }$ and $g\in {\mathcal {F}}_{\tau \zeta }$ there is some $\alpha <\mu $ , $i\le j \le l\le \zeta ^{p_{\alpha }}$ and $h\in {\mathcal {F}}^{p_\alpha }_{j\zeta ^{p_{\alpha }}}$ such that $k_\alpha (i)=\nu $ , $k_\alpha (\;j)=\tau $ , $f\in {\mathcal {F}}^{p_\alpha }_{ij}$ and $g=F^{-1}\cdot F^{p_{\alpha }} \cdot h$ . Then $g\cdot f = F^{-1}\cdot F^{p_{\alpha }} \cdot (h\cdot f)$ , where $h\cdot f \in {\mathcal {F}}^{p_\alpha }_{i\zeta ^{p_{\alpha }}} \subseteq {\mathcal {F}}_{\nu \tau }$ , and hence $g\cdot f\in {\mathcal {F}}_{\nu \zeta }$ .

Similarly, if $\zeta =\zeta '$ , $\nu \le \tau <\zeta $ and $f\in {\mathcal {F}}_{\nu \zeta }$ there is some $\alpha <\mu $ , $i\le j \le l\le \zeta ^{p_{\alpha }}$ , $h\in {\mathcal {F}}^{p_\alpha }_{ij}$ and $g\in {\mathcal {F}}^{p_\alpha }_{j \zeta ^{p_{\alpha }}}$ such that $k_\alpha (i)=\nu $ , $k_\alpha (\;j)=\tau $ , and $f=(F^{-1}\cdot F^{p_{\alpha }}\cdot g)\cdot h$ .

Next, we show directedness at limits.

Suppose $\nu $ , $\tau <\xi <\zeta '$ , $\xi $ is a limit ordinal, $f\in {\mathcal {F}}_{\nu \xi }$ and $g\in {\mathcal {F}}_{\tau \xi }$ . Let $\alpha <\mu $ be such that there are $i\le j< l\le \zeta ^{p_{\alpha }}$ with $k_\alpha (i)=\nu $ , $k_\alpha (\;j)=\tau $ , $f\in {\mathcal {F}}^{p_\alpha }_{il}$ and $g\in {\mathcal {F}}^{p_\alpha }_{jl}$ .

If l is a successor then ${\mathcal {F}}^p_{l-1l}$ is a singleton, say $\{ h_{l-1}\}$ (as otherwise $k_{\alpha \beta }(l)=k_{\alpha \beta }(l-1)+1$ for all $\beta \in (\alpha ,\mu )$ and hence $k_\alpha (l)=\xi $ is a successor). Furthermore, as $S^p$ is a SMS there are $f'\in {\mathcal {F}}^{p_{\alpha }}_{il-1}$ and $g'\in {\mathcal {F}}^{p_{\alpha }}_{jl-1}$ such that $f=h_{l-1}\cdot f'$ and $g=h_{l-1}\cdot g'$ . But then $k_\alpha (l-1) \in [\nu \cup \tau ,\xi )$ and $f'\in {\mathcal {F}}_{\nu k_\alpha (l-1)}$ , $g'\in {\mathcal {F}}_{\tau k_\alpha (l-1)}$ and $h_{l-1}\in {\mathcal {F}}_{k_\alpha (l-1)\xi }$ .

If l is a limit, by the definition of small SMS, there is some $m\in [\max (\{ i,j\}), l)$ , and $h\in {\mathcal {F}}^{p_{\alpha }}_{ml}$ , $f'\in {\mathcal {F}}^{p_{\alpha }}_{im}$ and $g'\in {\mathcal {F}}^{p_{\alpha }}_{jm}$ such that $f = h \cdot f'$ and $g= h\cdot g'$ . Hence $f'\in {\mathcal {F}}_{\nu k_\alpha (m)}$ , $g'\in {\mathcal {F}}_{\tau k_\alpha (l-1)}$ and $h\in {\mathcal {F}}_{k_\alpha (m)\xi }$ .

Now suppose that $\nu $ , $\tau <\zeta =\zeta '$ , $f\in {\mathcal {F}}_{\nu \zeta }$ and $g\in {\mathcal {F}}_{\tau \zeta }$ . Then there is some $\alpha <\mu $ and $i\le j\le \zeta ^{p_{\alpha }}$ with $k_\alpha (i)=\nu $ , $k_\alpha (\;j)=\tau $ , and $k_\alpha (\zeta ^{p_{\alpha }})=\xi <\zeta $ , and some maps $f'\in {\mathcal {F}}_{i\zeta ^{p_{\alpha }}}$ and $g'\in {\mathcal {F}}_{j\zeta ^{p_{\alpha }}}$ with $f=F^{-1}\cdot F^{p_{\alpha }}\cdot f'$ and $g=F^{-1}\cdot F^{p_{\alpha }}\cdot g'$ . As $F^{-1}\cdot F^{p_{\alpha }}\in {\mathcal {F}}_{\xi \zeta }$ we are done.

We have now shown that $\langle \langle \theta _\xi | \xi \le \zeta \rangle , \langle {\mathcal {F}}_{\nu \xi }| \nu \le \xi \le \zeta \rangle \rangle $ has all of the properties of a small SMS, except possibly the property that if $\xi \le \zeta $ is a limit ordinal then $\mathrm {cf}(\theta _\xi )<\kappa $ . In order to guarantee this last property, in general, we must add additional levels to ${\mathcal E}$ .

In order to see why consider the example of some $\langle \alpha _l| l<\omega \rangle $ in which for all $l<\omega $ we have $k_{\alpha _l}(i) = i$ for all $i<l$ but $\omega = k_{\alpha _l}(l)$ and $\mathrm {cf}(\theta ^{p_{\alpha _l}}_l)=\kappa $ . We then need to insert a ‘new’ $\omega $ th level in order to ensure that the $\theta $ of the $\omega $ th level of the lower bound we are constructing does not have cofinality $\kappa $ .

This interleaving of additional levels is not hard to do, but writing down the details formally, as we do in the following four paragraphs, is a little involved.

So, set $\zeta ^*=\zeta +1$ if $\mathrm {cf}(\theta _{x_\zeta })=\kappa $ , and otherwise let $\zeta ^*=\zeta $ . For $\alpha <\mu $ we define $k^*_\alpha :\zeta ^{p_{\alpha }}\longrightarrow \zeta ^*$ as follows. For each $i\le \zeta ^{p_{\alpha }}$ let $\xi _i$ be the greatest limit ordinal such that $\xi _i\le k_\alpha (i)$ if it exists and undefined otherwise. If $\xi _i$ is defined and $\mathrm {cf}(\theta _{\xi _i})=\kappa $ then $k^*_\alpha (i)=k_\alpha (i)+1$ . Otherwise we take $k^*_\alpha (i)=k_\alpha (i)$ .

For $\xi \le \zeta ^*$ such that $\xi \in \bigcup \{ \mathrm {rge}\;{k^*_\alpha }| \alpha <\mu \}$ we let $\theta ^*_\zeta = \theta ^{p_{\alpha }}_j$ for any pair $(\alpha ,j)$ such that $k^*_{\alpha }(\;j)=\xi $ . If $\nu \le \xi \le \zeta ^*$ and $\nu $ , $\xi \in \bigcup \{ \mathrm {rge}\;{k^*_\alpha }| \alpha <\mu \}$ let $\alpha <\mu $ be such that $\nu $ , $\xi \in \mathrm {rge}\;{k^*_\alpha }$ , let $\nu '= k_\alpha ((k^*_{\alpha })^{-1}(\nu )$ and $\xi '= k_\alpha ((k^*_{\alpha })^{-1}(\xi )$ and set ${\mathcal {F}}^*_{\nu \xi }= {\mathcal {F}}_{\nu '\xi '}$ .

Now suppose otherwise, that $\xi \le \zeta ^*$ but $\xi \notin \bigcup \{ \mathrm {rge}\;{k^*_\alpha }| \alpha <\mu \}$ , hence $\xi $ is a limit ordinal. Let $\alpha ^*$ be least such that $\xi \in \mathrm {rge}\;(k_{\alpha ^*})$ . For all $\alpha \in [\alpha ^*,\mu )$ and all $f\in {\mathcal {F}}^{p_\alpha }_{ij}$ , where $k_\alpha (\;j)=\xi $ let $\gamma _f=\mathrm {ssup}(\mathrm {rge}(\;f))$ . For all such f we have $\gamma _f<\theta _\xi $ . Hence, as $\mathrm {cf}(\theta _\xi )=\kappa $ , we have $\mathrm {ssup}(\{ \gamma _f| \alpha \in [\alpha ^*,\mu )\&\ k_\alpha (\;j)=\xi \&\ f\in {\mathcal {F}}^{p_\alpha }_{ij}\})<\theta _\xi $ .

So we can choose $\theta ^*_\xi $ such that $\mathrm {ssup}(\{ \gamma _f| \alpha \in [\alpha ^*,\mu )\&\ k_\alpha (\;j)=\xi \&\ f\in {\mathcal {F}}^{p_\alpha }_{ij}\})<\theta ^*_\xi <\theta _\xi $ and $\mathrm {cf}(\theta ^*_\xi )<\kappa $ . We set ${\mathcal {F}}^*_{\xi \xi +1} = \{ \mathrm {id}\}$ and define the other sets of maps in the obvious way using composition to ensure the last clause in the definition of small SMS holds: e.g., for $\nu \in \bigcup \{ \mathrm {rge}{k^*_\alpha }| \alpha <\mu \}$ if $\nu <\xi $ we take ${\mathcal {F}}^*_{\nu \xi +1}=\{ \mathrm {id}\cdot f| f\in {\mathcal {F}}^*_{\nu \xi }\}$ and if $\xi <\nu $ we take ${\mathcal {F}}^*_{\xi \nu }=\{ f\cdot \mathrm {id}| f\in {\mathcal {F}}^*_{\xi +1\nu }\}$ .

Now set $S=\langle \langle \theta ^*_\xi | \xi \le \zeta ^*\rangle , \langle {\mathcal {F}}^*_{\nu \xi }| \nu \le \xi \le \zeta ^*\rangle \rangle $ . It is clear that S is an small SMS.

We also define $\mathcal {N}=\{ N| \exists \alpha <\mu N\in \mathcal {N}^{p_\alpha }\}$ and set $p=(S,F,\mathcal {N})$ .

In order to finish the proof we must show that $p\in \mathbb {P}$ and that $p\le p_\alpha $ for each $\alpha <\mu $ .

For the former all that is left to show is that $\mathcal {N}$ is as required.

Suppose $M\in \mathcal {N}$ . Let $\alpha <\mu $ be such that $M\in \mathcal {N}^{p_\alpha }$ ., and let $i^M\le \zeta ^{p_\alpha }$ and $f\in {\mathcal {F}}^{p_\alpha }_{i\zeta ^{p_\alpha }}$ witness this. Then M fits $F^{p_\alpha } \cdot f $ , $=F\cdot (F^{-1}\cdot F^{p_\alpha } \cdot f )$ and the uniqueness of $(k^*_\alpha (i^M), F^{-1}\cdot F^{p_\alpha } \cdot f)$ follows from Lemma 3.2.

Suppose $i'\le j'\le k^*_\alpha (i^M)$ and $f\in {\mathcal {F}}_{i'j'}$ . If $i'\in \bigcup \{ \mathrm {rge}(k^*_\beta )| \beta \in (\alpha ,\mu )\}$ let $i=i'$ and otherwise let $i=i'+1$ . Define j similarly. Then there is some $\beta \in (\alpha ,\mu )$ and $\overline {i}$ , $\overline {j}\le \zeta ^{p_\beta }$ such that $k(\overline {i})=i$ , $k(\overline {j})=j$ and $f\in {\mathcal {F}}^{p_\beta }_{\overline {i}\overline {j}}$ . As $M\in {\mathcal N}^{p_\beta }$ we have that $f\in \overline {M}$ .

Similarly, if $N\in {\mathcal N}$ , $\gamma <\delta ^N$ , $g\in (\theta _{i^N})^\gamma $ , $i^N\le i^M$ and $g\in \overline {N}$ there is some $\beta \in (\alpha ,\mu )$ such that $N\in {\mathcal N}^{p_\beta }$ and $i^N\in \mathrm {rge}(k^*_\beta )$ and hence $g\in \overline {M}$ .

Now let $\alpha <\mu $ . We check that $p\le p_\alpha $ .

We take the witnessing ‘k’ to be $k^*_\alpha $ . The three bulleted properties of $k^*_\alpha $ are immediate from the definition of $k^*_\alpha $ and that of S. We take $f^{p_\alpha p}$ to be $F^{-1}\cdot F^{p_\alpha }$ , making it immediate that $F^{p_\alpha } = F\cdot f^{p_\alpha p}$ . It is also immediate from the definition of ${\mathcal N}$ that ${\mathcal N}^{p_\alpha }\subseteq {\mathcal N}$ .

Finally, if $N\in {\mathcal N}$ is witnessed by $(k^*_\alpha (i),f^{p_\alpha p}\cdot g)$ for some $g\in {\mathcal {F}}^{p_\alpha }_{i\zeta ^{p_\alpha }}$ then we must show that $N\in {\mathcal N}^p$ .

Let $\beta \in [\alpha ,\mu )$ be such that $N \in {\mathcal N}^{p_\beta }$ . By the definition of $k^*_\alpha $ and the fact that $k^*_\alpha = k^*_\beta \cdot k_{\alpha \beta }$ , and since $F^{p_\alpha } = F^{p_\beta }\cdot f^{p_\alpha p_\beta }$ , we have that this is witnessed by $(k_{\alpha \beta }(i),f^{p_\alpha p_\beta }\cdot g)$ . As $p_\beta \le p_\alpha $ we have (by the final property in the list of requirements defining $\le _{\mathbb {P}}$ ) that $N\in {\mathcal N}^{p_\alpha }$ (and $(i,g)$ witnesses that N fits $F^g\cdot g$ ). $\blacktriangle _{6.1}$

Proof. Define q as follows. Let $\zeta ^q=\zeta ^p+1$ .

If $\zeta \notin \mathrm {rge}(F^p)$ let $\gamma =\mathrm {otp}(\mathrm {rge}(F^p)\cap \zeta ))$ . Define $f_{\zeta ^p}:\theta ^p_{\zeta ^p}\longrightarrow \theta $ by $f_{\zeta ^p}\upharpoonright \gamma =\mathrm {id}$ and $f_{\zeta ^p}(\gamma +\xi )=\gamma +1+\xi $ for $\xi \in \mathrm {otp}(\theta ^p_{\zeta ^p}\setminus \gamma )$ . Define $F^q:\theta \longrightarrow \lambda $ by $F^q(\xi )=F^p(\xi )$ for $\xi <\gamma $ , $F^q(\gamma )=\zeta $ , $F^q(\gamma +1+\xi ) =F^p(\gamma +\xi )$ for $\xi \in \mathrm {otp}(\theta ^p_{\zeta ^p}\setminus \gamma )$ and $F^q(\theta ^p_{\zeta ^p} + \xi )= \mathrm {ssup}(F^p``\theta ^p_{\zeta ^p})+\xi $ for $\xi \in \mathrm {otp}(\theta \setminus (\gamma +1+\mathrm {otp}(\theta ^p_{\zeta ^p} \setminus \gamma )))$ .

If, on the other hand, $\zeta \in \mathrm {rge}(F^p)$ let $f_{\zeta ^p}=\mathrm {id}$ and let $F^q:\theta \longrightarrow \lambda $ be given by $F^q(\xi )=F^p(\xi )$ for $\xi <\theta ^p_{\zeta ^p}$ and $F^q(\theta ^p_{\zeta ^p} + \xi )= \mathrm {ssup}(F^p``\theta ^p_{\zeta ^p})+\xi $ for $\xi \in \mathrm {otp}(\theta \setminus \theta ^p_{\zeta ^p})$ .

Finally, let $S^q=\langle \langle \theta ^p_{i} | i\le \zeta ^p \rangle \frown \theta , \langle {\mathcal {F}}^p_{ij} | i\le j\le \zeta ^p \rangle \rangle $

$$ \begin{align*} \hskip7pc \langle \{ f_{\zeta^p} \cdot f | f\in {\mathcal{F}}^p_{i\zeta^p} \} | i\le \zeta^p\rangle \rangle ,\end{align*} $$

${\mathcal {F}}^*=\{ (F^q\cdot f, \theta ^{q}_i) | i\le \zeta ^p+1 \&\ f\in {\mathcal {F}}^q_{i\zeta ^p+1} \}$ and ${\mathcal N}^q={\mathcal N}^p$ . $\blacktriangle _{7.1}$

Proof. Let G be $\mathbb {P}$ -generic. Define two relations $\sim $ and $<$ on the set $\{ (p,i)| p\in G\&\ i\le \zeta ^p\}$ , by $(p,i)\sim (q,j)$ if there is some $r\in G$ with $r\le p$ , q witnessed by $k:\zeta ^p\longrightarrow \zeta ^r$ and $l:\zeta ^q\longrightarrow \zeta ^r$ and $k(i)=l(\;j)$ and, similarly, $(p,i) < (q,j)$ if there is some $r\in G$ with $r\le p$ , q witnessed by $k:\zeta ^p\longrightarrow \zeta ^r$ and $l:\zeta ^q\longrightarrow \zeta ^r$ and $k(i)<l(\;j)$ . Clearly $\sim $ is an equivalence relation and $<$ is a linear ordering which respects $\sim $ , inducing a linear ordering on the $\sim $ -equivalence classes ${\mathcal E}=\{ [(p,i)]_\sim | p\in G\&\ i\le \zeta ^p\}$ .

For $x=[(p,i)]_\sim \in {\mathcal E}$ let $\theta _x = \theta ^p_i$ . Note, by the definition of $\sim $ and of $\le _{\mathbb {P}}$ , that $\theta _x$ is independent of the choice of the pair $(p,i)$ . Note, also, that $x = [ p,i)]_{\sim } < y = [(q,j)]_{\sim }$ if and only if $\theta _x<\theta _y<\kappa ^{+}$ . Consequently, $<$ well-orders ${\mathcal E}$ and, by Lemma 7.1, it does so in order-type $\kappa ^{+}$ , since $\{ \theta _x| x\in {\mathcal E}\}$ is unbounded in $\kappa ^{+}$ . Let $\{ x_{\alpha }| \alpha <\kappa ^{+}\}$ be the $<$ -increasing ordering of ${\mathcal E}$ .

For x, $y\in {\mathcal E}$ define ${\mathcal {F}}_{xy}=\bigcup \{ f\in {\mathcal {F}}^p_{ij}| p\in G \&\ x=[(p,i)]_\sim \&\ y=[(p,j)]_\sim \}$ . Let ${\mathcal {F}}_x= \{ F^p\cdot f| p\in G\&\ x=[(p,i)]_\sim \&\ f\in {\mathcal {F}}^p_{i\zeta ^p} \}$ .

For $\alpha <\kappa ^{+}$ let $\theta _\alpha =\theta _{x_\alpha }$ . Let $\theta _{\kappa ^{+}}=\lambda $ . For $\alpha \le \beta <\kappa ^{+}$ let ${\mathcal {F}}_{\alpha \beta }={\mathcal {F}}_{x_{\alpha }x_{\beta }}$ . For $\alpha <\kappa ^{+}$ let ${\mathcal {F}}_{\alpha \kappa ^{+}}= {\mathcal {F}}_{x_\alpha }$ .

It remains to check that $M=\langle \langle \theta _\alpha | \alpha \le \kappa ^{+}\rangle , \langle {\mathcal {F}}_{\alpha \beta }| \alpha \le \beta \le \kappa ^{+}\rangle \rangle $ is an $(\kappa ^{+},1)$ -simplified morass.

Given $\alpha <\kappa ^{+}$ let $p\in G$ be such that for some $i<\zeta ^p$ one has $x_\alpha =[(p,i)]_\sim $ and $x_{\alpha +1}=[(p,i+1)]_\sim $ . Then, by the definition of $\mathbb {P}$ , ${\mathcal {F}}^p_{ii+1}$ is either a singleton (consisting of a order-preserving map) or an amalgamation pair, and by the definition of $\le _{\mathbb {P}}$ , one must then have that ${\mathcal {F}}_{\alpha \alpha +1}$ is a singleton (consisting of a order-preserving map) or an amalgamation pair.