# Graphs represented by Ext

Mohsen Asgharzadeh  Mohsen Asgharzadeh, Hakimiyeh, Tehran, Iran.
<EMAIL_ADDRESS>,  Mohammad Golshani  Mohammad Golshani, School of
Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box:
19395–5746, Tehran, Iran<EMAIL_ADDRESS>http://math.ipm.ac.ir/
golshani/ and  Saharon Shelah  Einstein Institute of Mathematics
Edmond J. Safra Campus, Givat Ram
The Hebrew University of Jerusalem
Jerusalem, 91904, Israel
and
Department of Mathematics
Hill Center - Busch Campus
Rutgers, The State University of New Jersey
110 Frelinghuysen Road
Piscataway, NJ 08854-8019 USA<EMAIL_ADDRESS>http://shelah.logic.at

###### Abstract.

This paper opens and discusses the question originally due to Daniel Herden,
who asked for which graph $(\mu,R)$ we can find a family
$\\{\mathbb{G}_{\alpha}:\alpha<\mu\\}$ of abelian groups such that for each
$\alpha,\beta\in\mu$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{G}_{\beta})=0$ iff $(\alpha,\beta)\in
R$.

In this regard, we present four results. First, we give a connection to
Quillen’s small object argument which helps ${\rm Ext}$ vanishes and uses to
present a useful criteria to the question. Suppose
$\lambda=\lambda^{\aleph_{0}}$ and $\mu=2^{\lambda}$. We apply Jensen’s
diamond principle along with the criteria to present $\lambda$-free abelian
groups representing bipartite graphs. Third, we use a version of black box to
construct in ZFC, a family of $\aleph_{1}$-free abelian groups representing
bipartite graphs. Finally, applying forcing techniques, we present a
consistent positive answer for general graphs.

###### Key words and phrases:

Abelian groups; almost-free modules; Ext-groups; forcing; graph theory; set
theoretic methods in algebra; vanishing of Ext.

###### 2020 Mathematics Subject Classification:

Primary: 03C60; 20A15 Secondary:13L05.

The second author’s research has been supported by a grant from IPM (No.
1400030417) and Iran National Science Foundation (INSF) (No. 98008254). The
third author’s research partially supported by NSF grant no: DMS 1833363. This
is publication 1217 of third author.

###### Contents

  1. 1 Introduction
  2. 2 Preliminary notation
  3. 3 A realization criteria
  4. 4 From diamond to Ext-graph of quite free groups
  5. 5 Representing a bipartite graph by Ext in ZFC
  6. 6 Representing a general graph by Ext

##  1\. Introduction

The vanishing and non-vanishing properties of ${\rm Ext}(-,\sim)$ are useful
tools, see for instance the book [18]. Here, we assume the objects $-,\sim$
are not necessarily noetherian, so the corresponding Ext-family becomes more
mysterious. Despite its ubiquity, there is very little known about the
correspondence between graph theory and the Ext-family. Our aim in this paper
is to present a sample of such connection by coding graphs using the vanishing
property of Ext of a family of almost free abelian groups.

Recall the following achievements from literature. In his seminal paper [13],
Shelah proved that freeness of Whitehead groups, that is an abelian
$\mathbb{G}$ satisfying the vanishing property ${\rm
Ext}(\mathbb{G},\mathbb{Z})=0$, is undecidable in ZFC. After this, there has
been a considerable amount of work to understand set theoretical methods in
algebra. Göbel and Shelah [6] introduced a method to construct splitters, that
is groups $\mathbb{G}$ satisfying ${\rm Ext}(\mathbb{G},\mathbb{G})=0$. They
applied their method to prove the existence of enough projective and injective
objects in the rational cotorsion pairs. Cotorsion pairs were introduced by
Salce [12] in 1979. Combining with the splitters, this theory has a lot of
applications, not only in group theory but also in the theory of rings and
modules. For instance, see the book [7]. Göbel, Shelah and Wallutis, proved in
[5] that any poset embeds into the lattice of cotorsion pairs of abelian
groups. To be more explicit, let $I$ be any set, and look at the power set
$\mathcal{P}(I)$ of $I$. For any $X\in\mathcal{P}(I)$ they construct
$\aleph_{1}$-free abelian groups $\mathbb{G}_{X},\mathbb{H}^{X}$ such that for
all $X,Y\subseteq I,$

${\rm Ext}(\mathbb{G}_{Y},\mathbb{H}^{X})=0\Longleftrightarrow Y\subseteq
X\quad(\ast).$

Also, there are some restrictions on the size of Ext-groups. As a sample,
suppose $\mathbb{G}$ is countable torsion-free, then ${\rm
Ext}(\mathbb{G},\mathbb{Z})$ is divisible and hence determined up to
isomorphism by its torsion-free rank and its $p$-rank. Shelah and Strüngmann
proved that the $p$-rank of ${\rm Ext}(\mathbb{G},\mathbb{Z})$ is either
countable or $2^{\aleph_{0}}$. For this and its generalization, see [17].

######  Definition 1.1.

Given a directed graph $G=(\mu,R)$, we say that $G$ is realized as an Ext-
graph of groups, provided that there exists a family of groups
$\\{\mathbb{G}_{\alpha}:\alpha<\mu\\}$ such that for each $\alpha,\beta\in\mu$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{G}_{\beta})=0$ iff $(\alpha,\beta)\in
R$.

Given the success of representing a wide range of rings as endomorphism rings
of abelian groups (see [2] and [7]), and useful constructions of a rigid
system consisting of $2^{\kappa}$ abelian groups (see [3]), it is a natural
and interesting question of what can be represented as extension groups of
abelian groups. In fact, Daniel Herden asked the following question:

######  Question 1.2.

Which graphs $(\mu,R)$ can be realized as an Ext-graph of abelian groups.

Our aim in this paper is to partially answer Question 1.2.

The organization of the paper is as follows. Section 2 contains the
preliminaries and basic notations that we need. In Section 3 we apply some
ideas similar to Quillen’s small object argument from model category (see [4,
Theorem 2.1.14]) to present a general criteria for representing bipartite
graphs as an Ext-graph of $\lambda$-free abelian groups:

Theorem (A). Let $\mu_{1},\mu_{2}\leq 2^{\lambda}$ be such that ${\rm
cf}(\mu_{2})>\lambda$ and let $R\subseteq\mu_{1}\times\mu_{2}$. Suppose the
following assumptions are satisfied:

  1. (a)

$\bar{\mathbb{G}}=\langle\mathbb{G}_{\alpha}:\alpha<\mu_{1}\rangle$ and
$\bar{\mathbb{G}}^{\iota}=\langle\mathbb{G}^{\iota}_{\alpha}:\alpha<\mu_{1}\rangle$
for $\iota=1,2$, are sequences of abelian groups,

  2. (b)

for each $\alpha$, $\mathbb{G}_{\alpha}$ is an $\aleph_{1}$-free abelian group
of cardinality $\lambda$ and
$\mathbb{G}_{\alpha}=\mathbb{G}^{2}_{\alpha}/\mathbb{G}^{1}_{\alpha},$ where
$\mathbb{G}^{1}_{\alpha}\subseteq\mathbb{G}^{2}_{\alpha}$ are free abelian
groups of cardinality $\lambda$,

  3. (c)

if $\alpha<\mu_{1}$ and $\mathbb{L}$ is constructible by
$\\{\mathbb{G}_{\gamma}:\gamma\in\mu_{1}\setminus\alpha\\}$ over
$\mathbb{G}^{1}_{\alpha}$ then, there is no homomorphism ${\bf g}$ from
$\mathbb{G}^{2}_{\alpha}$ into $\mathbb{L}$ extending ${\rm
id}_{\mathbb{G}^{1}_{\alpha}}$.

Then there exists a sequence
$\bar{\mathbb{K}}=\langle\mathbb{K}_{\beta}:\beta<\mu_{2}\rangle$ equipped
with the following three properties:

  1. $(\alpha)$

$\mathbb{K}_{\beta}$ is an $\aleph_{1}$-free abelian group of cardinality
$2^{\lambda}$ for each $\beta<\mu_{2}$,

  2. $(\beta)$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0$ iff $\alpha R\beta$,

  3. $(\gamma)$

if every $\mathbb{G}_{\alpha}$ is $\lambda$-free then every
$\mathbb{K}_{\alpha}$ is $\lambda$-free as well.

Theorem (A) is one of the main technical results of the paper and plays an
essential role in the sequel. Given cardinals $\mu<\lambda$ with $\mu$ regular
we set

$S^{\lambda}_{\mu}=\\{\alpha<\lambda:{\rm cf}(\alpha)=\mu\\}.$

In Section 4 we apply Jensen’s diamond principle $\diamondsuit_{S}$ along with
Theorem (A), and show the following:

Theorem (B). Let $S\subseteq S^{\lambda}_{\aleph_{0}}$ be a non-reflecting
stationary subset of $\lambda$ and suppose $\diamondsuit_{S}$ holds. Suppose
$\lambda=\lambda^{\aleph_{0}},\mu=2^{\lambda}$ and let
$R\subseteq\mu\times\mu$ be a relation. Then there are sequences
$\langle\mathbb{G}_{\alpha}:\alpha<\mu\rangle$ and
$\langle\mathbb{K}_{\alpha}:\alpha<\mu\rangle$ of $\lambda$-free abelian
groups such that for all $\alpha<\mu,|\mathbb{G}_{\alpha}|=\lambda,$
$|\mathbb{K}_{\alpha}|=2^{\lambda}$ and for all $\alpha,\beta<\mu,$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0\iff\alpha R\beta.$

The new advantage we have is that we work with $\lambda$-free abelian groups
with a control on their size. Recall that Jensen’s diamond principle is a kind
of prediction principle whose truth is independent of ZFC.

In Section 5, we descend from Section 4 to the ordinary ZFC set theory and as
another application of Theorem (A), we prove the following theorem, where
instead of using the diamond principle we use some variant of “Shelah’s black
box”:

Theorem (C). Let $\lambda=\lambda^{\aleph_{0}},\mu=2^{\lambda}$ and let
$R\subseteq\mu\times\mu$ be a relation. Then there are families
$\langle\mathbb{G}_{\alpha}:\alpha<\mu\rangle,$ and
$\langle\mathbb{K}_{\alpha}:\alpha<\mu\rangle$ of $\aleph_{1}$-free abelian
groups equipped with the following properties:

  1. (1)

for all $\alpha<\mu,\mathbb{G}_{\alpha}$ has size $\lambda$ and
$\mathbb{K}_{\alpha}$ has size $2^{\lambda},$

  2. (2)

for all $\alpha,\beta<\mu,$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0\iff\alpha R\beta.$

Here, we lose the $\lambda$-freeness from Theorem (B), the groups are just
$\aleph_{1}$-free, and this is the price that Theorem (B) should pay to be in
ZFC. The black boxes were introduced by Shelah in [14], where he proved that
they can be considered as a general method to generate a class of diamond-like
principles provable in ZFC. In particular, Question 1.2 has a positive answer
for the case of bipartite graphs, where:

######  Definition 1.3.

A graph $(\mu,R)$ is called bipartite if the vertex set can be decomposed as
$V_{1}\cup V_{2}$ such that all edges go between $V_{1}$ and $V_{2}$.

Concerning Theorem (C), we can realize bipartite graphs as the Ext-graph of
$\aleph_{1}$-free abelian groups. Nevertheless, it is easy to see that
bipartite graphs fit in the situation of $(\ast)$. In particular, we recover
the main result of [5] by a new argument, see Lemma 5.7. It may be worth to
mention that Theorem (C) slightly improves [5] via computing the size of
objects, namely $|\mathbb{G}_{\alpha}|=\lambda$ and
$|\mathbb{K}_{\alpha}|=2^{\lambda}$ for all $\alpha<\mu$.

In the final section we prove the following theorem:

Theorem (D). Suppose GCH holds and the pair $(S,R)$ is a graph where
$R\subseteq S\times S$, and let $\lambda>|S|$ be an uncountable regular
cardinal. Then there exists a cardinal preserving generic extension of the
universe, and there is a family $\\{\mathbb{G}_{s}:s\in S\\}$ of
$\lambda$-free abelian groups such that

${\rm Ext}(\mathbb{G}_{s},\mathbb{G}_{t})=0\iff sRt.$

The strategy of the proof of Theorem (D) is given in Discussion 6.2. However,
there are some details to be checked, and this is our task in §6. Here, the
new advantage we have is that we work with a general graph and also we rely on
forcing techniques. This theorem gives a consistent positive answer to
Question 1.2. Despite this, we think Herden’s question has a positive answer
in ZFC. Namely, we present the following conjecture:

######  Conjecture 1.4.

Any graph can be realized as an Ext-graph of groups.

We hope our results will shed more light on interplay between homological
algebra and graph theory.

For all unexplained definitions from algebra see the books by Eklof-Mekler [2]
and Göbel-Trlifaj [7]. Also, for unexplained definitions from the theory of
forcing see the books of Jech [9] and Kunen [10].

##  2\. Preliminary notation

In this section, we set out our notation and discuss some facts that will be
used throughout the paper and refer to the book of Eklof and Mekler [2] for
more information. We restrict our discussion to the category Mod-$\mathbb{Z}$
of abelian groups, though most of the notions and results can be extended to
module categories over more general rings. For abelian groups $\mathbb{G}$ and
$\mathbb{H}$, we set ${\rm Ext}(\mathbb{G},\mathbb{H}):={\rm
Ext}^{1}_{\mathbb{Z}}(\mathbb{G},\mathbb{H})$ and similarly, ${\rm
Hom}(\mathbb{G},\mathbb{H}):={\rm Hom}_{\mathbb{Z}}(\mathbb{G},\mathbb{H})$.
We need the following well-known fact (see e.g. the book [18, Page 77]):

######  Fact 2.1.

(Baer and Yoneda) Let
$\zeta_{i}:=0\longrightarrow\mathbb{B}\stackrel{{\scriptstyle
g_{i}}}{{\longrightarrow}}\mathbb{C}_{i}\stackrel{{\scriptstyle
f_{i}}}{{\longrightarrow}}\mathbb{A}\longrightarrow 0$ be two short exact
sequences of abelian groups. We say $\zeta_{1}$ is equivalent to $\zeta_{2}$
if there is a commutative diagram:

$\begin{CD}\zeta_{2}=0@>{}>{}>\mathbb{B}@>{g_{2}}>{}>\mathbb{C}_{2}@>{f_{2}}>{}>\mathbb{A}@>{}>{}>0\\\
=@A{}A{}A\cong @A{}A{}A=@A{}A{}A\\\
\zeta_{1}=0@>{}>{}>\mathbb{B}@>{g_{1}}>{}>\mathbb{C}_{1}@>{f_{1}}>{}>\mathbb{A}@>{}>{}>0\\\
\end{CD}$

Indeed, this is an equivalent relation, and there is a 1-1 correspondence
between the equivalent class of these short exact sequences and ${\rm
Ext}(\mathbb{A},\mathbb{B})$. In addition, $[\zeta_{1}]=0\in{\rm
Ext}(\mathbb{A},\mathbb{B})$ iff $\zeta_{1}$ splits.

######  Definition 2.2.

An abelian group $\mathbb{G}$ is called $\aleph_{1}$-free if every subgroup of
$\mathbb{G}$ of cardinality $<\aleph_{1}$, i.e., every countable subgroup, is
free. More generally, an abelian group $\mathbb{G}$ is called $\lambda$-free
if every subgroup of $\mathbb{G}$ of cardinality $<\lambda$ is free.

######  Definition 2.3.

Let $\kappa$ be a regular cardinal. An abelian group $\mathbb{G}$ is said to
be strongly $\kappa$-free if there is a set $\mathcal{S}$ of
$<\kappa$-generated free subgroups of $\mathbb{G}$ containing 0 such that for
any subset $S$ of $\mathbb{G}$ of cardinality $<\kappa$ and any
$\mathbb{N}\in\mathcal{S}$, there is $\mathbb{L}\in\mathcal{S}$ such that
$S\cup\mathbb{N}\subset\mathbb{L}$ and $\mathbb{L}/\mathbb{N}$ is free.

Also, by a club subset of an uncountable regular cardinal $\kappa$ we mean a
closed and unbounded subset of $\kappa$.

######  Definition 2.4.

Suppose $\kappa$ is an uncountable regular cardinal. Let
${\mathscr{D}}_{\kappa}$ denote the club filter on $\kappa$, i.e.,

${\mathscr{D}}_{\kappa}=\\{A\subseteq\kappa$: $A$ contains a club subset of
$\kappa$}.

Let also $\mathcal{P}(\kappa)/{\mathscr{D}}_{\kappa}$ denote the resulting
quotient Boolean algebra.

It is easily seen that ${\mathscr{D}}_{\kappa}$ is a normal $\kappa$-complete
filter on $\kappa$ and that it is closed under diagonal intersections, i.e.,
if $A_{i}\in{\mathscr{D}}_{\kappa}$, for $i<\kappa$, then their diagonal
intersection

$\bigtriangleup_{i<\kappa}A_{i}=\\{\xi<\kappa:\forall i<\xi,\xi\in A_{i}\\}$

is also in ${\mathscr{D}}_{\kappa}$. This can be used to prove the following
easy lemma.

######  Lemma 2.5.

Suppose $\kappa$ is a regular uncountable cardinal, $\delta\leq\kappa$ and let
$\Gamma_{i}\in{\mathscr{P}}(\kappa)/{\mathscr{D}}_{\kappa}$ for $i<\delta$.
Then in the Boolean Algebra ${\mathscr{P}}(\kappa)/{\mathscr{D}}_{\kappa},$
the sequence $\\{\Gamma_{i}:i<\delta\\}$ has a lub (least upper bound)
$\Gamma$.

###### Proof.

For each $i<\delta$ let $A_{i}\subseteq\kappa$ be such that
$\Gamma_{i}=A_{i}/{\mathscr{D}}_{\kappa}$. If $\delta<\kappa,$ then
$\Gamma=A/{\mathscr{D}}_{\kappa}$ is as required where
$A=\bigcup\limits_{i<\delta}A_{i}$, and if $\delta=\kappa$, then
$\Gamma=A/{\mathscr{D}}_{\kappa}$ is as required where
$A=\bigtriangleup_{i<\kappa}A_{i}$ is the diagonal intersection of the sets
$A_{i},i<\kappa.$ ∎

The following definition plays an important role in the sequel.

######  Definition 2.6.

Let $\kappa$ be a regular cardinal. If $\mathbb{G}$ is a
$\leq\kappa$-generated abelian group, a $\kappa$-filtration of $\mathbb{G}$ is
a sequence $\\{\mathbb{G}_{\nu}:\nu<\kappa\\}$ of subgroups of $\mathbb{G}$
whose union is $\mathbb{G}$ such that for all $\nu<\kappa:$

  1. $(a)$

$\mathbb{G}_{\nu}$ is a $<\kappa$-generated subgroup of $\mathbb{G}$;

  2. $(b)$

if $\mu<\nu$, then $\mathbb{G}_{\mu}\subset\mathbb{G}_{\nu}$;

  3. $(c)$

if $\nu$ is a limit ordinal, then
$\mathbb{G}_{\nu}=\bigcup_{\mu<\nu}\mathbb{G}_{\mu}$ i.e., the sequence is
continuous.

It is easily seen that if $\\{\mathbb{G}_{\nu}:\nu<\kappa\\}$ and
$\\{\mathbb{H}_{\nu}:\nu<\kappa\\}$ are two $\kappa$-filtrations of a group
$\mathbb{G}$, then the set

$\\{\nu<\kappa:\mathbb{G}_{\nu}=\mathbb{H}_{\nu}\\}$

contains a club subset of $\kappa,$ in particular, modulo the club filter
${\mathscr{D}}_{\kappa},$ the choice of the $\kappa$-filtration does not
matter. This observation makes the following definition well-defined.

######  Definition 2.7.

Let $\lambda$ be an uncountable regular cardinals.

  1. (1)

If $\mathbb{G}$ is an abelian group of cardinality $\lambda$ and
$\langle\mathbb{G}_{\alpha}:\alpha<\lambda\rangle$ is a filtration of
$\mathbb{G}$, then

$\Gamma(\mathbb{G},\bar{\mathbb{G}})=\\{\delta<\lambda:\mathbb{G}/\mathbb{G}_{\delta}\text{
is not }\lambda\text{-free}\\}.$

  2. (2)

Let

$\Gamma(\mathbb{G})=\Gamma(\mathbb{G},\bar{\mathbb{G}})/{\mathscr{D}}_{\lambda}$

for some (and hence every) filtration $\bar{\mathbb{G}}$ of $\mathbb{G}.$

We recall that $\Gamma(\mathbb{G})$ is called the $\Gamma$-invariant of the
group $\mathbb{G}$, and refer to [2, §IV.1] for more details and properties of
this invariant. The following lemma gives a combinatorial characterization for
$\lambda$-free groups to be free.

######  Lemma 2.8.

([2, Ch IV, Proposition 1.7]) Let $\lambda$ be an uncountable regular
cardinals and let $\mathbb{G}$ be a $\lambda$-free abelian group of
cardinality $\lambda.$ The following are equivalent:

  1. (1)

$\mathbb{G}$ is free,

  2. (2)

$\mathbb{G}$ has a filtration
$\langle\mathbb{G}_{\alpha}:\alpha<\lambda\rangle$ such that for all
$\alpha<\lambda,\mathbb{G}_{\alpha+1}/\mathbb{G}_{\alpha}$ is free,

  3. (3)

$\Gamma(\mathbb{G})=\emptyset/\mathcal{D}_{\lambda}$.

These mean that $\Gamma(\mathbb{G},\bar{\mathbb{G}})$ is non-stationary for
some (and hence every) filtration
$\langle\mathbb{G}_{\alpha}:\alpha<\lambda\rangle$ of $\mathbb{G}.$

##  3\. A realization criteria

Our main result in this section is Theorem 3.5. Let us start by some lemmas
and definitions.

######  Lemma 3.1.

(See [2, Ex. IV.22]) If $\mathbb{G}$ is the union of a continuous chain
$\\{\mathbb{G}_{\alpha}:\alpha\leq\beta\\}$ of abelian groups such that
$\mathbb{G}_{0}$ and $\mathbb{G}_{\alpha+1}/\mathbb{G}_{\alpha}$ are
$\aleph_{1}$-free for all $\alpha+1<\beta$, then $\mathbb{G}$ is
$\aleph_{1}$-free.

More generally, the following holds:

######  Lemma 3.2.

(See [3, Pages 112-113]) Let $\kappa<\lambda$ be infinite cardinals.

  1. $(i)$

$\lambda$-free implies $\kappa$-free.

  2. $(ii)$

Subgroups and direct sums of $\kappa$-free are $\kappa$-free

  3. $(iii)$

Extension of $\kappa$-free group is $\kappa$-free.

  4. $(iv)$

Let $0=\mathbb{G}_{0}\subseteq\ldots\subseteq\mathbb{G}_{i}\subseteq\ldots$ be
a smooth chain of groups with union $\mathbb{G}$ such that all
$\mathbb{G}_{i}/\mathbb{G}_{i+1}$ are $\kappa$-free. Then $\mathbb{G}$ is
$\kappa$-free.

######  Definition 3.3.

Let ${\mathscr{G}}$ be a set or class of abelian groups.

  1. (1)

We say $\bar{\mathbb{L}}$ is a construction by ${\mathscr{G}}$ over
$\mathbb{G}$ when:

    1. $(a)$

$\bar{\mathbb{L}}=\langle\mathbb{L}_{\varepsilon}:\varepsilon\leq\varepsilon(*)\rangle$
is a $\subseteq$-increasing and continuous sequence of abelian groups,

    2. $(b)$

$\mathbb{L}_{0}=\mathbb{G}$,

    3. $(c)$

for every
$\varepsilon<\varepsilon(*),\mathbb{L}_{\varepsilon+1}/\mathbb{L}_{\varepsilon}$
is free or is isomorphic to some member of ${\mathscr{G}}$.

  2. (2)

Omitting “over $\mathbb{G}$” means for $\mathbb{G}=\\{0\\}$.

  3. (3)

We say $\mathbb{L}$ is constructible by ${\mathscr{G}}$ (over $\mathbb{G}$)
when for some
$\bar{\mathbb{L}}=\langle\mathbb{L}_{\varepsilon}:\varepsilon\leq\varepsilon(*)\rangle$,
$\bar{\mathbb{L}}$ is a construction by ${\mathscr{G}}$ (over $\mathbb{G}$)
and $\mathbb{L}=\mathbb{L}_{\varepsilon(*)}$.

######  Notation 3.4.

Suppose we have the following data of abelian groups and homomorphisms:

$\begin{CD}\mathbb{A}\\\ @A{}A{}A\\\ \mathbb{B}@>{}>{}>\mathbb{C},\\\
\end{CD}$

We denote the corresponding pushout by
$\mathbb{A}\oplus_{\mathbb{B}}\mathbb{C}$.

The next result gives sufficient conditions for representing bipartite graphs
using the functor Ext.

######  Theorem 3.5.

Let $\mu_{1},\mu_{2}\leq 2^{\lambda}$ be such that ${\rm cf}(\mu_{2})>\lambda$
and let $R\subseteq\mu_{1}\times\mu_{2}$. Suppose the following assumptions
are satisfied:

  1. $\boxtimes^{1}_{\lambda,\mu_{1}}$

    1. (a)

$\bar{\mathbb{G}}=\langle\mathbb{G}_{\alpha}:\alpha<\mu_{1}\rangle$ and
$\bar{\mathbb{G}}^{\iota}=\langle\mathbb{G}^{\iota}_{\alpha}:\alpha<\mu_{1}\rangle$
for $\iota=1,2$, are sequences of abelian groups,

    2. (b)

for each $\alpha$, $\mathbb{G}_{\alpha}$ is an $\aleph_{1}$-free abelian group
of cardinality $\lambda$ and
$\mathbb{G}_{\alpha}=\mathbb{G}^{2}_{\alpha}/\mathbb{G}^{1}_{\alpha},$ where
$\mathbb{G}^{1}_{\alpha}\subseteq\mathbb{G}^{2}_{\alpha}$ are free abelian
groups cardinality $\lambda$,

    3. (c)

if $\alpha<\mu_{1}$ and $\mathbb{L}$ is constructible by
$\\{\mathbb{G}_{\gamma}:\gamma\in\mu_{1}\setminus\alpha\\}$ over
$\mathbb{G}^{1}_{\alpha}$, then there is no homomorphism ${\bf
g}:\mathbb{G}^{2}_{\alpha}\to\mathbb{L}$ extending ${\rm
id}_{\mathbb{G}^{1}_{\alpha}}$:

$\textstyle{\mathbb{G}^{1}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{{\rm
id}}$$\textstyle{\mathbb{G}^{2}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nexists{\bf
g}}$$\textstyle{\mathbb{G}^{1}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\textstyle{\mathbb{L}}$

Then there exists a sequence
$\bar{\mathbb{K}}=\langle\mathbb{K}_{\beta}:\beta<\mu_{2}\rangle$ equipped
with the following three properties:

  1. $\boxtimes^{2}_{\lambda,\mu_{1}}$

    1. $(\alpha)$

$\mathbb{K}_{\beta}$ is an $\aleph_{1}$-free abelian group of cardinality
$2^{\lambda}$ for each $\beta<\mu_{2}$,

    2. $(\beta)$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0$ iff $\alpha R\beta$,

    3. $(\gamma)$

if every $\mathbb{G}_{\alpha}$ is $\lambda$-free then every
$\mathbb{K}_{\alpha}$ is $\lambda$-free as well.

###### Proof.

Let $\langle{\mathscr{U}}_{\varepsilon}:\varepsilon<2^{\lambda}\rangle$ be a
partition of $2^{\lambda}$ such that
${\mathscr{U}}_{\varepsilon}\subseteq[\varepsilon,2^{\lambda})$ has
cardinality $2^{\lambda}$. We claim that there are sequences
$\bar{\mathbb{K}}_{\varepsilon}$, $\bar{\mathbf{H}}_{\varepsilon,\beta}$ and
$\bar{\mathbf{h}}^{*}_{\varepsilon,\beta}$, for $\varepsilon\leq 2^{\lambda}$
and $\beta<\mu_{2}$ with the following properties:

  1. $\oplus_{\varepsilon}$

    1. (a)

$\bar{\mathbb{K}}_{\varepsilon}=\langle\mathbb{K}_{\varepsilon,\beta}:\beta<\mu_{2}\rangle$
is a sequence of abelian groups,

    2. (b)

for each $\beta<\mu_{2}$ the sequence
$\langle\mathbb{K}_{\zeta,\beta}:\zeta\leq 2^{\lambda}\rangle$ is
$\subseteq$-increasing and continuous,

    3. (c)

$\mathbb{K}_{\varepsilon,\beta}$ has cardinality $\leq 2^{\lambda}$,

    4. (d)

$\bar{\bf
H}_{\varepsilon,\beta}=\langle(\alpha_{\beta,\zeta},h_{\beta,\zeta}):\zeta\in{\mathscr{U}}_{\varepsilon}\rangle$
lists all the pairs $(\alpha,h)$ where $\alpha<\mu_{1},~{}\alpha R\beta$ and
$h\in{\rm Hom}(\mathbb{G}^{1}_{\alpha},\mathbb{K}_{\varepsilon,\beta})$,

    5. (e)

$\bar{\bf h}^{*}_{\varepsilon,\beta}=\langle
h^{*}_{\beta,\zeta}:\zeta\in{\mathscr{U}}_{\varepsilon}\rangle$,

    6. (f)

if $\varepsilon\leq\zeta\in{\mathscr{U}}_{\varepsilon}$, then
$h^{*}_{\beta,\zeta}\in{\rm
Hom}(\mathbb{G}^{2}_{\alpha_{\beta,\zeta}},\mathbb{K}_{\zeta+1,\beta})$
extends $h_{\beta,\zeta}$. The property is conveniently summarized by the
subjoined diagram:

$\textstyle{\mathbb{G}^{1}_{\alpha_{\beta,\zeta}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{h_{\beta,\zeta}}$$\textstyle{\mathbb{G}^{2}_{\alpha_{\beta,\zeta}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h^{*}_{\beta,\zeta}}$$\textstyle{\mathbb{K}_{\zeta,\beta}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\textstyle{\mathbb{K}_{\zeta+1,\beta}}$

We proceed by a double induction on $\zeta$ and $\beta$ to construct such a
sequence. For $\zeta=0$ and for all $\beta<\mu_{2}$ set

$\mathbb{K}_{0,\beta}:=\bigoplus\\{\mathbb{G}^{1}_{\alpha}:\alpha<\mu_{1},\neg(\alpha
R\beta)\\}.$

By Fact 3.2(ii), $\mathbb{K}_{0,\beta}$ is free, and so $\lambda$-free.

For $\zeta$ a limit ordinal and $\beta<\mu_{2}$, we set

$\mathbb{K}_{\zeta,\beta}=\bigcup\limits_{\varepsilon<\zeta}\mathbb{K}_{\varepsilon,\beta}.$

Now suppose that the groups $K_{\varepsilon,\gamma}$ are defined for all
$\varepsilon<\zeta+1$ and $\gamma<\mu_{2}$. We define
$\mathbb{K}_{\zeta+1,\beta}$ for $\beta<\mu_{2}$ as follows.

Let $\bar{\bf
H}_{\varepsilon,\beta}=\langle(\alpha_{\beta,\zeta},h_{\beta,\zeta}):\zeta\in{\mathscr{U}}_{\varepsilon}\rangle$
be as in clause (d). We look at the following diagram

$\begin{CD}\mathbb{K}_{\zeta,\beta}\\\ h_{\beta,\zeta}@A{}A{}A\\\
\mathbb{G}^{1}_{\alpha_{\beta,\zeta}}@>{\subseteq}>{}>\mathbb{G}^{2}_{\alpha_{\beta,\zeta}},\\\
\end{CD}$

and set

$\mathbb{K}_{\zeta+1,\beta}:=\mathbb{K}_{\zeta,\beta}\oplus_{\mathbb{G}^{1}_{\alpha_{\beta,\zeta}}}\mathbb{G}^{2}_{\alpha_{\beta,\zeta}}.$

In particular, there is a homomorphism $f$ induced from $h_{\beta,\zeta}$
which commutes the following diagram:

$\begin{CD}0@>{}>{}>\mathbb{K}_{\zeta,\beta}@>{g}>{}>\mathbb{K}_{\zeta+1,\beta}@>{}>{}>\frac{\mathbb{G}^{2}_{\alpha_{\beta,\zeta}}}{\mathbb{G}^{1}_{\alpha_{\beta,\zeta}}}@>{}>{}>0\\\
h_{\beta,\zeta}@A{}A{}Af@A{}A{}A=@A{}A{}A\\\
0@>{}>{}>\mathbb{G}^{1}_{\alpha_{\beta,\zeta}}@>{}>{}>\mathbb{G}^{2}_{\alpha_{\beta,\zeta}}@>{}>{}>\frac{\mathbb{G}^{2}_{\alpha_{\beta,\zeta}}}{\mathbb{G}^{1}_{\alpha_{\beta,\zeta}}}@>{}>{}>0\\\
\end{CD}$

We set $h^{*}_{\beta,\zeta}:=f$. By a diagram chasing, $h^{*}_{\beta,\zeta}$
extends $h_{\beta,\zeta}$, and recall that
$g:\mathbb{K}_{\zeta,\beta}\hookrightarrow\mathbb{K}_{\zeta+1,\beta}$ is an
embedding. This completes the inductive construction and hence proves the
existence of the sequences claimed to exist above, i.e., the construction of
$\oplus_{\varepsilon}$ is now complete. Suppose now that every group
$\mathbb{G}_{\alpha}$ is $\lambda$-free. By Fact 3.2(iii) it follows that
$\mathbb{K}_{\zeta,\beta}$ is $\lambda$-free as well.

For $\beta<\mu_{2}$ let

$\mathbb{K}_{\beta}:=\mathbb{K}_{2^{\lambda},\beta}.$

In view of Lemma 3.1, the above short exact sequence and the hypotheses, each
$\mathbb{K}_{\beta}$ is an $\aleph_{1}$-free abelian group of size
$2^{\lambda}.$ Recall that $\lambda$-freeness of $\mathbb{K}_{\alpha}$ follows
by $\lambda$-freeness of $\mathbb{G}_{\alpha}$, as desired by
$\boxtimes^{2}_{\lambda,\mu_{1}}(\gamma)$.

Let us check the item $\boxtimes^{2}_{\lambda,\mu_{1}}(\beta)$. First, suppose
that $\alpha R\beta$ and let $h\in{\rm
Hom}(\mathbb{G}^{1}_{\alpha},\mathbb{K}_{\beta})$. Since
$|\mathbb{G}^{1}_{\alpha}|\leq\lambda<{\rm cf}(\mu_{2})$, for some
$\varepsilon<2^{\lambda}$ we have ${\rm Rang}(h)\subseteq
K_{\varepsilon,\beta}$, and hence we can assume that $h\in{\rm
Hom}(\mathbb{G}^{1}_{\alpha},\mathbb{K}_{\varepsilon,\beta})$. By the
definition of $\bar{\bf H}_{\varepsilon,\beta},$ for some
$\zeta\in{\mathscr{U}}_{\varepsilon}$ we have
$(\alpha,h)=(\alpha_{\beta,\zeta},h_{\beta,\zeta})$. Then by clause (f) of the
construction, $h^{*}_{\beta,\zeta}\in{\rm
Hom}(\mathbb{G}^{2}_{\alpha},\mathbb{K}_{\zeta+1,\beta})$ extends
$h_{\beta,\zeta}$. There is a natural embedding map from ${\rm
Hom}(\mathbb{G}^{2}_{\alpha},\mathbb{G}_{\zeta+1,\beta})$ into ${\rm
Hom}(\mathbb{G}^{2}_{\alpha},\mathbb{K}_{\beta})$, so without loss of
generality, we may and do assume that

$h^{*}_{\beta,\zeta}\in{\rm Hom}(\mathbb{G}^{2}_{\alpha},\mathbb{K}_{\beta}).$

Now, we look at the exact sequence

$(+):=0\longrightarrow\mathbb{G}^{1}_{\alpha}\longrightarrow\mathbb{G}^{2}_{\alpha}\longrightarrow\mathbb{G}_{\alpha}\longrightarrow
0.$

Applying ${\rm Hom}(-,\mathbb{K}_{\beta})$ to $(+)$, it induces the following
long exact sequence

${\rm
Hom}(\mathbb{G}^{2}_{\alpha},\mathbb{K}_{\beta})\stackrel{{\scriptstyle{\bf
f}}}{{\longrightarrow}}{\rm
Hom}(\mathbb{G}^{1}_{\alpha},\mathbb{K}_{\beta})\longrightarrow{\rm
Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})\longrightarrow{\rm
Ext}(\mathbb{G}^{2}_{\alpha},\mathbb{K}_{\beta})=0,$

where the last vanishing ${\rm
Ext}(\mathbb{G}^{2}_{\alpha},\mathbb{K}_{\beta})=0$ holds as
$\mathbb{G}^{2}_{\alpha}$ is free. From this,

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=\frac{{\rm
Hom}(\mathbb{G}^{1}_{\alpha},\mathbb{K}_{\beta})}{{\rm Rang}(f)}.$

Thus, ${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0$ if and only if
${\bf f}$ is surjective. But as we observed above, ${\bf f}$ is onto, and
hence we conclude that ${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0$.

Let $(\alpha,\beta)\in(\mu_{1}\times\mu_{2})\setminus R$. We need to show
${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})\neq 0$. Suppose on the
contrary that ${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0$, and
search for a contradiction. Since $\mathbb{G}^{1}_{\alpha}$ is a direct
summand of $\mathbb{K}_{0,\beta}$, we have $\mathbb{G}^{1}_{\alpha}\oplus
X=\mathbb{K}_{0,\beta}$. Let
$\rho:\mathbb{G}^{1}_{\alpha}\to\mathbb{G}^{1}_{\alpha}\oplus X$ be the
natural map, and look at

$\begin{CD}g^{1}_{\alpha}:=\mathbb{G}^{1}_{\alpha}@>{\rho}>{}>\mathbb{G}^{1}_{\alpha}\oplus
X@>{=}>{}>\mathbb{K}_{0,\beta}@>{\subseteq}>{}>\mathbb{K}_{\beta}\end{CD}$

So by our assumption, there is
$g^{2}_{\alpha}:\mathbb{G}^{2}_{\alpha}\rightarrow\mathbb{K}_{\beta}$
extending $g^{1}_{\alpha}$:

$\textstyle{\mathbb{G}^{1}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{g^{1}_{\alpha}}$$\textstyle{\mathbb{G}^{2}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\exists
g^{2}_{\alpha}}$$\textstyle{\mathbb{K}_{\beta}}$

We now show that there exists $u\subseteq 2^{\lambda}$ equipped with:

  1. (a)

$|u|=\lambda$,

  2. (b)

$0,\alpha,\beta\in u$,

  3. (c)

$\varepsilon\in u\Rightarrow{\rm
Rang}(h_{\beta,\varepsilon})\subseteq\sum\limits_{\zeta\in
u\cap\varepsilon}{\rm Rang}(h^{*}_{\beta,\zeta})+\sum\limits_{\gamma\in
u}\mathbb{G}^{1}_{\gamma}$,

  4. (d)

${\rm Rang}(g^{2}_{\alpha})\subseteq\sum\limits_{\varepsilon\in u}{\rm
Rang}(h^{*}_{\beta,\varepsilon})+\sum\limits_{\gamma\in
u}\mathbb{G}^{1}_{\gamma}$.

Indeed, by induction on $i<\lambda$ we define an increasing and continuous
sequence $\langle u_{i}:i<\lambda\rangle$ of subsets of $2^{\lambda}$ such
that for each $i<\lambda$, the following properties are valid:

  1. $\bullet$

$|u_{i}|\leq\lambda$,

  2. $\bullet$

$0,\alpha,\beta\in u_{0}$,

  3. $\bullet$

If $i$ is limit ordinal, then $u_{i}=\bigcup_{j<i}u_{j}$,

  4. $\bullet$

${\rm Rang}(g^{2}_{\alpha})\subseteq\sum\limits_{\varepsilon\in u_{0}}{\rm
Rang}(h^{*}_{\beta,\varepsilon})+\sum\limits_{\gamma\in
u_{0}}\mathbb{G}^{1}_{\gamma}$,

  5. $\bullet$

If $\varepsilon\in u_{i}$ then

${\rm Rang}(h_{\beta,\varepsilon})\subseteq\sum\limits_{\zeta\in
u_{i+1}\cap\varepsilon}{\rm Rang}(h^{*}_{\beta,\zeta})+\sum\limits_{\gamma\in
u_{i+1}}\mathbb{G}^{1}_{\gamma}.$

Set $u:=\bigcup\limits_{i<\lambda}u_{i}$. It is easily seen that $u$ satisfies
the required properties. Let $\varepsilon(*)={\rm otp}(u)$, where ${\rm
otp}(u)$ denotes the order type of the set $u$ and let
$\langle\gamma_{\varepsilon}:\varepsilon<\varepsilon(*)\rangle$ be the
increasing enumeration of $u$.

Set

${\mathscr{G}}:=\langle\mathbb{G}_{\gamma}:\gamma\in u\setminus\alpha\rangle.$

We now define a construction
$\bar{\mathbb{L}}:=\langle\mathbb{L}_{\varepsilon}:\varepsilon\leq\varepsilon(*)\rangle$
by ${\mathscr{G}}$ over $\mathbb{L}_{0}=\mathbb{G}_{\alpha}^{1}$. To achieve
this, we set

$\mathbb{L}_{\epsilon+1}:=\sum\limits_{\zeta<\epsilon}{\rm
Rang}(h^{*}_{\beta,\zeta})+\sum\limits_{\gamma\in u}\mathbb{G}^{1}_{\gamma}.$

Let $A:={\rm coker}(h^{*}_{\beta,\zeta})$. Let put this into the previous
diagram and obtain the following:

$\begin{CD}00\\\ @A{}A{}A@A{}A{}A\\\ A@>{=}>{}>A\\\ \pi @A{}A{}A\pi
@A{}A{}A\\\
0@>{}>{}>\mathbb{K}_{\zeta,\beta}@>{}>{}>\mathbb{K}_{\zeta+1,\beta}@>{}>{}>\mathbb{G}_{\alpha_{\beta,\epsilon}}@>{}>{}>0\\\
h_{\beta,\zeta}@A{}A{}Ah^{*}_{\beta,\zeta}@A{}A{}A=@A{}A{}A\\\
0@>{}>{}>\mathbb{G}^{1}_{\alpha_{\beta,\zeta}}@>{}>{}>\mathbb{G}^{2}_{\alpha_{\beta,\zeta}}@>{}>{}>\mathbb{G}_{\alpha_{\beta,\epsilon}}@>{}>{}>0\\\
\end{CD}$

This yields that:

$\begin{CD}0@>{}>{}>{\rm
Rang}(h^{*}_{\beta,\zeta})@>{}>{}>\mathbb{K}_{\zeta+1,\beta}@>{}>{}>A@>{}>{}>0\\\
f_{1}@A{}A{}Af_{2}@A{}A{}A=@A{}A{}A\\\ 0@>{}>{}>{\rm
Rang}(h_{\beta,\zeta})@>{}>{}>\mathbb{K}_{\zeta,\beta}@>{}>{}>A@>{}>{}>0,\\\
\end{CD}$

where $f_{i}$ are the natural inclusions. Now, we use the ${\rm Ker}-{\rm
coker}$ exact sequence:

$0\to{\rm Ker}(f_{1})\to{\rm Ker}(f_{2})\to{\rm Ker}(=)\to{\rm
coker}(f_{1})\to{\rm coker}(f_{2})\to{\rm coker}(=)\to 0$

Recall that ${\rm Ker}(=)={\rm coker}(=)=0$, ${\rm coker}(f_{1})=\frac{{\rm
Rang}(h^{*}_{\beta,\zeta})}{{\rm Rang}(h_{\beta,\zeta})}$ and ${\rm
coker}(f_{2})=\mathbb{G}_{\alpha_{\beta,\epsilon}}$. So,

${\rm Rang}(h^{*}_{\beta,\zeta})/{\rm
Rang}(h_{\beta,\zeta})\cong\mathbb{G}_{\alpha_{\beta,\epsilon}}\quad(+)$

For each $\epsilon<\epsilon(\star)$ we have

$\frac{\mathbb{L}_{\epsilon+1}}{\mathbb{L}_{\epsilon}}\cong{\rm
Rang}(h^{*}_{\beta,\zeta})/{\rm
Rang}(h_{\beta,\zeta})\cong\mathbb{G}_{\alpha_{\beta,\epsilon}}\in{\mathscr{G}}.$

In view of Definition 3.3, $\bar{\mathbb{L}}$ is indeed a construction. Recall
from $(d)$ that ${\rm
Rang}(g^{2}_{\alpha})\subset\mathbb{L}_{\varepsilon(*)}$. Hence
$g^{2}_{\alpha}:\mathbb{G}^{2}_{\alpha}\to\mathbb{L}_{\varepsilon(*)}$ extends
the identity function over $\mathbb{G}^{1}_{\alpha}$. This is in contradiction
with $\boxtimes^{1}_{\lambda,\mu_{1}}(c)$. ∎

##  4\. From diamond to Ext-graph of quite free groups

In this section we are going to prove Theorem (B) from the introduction.

######  Notation 4.1.

Let $\mathbb{G}$ be a reduced torsion free abelian group, e.g., a free group.
The notation ${\mathbb{G}}$ stands for the $\mathbb{Z}$-adic completion of
$\mathbb{G}$.

######  Remark 4.2.

i) $\mathbb{Z}$-adic topology of a free abelian group is Hausdorff.

ii) For example, each element of $\widehat{\mathbb{\mathbb{Z}}}$ can be
represented as $\Sigma_{n}k_{n}$ where $n!|k_{n}$.

######  Discussion 4.3.

Recall that for a stationary set $S\subseteq\lambda,$ Jensen’s diamond
$\diamondsuit_{\lambda}(S)$ asserts the existence of a sequence $\langle
S_{\alpha}:\alpha\in S\rangle$ such that for every $X\subseteq\lambda$ the set
$\\{\alpha\in S:X\cap\alpha=S_{\alpha}\\}$ is stationary.

For simplicity of the reader we cite the following result:

######  Fact 4.4.

If $\diamondsuit_{\lambda}(S)$ holds, then there is a decomposition
$S=\bigcup_{\beta<\lambda}S_{\beta}$ such that
$\diamondsuit_{\lambda}(S_{\beta})$ holds for all $\beta<\lambda$.

###### Proof.

See [7, Theorem 9.1.17]. ∎

The following lemma plays a key role in our construction.

######  Lemma 4.5.

Let $\iota\in\\{1,2\\}$. Suppose $\mathbb{G}^{\iota}_{n}$ are free abelian
groups of the same size such that for all $n<\omega$ the following conditions
are satisfied:

  1. (1)

$\mathbb{G}^{1}_{n}\subseteq\mathbb{G}^{2}_{n},$

  2. (2)

$\mathbb{G}^{\iota}_{n}\subseteq\mathbb{G}^{\iota}_{n+1}$,

  3. (3)

$\mathbb{G}^{\iota}_{n+1}/\mathbb{G}^{\iota}_{n}$ is free,

  4. (4)

$\mathbb{G}^{2}_{n+1}/(\mathbb{G}^{1}_{n+1}+\mathbb{G}^{2}_{n})$ is free,

  5. (5)

$\mathbb{G}^{2}_{n+1}=\left(\mathbb{G}^{1}_{n+1}\oplus_{\mathbb{G}^{1}_{n}}\mathbb{G}^{2}_{n}\right)\oplus\bigoplus\limits_{m<\omega}y_{m}\mathbb{Z},$

  6. (6)

$\bar{\mathbb{L}}=\langle\mathbb{L}_{\varepsilon}:\varepsilon\leq\varepsilon(*)\rangle$
is a construction,

  7. (7)

$\mathbb{L}_{\varepsilon}$ and
$\mathbb{L}_{\varepsilon+1}/\mathbb{L}_{\varepsilon}$ are $\aleph_{1}$-free.

Let $\mathbb{G}^{\iota}=\bigcup\limits_{n<\omega}\mathbb{G}^{\iota}_{n},$ so
that $\mathbb{G}^{1}\subseteq\mathbb{G}^{2}$ are free abelian groups and let
$\mathbb{f}\in{\rm Hom}(\mathbb{G}^{2},\mathbb{L}_{\varepsilon(*)})$. Then
there are free abelian groups
$\dot{\mathbb{G}}^{1}\subseteq\dot{\mathbb{G}}^{2}$ such that:

  1. (a)

$\mathbb{G}^{1}\subseteq\dot{\mathbb{G}}^{1}$,

  2. (b)

$\mathbb{G}^{2}\subseteq\dot{\mathbb{G}}^{2}$,

  3. (c)

$\mathbb{G}^{1}=\mathbb{G}^{2}\cap\dot{\mathbb{G}}^{1}$,

  4. (d)

$\dot{\mathbb{G}}^{\iota}/{\mathbb{G}}^{\iota}$ is free,

  5. (e)

$\dot{\mathbb{G}}^{2}/(\dot{\mathbb{G}}^{1}+\mathbb{G}^{2})$ is torsion free,

  6. (f)

for all $n<\omega$,
$\dot{\mathbb{G}}^{2}/(\dot{\mathbb{G}}^{1}+\mathbb{G}^{2}_{n})$ is free,

  7. (g)

there are no $\mathbb{\dot{f}}$ and
$\langle\dot{\mathbb{L}}_{\varepsilon}:\varepsilon\leq\varepsilon(*)\rangle$
equipped with the following properties:

    1. $(g1)$

$\langle\dot{\mathbb{L}}_{\varepsilon}:\varepsilon\leq\varepsilon(*)\rangle$
is a nice construction over $\dot{\mathbb{G}}^{1},$

    2. $(g2)$

$\mathbb{\dot{f}}\in{\rm
Hom}(\dot{\mathbb{G}^{2}},\dot{\mathbb{L}}_{\varepsilon(*)})$,

    3. $(g3)$

$\dot{\mathbb{f}}$ extends $\mathbb{f}\cup{\rm id}_{\dot{\mathbb{G}}^{1}}$.
This property is conveniently summarized by the subjoined diagram:

$\textstyle{\dot{\mathbb{L}}_{\varepsilon(*)}}$$\textstyle{\dot{\mathbb{G}}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{\subseteq}$$\textstyle{\dot{\mathbb{G}}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nexists\dot{\mathbb{f}}}$$\textstyle{\mathbb{G}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{\subseteq}$$\textstyle{\mathbb{G}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{{\mathbb{f}}}$

    4. $(g4)$

$\mathbb{L}_{\varepsilon}\subseteq\dot{\mathbb{L}}_{\varepsilon}$ for
$\varepsilon\leq\varepsilon(*),$

    5. $(g5)$

$\mathbb{L}_{\varepsilon}=\mathbb{L}_{\varepsilon+1}\cap\dot{\mathbb{L}}_{\varepsilon},$

    6. $(g6)$

$\dot{\mathbb{L}}_{\varepsilon+1}/(\mathbb{L}_{\varepsilon+1}+\dot{\mathbb{L}}_{\varepsilon})$
is free.

###### Proof.

Set
$\dot{\mathbb{G}}^{1}:=\mathbb{G}^{1}\oplus\bigoplus\limits_{n<\omega}z_{n}\mathbb{Z}$,
where $\bigoplus\limits_{n<\omega}z_{n}\mathbb{Z}$ is a free abelian group
with a base given by $\\{z_{n}:n<\omega\\}$. To define $\dot{\mathbb{G}}^{2},$
let us first set

$\mathbb{G}^{2,\ast}:=\dot{\mathbb{G}}^{1}\oplus_{\mathbb{G}^{1}}\mathbb{G}^{2}.$

For any infinite sequence $\vec{a}=\langle
a_{n}:n<\omega\rangle\in\prod_{\omega}2={}^{\omega}2$, we look at the
following

$\mathbb{G}^{2,\vec{a}}:=\langle\mathbb{G}^{2,\ast}\cup\\{\sum\limits_{n<\omega}n!(y_{n}+a_{n}z_{n})\\}\rangle,$

i.e., the subgroup of $\widehat{\mathbb{G}^{2,\ast}}$ generated by
$\mathbb{G}^{2,\ast}$ and the distinguished element

$\sum\limits_{n<\omega}n!(y_{n}+a_{n}z_{n})\in\widehat{\mathbb{G}^{2,\ast}}.$

It is routine to see that the properties $(a)-(f)$ are satisfied with
$\mathbb{G}^{2,\vec{a}}$. We are going to show that
$\dot{\mathbb{G}}^{2}=\mathbb{G}^{2,\vec{a}}$ satisfies in the remaining
property $(g)$ for a suitable choice of $\vec{a}$. Suppose on the way of
contradiction that for each $\vec{a}=\langle
a_{n}:n<\omega\rangle\in{}^{\omega}2$ there is a counterexample
$\mathbb{\dot{f}}_{\vec{a}}$ and
$\langle\dot{\mathbb{L}}^{\vec{a}}_{\varepsilon}:\varepsilon\leq\varepsilon(*)\rangle$
to clause $(g)$. In particular, $\mathbb{\dot{f}}_{\vec{a}}\in{\rm
Hom}(\mathbb{G}^{2,\vec{a}},\dot{\mathbb{L}}_{\varepsilon(*)})$, and it
extends $\mathbb{f}\cup{\rm id}_{\dot{\mathbb{G}}^{1}}$.

For $\vec{a}\in{}^{\omega}2$ and $m<\omega$ set

$(+)\quad\quad\quad y^{\vec{a}}_{m}:=\sum\limits_{n\geq
m}\frac{n!}{m!}(y_{n}+a_{n}z_{n}).$

  1. Claim

(A): Let $\vec{a}$, $\vec{b}$ and $m$ be such that

    1. $\bullet$

$m<\omega$,

    2. $\bullet$

$a_{\ell}=b_{\ell}$ for all $\ell\leq m$,

    3. $\bullet$

$\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{\ell})=\mathbb{\dot{f}}_{\vec{b}}(y^{\vec{b}}_{\ell})$
for all $\ell\leq n$.

Then $\vec{a}=\vec{b}$.

To prove the claim, we proceed by induction on $\ell<\omega$ and show that:

  1. (i)

$a_{\ell}=b_{\ell}$,

  2. (ii)

$\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{\ell})=\mathbb{\dot{f}}_{\vec{b}}(y^{\vec{b}}_{\ell})$.

The case $\ell\leq m$ follows from the assumption, so we may assume that
$\ell>m$. Now suppose that $\ell=k+1$ and the result is true for all natural
numbers less or equal to $k$. Let us evaluate $\mathbb{\dot{f}}_{\vec{a}}$ at
$y^{\vec{a}}_{m}$ from $(+)$, and recall that $\mathbb{\dot{f}}_{\vec{a}}$
extends $\mathbb{f}\cup{\rm id}_{\dot{\mathbb{G}}^{1}}$. It turns out that:

$\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{m})=\sum\limits_{n\geq
m}(\frac{n!}{m!}\mathbb{\dot{f}}_{\vec{a}}(y_{n})+a_{n}\mathbb{\dot{f}}_{\vec{a}}(z_{n}))=\sum\limits_{n\geq
m}\frac{n!}{m!}(y_{n}+a_{n}z_{n}).$

It immediately follows that for all $i<\omega$

$(\ast)\quad\quad\quad\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{i})=(i+1)\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{i+1})-(y_{i}+a_{i}z_{i}).$

We apply $(\ast)$ at level $i=k$ and combine it with inductive hypothesis to
observe that

$\begin{array}[]{clcr}0&=\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{k})-\mathbb{\dot{f}}_{\vec{b}}(y^{\vec{b}}_{k})\\\
&=\ell\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{\ell})-(y_{\ell-1}+a_{\ell-1}z_{\ell-1})-\ell\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{\ell})+(y_{\ell-1}+a_{\ell-1}z_{\ell-1})\\\
&=\ell(\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{\ell})-\mathbb{\dot{f}}_{\vec{b}}(y^{\vec{b}}_{\ell})).\end{array}$

Since $\dot{\mathbb{L}}_{\varepsilon(*)}$ is torsion-free, it follows that
$\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{\ell})-\mathbb{\dot{f}}_{\vec{b}}(y^{\vec{b}}_{\ell})=0$,
so (ii) is valid. In order to show (i) we apply $(\ast)$ at level $i=\ell$:

$0=\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{\ell})-\mathbb{\dot{f}}_{\vec{b}}(y^{\vec{b}}_{\ell})=(\ell+1)(\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{\ell+1})-\mathbb{\dot{f}}_{\vec{b}}(y^{\vec{b}}_{\ell+1}))-(a_{\ell}-b_{\ell})z_{\ell},$

i.e.,

$(a_{\ell}-b_{\ell})z_{\ell}=(\ell+1)(\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{\ell})-\mathbb{\dot{f}}_{\vec{b}}(y^{\vec{b}}_{\ell})).$

It follows that $(\ell+1)|(a_{\ell}-b_{\ell})$. Since
$a_{i},b_{i}\in\\{0,1\\}$, we must have $a_{\ell}-b_{\ell}=0$. This completes
the proof of the desired claim. Let $\vec{a}$ and $\vec{b}$ be distinct in
${}^{\omega}2$ such that $a_{0}=b_{0}$ and
$\mathbb{\dot{f}}_{\vec{a}}(y^{\vec{a}}_{0})=\mathbb{\dot{f}}_{\vec{b}}(y^{\vec{b}}_{0})$.
This contradicts Claim(A), so we are done. ∎

######  Remark 4.6.

Adopt the notation of Lemma 4.5, and only replace “free” by
“$\aleph_{1}$-free”. Then the same claim, as Lemma 4.5 indicates, is valid. We
leave its straightforward modification to the reader.

Recall that $S^{\lambda}_{\aleph_{0}}=\\{\alpha<\lambda:{\rm
cf}(\alpha)=\aleph_{0}\\}$. The main point in the development of Theorem B) is
to introduce $\boxtimes_{\lambda,\mu}^{1}$ from Theorem 3.5. Here, we present
a variation of it:

######  Theorem 4.7.

Let $S\subseteq S^{\lambda}_{\aleph_{0}}$ be stationary non-reflecting and
suppose $\diamondsuit_{S}$ holds. Suppose one of the followings:

  1. $(1)$

$\mu=\lambda={\rm cf}(\lambda)>\aleph_{0}$, or

  2. $(2)$

$\mu=2^{\lambda},\lambda={\rm cf}(\lambda)>\aleph_{0}$.

Let $\iota=1,2$. Then there are sequences
$\bar{\mathbb{G}}^{\iota}=\langle\mathbb{G}^{\iota}_{\alpha}:\alpha<\mu\rangle$
of abelian groups such that the following assertions are valid:

  1. (i)

$\mathbb{G}^{1}_{\alpha}\subseteq\mathbb{G}^{2}_{\alpha}$ are free abelian
groups cardinality $\lambda$,

  2. (ii)

for each $\alpha$,
$\mathbb{G}_{\alpha}:=\mathbb{G}^{2}_{\alpha}/\mathbb{G}^{1}_{\alpha}$ is a
strongly $\lambda$-free abelian group of cardinality $\lambda$,

  3. (iii)

if $\alpha<\mu$ and $\mathbb{L}$ is constructible by
$\\{\mathbb{G}_{\gamma}:\gamma\in\mu\setminus\alpha\\}$ over
$\mathbb{G}^{1}_{\alpha}$, then there is no homomorphism ${\bf
g}:\mathbb{G}^{2}_{\alpha}\to\mathbb{L}$ extending ${\rm
id}_{\mathbb{G}^{1}_{\alpha}}$.

###### Proof.

(1): Since $S$ is non-reflecting, given any limit ordinal $\delta<\lambda,$
there exists a club $C_{\delta}\subseteq\delta$ such that $S\cap
C_{\delta}=\emptyset.$ Furthermore as $\diamondsuit_{S}$ holds, and in the
light of Fact 4.4, we can find a sequence $\langle
S_{\varepsilon}:\varepsilon<\lambda\rangle$ of subsets of $S$ such that the
following two properties are satisfied:

  * •

the sets $S_{\varepsilon}$ are pairwise almost disjoint, i.e., for all
$\varepsilon<\zeta<\lambda,S_{\varepsilon}\cap S_{\zeta}$ is a bounded subset
of $\lambda,$

  * •

$\diamondsuit_{S_{\varepsilon}}$ holds for $\varepsilon<\lambda$.

For $\varepsilon<\lambda$ let

$\langle(f^{\varepsilon}_{\alpha},\mathbb{H}^{1,\varepsilon}_{\alpha},\mathbb{H}^{2,\varepsilon}_{\alpha},\bar{\mathbb{L}}_{\alpha}):\alpha\in
S_{\varepsilon}\rangle$

be such that:

  * •

$f^{\varepsilon}_{\alpha}:\alpha\to\alpha$ is a function,

  * •

$\mathbb{H}^{1,\varepsilon}_{\alpha}\subseteq\mathbb{H}^{2,\varepsilon}_{\alpha}$
are abelian groups,

  * •

the universe of $\mathbb{H}^{2,\varepsilon}_{\alpha}$ is $\alpha$,

  * •

$\bar{\mathbb{L}}_{\alpha}=\langle\mathbb{L}^{\alpha}_{\varepsilon}:\varepsilon\leq\alpha\rangle$
is a construction such that $\mathbb{L}_{\alpha}$ has universe $\alpha,$

  * •

each $\mathbb{L}_{\varepsilon}$ and
$\mathbb{L}_{\varepsilon+1}/\mathbb{L}_{\varepsilon}$ is $\aleph_{1}$-free,

  * •

for any tuple $(f,\mathbb{H}^{1},\mathbb{H}^{2},\bar{\mathbb{L}})$, where
$f:\lambda\to\lambda$ is a function, $\mathbb{H}^{1}\subseteq\mathbb{H}^{2}$
are abelian groups where the universe of $\mathbb{H}^{2}$ is $\lambda$ and
$\bar{\mathbb{L}}=\langle\mathbb{L}_{\varepsilon}:\varepsilon\leq\lambda\rangle$
is a construction such that the universe of $\mathbb{L}_{\lambda}$ is
$\lambda$ and $\mathbb{L}_{\varepsilon}$ and
$\mathbb{L}_{\varepsilon+1}/\mathbb{L}_{\varepsilon}$ are $\aleph_{1}$-free,
then the following set

$\\{\alpha\in
S_{\varepsilon}:(f\restriction\alpha,\mathbb{H}^{1}\restriction\alpha,\mathbb{H}^{2}\restriction\alpha,\bar{\mathbb{L}}\restriction\alpha)=(f^{\varepsilon}_{\alpha},\mathbb{H}^{1,\varepsilon}_{\alpha},\mathbb{H}^{2,\varepsilon}_{\alpha},\bar{\mathbb{L}}_{\alpha})\\}$

is stationary.

Given $\varepsilon<\lambda$ by induction on $\alpha<\lambda$, we choose the
sequences
$\bar{\mathbb{G}}^{\iota}_{\varepsilon}=\langle\mathbb{G}^{\iota}_{\alpha,\varepsilon}:\alpha<\lambda\rangle$,
for $\iota=1,2$ such that:

  1. $\circledast$

    1. $(a)$

$\bar{G}^{\iota}_{\varepsilon}$ is an increasing and continuous sequence of
abelian groups
$\mathbb{G}^{\iota}_{\alpha,\varepsilon}=(G^{\iota}_{\alpha,\varepsilon},+,0)$.

    2. $(b)$

$\mathbb{G}^{\iota}_{0,\varepsilon}=\\{0\\}$ and for $\alpha<\lambda,$ the
universe of $\mathbb{G}^{2}_{\alpha,\varepsilon}$, namely
$G^{2}_{\alpha,\varepsilon}$ is an ordinal
$\gamma_{\alpha,\varepsilon}<\lambda$.

    3. $(c)$

The following holds:

      1. $(c1):$

$\mathbb{G}^{1}_{\alpha,\varepsilon}\subseteq\mathbb{G}^{2}_{\alpha,\varepsilon}$
are free.

      2. $(c2):$

If $\alpha<\beta$ then
$\mathbb{G}^{2}_{\alpha,\varepsilon}\cap\mathbb{G}^{1}_{\beta,\varepsilon}=\mathbb{G}^{1}_{\alpha,\varepsilon}$.

      3. $(c3):$

If $\alpha<\beta$ and $\alpha\notin S_{\varepsilon}$, then
$\mathbb{G}^{2}_{\beta}/(\mathbb{G}^{1}_{\beta}+\mathbb{G}^{2}_{\alpha})$ is
free.

    4. $(d)$

If $\alpha<\beta$ and $\alpha\notin S_{\varepsilon}$, then
$\mathbb{G}^{1}_{\beta,\varepsilon}/\mathbb{G}^{1}_{\alpha,\varepsilon}$ and
$\mathbb{G}^{2}_{\beta,\varepsilon}/\mathbb{G}^{2}_{\alpha,\varepsilon}$ are
free.

    5. $(e)$

if $\delta\in S_{\varepsilon},$ then there are no $\dot{\mathbb{f}}$ and
$\langle\dot{\mathbb{L}}_{\zeta}:\zeta\leq\delta\rangle$ equipped with the
following properties:

      1. $(e1)$

$\langle\dot{\mathbb{L}}_{\zeta}:\zeta\leq\delta\rangle$ is a construction
over $\mathbb{G}_{\delta+1,\varepsilon}^{1},$

      2. $(e2)$

$\dot{\mathbb{f}}\in{\rm
Hom}(\mathbb{G}^{2}_{\delta+1,\varepsilon},\dot{\mathbb{L}}_{\delta})$,

      3. $(e3)$

$\dot{\mathbb{f}}$ extends $\mathbb{f}^{\varepsilon}_{\delta}\cup{\rm
id}_{\mathbb{G}_{\delta+1,\varepsilon}^{1}}$. This means that
$\mathbb{G}^{2}_{\delta,\varepsilon}\subseteq\mathbb{H}^{2,\varepsilon}_{\delta}$,
and also:

$\textstyle{\dot{\mathbb{L}}_{\delta}}$$\textstyle{\mathbb{G}^{1}_{\delta+1,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{\subseteq}$$\textstyle{\mathbb{G}^{2}_{\delta+1,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nexists\dot{\mathbb{f}}}$$\textstyle{\mathbb{G}^{1}_{\delta,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{\subseteq}$$\textstyle{\mathbb{G}^{2}_{\delta,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{\mathbb{f}^{\varepsilon}_{\delta}}$

      4. $(e4)$

$\mathbb{L}^{\delta}_{\zeta}\subseteq\dot{\mathbb{L}}_{\zeta}$ for
$\zeta\leq\delta,$

      5. $(e5)$

$\mathbb{L}^{\delta}_{\zeta}=\mathbb{L}^{\delta}_{\zeta+1}\cap\dot{\mathbb{L}}_{\zeta},$

      6. $(e6)$

$\dot{\mathbb{L}}_{\zeta+1}/(\mathbb{L}^{\delta}_{\zeta+1}+\dot{\mathbb{L}}_{\zeta})$
is free.

For $\alpha=0,$ set
$\mathbb{G}^{1}_{0,\varepsilon}=\mathbb{G}^{2}_{0,\varepsilon}=\\{0\\}.$ For
limit ordinal $\delta,$ set
$\mathbb{G}^{\iota}_{\delta,\varepsilon}=\bigcup\limits_{\alpha<\delta}\mathbb{G}^{\iota}_{\alpha,\varepsilon}$.
Let us show that items (c) and (d) continue to hold. By the induction
hypothesis and for all $\alpha<\beta<\delta$ with $\alpha\notin S$,
$\mathbb{G}^{\iota}_{\beta,\varepsilon}/\mathbb{G}^{\iota}_{\alpha,\varepsilon}$
and $\mathbb{G}^{2}_{\beta}/(\mathbb{G}^{1}_{\beta}+\mathbb{G}^{2}_{\alpha})$
are free, and since we have a club $C_{\delta}$ of $\delta$ which is disjoint
to $S,$ it immediately follows that the groups
$\mathbb{G}^{\iota}_{\delta,\varepsilon}$,
$\mathbb{G}^{\iota}_{\delta,\varepsilon}/\mathbb{G}^{\iota}_{\alpha,\varepsilon}$
and
$\mathbb{G}^{2}_{\delta,\varepsilon}/(\mathbb{G}^{1}_{\delta,\varepsilon}+\mathbb{G}^{2}_{\alpha,\varepsilon})$
are free for all $\alpha<\delta$ with $\alpha\notin S$.

Let $\iota=1,2$. Now suppose that $\delta<\lambda$ and we have defined the
groups $\mathbb{G}^{\iota}_{\alpha,\varepsilon}$ and ordinals
$\alpha\leq\delta$. We would like to define the groups
$\mathbb{G}^{1}_{\delta+1,\varepsilon}$ and
$\mathbb{G}^{2}_{\delta+1,\varepsilon}$ so that the (a)-(e) continue to hold.

In the case $\delta\notin S_{\varepsilon},$ we set

$\mathbb{G}^{1}_{\delta+1,\varepsilon}:=\mathbb{G}^{1}_{\delta,\varepsilon}$

and

$\mathbb{G}^{2}_{\delta+1,\varepsilon}:=(\mathbb{G}^{1}_{\delta+1,\varepsilon}\oplus_{\mathbb{G}^{1}_{\delta+1,\varepsilon}}\mathbb{G}^{2}_{\delta,\varepsilon})\oplus\bigoplus\limits_{n<\omega}y_{\delta,n}\mathbb{Z}.$

It is not difficult to show that items (a)-(d) continue to hold and there is
nothing to prove for case (e).

Now suppose that $\delta\in S_{\varepsilon}$. We define the groups
$\mathbb{G}^{1}_{\delta+1,\varepsilon}$ and
$\mathbb{G}^{2}_{\delta+1,\varepsilon}$ such that items (a)-(e) above continue
to hold, and further we have:

  1. $(f)$

$\mathbb{G}^{2}_{\delta+1,\varepsilon}/(\mathbb{G}^{1}_{\delta,\varepsilon}+\mathbb{G}^{2}_{\delta,\varepsilon})$
is not free,

  2. $(g)$

if $\gamma\in\delta\backslash S_{\varepsilon}$, then
$\mathbb{G}^{2}_{\delta+1,\varepsilon}/(\mathbb{G}^{1}_{\delta,\varepsilon}+\mathbb{G}^{2}_{\gamma,\varepsilon})$
is free.

As $\delta\in S_{\varepsilon}$, we have ${\rm cf}(\delta)=\aleph_{0}$, so let
$\langle\gamma_{\delta,n}:n<\omega\rangle$ be an increasing sequence of
successor ordinals $<\delta$ with limit $\delta$.

For any $n<\omega,\gamma_{\delta,n}\notin S_{\varepsilon},$ it is easily seen
that the abelian groups
$\mathbb{G}^{1}_{\gamma_{\delta,n},\varepsilon}\subseteq\mathbb{G}^{2}_{\gamma_{\delta,n},\varepsilon}$
satisfy the hypotheses of Lemma 4.5, hence by the lemma, we can find the
groups
$\mathbb{G}^{1}_{\delta+1,\varepsilon}\subseteq\mathbb{G}^{2}_{\delta+1,\varepsilon}$
such that there for all constructions
$\mathbb{L}\supseteq\mathbb{G}^{1}_{\delta+1,\varepsilon}$ as described in
(e), there is no $\mathbb{f}\in{\rm
Hom}(\mathbb{G}^{2}_{\delta+1,\varepsilon},\mathbb{L})$ such that
$\mathbb{f}\supseteq\mathbb{f}^{\varepsilon}_{\delta}\cup{\rm
id}_{\mathbb{G}^{1}_{\delta+1,\varepsilon}}$.

This finishes our inductive construction. For $\iota=1,2$ and
$\varepsilon<\lambda$, we define:

$\mathbb{G}^{\iota}_{\varepsilon}=\bigcup\limits_{\alpha<\lambda}\mathbb{G}^{\iota}_{\alpha,\varepsilon}.$

Set also

  * (h1):

$\mathbb{G}_{\alpha,\varepsilon}=\mathbb{G}^{2}_{\alpha,\varepsilon}/\mathbb{G}^{1}_{\alpha,\varepsilon}$,

  * (h2):

$\mathbb{G}_{\varepsilon}=\bigcup\limits_{\alpha<\lambda}\mathbb{G}_{\alpha,\varepsilon},$

  * (h3):

$\bar{\mathbb{G}}=\langle\mathbb{G}_{\varepsilon}:\varepsilon<\lambda\rangle.$

Let us show that the properties.

$(i)$: This is clear.

$(ii)$: We apply the properties taken from items (c) and (d) of the
construction, along with freeness
$\mathbb{G}^{1}_{\varepsilon}\subseteq\mathbb{G}^{2}_{\varepsilon}$ to deduce
$\mathbb{G}_{\varepsilon}$ is strongly $\lambda$-free as witnessed by the
sequence

$\mathcal{S}_{\varepsilon}=\\{\mathbb{G}_{\alpha,\varepsilon}:\alpha\in\lambda\setminus
S_{\varepsilon}\\},$

for more details, see [2, IV.1.11].

$(iii)$: Suppose by contradiction that there are $\varepsilon<\lambda$, a
construction
$\bar{\mathbb{L}}=\langle\mathbb{L}_{\varepsilon}:\varepsilon\leq\lambda\rangle$
by $\\{\mathbb{G}_{\alpha}:\alpha\in\lambda\setminus\varepsilon\\}$ over
$\mathbb{G}^{1}_{\varepsilon}$ and there is a homomorphism ${\bf g}$ from
$\mathbb{G}^{2}_{\varepsilon}$ into $\mathbb{L}_{\lambda}$ which extends ${\rm
id}_{\mathbb{G}^{1}_{\varepsilon}}$:

$\textstyle{\mathbb{G}^{1}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{{\rm
id}}$$\textstyle{\mathbb{G}^{2}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\bf
g}}$$\textstyle{\mathbb{G}^{1}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\textstyle{\mathbb{L}}$

Without loss of generality we can assume that $\mathbb{L}_{\lambda}$ has size
$\lambda$ and that its universe is $\lambda$. The set

$E=\\{\delta<\lambda:{\bf g}\restriction\delta:\delta\to\delta$ and
$\bar{\mathbb{L}}\restriction\delta=\bar{\mathbb{L}}_{\delta}\restriction\delta$
and $\mathbb{L}_{\lambda}\restriction\delta$ is a subgroup of $\mathbb{L}\\}$

is a club, thus we can find some $\delta\in E\cap S_{\varepsilon}$ such that:

  * (i1):

${\bf g}\restriction\delta=f^{\varepsilon}_{\delta}$,

  * (i2):

$\mathbb{L}_{\kappa}\restriction\delta=\mathbb{H}^{2,\varepsilon}_{\delta}$,

  * (i3):

$\mathbb{H}^{1,\varepsilon}_{\delta}=\mathbb{G}^{1}_{\delta,\varepsilon},$

  * (i4):

$\mathbb{G}^{2}_{\varepsilon}\cap\delta=\mathbb{G}^{2}_{\delta,\varepsilon}.$

Now note that $f={\bf
g}\restriction\mathbb{G}^{2}_{\delta+1,\varepsilon}:\mathbb{G}^{2}_{\delta+1,\varepsilon}\to\mathbb{L}_{\lambda}$
is such that ${f}^{\varepsilon}_{\delta}\cup{\rm
id}_{\mathbb{G}_{\delta+1,\varepsilon}^{1}}\subseteq{f}$, which is in
contradiction with clause (e) of the construction.

$(2)$: The proof is similar to the proof of (1), this time, we find the
following family

$\langle S_{\varepsilon}:\varepsilon<2^{\lambda}\rangle$

of almost disjoint subsets of $\lambda$ such that $\Diamond_{S_{\varepsilon}}$
holds for all $\varepsilon$. By [1], such a sequence exists. ∎

Now, we are ready to prove Theorem (B):

######  Theorem 4.8.

Let $S\subseteq S^{\lambda}_{\aleph_{0}}$ be stationary non-reflecting and
suppose $\diamondsuit_{S}$ holds. Suppose one of the followings:

  1. $(1)$

$\mu=\lambda={\rm cf}(\lambda)>\aleph_{0}$, or

  2. $(2)$

$\mu=2^{\lambda},\lambda={\rm cf}(\lambda)>\aleph_{0}$.

Then there are sequences $\langle\mathbb{G}_{\alpha}:\alpha<\mu\rangle$ and
$\langle\mathbb{K}_{\alpha}:\alpha<\mu\rangle$ of $\lambda$-free abelian
groups such that for all $\alpha<\mu,|\mathbb{G}_{\alpha}|=\lambda,$
$|\mathbb{K}_{\alpha}|=2^{\lambda}$ and for all $\alpha,\beta<\mu,$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0\Leftrightarrow\alpha
R\beta.$

###### Proof.

This follows from Theorem 3.5 and Theorem 4.7. ∎

##  5\. Representing a bipartite graph by Ext in ZFC

In this section we show that it is possible to remove the diamond principle
from the construction of Section 4 and get ZFC result. The main result is
Theorem 5.5. This answers Herden’s question for the case of bipartite graphs.
We will do this by using a simple version of Shelah’s black box. In this case,
the groups $\mathbb{G}_{\alpha}$ that we construct are not $\lambda$-free, but
just $\aleph_{1}$-free.

######  Notation 5.1.

Let $\chi$ be be infinite cardinal. By ${\mathscr{H}}(\chi)$ we mean the
collection of sets of hereditary cardinality less than $\chi$.

Let us start by stating the version of the black box we are using in this
paper.

######  Theorem 5.2.

Let $\chi$, $\lambda$ and $\mu$ be infinite cardinals such that
$\lambda=\mu^{+},\mu^{\aleph_{0}}=\mu$, and $E_{0},\ldots,E_{m-1}$ are
pairwise disjoint stationary subsets of $\lambda$ consisting of ordinals of
cofinality $\omega$, and $\chi>\lambda$. Let $N$ be an expansion in a
countable language of $({\mathscr{H}}(\chi),\in,\lhd,\lambda)$ where $\lhd$ is
a well ordering of ${\mathscr{H}}(\chi)$. Then there is a family of countable
sets $\\{(M_{i},X_{i}):i\in I\\}$ such that the following properties hold:

  1. (a)

$M_{i}\prec N$ and $X_{i}\subset\lambda$.

  2. (b)

Let $\delta(i):=\sup(M_{i}\cap\lambda)$. If $\delta(i)=\delta(j)$, then
$(M_{i},X_{i})\cong(M_{j},X_{j})$ and $M_{i}\cap M_{j}\cap\lambda$ is an
initial segment of $M_{i}\cap\lambda$.

  3. (c)

For all $X\subset\lambda$, all $\ell<m$, the following set

$\\{\delta\in E_{\ell}:\exists i\emph{ such that }\delta(i)=\delta\emph{ and
}(M_{i},X_{i})\equiv_{M_{i}\cap\lambda}(N,X)\\}$

is stationary in $\lambda$.

###### Proof.

See [2, Page 444]. ∎

######  Notation 5.3.

Given a torsion-free group $\mathbb{G}$ and a subgroup
$\mathbb{H}\subseteq\mathbb{G}$. By $\mathbb{H}_{\ast}$ we mean the pure-
closure, i.e., the smallest pure subgroup of $\mathbb{G}$ containing
$\mathbb{H}$. In fact,

$\mathbb{H}_{\ast}=\\{g\in\mathbb{G}:ng\in\mathbb{H}\emph{ for some nonzero
}n\in\mathbb{Z}\\}.$

The next task is to construct $\boxtimes_{\lambda,\mu}^{1}$ from Theorem 3.5
in ZFC:

######  Theorem 5.4.

Adopt one of the following assumptions:

  1. (1)

If $\lambda=\lambda^{\aleph_{0}},\mu=\lambda$,

  2. (2)

If $\lambda=\lambda^{\aleph_{0}},\mu=2^{\lambda}$.

Then $\boxtimes^{1}_{\lambda,\mu}$ from Lemma 3.5 holds.

###### Proof.

(1): By a result of Solovay (see [2, Corollary II.4.9]) we can find a
partition $\langle S_{\varepsilon}:\varepsilon<\lambda\rangle$ of $\lambda$
into $\lambda$ many disjoint stationary sets. Let

$\mathbb{C}_{\varepsilon}:=\bigoplus\limits_{\nu<\lambda,n<\omega}(y_{\varepsilon,\nu,n}\mathbb{Z}\oplus
z_{\varepsilon,\nu,n}\mathbb{Z})$

and recall that its $\mathbb{Z}$-adic completion is denoted by
$\widehat{\mathbb{C}_{\varepsilon}}$. Clearly,
$\widehat{\mathbb{C}_{\varepsilon}}$ has cardinality $\lambda,$ so we identify
it with $\lambda$. Let also

$Y_{\varepsilon}:=\\{y_{\varepsilon,\nu,n}:\nu<\lambda,n<\omega\\}\cup\\{z_{\varepsilon,\nu,n}:\nu<\lambda,n<\omega\\}.$

Set $\chi=\lambda^{+3}$. The initial structure for the Black box,
corresponding to the stationary set $S_{\varepsilon}$ is as follows:

$N_{\varepsilon}:=({\mathscr{H}}(\chi),\in,\lhd,\lambda,\widehat{C_{\varepsilon}},Y_{\varepsilon}),$

where $\widehat{C_{\varepsilon}}$ denotes the 3-ary relation on $\lambda$
which is the graph of the addition operation on the group
$\widehat{\mathbb{C}_{\varepsilon}}$. We take the first bijection
$\mathbb{g}_{\varepsilon}:\lambda\times\lambda\to\lambda$ with respect to
$\lhd$, and use it to identify each $X\subseteq\lambda$ with a subset of
$\widehat{C_{\varepsilon}}\times\widehat{C_{\varepsilon}}.$ Let
$\\{(M^{\varepsilon}_{i},X^{\varepsilon}_{i}):i\in I_{\varepsilon}\\}$ be as
in the statement of Theorem 5.2 when $m=1$ and $E_{0}=S_{\varepsilon}$, and
note that for each $i\in I_{\varepsilon}$ and $\varepsilon<\lambda,$
$\mathbb{g}_{\varepsilon}\in M^{\varepsilon}_{i}$.

Let $\iota=1,2$. We proceed as in the previous section and for a given
$\varepsilon<\lambda$, by induction on $\alpha<\lambda$, we choose the
sequences
$\bar{\mathbb{G}}^{\iota}_{\varepsilon}=\langle\mathbb{G}^{\iota}_{\alpha,\varepsilon}:\alpha<\lambda\rangle$
equipped with the following five items:

  1. $(a)$

$\bar{\mathbb{G}}^{\iota}_{\varepsilon}$ is an increasing and continuous
sequence of abelian groups
$\mathbb{G}^{\iota}_{\alpha,\varepsilon}=(G^{\iota}_{\alpha,\varepsilon},+,0)$.

  2. $(b)$

$\mathbb{G}^{\iota}_{0,\varepsilon}=\\{0\\}$ and for $\alpha<\lambda,$ the
universe of $\mathbb{G}^{2}_{\alpha,\varepsilon}$, namely
$G^{2}_{\alpha,\varepsilon}$ is an ordinal
$\gamma_{\alpha,\varepsilon}<\lambda$.

  3. $(c)$

The following three properties hold:

    1. $(c1)$:

$\mathbb{G}^{1}_{\alpha,\varepsilon}\subseteq\mathbb{G}^{2}_{\alpha,\varepsilon}$
are $\aleph_{1}$-free,

    2. $(c2)$:

if $\alpha<\beta$ then
$\mathbb{G}^{2}_{\alpha,\varepsilon}\cap\mathbb{G}^{1}_{\beta,\varepsilon}=\mathbb{G}^{1}_{\alpha,\varepsilon}$,

    3. $(c3)$:

if $\alpha<\beta$ and $\alpha\notin S_{\varepsilon}$, then
$\mathbb{G}^{2}_{\beta}/(\mathbb{G}^{1}_{\beta}+\mathbb{G}^{2}_{\alpha})$ is
$\aleph_{1}$-free.

  4. $(d)$

If $\alpha<\beta$ and $\alpha\notin S_{\varepsilon}$, then
$\mathbb{G}^{1}_{\beta,\varepsilon}/\mathbb{G}^{1}_{\alpha,\varepsilon}$ and
$\mathbb{G}^{2}_{\beta,\varepsilon}/\mathbb{G}^{2}_{\alpha,\varepsilon}$ are
$\aleph_{1}$-free.

  5. $(e)$

If $\delta\in S_{\varepsilon},$ then there are no $f$ and
$\langle\dot{\mathbb{L}}_{\zeta}:\zeta\leq\delta\rangle$ equipped with the
following properties:

    1. $(e1)$

$\langle\dot{\mathbb{L}}_{\zeta}:\zeta\leq\delta\rangle$ is a construction
over $\mathbb{G}_{\delta+1,\varepsilon}^{1},$

    2. $(e2)$

$f\in{\rm
Hom}(\mathbb{G}^{2}_{\delta+1,\varepsilon},\dot{\mathbb{L}}_{\delta})$,

    3. $(e3)$

$f$ extends $f^{\varepsilon}_{\delta,i}\cup{\rm
id}_{\mathbb{G}^{1}_{\delta+1,\varepsilon}}$ where $i\in I_{\varepsilon}$ and

$\langle
f^{\varepsilon}_{\delta,i},\bar{\mathbb{L}}^{\varepsilon}_{\delta,i}=\langle\mathbb{L}^{\varepsilon}_{\delta,i,\zeta}:\zeta\leq\delta\rangle\rangle$

is coded by $X^{\varepsilon}_{i}$, under the identification given by
$\mathbb{g}_{\varepsilon}$. Also, $f^{\varepsilon}_{\delta,i}$ is in ${\rm
Hom}(\mathbb{G}^{2}_{\delta,\varepsilon},\mathbb{L}^{\varepsilon}_{\delta,i,\delta})$
and $\bar{\mathbb{L}}^{\varepsilon}_{\delta,i}$ is a construction:

$\textstyle{\dot{\mathbb{L}}_{\delta}}$$\textstyle{\mathbb{G}^{1}_{\delta+1,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{\subseteq}$$\textstyle{\mathbb{G}^{2}_{\delta+1,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nexists{f}}$$\textstyle{\mathbb{G}^{2}_{\delta,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{f}^{\varepsilon}_{\delta,i}}$$\scriptstyle{\subseteq}$

    4. $(e4)$

$\mathbb{L}^{\delta}_{\zeta}\subseteq\dot{\mathbb{L}}_{\zeta}$ for
$\zeta\leq\delta,$

    5. $(e5)$

$\mathbb{L}^{\delta}_{\zeta}=\mathbb{L}^{\delta}_{\zeta+1}\cap\dot{\mathbb{L}}_{\zeta},$

    6. $(e6)$

$\dot{\mathbb{L}}_{\zeta+1}/(\mathbb{L}^{\delta}_{\zeta+1}+\dot{\mathbb{L}}_{\zeta})$
is free.

For $\alpha=0,$ set
$\mathbb{G}^{1}_{0,\varepsilon}=\mathbb{G}^{2}_{0,\varepsilon}=\\{0\\}.$ For
the limit ordinal $\delta,$ we set
$\mathbb{G}^{\iota}_{\delta,\varepsilon}:=\bigcup\limits_{\alpha<\delta}\mathbb{G}^{\iota}_{\alpha,\varepsilon}$.
Let us show that items (c) and (d) continue to hold. By the induction
hypothesis and for all $\alpha<\beta<\delta$ with $\alpha\notin S$,
$\mathbb{G}^{\iota}_{\beta,\varepsilon}/\mathbb{G}^{\iota}_{\alpha,\varepsilon}$
and $\mathbb{G}^{2}_{\beta}/(\mathbb{G}^{1}_{\beta}+\mathbb{G}^{2}_{\alpha})$
are $\aleph_{1}$-free, and since we have a club $C_{\delta}$ of $\delta$ which
is disjoint to $S,$ it immediately follows from Lemma 3.1 that the groups
$\mathbb{G}^{\iota}_{\delta,\varepsilon}$,
$\mathbb{G}^{\iota}_{\delta,\varepsilon}/\mathbb{G}^{\iota}_{\alpha,\varepsilon}$
and
$\mathbb{G}^{2}_{\delta,\varepsilon}/(\mathbb{G}^{1}_{\delta,\varepsilon}+\mathbb{G}^{2}_{\alpha,\varepsilon})$
are $\aleph_{1}$-free for all $\alpha<\delta$ with $\alpha\notin
S_{\varepsilon}$.

Now suppose that $\delta<\lambda$ and we have defined the groups
$\mathbb{G}^{\iota}_{\alpha,\varepsilon}$ for $\iota=1,2$ and ordinals
$\alpha\leq\delta$. We would like to define the groups
$\mathbb{G}^{1}_{\delta+1,\varepsilon}$ and
$\mathbb{G}^{2}_{\delta+1,\varepsilon}$ so that the (a)-(e) continue to hold.

If $\delta\notin S_{\varepsilon},$ then we set

$\mathbb{G}^{1}_{\delta+1,\varepsilon}:=\mathbb{G}^{1}_{\delta,\varepsilon}$

and

$\mathbb{G}^{2}_{\delta+1,\varepsilon}:=(\mathbb{G}^{1}_{\delta+1,\varepsilon}\oplus_{\mathbb{G}^{1}_{\delta+1,\varepsilon}}\mathbb{G}^{2}_{\delta,\varepsilon})\oplus\bigoplus\limits_{n<\omega}y_{\varepsilon,\delta,n}\mathbb{Z}.$

We leave to the reader to check that the items presented from (a) to (d) all
are valid, and recall that there is nothing to prove for case (e).

Now suppose that $\delta\in S_{\varepsilon}$. We define the groups
$\mathbb{G}^{1}_{\delta+1,\varepsilon}$ and
$\mathbb{G}^{2}_{\delta+1,\varepsilon}$ such that items (a)-(e) above continue
to hold, and further we have:

  1. $(f)$

$\mathbb{G}^{2}_{\delta+1,\varepsilon}/(\mathbb{G}^{1}_{\delta,\varepsilon}+\mathbb{G}^{2}_{\delta,\varepsilon})$
is not free,

  2. $(g)$

if $\gamma\in\delta\backslash S_{\varepsilon}$, then
$\mathbb{G}^{2}_{\delta+1,\varepsilon}/(\mathbb{G}^{1}_{\delta,\varepsilon}+\mathbb{G}^{2}_{\gamma,\varepsilon})$
is $\aleph_{1}$-free.

As $\delta\in S_{\varepsilon}$, we have ${\rm cf}(\delta)=\aleph_{0}$, so let
$\langle\gamma_{\delta,n}:n<\omega\rangle$ be an increasing sequence of
successor ordinals $<\delta$ with limit $\delta$.

Let $n<\omega$ be such that $\gamma_{\delta,n}\notin S_{\varepsilon}$. It
turns out that the abelian groups
$\mathbb{G}^{1}_{\gamma_{\delta,n},\varepsilon}\subseteq\mathbb{G}^{2}_{\gamma_{\delta,n},\varepsilon}$
are suited well in the hypotheses of Lemma 4.5.

The notation $\Sigma_{\delta,\varepsilon}$ stands for the following set:

$\left\\{i\in I_{\varepsilon}:\delta(i)=\delta\emph{ and
}X^{\varepsilon}_{i}\emph{ codes }\langle
f^{\varepsilon}_{\delta,i},\bar{\mathbb{L}}^{\varepsilon}_{\delta,i}=\langle\mathbb{L}^{\varepsilon}_{\delta,i,\zeta}:\zeta\leq\delta\rangle\rangle\emph{
as in }(e3)\right\\}.$

Given any $i\in\Sigma_{\delta,\varepsilon}$, and according to Lemma 4.5, we
can find the $\aleph_{1}$-free groups
$\mathbb{G}^{1,i}_{\delta+1,\varepsilon}\subseteq\mathbb{G}^{2,i}_{\delta+1,\varepsilon}$
such that there for all constructions
$\langle\dot{\mathbb{L}}_{\zeta}:\zeta\leq\delta\rangle$ over
$\mathbb{G}^{1,i}_{\delta+1,\varepsilon}$ as in item (e), if we set
$\mathbb{L}=\dot{\mathbb{L}}_{\delta}$, then there is no $\mathbb{f}\in{\rm
Hom}(\mathbb{G}^{2,i}_{\delta+1,\varepsilon},\mathbb{L})$ such that
$\mathbb{f}\supseteq f^{\varepsilon}_{\delta,i}\cup{\rm
id}_{\mathbb{G}^{1,i}_{\delta+1,\varepsilon}}$. For $\iota=1,2$ we look at

$\mathbb{G}^{\iota}_{\delta+1,\varepsilon}:=\langle\mathbb{G}^{\iota}_{\delta,\varepsilon}\cup\bigcup\\{\mathbb{G}^{\iota,i}_{\delta+1,\varepsilon}:i\in\Sigma_{\delta,\varepsilon}\\}\rangle_{\ast},$

i.e., the pure closure of
$\langle\mathbb{G}^{\iota}_{\delta,\varepsilon}\cup\bigcup\\{\mathbb{G}^{\iota,i}_{\delta+1,\varepsilon}:i\in\Sigma_{\delta,\varepsilon}\\}\rangle$
in $\widehat{\mathbb{C}_{\varepsilon}}$. Let us show that the hypothesis
(a)-(e) hold. We first show that the group
$\frac{\mathbb{G}^{2}_{\delta+1,\varepsilon}}{\mathbb{G}^{2}_{\alpha,\varepsilon}}$
is $\aleph_{1}$-free, provided that $\alpha$ is a successor ordinal. Let
$\mathbb{K}$ be any countable subgroup of
$\mathbb{G}^{2}_{\delta+1,\varepsilon}$. We are going to show that
$\frac{\mathbb{K}+\mathbb{G}^{2}_{\delta+1,\varepsilon}}{\mathbb{G}^{2}_{\alpha,\varepsilon}}$
is free. There is an $\omega$-sequence $\\{i_{m}:m\in\omega\\}$ together with
a countable subgroup $\mathbb{I}\subset\mathbb{G}^{2}_{\delta,\varepsilon}$
such that $\mathbb{K}$ is the subgroup generated by $\mathbb{I}$ together with
some countable subset $\\{w_{n,{i_{m}}}:n,m\in\omega\\}$ of
$\bigcup\limits_{i\in\Sigma_{\delta,\varepsilon}}\mathbb{G}^{\iota,i}_{\delta+1,\varepsilon}$.
We can assume that for all $n,m$,
$y_{\varepsilon,\alpha_{n,{i_{m}}}}\in\mathbb{I}$. Choose an increasing
sequence of ordinals $\\{\alpha_{k}:k<\omega\\}$ with limit $\delta$ such that
$\alpha_{0}=\alpha$ and for all $m\in\omega$ and all but finitely many of $n$,
$\alpha_{n,i_{m}}\in\\{\alpha_{k}:k<\omega\\}$. Notice that for any successor
ordinal $\gamma<\delta$

$\mathbb{I}/(\mathbb{I}\cap\mathbb{G}^{2}_{\gamma,\varepsilon})\cong(\mathbb{I}+\mathbb{G}^{2}_{\gamma,\varepsilon})/\mathbb{G}^{2}_{\gamma,\varepsilon}$

which is free by the induction hypothesis. So for such $\gamma$,
$\mathbb{I}\cap\mathbb{G}^{2}_{\gamma,\varepsilon}$ is a direct summand of
$\mathbb{I}$. Inductively choose subgroups $\mathbb{I}_{k}$ so that:

$\mathbb{I}\cap\mathbb{G}^{2}_{{\alpha_{k}+1},\varepsilon}\oplus\mathbb{I}_{k}=\mathbb{I}\cap\mathbb{G}^{2}_{{\alpha_{k+1}},\varepsilon}$

for all $k$. Hence

$\mathbb{I}=\bigoplus_{k}\mathbb{I}_{k}\oplus\bigoplus\mathbb{Z}y_{\varepsilon,\alpha_{k},n}.$

In view of Theorem 5.2(b), we are able to choose $\\{n(m):m<\omega\\}$ so
that:

  * •

the collections $\\{\alpha_{n,i_{m}}:n(m)<n\\}$ are pairwise disjoint.

  * •

$\\{\alpha_{n,i_{m}}:n(m)<n\\}\subset\\{\alpha_{k}:k<\omega\\}$.

We observe that
$\frac{\mathbb{K}+\mathbb{G}^{2}_{\delta,\varepsilon}}{\mathbb{G}^{2}_{\delta,\varepsilon}}$
is isomorphic to the direct sum of $\bigoplus_{k}\mathbb{I}_{k}$ together with
the group freely generated by

$\\{w_{n,{i_{m}}}:n(m)\leq n\emph{ and }m\in\omega\\}$

and

$\\{y_{\varepsilon,\alpha_{k},n}:\forall m<\omega\text{~{} and~{} }n(m)\leq
n,\alpha_{k}\neq\alpha_{n,i_{m}}\\}.$

From this, the claim follows. By a similar argument, the group
$\frac{\mathbb{G}^{1}_{\delta+1,\varepsilon}}{\mathbb{G}^{1}_{\alpha,\varepsilon}}$
is $\aleph_{1}$-free, provided that $\alpha$ is a successor ordinal.

In the same vein, we also observe that the group
$\mathbb{G}^{1}_{\delta+1,\varepsilon}$ is $\aleph_{1}$-free and the
hypotheses (a)-(e) continue to hold.

The rest of the argument is similar to Theorem 4.7. Let us elaborate the main
idea of the proof. For $\iota=1,2$ and $\varepsilon<\lambda$, we define:

$\mathbb{G}^{\iota}_{\varepsilon}=\bigcup\limits_{\alpha<\lambda}\mathbb{G}^{\iota}_{\alpha,\varepsilon}.$

Set also

  * (h1):

$\mathbb{G}_{\alpha,\varepsilon}=\mathbb{G}^{2}_{\alpha,\varepsilon}/\mathbb{G}^{1}_{\alpha,\varepsilon}$,

  * (h2):

$\mathbb{G}_{\varepsilon}=\bigcup\limits_{\alpha<\lambda}\mathbb{G}_{\alpha,\varepsilon},$

  * (h3):

$\bar{\mathbb{G}}=\langle\mathbb{G}_{\varepsilon}:\varepsilon<\lambda\rangle.$

Let us show that $\boxtimes^{1}_{\lambda,\mu_{1}}$ is satisfied. By items (c)
and (d) of the construction,
$\mathbb{G}^{1}_{\varepsilon}\subseteq\mathbb{G}^{2}_{\varepsilon}$ are
$\aleph_{1}$-free and $\mathbb{G}_{\varepsilon}$ is $\aleph_{1}$-free as well.
Let us now show that $\boxtimes^{1}_{\lambda,\mu_{1}}(c)$ is satisfied as
well. Suppose by contradiction that there are $\varepsilon<\lambda$, a
construction $\mathbb{L}$ by
$\\{\mathbb{G}_{\alpha}:\alpha\in\lambda\setminus\varepsilon\\}$ over
$\mathbb{G}^{1}_{\varepsilon}$, witnessed by
$\langle\mathbb{L}_{\varepsilon}:\varepsilon\leq\lambda\rangle,$ and there is
a homomorphism ${\bf g}$ from $\mathbb{G}^{2}_{\varepsilon}$ into $\mathbb{L}$
which extends ${\rm id}_{\mathbb{G}^{1}_{\varepsilon}}$:

$\textstyle{\mathbb{G}^{1}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{=}$$\textstyle{\mathbb{G}^{2}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\bf
g}}$$\textstyle{\mathbb{G}^{1}_{\alpha}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\textstyle{\mathbb{L}}$

Without loss of generality we can assume that $\mathbb{L}$ has size $\lambda$
and that its universe is $\lambda$. Let $X\subseteq\lambda$ be codes
$\langle\bf
g,\langle\mathbb{L}_{\varepsilon}:\varepsilon\leq\lambda\rangle\rangle$. The
set

$E:=\\{\delta<\lambda:{\bf g}\restriction\delta:\delta\to\delta$,
$X\cap\delta$ codes $\langle{\bf
g}\restriction\delta,\langle\mathbb{L}_{\zeta}:\zeta\leq\delta\rangle\rangle$
and $\mathbb{L}\restriction\delta\subseteq_{group}\mathbb{L}\\}$

is a club, thus we can find some $\delta\in E\cap S_{\varepsilon}$ and some
$i$ with $\delta(i)=i$ such that
$(M^{\varepsilon}_{i},X^{\varepsilon}_{i})\equiv_{M^{\varepsilon}_{i}\cap\lambda}(N_{\varepsilon},X)$.
It then follows that
$M^{\varepsilon}_{i}\cap\mathbb{G}^{2}_{\varepsilon}=M^{\varepsilon}_{i}\cap\mathbb{G}^{2,i}_{\delta,\varepsilon}$,
and since
$(M^{\varepsilon}_{i},X^{\varepsilon}_{i})\equiv_{M^{\varepsilon}_{i}\cap\lambda}(N_{\varepsilon},X)$,
we can easily observe that $X^{\varepsilon}_{i}$ codes

$\langle{\bf g}\restriction
M^{\varepsilon}_{i}\cap\mathbb{G}^{2,i}_{\delta,\varepsilon},\langle\mathbb{L}_{\zeta}\cap
M^{\varepsilon}_{i}:\zeta\leq\lambda\rangle\rangle.$

Now by elementarily and the choice of $\bf g$,

$(M_{i}^{\varepsilon},X^{\varepsilon}_{i})\models$“$X^{\varepsilon}_{i}$ codes
a homomorphism $f^{\varepsilon}_{\delta,i}$ from
$\mathbb{G}^{2,i}_{\delta,\varepsilon}$.

It follows that $i\in\Sigma_{\delta,\varepsilon},$ and hence by our
construction, there is no $\mathbb{f}\in{\rm
Hom}(\mathbb{G}^{2,i}_{\delta+1,\varepsilon},\mathbb{L})$ such that
$\mathbb{f}$ extends $f^{\varepsilon}_{\delta,i}\cup{\rm
id}_{\mathbb{G}^{1,i}_{\delta+1,\varepsilon}}$. This property is conveniently
summarized by the subjoined diagram:

$\textstyle{\mathbb{L}}$$\textstyle{\mathbb{G}^{1,i}_{\delta+1,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{\subseteq}$$\textstyle{\mathbb{G}^{2,i}_{\delta+1,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nexists{\mathbb{f}}}$$\textstyle{\mathbb{G}^{1,i}_{\delta,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{\subseteq}$$\textstyle{\mathbb{G}^{2,i}_{\delta,\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{f^{\varepsilon}_{\delta,i}}$

This is not possible, as $\bf g$ is such an extension. This shows that there
is no homomorphism ${\bf g}$ as above and the result follows.

$(2)$: This is similar to the proof of (1). Take a sequence $\langle
S_{\varepsilon}:\varepsilon<2^{\lambda}\rangle$ of almost disjoint subsets of
$\lambda$ such that each $S_{\varepsilon}$ is stationary and proceed as
before. ∎

Now, we are ready to prove Theorem (C) from §1:

######  Theorem 5.5.

Let $\lambda=\lambda^{\aleph_{0}},\mu=2^{\lambda}$ and let
$R\subseteq\mu\times\mu$ be a relation. Then there are families
$\langle\mathbb{G}_{\alpha}:\alpha<\mu\rangle,$ and
$\langle\mathbb{K}_{\alpha}:\alpha<\mu\rangle$ of $\aleph_{1}$-free abelian
groups such that:

  1. (1)

for all $\alpha<\mu,\mathbb{G}_{\alpha}$ has size $\lambda$ and
$\mathbb{K}_{\alpha}$ has size $2^{\lambda},$

  2. (2)

for all $\alpha,\beta<\mu,$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0\iff\alpha R\beta.$

###### Proof.

This follows using Theorem 3.5 and Theorem 5.4. ∎

Let us close this section by showing the relation between Theorem 5.5 and the
result of Göbel, Shelah and Wallutis [5] stated in the introduction.

######  Lemma 5.6.

Let $\mu$ be an infinite cardinal. Then (1) implies (2), where:

  1. (1)

There are abelian groups
$\langle\mathbb{G}_{X},\mathbb{H}^{X}:X\subseteq\mu\rangle$ such that for all
$X,Y\subseteq\mu,$

${\rm Ext}(\mathbb{G}_{Y},\mathbb{H}^{X})=0\iff Y\subseteq X.$

  2. (2)

If $(\mu,R)$ is a bipartite graph, then there are families
$\langle\mathbb{G}_{\alpha}:\alpha<\mu\rangle,$ and
$\langle\mathbb{K}_{\alpha}:\alpha<\mu\rangle$ of abelian groups such that for
all $\alpha,\beta<\mu,$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0\iff\alpha R\beta.$

###### Proof.

Suppose (1) holds as witnessed by the sequence
$\langle\mathbb{G}_{X},\mathbb{H}^{X}:X\subseteq\mu\rangle$ and let $(\mu,R)$
be a bipartite graph. Given any $\alpha,\beta<\mu$ set
$Y_{\alpha}=\\{\alpha\\}$ and $X_{\beta}=\\{\alpha<\mu:\alpha R\beta\\}$. Now
define

  1. $i)$

$\mathbb{G}_{\alpha}:=\mathbb{G}_{Y_{\alpha}}$ and

  2. $ii)$

$\mathbb{K}_{\alpha}:=\mathbb{H}^{X_{\alpha}}$.

This is now straightforward to see

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0\iff Y_{\alpha}\subseteq
X_{\beta}\iff\alpha R\beta.$

Thus, the family
$\langle\mathbb{G}_{\alpha},\mathbb{K}_{\alpha}:\alpha<\mu\rangle$ is as
required. ∎

The converse of the above lemma also holds in the following sense:

######  Lemma 5.7.

Let $\mu$ be an infinite cardinal and set $\lambda=2^{\mu}$. Then (1) implies
(2), where:

  1. (1)

If $(\lambda,R)$ is a bipartite graph, then there are families
$\langle\mathbb{G}_{\alpha}:\alpha<\lambda\rangle,$ and
$\langle\mathbb{K}_{\alpha}:\alpha<\lambda\rangle$ of abelian groups such that
for all $\alpha,\beta<\lambda,$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0\iff\alpha R\beta.$

  2. (2)

There are abelian groups
$\langle\mathbb{G}_{X},\mathbb{H}^{X}:X\subseteq\mu\rangle$ such that for all
$X,Y\subseteq\mu,$

${\rm Ext}(\mathbb{G}_{Y},\mathbb{H}^{X})=0\iff Y\subseteq X.$

###### Proof.

Suppose (1) holds. Let $\langle X_{\alpha}:\alpha<\lambda\rangle$ and $\langle
Y_{\beta}:\beta<\lambda\rangle$ be two enumerations of $\mathcal{P}(\mu)$. Let
us define the bipartite graph $(\lambda,R)$ as

$\alpha R\beta\iff Y_{\alpha}\subseteq X_{\beta}.$

By (1), there exists a family
$\langle\mathbb{G}_{\alpha},\mathbb{K}_{\alpha}:\alpha<\lambda\rangle,$ of
abelian groups such that for all $\alpha,\beta<\lambda,$

${\rm Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0\iff\alpha R\beta.$

For $X,Y\subseteq\mu$, let $\alpha,\beta<\lambda$ be such that $X_{\beta}=X$
and $Y_{\alpha}=Y$ and set

  1. $\bullet$

$\mathbb{G}_{Y}:=\mathbb{G}_{\alpha}$;

  2. $\bullet$

$\mathbb{H}^{X}:=\mathbb{K}_{\beta}$.

Then

${\rm Ext}(\mathbb{G}_{Y},\mathbb{H}^{X})=0\iff{\rm
Ext}(\mathbb{G}_{\alpha},\mathbb{K}_{\beta})=0\iff\alpha R\beta\iff Y\subseteq
X.$

The lemma follows. ∎

##  6\. Representing a general graph by Ext

In this section we consider general graphs and discuss if they can be
represented by Ext as before. We do not know the result in ZFC, but we show
the following consistency result which shows that there are no restrictions on
such graphs in ZFC, as promised by Theorem (D) from the introduction:

######  Theorem 6.1.

Suppose GCH holds and the pair $(S,R)$ is a graph where $R\subseteq S\times
S$, and let $\lambda>|S|$ be an uncountable regular cardinal. Then there
exists a cardinal preserving generic extension of the universe, and there is a
family $\\{\mathbb{G}_{s}\\}_{s\in S}$ of $\lambda$-free abelian groups such
that

${\rm Ext}(\mathbb{G}_{s},\mathbb{G}_{t})=0\iff sRt.$

The rest of this section is devoted to the proof of the above theorem. Before
we go into the details, let us sketch the idea of the proof:

######  Discussion 6.2.

We first define a forcing notion ${\mathbb{P}}_{*}$ which adds a sequence
$\langle\mathbb{G}_{s}:s\in S\rangle$ of $\lambda$-free abelian groups of size
$\lambda$ such that for each $s,t\in S$ if $(s,t)\notin R,$ then for some
abelian group $\mathbb{H}_{s,t}$ of size $\lambda$ there exists an exact
sequence

$0\longrightarrow\mathbb{G}_{s}\longrightarrow\mathbb{H}_{s,t}\longrightarrow\mathbb{G}_{t}\longrightarrow
0$

which does not split. We apply this along with Fact 2.1 to deduce that ${\rm
Ext}(\mathbb{G}_{s},\mathbb{G}_{t})\neq 0.$ Then working in the generic
extension by ${\mathbb{P}}_{*},$ we define a cardinal preserving
$\lambda$-support iteration forcing notion of length $\lambda^{+}$, which
makes ${\rm Ext}(\mathbb{G}_{s},\mathbb{G}_{t})=0$ for all $s,t\in S$ with
$sRt.$ This is done by adding a splitter for any exact sequence

$0\longrightarrow\mathbb{G}_{s}\longrightarrow\mathbb{H}\longrightarrow\mathbb{G}_{t}\longrightarrow
0,$

where $\mathbb{H}$ is an abelian group of size $\lambda.$ By using a suitable
book-keeping argument, we make sure that at the end all such exact sequences
are considered for all pairs $(s,t)\in R.$ We also show that the exact
sequence

$0\longrightarrow\mathbb{G}_{s}\longrightarrow\mathbb{H}_{s,t}\longrightarrow\mathbb{G}_{t}\longrightarrow
0$

still fails to split after the iteration, which will complete the proof.

Let us now go into the details of the proof. Recall from Notation 5.1 that
${\mathscr{H}}(\chi)$ is the collection of sets of hereditary cardinality less
than $\chi$.

######  Notation 6.3.

Let $\Phi:\lambda^{+}\to{\mathscr{H}}(\lambda^{+})$ be such that
$\Phi^{-1}[x]\subseteq\lambda^{+}$ is unbounded for all
$x\in{\mathscr{H}}(\lambda^{+})$.

The existence of $\Phi$ follows by the GCH assumption. We will use $\Phi$ as
our book-keeping function. Here, we define the forcing notion
${\mathbb{P}}_{*}.$

######  Definition 6.4.

Let $\lambda=\text{ cf}(\lambda)>\aleph_{0}$ be an uncountable regular
cardinal, and let $S$ be a set of cardinality $<\lambda$. Finally, let
$R\subseteq S\times S$.

  * (a)

The forcing notion ${\mathbb{P}}_{*}$ consists of conditions

$p=\langle\langle\mathbb{G}^{p}_{s,\beta}:s\in
S,\beta\leq\alpha_{p}\rangle,E_{p},\langle\mathbf{x}^{p}_{s,t,\beta}:(s,t)\notin
R,\beta\leq\alpha_{p}\rangle\rangle,$

where

    1. (1)

$\alpha_{p}<\lambda$ is an ordinal,

    2. (2)

for each $s\in S,$
$\langle\mathbb{G}^{p}_{s,\beta}:\beta\leq\alpha_{p}\rangle$ is an increasing
and continuous sequence of free Abelian groups from ${\mathscr{H}}(\lambda)$,

    3. (3)

$E_{p}\subseteq(\alpha_{p}+1)\cap S^{\lambda}_{\aleph_{0}}$ does not reflect,

    4. (4)

$\mathbb{G}^{p}_{s,\beta}/\mathbb{G}^{p}_{s,\gamma}$ is free when
$\gamma<\beta<\alpha_{p},\gamma\notin E_{p},$

    5. (5)

if $(s,t)\notin R$, then

$\begin{CD}\mathbf{x}^{p}_{s,t,\beta}:=0@>{}>{}>\mathbb{G}^{p}_{t,\beta}@>{f^{p}_{s,t,\beta}}>{}>\mathbb{H}^{p}_{s,t,\beta}@>{g^{p}_{s,t,\beta}}>{}>\mathbb{G}^{p}_{s,\beta}@>{}>{}>0\end{CD}$

is an exact sequence, all increasing with $\beta\leq\alpha_{p}$, from
${\mathscr{H}}(\lambda)$:

$\begin{CD}\mathbf{x}^{p}_{s,t,\alpha_{p}}:=0@>{}>{}>\mathbb{G}^{p}_{t,\alpha_{p}}@>{f^{p}_{s,t,\alpha_{p}}}>{}>\mathbb{H}^{p}_{s,t,\alpha_{p}}@>{g^{p}_{s,t,\alpha_{p}}}>{}>\mathbb{G}^{p}_{s,\alpha_{p}}@>{}>{}>0\\\
\subseteq @A{}A{}A\subseteq @A{}A{}A\subseteq @A{}A{}A\\\
\vdots\vdots\vdots\\\ \subseteq @A{}A{}A\subseteq @A{}A{}A\subseteq
@A{}A{}A\\\
\mathbf{x}^{p}_{s,t,1}:=0@>{}>{}>\mathbb{G}^{p}_{t,1}@>{f^{p}_{s,t,1}}>{}>\mathbb{H}^{p}_{s,t,1}@>{g^{p}_{s,t,1}}>{}>\mathbb{G}^{p}_{s,1}@>{}>{}>0\\\
\subseteq @A{}A{}A\subseteq @A{}A{}A\subseteq @A{}A{}A\\\
\mathbf{x}^{p}_{s,t,0}:=0@>{}>{}>\mathbb{G}^{p}_{t,0}@>{f^{p}_{s,t,0}}>{}>\mathbb{H}^{p}_{s,t,0}<EMAIL_ADDRESS>\end{CD}$

  * (b)

Given $p,q\in{\mathbb{P}}_{*},$ let $p\leq q$ ($q$ is stronger than $p$) when

    1. (1)

$\alpha_{p}\leq\alpha_{q},$

    2. (2)

for all $s\in S$ and $\beta\leq\alpha_{p},$
$\mathbb{G}^{q}_{s,\beta}=\mathbb{G}^{p}_{s,\beta}$,

    3. (3)

$E^{q}\cap(\alpha_{p}+1)=E^{p}$,

    4. (4)

If $(s,t)\notin R$ and $\beta\leq\alpha_{p}$, then
$\mathbf{x}^{q}_{s,t,\beta}=\mathbf{x}^{p}_{s,t,\beta}.$

For simplicity of the reader we recall:

######  Definition 6.5.

Let ${\mathbb{P}}$ be a forcing notion and $\kappa$ be a regular cardinal.

  1. (a)

${\mathbb{P}}$ is called $\kappa$-distributive if the intersection of less
than $\kappa$-many dense open subsets of ${\mathbb{P}}$ is dense, or
equivalently, forcing with ${\mathbb{P}}$ does not add any new sequences of
ordinals of length less than $\kappa.$

  2. (b)

${\mathbb{P}}$ is called $\kappa$-strategically closed if for every
$\alpha<\kappa$, player II has a winning strategy in the following game:

     * $\Game_{{\mathbb{P}}}(\alpha)$

: the game has length $\alpha$ in which the players I and II take turns to
play conditions from ${\mathbb{P}}$ for $\alpha$-many moves, with player I
playing at odd stages and player II at even stages (including all limit
stages). II must play $1_{{\mathbb{P}}}$ at move zero. Let $p_{\beta}$ b the
condition played at move $\beta$; the player who played $p_{\beta}$ loses
immediately unless $p_{\beta}\geq p_{\gamma}$ for all $\gamma<\beta.$ If
neither player loses at any stage $\beta<\alpha,$ then player II wins.

  3. (c)

${\mathbb{P}}$ is called $\kappa$-c.c. if any antichain
$A\subseteq{\mathbb{P}}$ has size less than $\kappa.$

It is evident that if ${\mathbb{P}}$ is $\kappa$-strategically closed, then it
is $\kappa$-distributive.

######  Lemma 6.6.

Adopt the above notation. Then ${\mathbb{P}}_{*}$ is $\lambda$-distributive
and $\lambda^{+}$-c.c. In particular, forcing with ${\mathbb{P}}_{*}$ does not
add any new sequences of ordinals of size less than $\lambda$ and it preserves
all cardinals.

###### Proof.

Let us first show that ${\mathbb{P}}_{*}$ is $\lambda$-distributive. Thus
suppose that $\mu<\lambda$ and let $f:\mu\to V$ be a function in the generic
extension by ${\mathbb{P}}_{*}.$ We will show that $f$ is in $V$. Let
$\dot{f}$ be a ${\mathbb{P}}_{*}$-name for $f$ and let $p\in{\mathbb{P}}_{*}$
be such that

$p\Vdash\text{``}\dot{f}:\mu\to\check{V}\text{''}.$

By induction on $\xi\leq\mu$ we define a sequence $\langle
C_{\xi}:\xi\leq\mu\rangle$ of closed subsets of $\lambda$ and an increasing
sequence $\langle p_{\xi}:\xi\leq\mu\rangle$ of conditions in
${\mathbb{P}}_{*}$ such that:

  1. (1)

$C_{\xi}$ is a closed subset of $\alpha_{p_{\xi}}+1$ with $\alpha_{p_{\xi}}\in
C$, $C_{\xi}\cap\alpha_{p_{\xi}}$ is unbounded in $\alpha_{p_{\xi}}$ and
$C_{\xi}\cap E_{p_{\xi}}=\emptyset,$

  2. (2)

$C_{\xi+1}\cap\alpha_{p_{\xi}}+1=C_{\xi},$

  3. (3)

if $\xi\leq\mu$ is a limit ordinal, then
$C_{\xi}\cap\alpha_{p_{\xi}}=\bigcup\limits_{\zeta<\xi}C_{p_{\zeta}}$,

  4. (4)

$p_{0}=p,$

  5. (5)

for each $\xi<\mu,p_{\xi+1}$ decides $\dot{f}(\xi)$, say
$p_{\xi+1}\Vdash$“$\dot{f}(\xi)=a_{\xi}$”.

  6. (6)

If $\xi\leq\mu$ is a limit ordinal and $\langle p_{\zeta}:\zeta<\xi\rangle$ is
defined, then $p_{\xi}$ is the least upper bound of the sequence $\langle
p_{\zeta}:\zeta<\xi\rangle$ defined as follows:

    1. (a)

$\alpha_{p_{\xi}}=\sup\limits_{\zeta<\xi}\alpha_{\zeta}$,

    2. (b)

$E_{p_{\xi}}=\big{(}\bigcup\limits_{\zeta<\xi}E_{p_{\zeta}}\cup\\{\alpha_{p_{\xi}}\\}\big{)}\cap
S^{\lambda}_{\aleph_{0}}$,

    3. (c)

for $s\in S$ and
$\beta<\alpha_{p_{\xi}},G^{p_{\xi}}_{s,\beta}=G^{p_{\zeta}}_{s,\beta}$ for
some, and hence all, $\zeta<\xi$ with $\beta<\alpha_{p_{\zeta}}$,

    4. (d)

if $(s,t)\notin R$ and $\beta<\alpha_{p_{\xi}}$, then
$\mathbb{x}^{p_{\xi}}_{s,t,\beta}=\mathbb{x}^{p_{\zeta}}_{s,t,\beta}$ for
some, and hence all, $\zeta<\xi$ with $\beta<\alpha_{p_{\zeta}}$,

    5. (e)

for $s\in S$,
$G^{p_{\xi}}_{s,\alpha_{p_{\xi}}}=\bigcup\limits_{\zeta<\xi}G^{p_{\zeta}}_{s,\alpha_{p_{\zeta}}}$,

    6. (f)

if $(s,t)\notin R$, then
$\mathbb{x}^{p_{\xi}}_{s,t,\alpha_{p_{\xi}}}=\bigcup\limits_{\zeta<\xi}\mathbb{x}^{p_{\zeta}}_{s,t,\alpha_{p_{\zeta}}}$,
which is defined in the natural way.

Given $\xi\leq\mu$, let us show that $p_{\xi}$ as defined above is a
condition, as then it is clear that $p_{\xi}$ extends all $p_{\zeta}$’s , for
$\zeta<\xi.$ Items (1) and (5) from Definition 6.4 are clearly satisfied.

In order to see clause (2), it suffices to show that for each $s\in S,$
$G^{p_{\xi}}_{s,\alpha_{p_{\xi}}}$ is free. We have
$C_{\xi}\cap\alpha_{p_{\xi}}$ is a club of $\alpha_{p_{\xi}}$ and

$G^{p_{\xi}}_{s,\alpha_{p_{\xi}}}=\bigcup\limits_{\gamma\in
C_{\xi}\cap\alpha_{p_{\xi}}}G^{p_{\xi}}_{s,\gamma}.$

Furthermore, the sequence $\langle G^{p_{\xi}}_{s,\gamma}:\gamma\in
C_{\xi}\rangle$ is an increasing and continuous sequence of free groups, such
that if $\gamma<\beta$ are successive points in $C_{\xi},$ then
$G^{p_{\xi}}_{s,\beta}/G^{p_{\xi}}_{s,\gamma}$ is free. It follows that
$G^{p_{\xi}}_{s,\alpha_{p_{\xi}}}$ is free as required.

The set $E_{p_{\xi}}$ does not reflect as witnessed by $C_{\xi}$, hence clause
(3) is satisfied.

To show that clause (4) holds, let $\gamma<\beta<\alpha_{p_{\xi}}$ with
$\gamma\notin E_{p_{\xi}}.$ Let $\zeta<\xi$ be such that
$\beta<\alpha_{p_{\zeta}}$. This implies $\gamma\notin E_{p_{\zeta}}$, and
consequently

$G^{p_{\xi}}_{s,\beta}/G^{p_{\xi}}_{s,\gamma}=G^{p_{\zeta}}_{s,\beta}/G^{p_{\zeta}}_{s,\gamma}$

is free.

Let $q=p_{\mu},$ and let $h=\langle a_{\xi}:\xi<\mu\rangle$. Then $h\in V$ and
$q\Vdash$“$\dot{f}=h$”. As $p$ was arbitrary, we are done.

Now as $\lambda^{<\lambda}=\lambda$ and
${\mathbb{P}}_{*}\subseteq{\mathscr{H}}(\lambda)$, we have
$|{\mathbb{P}}_{*}|=\lambda$ and hence it clearly satisfies the
$\lambda^{+}$-c.c. ∎

######  Remark 6.7.

The above proof shows that the forcing notion ${\mathbb{P}}_{*}$ is indeed
$\lambda$-strategically closed.

######  Lemma 6.8.

Suppose $\mathbf{G}_{*}\subseteq{\mathbb{P}}_{*}$ is generic over $V$. Then
$E:=\bigcup\limits_{p\in\mathbf{G}_{*}}E_{p}$ is a non-reflecting stationary
subset of $\lambda.$

###### Proof.

This is standard, so we just sketch the proof. Let $p\in{\mathbb{P}}_{*}$ and
suppose that

$p\Vdash``\dot{C}\subseteq\lambda\emph{ is a club of }\lambda.^{\prime\prime}$

Let $\chi>\lambda^{+}$ be large enough regular, $\lhd$ be a well-ordering of
${\mathscr{H}}(\chi)$ and let $M\prec({\mathscr{H}}(\chi),\in,\lhd)$ be an
elementary submodel of ${\mathscr{H}}(\chi)$ such that:

  1. (1)

$|M|<\lambda$,

  2. (2)

$p,\dot{C},{\mathbb{P}}_{*},\cdots\in M,$

  3. (3)

$M\cap\lambda=\delta$ for some $\delta\in S^{\lambda}_{\aleph_{0}}$.

By induction on $n<\omega$ we define an increasing sequence $\langle
p_{n}:n<\omega\rangle$ of conditions, together with a sequence $\langle
C_{n}:n<\omega\rangle$ satisfying the following:

  * (4)

$p\leq p_{0},$

  * (5)

$p_{n+1}$ decides $\dot{C}\cap\alpha_{p_{n}}$ to be $C_{n}$,

  * (6)

$p_{n+2}\Vdash$“$\dot{C}\cap(\alpha_{p_{n}},\alpha_{p_{n+1}})\neq\emptyset$”,

  * (7)

$\sup\limits_{n<\omega}\alpha_{p_{n}}=\delta.$

We are now ready to define

$q:=\langle\langle\mathbb{G}^{q}_{s,\beta}:s\in
S,\beta\leq\delta\rangle,E_{q},\langle\mathbf{x}^{q}_{s,t,\beta}:(s,t)\notin
R,\beta\leq\delta\rangle\rangle,$

as follows:

  1. (8)

for $s\in S$ and
$\beta<\delta,\mathbb{G}^{q}_{s,\beta}=\mathbb{G}^{p_{n}}_{s,\beta},$ where
$n$ is such that $\alpha_{p_{n}}>\beta$,

  2. (9)

for $s\in
S,\mathbb{G}^{q}_{\delta}=\bigcup\limits_{n<\omega}\mathbb{G}^{p_{n}}_{s,\alpha_{p_{n}}}$,

  3. (10)

$E_{q}=\bigcup\limits_{n<\omega}E_{p_{n}}\cup\\{\delta\\}$,

  4. (11)

if $(s,t)\notin R$ and $\beta<\delta,$
$\mathbf{x}^{q}_{s,t,\beta}=\mathbf{x}^{p_{n}}_{s,t,\beta}$, for some $n$ with
$\alpha_{p_{n}}>\beta$,

  5. (12)

$\mathbf{x}^{q}_{s,t,\delta}$ is the exact sequence that is the direct limit
of the sequence $\mathbf{x}^{p_{n}}_{s,t,\alpha_{p_{n}}}$.

It is easily seen that $q\in{\mathbb{P}}_{*}$ is well-defined and it forces
$\delta\in\dot{C}\cap\dot{E}$, which completes the proof. ∎

Recall the following easy fact:

######  Fact 6.9.

(See [3]) Any subgroup of a free abelian group is free.

For each $s\in S$ and $\beta<\lambda$, we set
$\mathbb{G}_{s,\beta}=\mathbb{G}^{p}_{s,\beta}$ for some (and hence any)
$p\in\mathbf{G}_{*}$ with $\alpha_{p}\geq\beta$. Also, we set

$\mathbb{G}_{s}:=\bigcup\limits_{\beta<\lambda}\mathbb{G}_{s,\beta}.$

Let us first show that $\mathbb{G}_{s}$ is not free.

######  Lemma 6.10.

Suppose $s\in S$. Then $\mathbb{G}_{s}$ is a non-free strongly $\lambda$-free
abelian group of size $\lambda.$

###### Proof.

It is clear from Definition 6.4(a)(2) that $\mathbb{G}_{s}$ is a strongly
$\lambda$-free abelian group of size $\lambda.$ Let us show that it is not
free. Recall that we are in $\mathbb{V},$ and let $\mathbb{A}$ be a
$\lambda$-free abelian group of size $\lambda$ which is not free. We may
assume that the universe of $\mathbb{A}$ is $\lambda.$ Let also $\langle
A_{\alpha}:\alpha<\lambda\rangle$ be a filtration of $\mathbb{A}$. It follows
that the set

$E_{\mathbb{A}}:=\\{\alpha<\lambda:\mathbb{A}/\mathbb{A}_{\alpha}\text{~{}in
not~{}}\lambda\text{-free}\\}$

is stationary. We apply an argument similar to the proof of Lemma 6.8, and we
are able to show $E_{\mathbb{A}}$ remains stationary in
$\mathbb{V}[\mathbb{G}_{*}]$. In particular, $\mathbb{A}$ remains non-free in
the generic extension $\mathbb{V}[\mathbb{G}_{*}]$.

For each $\gamma<\lambda$ let $D_{\gamma}$ be the set of all
$p\in{\mathbb{P}}_{*}$ such that there exists $\beta$ such that:

  * •

$\alpha_{p}=\beta+1>\gamma$,

  * •

$\mathbb{G}^{p}_{s,\beta+1}$ includes $\mathbb{A}_{\beta}$ as a subgroup.

It is easily seen that each set $D_{\gamma}$ is dense in ${\mathbb{P}}_{*}.$
Now for each $\gamma<\lambda$ pick some $p_{\gamma}\in
D_{\gamma}\cap\mathbb{G}_{*}$ and set
$\alpha_{p_{\gamma}}=\beta_{\gamma}+1>\gamma$.

Now we look at the following commutative diagram:

$\begin{CD}\ldots
@>{\subseteq}>{}>\mathbb{A}_{\beta_{\gamma}}@>{\subseteq}>{}>\mathbb{A}_{\beta_{\gamma+1}}@>{\subseteq}>{}>\ldots\\\
@V{\subseteq}V{}V@V{\subseteq}V{}V@V{}V{\subseteq}V\\\ \ldots
@>{\subseteq}>{}>\mathbb{G}^{p_{\gamma}}_{s,\beta_{\gamma+1}}@>{\subseteq}>{}>\mathbb{G}^{p_{\gamma+1}}_{s,\beta_{\gamma+1}+1}@>{\subseteq}>{}>\ldots\end{CD}$

Taking direct limits of these directed systems, lead us to a natural inclusion
map

$\mathbb{G}_{s}=\bigcup\limits_{\gamma<\lambda}\mathbb{G}^{p_{\gamma}}_{s,\beta_{\gamma}+1}\supseteq\bigcup\limits_{\gamma<\lambda}\mathbb{A}_{\beta_{\gamma}}=\mathbb{A}.$

As $\mathbb{A}$ is non-free, it follows from Fact 6.9 that $\mathbb{G}_{s}$ is
also non-free. This completes the proof. ∎

For each $(s,t)\in(S\times S)\setminus R$, we look at the following
commutative diagram of short exact sequences:

$\begin{CD}\vdots\vdots\vdots\\\ \subseteq @A{}A{}A\subseteq @A{}A{}A\subseteq
@A{}A{}A\\\
\mathbf{x}_{s,t,\beta}:=0@>{}>{}>\mathbb{G}_{t,\beta}@>{f_{s,t,\beta}}>{}>\mathbb{H}_{s,t,\beta}@>{g_{s,t,\beta}}>{}>\mathbb{G}_{s,\beta}@>{}>{}>0\\\
\subseteq @A{}A{}A\subseteq @A{}A{}A\subseteq @A{}A{}A\\\
\vdots\vdots\vdots\\\ \subseteq @A{}A{}A\subseteq @A{}A{}A\subseteq
@A{}A{}A\\\
\mathbf{x}_{s,t,1}:=0@>{}>{}>\mathbb{G}_{t,1}@>{f_{s,t,1}}>{}>\mathbb{H}_{s,t,1}@>{g_{s,t,1}}>{}>\mathbb{G}_{s,1}@>{}>{}>0\\\
\subseteq @A{}A{}A\subseteq @A{}A{}A\subseteq @A{}A{}A\\\
\mathbf{x}_{s,t,0}:=0@>{}>{}>\mathbb{G}_{t,0}@>{f_{s,t,0}}>{}>\mathbb{H}_{s,t,0}@>{g_{s,t,0}}>{}>\mathbb{G}_{s,0}@>{}>{}>0,\\\
\end{CD}$

where for each
$\beta<\lambda,\mathbf{x}_{s,t,\beta}=\mathbf{x}_{s,t,\beta}^{p}$, for some
and hence any $p\in\mathbb{G}_{*}$ with $\alpha_{p}\geq\beta.$ By taking the
corresponding inductive limit, we lead to the following short exact sequence

$\begin{CD}\mathbf{x}_{s,t}:=0@>{}>{}>\mathbb{G}_{t}@>{f_{s,t}}>{}>\mathbb{H}_{s,t}<EMAIL_ADDRESS>
In other words,
$\mathbf{x}_{s,t}:=\lim_{\beta\to\lambda}\mathbf{x}^{p}_{s,t,\beta}$. The next
lemma shows that $\mathbf{x}_{s,t}$ does not split.

######  Lemma 6.11.

Suppose $s_{*},t_{*}\in S$ and $(s_{*},t_{*})\notin R$. Then in
$V[\mathbf{G}_{*}]$ the exact sequence $\mathbf{x}_{s_{*},t_{*}}$ does not
split. In particular, ${\rm Ext}(\mathbb{G}_{s_{*}},\mathbb{G}_{t_{*}})\neq
0.$

###### Proof.

Suppose towards contradiction that the exact sequence
$\mathbf{x}_{s_{*},t_{*}}$ splits and let $p\in{\mathbb{P}}_{*}$ and
$\textstyle h$ $\textstyle{\tilde{\mkern-3.0mu}\mkern 3.0mu}{}$ be such that

$p\Vdash$“$\mathchoice{\oalign{$\displaystyle h$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle h$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}:\mathchoice{\oalign{$\displaystyle\mathbb{G}$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle\mathbb{G}$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle\mathbb{G}$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle\mathbb{G}$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{s_{*}}\to\mathchoice{\oalign{$\displaystyle\mathbb{H}$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle\mathbb{H}$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle\mathbb{H}$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle\mathbb{H}$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{s_{*},t_{*}}$ is such that
$\mathchoice{\oalign{$\displaystyle h$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle h$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}\circ\mathchoice{\oalign{$\displaystyle g$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle g$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle g$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle g$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{s_{*},t_{*}}={\rm
id}_{\mathchoice{\oalign{$\displaystyle\mathbb{G}$\crcr\vbox
to0.60275pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle\mathbb{G}$\crcr\vbox
to0.60275pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle\mathbb{G}$\crcr\vbox
to0.60275pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle\mathbb{G}$\crcr\vbox
to0.60275pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{s_{*}}}$”.

As in the proof of Lemma 6.8, we can find an extension $q\leq p$ such that

  1. (1)

$\alpha_{q}\in E_{q}\cap S^{\lambda}_{\aleph_{0}},$

  2. (2)

$q$ decides $\mathchoice{\oalign{$\displaystyle h$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle h$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}\restriction\mathbb{G}^{q}_{s_{*},\alpha_{q}},$ say

$q\Vdash\text{``}\mathchoice{\oalign{$\displaystyle h$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle h$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}\restriction\mathbb{G}^{q}_{s_{*},\alpha_{q}}=h_{*}\text{''}.$

We will need the following claim.

######  Claim 6.12.

Let

$\begin{CD}\mathbf{x}_{s_{*},t_{*},\alpha_{q}}:=0@>{}>{}>\mathbb{G}^{q}_{t_{*},\alpha_{q}}@>{f^{q}_{s_{*},t_{*},\alpha_{q}}}>{}>\mathbb{H}^{q}_{s_{*},t_{*},\alpha_{q}}@>{g_{s_{*},t_{*},\alpha_{q}}}>{}>\mathbb{G}^{q}_{s_{*},\alpha_{q}}@>{}>{}>0\\\
\end{CD}$

be given and suppose $h_{*}$ is an splitting of $g_{s_{*},t_{*},\alpha_{q}}$.
Then there exists an exact sequence $\mathbf{x}_{s_{*},t_{*}}$ that fits in
the following commutative diagram

$\begin{CD}\mathbf{x}_{s_{*},t_{*}}:=0@>{}>{}>\mathbb{G}^{\prime}_{t_{*}}@>{f^{\prime}}>{}>\mathbb{H}^{\prime}_{s_{*},t_{*}}@>{g^{\prime}}>{}>\mathbb{G}^{\prime}_{s_{*}}@>{}>{}>0\\\
@A{}A{}A@A{}A{}A@A{}A{}A\\\
\mathbf{x}_{s_{*},t_{*},\alpha_{q}}:=0@>{}>{}>\mathbb{G}^{q}_{t_{*},\alpha_{q}}@>{f^{q}_{s_{*},t_{*},\alpha_{q}}}>{}>\mathbb{H}^{q}_{s_{*},t_{*},\alpha_{q}}@>{g_{s_{*},t_{*},\alpha_{q}}}>{}>\mathbb{G}^{q}_{s_{*},\alpha_{q}}@>{}>{}>0\\\
\end{CD}$

such that there exists no splitting $h^{\prime}$ of $g^{\prime}$ extending
$h_{*}$. Furthermore, there is an extension $r$ of $q$ such that
$\alpha_{r}=\alpha_{q}+1$ and
$\mathbb{x}^{r}_{s_{*},t_{*},\alpha_{r}}=\mathbb{x}_{s_{*},t_{*}}$.

###### Proof.

Without loss of generality, either $\alpha_{p}$ is a cardinal, or
$|\alpha_{p}|<\alpha_{p}$ and $\alpha_{p}^{\omega}=\alpha_{p}$. Since ${\rm
cf}(\alpha_{p})=\omega$, there is an increasing sequence
$\langle\gamma_{m}:m<n\rangle$ which is cofinal in $\alpha_{p}$ with
$\gamma_{m}\notin E_{q}$. To simplicity, let $(s_{1},s_{2})=(t_{*},s_{*})$ and
for $\ell=1,2$ we set
$\mathbb{G}_{s_{\ell}}:=\mathbb{G}^{q}_{s_{\ell},\alpha_{q}}$. We have

$\mathbb{G}_{s_{\ell}}=\bigcup\limits_{n<\omega}\mathbb{G}^{q}_{s_{\ell},\gamma_{n}},$

where for each $n$, both $\mathbb{G}^{q}_{s_{\ell},\gamma_{n}}$ and
$\mathbb{G}^{q}_{s_{\ell},\gamma_{n+1}}/\mathbb{G}^{q}_{s_{\ell},\gamma_{n}}$
are free, so we can find a free basis $\vec{x}_{\ell}:=\langle
x_{\alpha}^{\ell}:\alpha<\alpha_{q}\rangle$ of $\mathbb{G}_{s_{\ell}}$ such
that $\vec{x}_{\ell}{\restriction}\gamma_{n}$ is a free basis of
$\mathbb{G}^{q}_{s_{\ell},\gamma_{n}}$. Without loss of generality, we may and
do assume that $f^{q}_{s,t,\alpha_{q}}$ is the natural inclusion morphism:

$\mathbb{G}_{{s_{1}}}\stackrel{{\scriptstyle
f^{q}_{s,t,\alpha_{q}}}}{{\subseteq}}\mathbb{H}^{q}_{{s_{2},s_{1}},\alpha_{q}},$

and $h_{\ast}(x^{2}_{\alpha})=x^{2}_{\alpha}$ for any
$x^{2}_{\alpha}\in\mathbb{G}_{s_{2}}$.

We define the abelian group $\mathbb{G}_{s_{\ell}}^{\prime}$ to be generated
by

$X_{\ell}:=\mathbb{G}_{s_{\ell}}\cup\\{z^{\ell}_{n}:n<\omega\\}$

freely except the equations below which $\mathbb{G}_{s_{\ell}}^{\prime}$
satisfy:

  * $(\ast)_{1}$

: $n!z^{\ell}_{n-1}=z^{\ell}_{n}+x^{\ell}_{\gamma_{n}}$.

Recall that $\langle x^{1}_{\alpha},x^{2}_{\alpha}:\alpha<\alpha_{q}\rangle$
is a free basis of $\mathbb{H}_{s_{1},s_{2}}$. We let
$\mathbb{H}_{s_{1},s_{2}}^{\prime}$be the abelian group generated by

$\mathbb{H}_{s_{1},s_{2}}\cup\\{z_{n}:n<\omega\\},$

freely except the equations below which $\mathbb{H}_{s_{1},s_{2}}$ should
satisfy:

  * $\odot_{1}$

: $n!z_{n-1}=z_{n}+x^{1}_{\gamma_{n}}+x^{2}_{\gamma_{n}}$.

First, we show that these new abelian groups
$\\{\mathbb{G}_{s_{1}}^{\prime},\mathbb{H}_{s_{1},s_{2}}^{\prime},\mathbb{G}_{s_{2}}^{\prime}\\}$
are free, by presenting their bases:

  * $(b)_{1}$

: The group $\mathbb{G}_{s_{1}}^{\prime}$ is free. Indeed, in view of
$(\ast)_{1}$ we observe that $z^{1}_{n}=n!z^{1}_{n-1}-x^{1}_{\gamma_{n}}.$ An
easy inductive argument on $\ell$ yields that $z^{1}_{n}\in\langle
x^{\ell}_{\gamma_{n}},z^{1}_{0}\rangle,$ and so the set

$B_{1}:=\\{x^{1}_{\alpha}:\alpha<\alpha_{q}\\}\cup\\{z^{1}_{0}\\}$

generates $\mathbb{G}_{s_{1}}^{\prime}$. Since there is no relation involved
in $B_{1}$ we deduce that it is a free base of $\mathbb{G}_{s_{1}}^{\prime}$.
Thus, $\mathbb{G}_{s_{1}}^{\prime}$ is free, as claimed.

  * $(b)_{2}$

: The group $\mathbb{G}_{s_{2}}^{\prime}$ is free. Indeed, in $(b)_{1}$
replace $B_{1}$ with

$B_{2}:=\\{x^{1}_{\alpha}:\alpha<\alpha_{q}\\}\cup\\{z^{2}_{0}\\}$

and conclude that $\mathbb{G}_{s_{2}}^{\prime}$ is free.

  * $(b)_{3}$

: The group $\mathbb{H}_{s_{1},s_{2}}^{\prime}$ is free. Indeed, in view of
$\odot_{1}$ the set

$B_{3}:=\\{x^{1}_{\alpha},x^{2}_{\alpha}:\alpha<\alpha_{q}\\}\cup\\{z_{0}\\}$

generates $\mathbb{H}_{s_{1},s_{2}}^{\prime}$. Since there is no relation
involved in elements of $B_{3}$ we deduce that it is a free base of
$\mathbb{H}_{s_{1},s_{2}}^{\prime}$. Thus, $\mathbb{H}_{s_{1},s_{2}}^{\prime}$
is free, as claimed.

Now, we are going to define a map $f^{\prime}$ (resp. $g^{\prime}$) extending
${\rm id}_{\mathbb{G}_{s_{1}}}$ (resp. $g=g_{s_{1},s_{2},\alpha_{q}})$ such
that the following data becomes an exact sequence

$\begin{CD}0@>{}>{}>\mathbb{G}_{s_{1}}^{\prime}@>{f^{\prime}}>{}>\mathbb{H}_{s_{1},s_{2}}^{\prime}<EMAIL_ADDRESS>\end{CD}$

Since we need $f^{\prime}$ extends $f$, we define
$f^{\prime}(x^{1}_{\alpha}):=x^{1}_{\alpha}$. In order to complete the
definition of $f^{\prime}$, we proceed by induction on $n$ to define
$f^{\prime}(z^{1}_{n})$. When $n=0$, we set

$f^{\prime}(z^{1}_{0}):=z_{1}-z_{0}-x^{2}_{\gamma_{0}}\quad(\dagger).$

Now suppose $f^{\prime}(z^{1}_{n-1})$ is defined. We use the equation
$n!z^{1}_{n-1}=z^{\ell}_{n}+x^{1}_{\gamma_{n}}$ and define

$f^{\prime}(z^{1}_{n}):=n!f^{\prime}(z^{1}_{n-1})-x^{1}_{\gamma_{n}}\quad(+)$

Also, the assignments $z_{n}\mapsto z^{2}_{n}$, $x^{1}_{\alpha}\mapsto 0$ and
$x^{2}_{\alpha}\mapsto x^{2}_{\alpha}$ define a map
$g^{\prime}:\mathbb{H}_{s_{1},s_{2}}^{\prime}\to\mathbb{G}_{s_{2}}^{\prime}$
which extends $g$. This is well-defined. Namely, it sends the relation

$n!z_{n-1}=z_{n}+x^{1}_{\gamma_{n}}+x^{2}_{\gamma_{n}}$

from $\odot_{1}$ into the relation
$n!z^{2}_{n-1}=z^{2}_{n}+x^{2}_{\gamma_{n}}$ from $(\ast)_{1}$.

Let us show that ${\rm Rang}(f^{\prime})\subseteq\ker(g^{\prime})$. To see
this, first note that $g^{\prime}(f^{\prime}(z^{1}_{0}))=0$. Now suppose by
induction that $g^{\prime}(f^{\prime}(z^{1}_{n-1}))=0$. Then

$g^{\prime}(f^{\prime}(z^{1}_{n}))\stackrel{{\scriptstyle(+)}}{{=}}n!g^{\prime}(f^{\prime}(z^{1}_{n-1}))n!-g^{\prime}(x^{1}_{\gamma_{n}})=-g^{\prime}(x^{1}_{\gamma_{n}})=0.$

We proved that

${\rm Rang}(f^{\prime})\subseteq\ker(g^{\prime})\quad(\star)$

To see the reverse inclusion, we revisit $\odot_{1}$ and note that
$\ker(g^{\prime})$ is generated by

$\mathbb{G}_{s_{1}}\cup\\{k_{n}:=n!z_{n-1}-z_{n}-x^{2}_{\gamma_{n}}\\}.$

As $g^{\prime}$ extends $g$, we have

$\mathbb{G}_{s_{1}}={\rm Rang}(f)\subseteq{\rm
Rang}(f^{\prime})\stackrel{{\scriptstyle(\star)}}{{\subseteq}}\ker(g^{\prime}).$

By induction on $n$, we show that $k_{n}\in{\rm Rang}(f^{\prime})$. Following
$(+)$ it is enough to deal with $n=0$, and in this case
$k_{0}=f^{\prime}(z_{0}^{1})$. In sum, we proved that

$\begin{CD}0@>{}>{}>\mathbb{G}_{s_{1}}^{\prime}@>{f^{\prime}}>{}>\mathbb{H}_{s_{1},s_{2}}^{\prime}@>{g^{\prime}}>{}>\mathbb{G}_{s_{2}}^{\prime}@>{}>{}>0\\\
\end{CD}$

is an exact sequence of free abelian groups.

Assume toward the contradiction that there exists $h^{\prime}\in{\rm
Hom}(\mathbb{G}_{s_{2}}^{\prime},\mathbb{H}_{s_{1},s_{2}}^{\prime})$ such that
$h^{\prime}\supseteq h_{\ast}$. Let
$z^{\ast}_{n}:=h^{\prime}(z^{2}_{n})\in\mathbb{H}_{s_{1},s_{2}}^{\prime}$. As

$\mathbb{G}_{s_{2}}^{\prime}\models
n!z^{2}_{n-1}=z_{n}^{2}+x^{2}_{\gamma_{n}},$

we have

  * $\odot_{2}$

: $\mathbb{H}_{s_{1},s_{2}}^{\prime}\models
n!z^{\ast}_{n-1}=z_{n}^{\ast}+h_{\ast}(x^{2}_{\gamma_{n}}).$

Subtract $\odot_{1}$-$\odot_{2}$ we get

  * $\odot_{3}$

:

$\mathbb{H}_{s_{1},s_{2}}^{\prime}\models
n!(z_{n-1}-z^{*}_{n-1})=(z_{n}-z^{*}_{n})+x^{1}_{\gamma_{n}}.$

Noting that $g^{\prime}(z_{n}-z^{\ast}_{n})=z^{2}_{n}-z^{2}_{n}=0$. From this,
$z_{n}-z^{*}_{n}=f^{\prime}(y_{n})$ for some
$y_{n}\in\mathbb{G}_{s_{1}}^{\prime}$. Recall that $f^{\prime}$ is injective.
Apply this along with $\odot_{3}$ and deduce that

$\mathbb{G}_{s_{1}}^{\prime}\models n!y_{n-1}=y_{n}+x^{1}_{\gamma_{n}}.$

Due to the uniqueness of solutions of $(\ast)_{1}$ we conclude that

$f^{\prime}(z^{1}_{n})=f^{\prime}(y_{n})=z_{n}-z_{n}^{\ast}=z_{n}-h^{\prime}(z^{2}_{n}).$

In other words, we determined $h^{\prime}$, that is:

$h^{\prime}(z^{2}_{n})=z_{n}-f^{\prime}(z^{1}_{n}).$

We substitute this from $(\dagger)$ and evaluate $g^{\prime}$ on the both
sides of that equation, then we observe that

$\begin{array}[]{clcr}z^{2}_{0}&=g^{\prime}(h^{\prime}(z^{2}_{0}))\\\
&=g^{\prime}(z_{0}-f^{\prime}(z^{1}_{0}))\\\
&=g^{\prime}(z_{0}-(z_{1}-z_{0}-x^{2}_{\gamma_{0}}))\\\
&=z_{0}^{2}-z_{1}^{2}+z_{0}^{2}+x^{2}_{\gamma_{0}}.\end{array}$

Thus

$z_{0}^{2}=z_{1}^{2}-x^{2}_{\gamma_{0}}.$

But recalling from $(\ast)_{1}$,

$z_{0}^{2}=z_{1}^{2}+x^{2}_{\gamma_{0}},$

which imply $x^{2}_{\gamma_{0}}=0,$ a contradiction. This contradiction shows
that a such $h^{\prime}$ does not exist.

Finally let $r$ be the extension of $q$ such that:

  * $r_{1})$

: $\alpha_{r}=\alpha_{q}+1$,

  * $r_{2})$

: $\mathbb{G}^{r}_{s,\alpha_{r}}=\mathbb{G}^{q}_{\alpha_{q}}$ if
$s\notin\\{s_{*},t_{*}\\}$,

  * $r_{3})$

: $\mathbf{x}^{r}_{s_{*},t_{*},\alpha_{r}}=\mathbf{x}_{s_{*},t_{*}},$ where
$\mathbf{x}_{s_{*},t_{*}}$ is the exact sequence defined above,

  * $r_{4})$

: $\mathbf{x}^{r}_{s,t,\alpha_{r}}=\mathbf{x}^{q}_{s,t,\alpha_{q}}$ for all
$(s,t)\notin R$ such that $s\neq s_{*}$ and $t\neq t_{*},$

  * $r_{5})$

: if $s=s_{*}$ and $t\neq t_{*},$ then the desired sequence
$\mathbf{x}^{r}_{s_{*},t,\alpha_{r}}$ is defined as follows

$\begin{CD}0@>{}>{}>\mathbb{G}^{r}_{t,\alpha_{r}}@>{f^{r}_{s_{*},t,\alpha_{r}}}>{}>\mathbb{H}^{r}_{s_{*},t,\alpha_{r}}@>{g^{r}_{s_{*},t,\alpha_{r}}}>{}>\mathbb{G}^{r}_{s_{*},\alpha_{r}}@>{}>{}>0,\\\
\end{CD}$

where

    * $(r_{5}.i)$

:
$\mathbb{H}^{r}_{s_{*},t,\alpha_{r}}=\frac{\mathbb{H}^{q}_{s_{*},t,\alpha_{q}}\oplus\bigoplus_{n<\omega}z_{n}\mathbb{Z}}{\langle
n!z_{n-1}-z_{n}-x^{2}_{\gamma_{n}}\rangle}$,

    * $(r_{5}.ii)$

: $f^{r}_{s_{*},t,\alpha_{r}}=f^{q}_{s_{*},t,\alpha_{q}}$,

    * ($r_{5}.iii)$

:
$g^{r}_{s_{*},t,\alpha_{r}}\restriction\mathbb{H}^{q}_{s_{*},t,\alpha_{q}}=g^{q}_{s_{*},t,\alpha_{q}}$
and $g^{r}_{s_{*},t,\alpha_{r}}(z_{n})=z^{2}_{n},$

  * $r_{6})$

: In the case $s\neq s_{*}$ and $t=t_{*},$ the proposed sequence
$\mathbf{x}^{r}_{s,t_{*},\alpha_{r}}$ is defined as follows

$\begin{CD}0@>{}>{}>\mathbb{G}^{r}_{t_{*},\alpha_{r}}@>{f^{r}_{s,t_{*},\alpha_{r}}}>{}>\mathbb{H}^{r}_{s,t_{*},\alpha_{r}}@>{g^{r}_{s,t_{*},\alpha_{r}}}>{}>\mathbb{G}^{r}_{s,\alpha_{r}}@>{}>{}>0,\\\
\end{CD}$

where

    * $(r_{6}.i)$

:
$\mathbb{H}^{r}_{s,t_{*},\alpha_{r}}=\frac{\mathbb{H}^{q}_{s,t_{*},\alpha_{q}}\oplus\bigoplus_{n<\omega}z_{n}\mathbb{Z}}{\langle
n!z_{n-1}-z_{n}-x^{1}_{\gamma_{n}}\rangle}$,

    * ($r_{6}.ii)$

:
$f^{r}_{s,t_{*},\alpha_{r}}\restriction\mathbb{G}^{q}_{t_{*},\alpha_{q}}=f^{q}_{s_{*},t,\alpha_{q}}$
and $f^{r}_{s,t_{*},\alpha_{r}}(z^{1}_{n})=0$,

    * $(r_{6}.iii)$

:
$g^{r}_{s_{*},t,\alpha_{r}}\restriction\mathbb{H}^{q}_{s_{*},t,\alpha_{q}}=g^{q}_{s_{*},t,\alpha_{q}}$
and $g^{r}_{s_{*},t,\alpha_{r}}(z_{n})=0.$

It is easily seen that $r$ as defined above is a condition extending $q$. ∎

Let us continue the proof of Lemma 6.11. By the way we defined $r$, see the
above itemized properties $r_{1}),\ldots,r_{6})$, we have

$r\Vdash$“there is no splitting map for $\mathchoice{\oalign{$\displaystyle
g$\crcr\vbox to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle g$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle g$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle g$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{s_{*},t_{*}}$ extending $h_{*}$”.

In order to see this, let $r^{\prime}\leq r$ be such that $r^{\prime}$ decides
$\mathchoice{\oalign{$\displaystyle h$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle h$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}\restriction\mathbb{G}^{r^{\prime}}_{s_{*},\alpha_{r}}$. Then

$r^{\prime}\Vdash\text{``}\mathchoice{\oalign{$\displaystyle h$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle h$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}\restriction\mathbb{G}^{r^{\prime}}_{s_{*},\alpha_{r}}=g^{\prime}\text{''}.$

This contradicts the choice of $g^{\prime}$. The argument of Lemma 6.11 is now
completed. ∎

Let us now work in the generic extension $V[\mathbf{G}_{*}]$. We define a
$\lambda$-support iteration

${\mathbb{P}}=\langle\langle{\mathbb{P}}_{\alpha}:\alpha\leq\lambda^{+}\rangle,\langle\dot{{\mathbb{Q}}}_{\beta}:\beta<\lambda^{+}\rangle\rangle$

of forcing notions which forces ${\rm Ext}(\mathbb{G}_{s},\mathbb{G}_{t})=0$
for all $s,t\in S$ with $sRt$. We first define the building blocks of this
iteration.

Suppose $W\supseteq V[\mathbf{G}_{*}]$ is a forcing extension of
$V[\mathbf{G}_{*}]$, $s,t\in S$ are such that $sRt$ and suppose that

$\begin{CD}\mathbf{x}:=0@>{}>{}>\mathbb{G}_{s}@>{f}>{}>\mathbb{H}@>{g}>{}>\mathbb{G}_{t}@>{}>{}>0\end{CD}$

is an exact sequence in $W$.

######  Definition 6.13.

The forcing notion ${\mathbb{Q}}^{W}_{s,t,\mathbf{x}}$ consists of partial
functions $q:\mathbb{G}_{t}\dashrightarrow\mathbb{H}$ with domain ${\rm
dom}(q)$ such that:

  1. (1)

${\rm dom}(q)=\mathbb{G}_{s,\gamma}$, where $\gamma\notin E$,

  2. (2)

$g\circ q={\rm id}_{{\rm dom}(q)}$.

${\mathbb{Q}}^{W}_{s,t,\mathbf{x}}$ is ordered by inclusion.

Thus, the forcing notion ${\mathbb{Q}}^{W}_{s,t,\mathbf{x}}$ aims to add a
splitter for the exact sequence $\mathbf{x}$. The next lemma shows that this
is indeed possible.

######  Lemma 6.14.

The forcing notion ${\mathbb{Q}}^{W}_{s,t,\mathbf{x}}$ is $\lambda$-closed,
$\lambda^{+}$-c.c., and forcing with it adds a function
$h:\mathbb{G}_{t}\to\mathbb{H}$ such that $g\circ h={\rm
id}_{\mathbb{G}_{t}}$.

###### Proof.

The fact that ${\mathbb{Q}}^{W}_{s,t,\mathbf{x}}$ is $\lambda$-closed follows
from a combination of Definition 6.13(1) along with Definition 6.4(4). This
allows us to take unions (or direct limits) at limit stages and still have a
condition. According to $\Delta$-system lemma, the forcing is
$\lambda^{+}$-c.c., and claimed. ∎

We are finally ready to define our iteration. Let $\beta<\lambda^{+}$ and
suppose that ${\mathbb{P}}_{\beta}$ is defined. If $\Phi(\beta)$ is a
${\mathbb{P}}_{*}\ast\dot{{\mathbb{P}}}_{\beta}$-name for a triple
$(s,t,\dot{\mathbf{x}}),$ where $s,t\in S,sRt$ and $\dot{\mathbf{x}}$ is a
name for an exact sequence

$\begin{CD}\mathbf{x}:=0@>{}>{}>\mathbb{G}_{s}@>{\dot{f}}>{}>\dot{\mathbb{H}}@>{\dot{g}}>{}>\mathbb{G}_{t}@>{}>{}>0,\end{CD}$

in ${\mathscr{H}}(\lambda^{+})$, then

$\Vdash_{{\mathbb{P}}_{\beta}}\text{``}\dot{{\mathbb{Q}}}_{\beta}=\dot{{\mathbb{Q}}}^{V^{{\mathbb{P}}_{*}\ast\dot{{\mathbb{P}}}_{\beta}}}_{s,t,\dot{\mathbf{x}}}\text{''.}$

Otherwise, let $\dot{{\mathbb{Q}}}_{\beta}$ be forced to be the trivial
forcing notion.

The next lemma follows from [16].

######  Lemma 6.15.

Work in $V[\mathbf{G}_{*}].$ The forcing notion
${\mathbb{P}}={\mathbb{P}}_{\lambda^{+}}$ is $\lambda$-complete and
$\lambda^{+}$-c.c.

It follows from the above lemma that forcing with ${\mathbb{P}}$ preserves all
cardinals and adds no new sequences of ordinals of length less than $\lambda$.
Suppose

$\mathbf{G}=\langle\langle\mathbf{G}_{\alpha}:\alpha\leq\lambda^{+}\rangle,\langle\mathbf{H}_{\beta}:\beta<\lambda^{+}\rangle\rangle$

is ${\mathbb{P}}$-generic over $V[\mathbf{G}_{*}]$.

######  Lemma 6.16.

Work in $V[\mathbf{G}_{*}\ast\mathbf{G}]$. If $s,t\in S$ and $sRt$, then ${\rm
Ext}(\mathbb{G}_{s},\mathbb{G}_{t})=0$.

###### Proof.

It suffices to show that any exact sequence

$\begin{CD}\mathbf{x}:=0@>{}>{}>\mathbb{G}_{s}@>{f}>{}>\mathbb{H}@>{g}>{}>\mathbb{G}_{t}@>{}>{}>0\end{CD}$

with $\mathbb{H}$ of size $\lambda$ splits. Let $\dot{\mathbf{x}}$ be a
${\mathbb{P}}_{*}\ast{\mathbb{P}}$-name for $\mathbf{x}$ which we may assume
that $\dot{\mathbf{x}}\in{\mathscr{H}}(\lambda^{+})$. Furthermore, we can find
some $\alpha<\lambda^{+}$ such that $\dot{\mathbf{x}}$ is a
${\mathbb{P}}_{*}\ast{\mathbb{P}}_{\alpha}$-name, and then by the choice of
$\Phi,$ we may also assume that $\Phi(\alpha)=\dot{\mathbf{x}}$. By definition
of forcing notion, we know

$\Vdash_{{\mathbb{P}}_{*}\ast{\mathbb{P}}_{\alpha+1}}\text{``}\dot{\mathbf{x}}\text{~{}splits~{}}\text{'',}$

and consequently

$\Vdash_{{\mathbb{P}}_{*}\ast{\mathbb{P}}}\text{``}\dot{\mathbf{x}}\text{~{}splits~{}}\text{''.}$

This completes the proof. ∎

Let $(s,t)\in(S\times S)\setminus R$. We now show that the iteration does not
add splitters for $\mathbf{x}_{s,t}$,

######  Lemma 6.17.

Suppose $(s,t)\in(S\times S)\setminus R$. Then the exact sequence

$\begin{CD}\mathbf{x}_{s,t}:=0@>{}>{}>\mathbb{G}_{s}@>{f_{s,t}}>{}>\mathbb{H}_{s,t}@>{g_{s,t}}>{}>\mathbb{G}_{t}@>{}>{}>0\end{CD}$

does not split in $V[\mathbf{G}_{*}\ast\mathbf{G}]$.

###### Proof.

Since the argument is similar to the proof of Lemma 6.11, we just present the
sketch of proof. Here, we work in $V$. Let us first combine Lemma 6.6 along
with Lemma 6.15, and note that the set of conditions of the form

$(p,\langle
q_{\alpha}:\alpha<\lambda^{+}\rangle)\in{\mathbb{P}}_{*}\ast{\mathbb{P}}_{\lambda^{+}}$

such that $p,q_{\alpha}\in V$, is dense in
${\mathbb{P}}_{*}\ast{\mathbb{P}}_{\lambda^{+}}$. Suppose towards
contradiction that the exact sequence $\mathbf{x}_{s,t}$ splits. This gives us

$(p,\langle
p_{\alpha}:\alpha<\lambda^{+}\rangle)\in{\mathbb{P}}_{*}\ast{\mathbb{P}}_{\lambda^{+}}$

and  $\textstyle h$ $\textstyle{\tilde{\mkern-3.0mu}\mkern 3.0mu}{}$
satisfying:

$(p,\langle
p_{\alpha}:\alpha<\lambda^{+}\rangle)\Vdash$“$\mathchoice{\oalign{$\displaystyle
h$\crcr\vbox to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle h$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}:\mathchoice{\oalign{$\displaystyle\mathbb{G}$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle\mathbb{G}$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle\mathbb{G}$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle\mathbb{G}$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{s}\to\mathchoice{\oalign{$\displaystyle\mathbb{H}$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle\mathbb{H}$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle\mathbb{H}$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle\mathbb{H}$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{s,t}$ is such that $\mathchoice{\oalign{$\displaystyle
h$\crcr\vbox to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle h$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}\circ\mathchoice{\oalign{$\displaystyle g$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle g$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle g$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle g$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{s,t}={\rm
id}_{\mathchoice{\oalign{$\displaystyle\mathbb{G}$\crcr\vbox
to0.60275pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle\mathbb{G}$\crcr\vbox
to0.60275pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle\mathbb{G}$\crcr\vbox
to0.60275pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle\mathbb{G}$\crcr\vbox
to0.60275pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{s}}$”.

As before, we can find an extension

$(p,\langle p_{\alpha}:\alpha<\lambda^{+}\rangle)\leq(q,\langle
q_{\alpha}:\alpha<\lambda^{+}\rangle)$

equipped with the following two properties:

  1. (1)

$\alpha_{q}\in E_{q}\cap S^{\lambda}_{\aleph_{0}},$

  2. (2)

$(q,\langle q_{\alpha}:\alpha<\lambda^{+}\rangle)$ decides
$\mathchoice{\oalign{$\displaystyle h$\crcr\vbox
to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle h$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}\restriction\mathbb{G}^{q}_{s,\alpha_{q}},$ say

$(q,\langle
q_{\alpha}:\alpha<\lambda^{+}\rangle)\Vdash\text{``}\mathchoice{\oalign{$\displaystyle
h$\crcr\vbox to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle h$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle h$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}\restriction\mathbb{G}^{q}_{s,\alpha_{q}}=h_{*}\text{''}.$

In view of Claim 6.12 we can find an extension $r$ of $q$ such that
$(r,\langle q_{\alpha}:\alpha<\lambda^{+}\rangle)$ extends $(q,\langle
q_{\alpha}:\alpha<\lambda^{+}\rangle)$ and also, it forces

$``\emph{there is no splitting map for }\mathchoice{\oalign{$\displaystyle
g$\crcr\vbox to0.86108pt{\hbox{$\displaystyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\textstyle g$\crcr\vbox
to0.86108pt{\hbox{$\textstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptstyle g$\crcr\vbox
to0.86108pt{\hbox{$\scriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}{\oalign{$\scriptscriptstyle g$\crcr\vbox
to0.86108pt{\hbox{$\scriptscriptstyle{\tilde{\mkern-3.0mu}\mkern
3.0mu}{}$}\vss}}}_{s,t}\emph{ extending }h_{*}.^{\prime\prime}$

We get a contradiction and the lemma follows. ∎

This completes the proof of Theorem 6.1.

## References

  * [1] Keith J. Devlin, _A note on the combinatorial principles_ $\Diamond(E)$, Proc. AMS 72 (1978), no. 1, 163-165.
  * [2] Paul C. Eklof and Alan Mekler, _Almost free modules: Set theoretic methods_ , Revised Edition, North–Holland Publishing Co., North–Holland Mathematical Library, 65, 2002\.
  * [3] Laszlo Fuchs, _Abelian groups_ , Springer Monographs in Mathematics. Springer, Cham, 2015.
  * [4] Mark Hovey, _Model categories_ , Mathematical Surveys and Monographs, 63 American Mathematical Society, Providence, RI, 1999.
  * [5] Rüdiger Göbel, Saharon Shelah, and Simone L. Wallutis, _On the lattice of cotorsion theories_ , J. Algebra 238 (2001), 292–313.
  * [6] Rüdiger Göbel and Saharon Shelah, _Cotorsion pairs and splitters_ , Trans. AMS, 352 (2000), 5357-5379.
  * [7] Rüdiger Göbel and Jan Trlifaj, _Approximations and endomorphism algebras of modules_ , Vols. 1, 2, de Gruyter Expositions in Mathematics, Vol. 41, Walter de Gruyter, Berlin, 2012.
  * [8] Daniel Herden, _Personal communication with Saharon Shelah_.
  * [9] Thomas Jech, _Set theory_ , The third millennium edition, revised and expanded. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2003.
  * [10] Kenneth Kunen, _Set theory. An introduction to independence proofs_ , Studies in Logic and the Foundations of Mathematics, 102. North-Holland Publishing Co., Amsterdam-New York, 1980.
  * [11] Alan Mekler, Andrzej Rosłanowski, and Saharon Shelah, _On the $p$-rank of Ext_, Israel J. Math., 112 (1999), 327–356.
  * [12] Luigi Salce, _Cotorsion theories for abelian groups_ , Symp. Math. 23 (1979), 11-32.
  * [13] Saharon Shelah, _Infinite abelian groups, Whitehead problem and some constructions_ , Israel Journal of Mathematics 18 (1974), 243-256.
  * [14] Saharon Shelah, _A combinatorial principle and endomorphism rings. I_ , On p-groups, Israel Journal of Mathematics 49 (1984), 239–-257, Proceedings of the 1980/1 Jerusalem Model Theory year.
  * [15] Saharon Shelah, _Categoricity of an abstract elementary class in two successive cardinals_ , Israel J. Math., 126 (2001), 29–128.
  * [16] Saharon Shelah, _Not collapsing cardinals $\leq\kappa$ in $(<\kappa)$-support iterations_, Israel J. Math., 136 (2003), 29–115.