# Finite presentation of finitely determined modules

Eero Hyry<EMAIL_ADDRESS>Faculty of Information Technology and
Communication Sciences, Tampere University, Kanslerinrinne 1 (Pinni B),
Tampere, 33100, Finland  and  Markus Klemetti<EMAIL_ADDRESS>Faculty of Information Technology and Communication Sciences, Tampere
University, Kanslerinrinne 1 (Pinni B), Tampere, 33100, Finland

(Date: January 19, 2023)

###### Abstract.

In this article we study certain notions of ‘tameness’ for the persistence
modules studied in topological data analysis. In particular, we show that
after adding infinitary points the so called finitely determined modules
become finitely presented.

###### Key words and phrases:

Persistence module, Finitely determined, Finitely presented

###### 2010 Mathematics Subject Classification:

55N31, 13E15.

## Introduction

This article is motivated by topological data analysis, which is a recent
field of mathematics studying the shape of data. One of the main methods of
topological data analysis is persistent homology. In persistent homology one
studies the data by associating a filtered topological space to it. By taking
homology with coefficients in a field, one obtains a diagram of vector spaces
and linear maps. This diagram is called a persistence module. In the standard
case, the filtration is indexed by $\mathbb{Z}$ or $\mathbb{R}$, but the
indexing set can by any poset. Carlsson and Zomorodian realized that one can
consider persistence modules indexed by $\mathbb{Z}^{n}$ as
$\mathbb{Z}^{n}$-graded modules over a polynomial ring of $n$ variables (see
[Carlsson]*p. 78, Thm. 1). This opened the way for methods of commutative
algebra and algebraic geometry in topological data analysis. However, it is
important to consider also more general indexing sets. More formally, a
persistence module indexed by a poset $\mathcal{C}$ with coefficients in a
field $k$ is a functor from $\mathcal{C}$, interpreted as a category, to the
category of $k$-vector spaces. For the sake of generality, instead of a field
$k$, we prefer in this article to work with any commutative ring $R$.
Following the terminology of representation theory, we call a functor
$\mathcal{C}\rightarrow R\text{-}\textbf{Mod}$ an $R\mathcal{C}$-module. In
this terminology, a persistence module is then a $k\mathcal{C}$-vector space.

Persistence modules need not be finitely presented. For computational reasons,
one has therefore introduced several notions of ‘tameness’ for them. In this
context, Miller defines in [Miller]*p. 24, Def. 4.1 an encoding of an
$R\mathcal{C}$-module $M$ by a poset $\mathcal{D}$ to be a poset morphism
$f\colon\mathcal{C}\rightarrow\mathcal{D}$ with an $R\mathcal{D}$-module $N$
such that the restriction $\operatorname{res}_{f}N\cong M$. We define in
[Hyry]*p. 22, Def. 4.1 an $R\mathcal{C}$-module $M$ to be _$S$ -determined_ if
there exists a subset $S\subseteq\mathcal{C}$ such that
$\operatorname{Supp}(M)\subseteq{\uparrow}S$, and for every $c\leq d$ in
$\mathcal{C}$ the implication

$S\cap{\downarrow}c=S\cap{\downarrow}d\ \Rightarrow\ M(c\leq d)\text{ is an
isomorphism}$

holds. For any $T\subseteq\mathcal{C}$, we use the usual notations

${\uparrow}T:=\\{c\in\mathcal{C}\mid t\leq c\text{ for some }t\in T\\}$

and

${\downarrow}T:=\\{c\in\mathcal{C}\mid c\leq t\text{ for some }t\in T\\}$

for the upset generated and the downset cogenerated by $T$, respectively. This
is a straightforward generalization of the notion of a ‘positively
$a$-determined’ $\mathbb{N}^{n}$-graded module, where $a\in\mathbb{N}^{n}$, as
defined in [Miller3]*p. 186, Def. 2.1. One can look at them as modules
determined by their restriction to the interval
$[0,a]\subseteq\mathbb{N}^{n}$. They are finitely generated. Positively
$a$-determined modules have been much studied by commutative algebraists. See
e.g. [Brun] and the references therein.

Suppose now that $S\subseteq\mathcal{C}$ is a finite set. We will consider the
set $\tilde{S}$ of all minimal upper bounds of the subsets of $S$. We are
going to define a functor $\alpha\colon\mathcal{C}\rightarrow\tilde{S}$ by
mapping an element of $\mathcal{C}$ to the unique minimal upper bound of the
elements of $S$ below it. In our main result, Theorem 2.5, we will prove that
$M$ is $S$-determined for some finite $S\subseteq\mathcal{C}$ if and only if
$\alpha$ is an encoding of $M$.

The definition given by Miller in [Miller3]*p. 186, Def. 2.1 includes also the
so called ‘finitely determined’ modules. They are further studied in the
context of topological data-analysis in [Miller]. Finitely determined modules
are $\mathbb{Z}^{n}$-graded modules fully determined by their restriction to
an interval $[a,b]\subseteq\mathbb{Z}^{n}$.They are not finitely presented in
general. However, as a consequence of Theorem 2.5 we can show in Theorem 4.7
that after adding infinitary points to $\mathbb{Z}^{n}$, finitely determined
modules in fact become finitely presented. We follow here an idea due to
Perling (see [Perling]*p. 16). We also show in Proposition 4.13 that our
terminology is compatible with that of admissible posets used in [Perling].

##  1\. Preliminaries

Throughout this article we use the terminology of category theory. We will
always assume that $\mathcal{C}$ is a small category and $R$ a commutative
ring. For any set $X$, we denote by $R[X]$ the free $R$-module generated by
$X$. An $R\mathcal{C}$-module is a functor from $\mathcal{C}$ to the category
of $R$-modules. A morphism between $R\mathcal{C}$-modules is a natural
transformation. For more details on $R\mathcal{C}$-modules, we refer to [Luck]
and [TomDieck].

Recall first that an $R\mathcal{C}$-module $M$ is called

  * •

_finitely generated_ if there exists an epimorphism

$\bigoplus_{i\in I}R[\operatorname{Mor}_{\mathcal{C}}(c_{i},-)]\rightarrow M,$

where $I$ is a finite set, and $c_{i}\in\mathcal{C}$ for all $i\in I$;

  * •

_finitely presented_ if there exists an exact sequence

$\bigoplus_{j\in
J}R[\operatorname{Mor}_{\mathcal{C}}(d_{j},-)]\rightarrow\bigoplus_{i\in
I}R[\operatorname{Mor}_{\mathcal{C}}(c_{i},-)]\rightarrow M\rightarrow 0,$

where $I$ and $J$ are finite sets, and $c_{i},d_{j}\in\mathcal{C}$ for all
$i\in I$ and $j\in J$.

See, for example, [Popescu].

Let $\varphi\colon S\rightarrow\mathcal{C}$ be a functor between small
categories. Recall that the _restriction_ $\operatorname{res}_{\varphi}\colon
R\mathcal{C}\text{-}\textbf{Mod}\rightarrow RS\text{-}\textbf{Mod}$ is the
functor defined by precomposition with $\varphi$, and the _induction_
$\operatorname{ind}_{\varphi}\colon RS\text{-}\textbf{Mod}\rightarrow
R\mathcal{C}\text{-}\textbf{Mod}$ is its left Kan extension along $\varphi$.
The induction is the left adjoint of the restriction. The counit of this
adjunction gives us for every $R\mathcal{C}$-module $M$ the _canonical
morphism_

$\mu_{M}\colon\operatorname{ind}_{\varphi}\operatorname{res}_{\varphi}M\rightarrow
M.$

More explicitly, for any $R\mathcal{C}$-module $M$ and $RS$-module $N$, we
have the pointwise formulas

$(\operatorname{res}_{\varphi}M)(s)=M(\varphi(s))\quad\text{and}\quad(\operatorname{ind}_{\varphi}N)(c)=\underset{(t,u)\in(\varphi/c)}{\operatorname{colim}}N(t)$

for all $s\in S$ and $c\in\mathcal{C}$. Here $(\varphi/c)$ denotes the slice
category. Its objects are pairs $(s,u)$, where $s\in S$ and
$u\colon\varphi(s)\rightarrow c$ is a morphism in $\mathcal{C}$. For
$(s,u),(t,v)\in\operatorname{Ob}(\varphi/c)$, a morphism
$(s,u)\rightarrow(t,v)$ is a morphism $f\colon s\rightarrow t$ in $S$ with
$v\varphi(f)=u$. We will typically assume that $S$ is a full subcategory of
$\mathcal{C}$ and that $\varphi$ is the inclusion functor. In this case, we
use the notations $\operatorname{res}_{S}$ and $\operatorname{ind}_{S}$
instead of $\operatorname{res}_{\varphi}$ and $\operatorname{ind}_{\varphi}$.
If $\mathcal{C}$ is also a poset, the latter formula yields

$(\operatorname{ind}_{S}N)(c)=\underset{t\in S,\ t\leq
c}{\operatorname{colim}}N(t).$

Let $\mathcal{C}$ be a small category and $S\subseteq\mathcal{C}$ a full
subcategory. An $R\mathcal{C}$-module $M$ is said to be _$S$ -generated_ if
the natural morphism

$\rho_{M}\colon\bigoplus_{s\in
S}M(s)[\operatorname{Mor}_{\mathcal{C}}(s,-)]\rightarrow M$

is an epimorphism. Here

$M(s)[\operatorname{Mor}_{\mathcal{C}}(s,-)]:=M(s)\otimes_{R}R[\operatorname{Mor}_{\mathcal{C}}(s,-)],$

where the tensor product is taken pointwise. Since the morphism $\rho_{M}$
factors through the canonical morphism $\mu_{M}$, we see that $M$ is
$S$-generated if and only if $\mu_{M}$ is an epimorphism.

Following [Djament]*p. 13, Prop. 2.14, we say that $M$ is _$S$ -presented_ if
it is $S$-generated and the following condition holds: Given an exact sequence
of $R\mathcal{C}$-modules

$0\rightarrow L\rightarrow N\rightarrow M\rightarrow 0,$

where $N$ is $S$-generated, then $L$ is $S$-generated. It is shown in
[Djament]*p. 13, Prop. 2.14, that $M$ is $S$-presented if and only if
$\mu_{M}$ is an isomorphism.

##  2\. Modules over strongly bounded posets

In order to prove Theorem 2.5, we need to recall some order theory. In the
following, $\mathcal{C}$ always denotes a poset.

######  Notation 2.1.

Let $S\subseteq\mathcal{C}$ be a finite subset. We denote the set of minimal
upper bounds of $S$ by $\operatorname{mub}(S)$. If $S$ is finite, we set

$\hat{S}:=\bigcup_{\emptyset\neq S^{\prime}\subseteq
S}\operatorname{mub}(S^{\prime}).$

In other words, $\hat{S}$ is the set of minimal upper bounds of non-empty
subsets of $S$.

We say that the poset $\mathcal{C}$ is _strongly bounded from above_ if every
finite $S\subseteq\mathcal{C}$ has a unique minimal upper bound in
$\mathcal{C}$. If $\mathcal{C}$ is strongly bounded from above, then $\hat{S}$
is finite. The condition of $\mathcal{C}$ being strongly bounded from above is
equivalent to $\mathcal{C}$ being a bounded join-semilattice. Also note that
if $\mathcal{C}$ is strongly bounded from above, then $\mathcal{C}$ is weakly
bounded from above and mub-complete, as defined in [Hyry]*p. 23, Def. 4.5,
Def. 4.6.

Let $\mathcal{C}$ be strongly bounded from above, and let
$S\subseteq\mathcal{C}$ be a finite set. From now on, we consider
$\operatorname{mub}(S)$ as an element of $\mathcal{C}$, and not as a (one
element) set. In particular, every element of $\hat{S}$ is then of the form
$\operatorname{mub}(S^{\prime})$, where $S^{\prime}\subseteq S$ is a non-empty
subset. Viewing $\mathcal{C}$ as a join-semilattice, we have the join-
operation

$a\vee b:=\operatorname{mub}(a,b):=\operatorname{mub}(\\{a,b\\}).$

Extending this operation to finite sets, we get an operation that coincides
with taking minimal upper bounds.

######  Lemma 2.2.

Let $\mathcal{C}$ be strongly bounded from above, and let
$S\subseteq\mathcal{C}$ be a finite subset. Then $\hat{\hat{S}}=\hat{S}$.

###### Proof.

An element $s\in\hat{\hat{S}}$ may be written as

$s=\operatorname{mub}(\operatorname{mub}(S_{1}),\ldots,\operatorname{mub}(S_{n})),$

where $S_{1},\ldots,S_{n}$ are (finite) non-empty subsets of $S$. Since the
join-operation is associative in join-semilattices, we see that

$s=\bigvee_{i=1}^{n}(\bigvee S_{i})=\bigvee(\bigcup_{i=1}^{n}S_{i}).$

This implies that $s=\operatorname{mub}(S_{1}\cup\cdots\cup S_{n})$, which
belongs to $\hat{S}$ by definition. ∎

Assume that $\mathcal{C}$ is strongly bounded from above. Then $\mathcal{C}$
has a minimum element $\min(\mathcal{C})=\operatorname{mub}(\emptyset)$. Let
$S\subseteq\mathcal{C}$ be a finite subset. Denote

$\tilde{S}:=\hat{S}\cup\\{\min(\mathcal{C})\\}.$

We define a poset morphism $\alpha_{S}\colon\mathcal{C}\rightarrow\tilde{S}$
by setting

$\alpha_{S}(c)=\operatorname{mub}(S\cap{\downarrow}c)$

for every $c\in\mathcal{C}$. In other words, $\alpha_{S}$ maps each
$c\in\mathcal{C}$ to the minimal upper bound of the elements of $S$ below it.
To show that $\alpha_{S}$ actually is a poset morphism, suppose that $c\leq d$
in $\mathcal{C}$. Then $S\cap{\downarrow}c\subseteq S\cap{\downarrow}d$, which
implies that $\alpha_{S}(c)\leq\alpha_{S}(d)$.

######  Proposition 2.3.

Let $\mathcal{C}$ be strongly bounded from above, and let
$S\subseteq\mathcal{C}$ be a finite subset. Then
$\alpha_{S}=\alpha_{\hat{S}}=\alpha_{\tilde{S}}$.

###### Proof.

Using Lemma 2.2, we first note that $\tilde{\hat{S}}=\tilde{S}$ and
$\tilde{\tilde{S}}=\tilde{S}$. Let $c\in\mathcal{C}$. We claim that

$\operatorname{mub}(S\cap{\downarrow}c)=\operatorname{mub}(\hat{S}\cap{\downarrow}c)=\operatorname{mub}(\tilde{S}\cap{\downarrow}c).$

The latter equation follows from the fact that for all subsets
$T\subseteq\mathcal{C}$, we have
$\operatorname{mub}(T)=\operatorname{mub}(T\cup\\{\min(\mathcal{C})\\})$. In
particular, $\operatorname{mub}(T)=\min(\mathcal{C})$, if $T=\emptyset$.

For the first equation, since $S\subseteq\hat{S}$, we have
$\operatorname{mub}(S\cap{\downarrow}c)\leq\operatorname{mub}(\hat{S}\cap{\downarrow}c)$.
On the other hand, $\hat{S}\cap{\downarrow}c$ is a subset of $\hat{S}$. Thus
$\operatorname{mub}(\hat{S}\cap{\downarrow}c)\in\hat{\hat{S}}=\hat{S}$, where
the equation follows from Lemma 2.2. By the definition of $\hat{S}$, we may
now write

$\operatorname{mub}(\hat{S}\cap{\downarrow}c)=\operatorname{mub}(s_{1},\ldots,s_{n}),$

where $s_{1},\ldots,s_{n}\in S$. Furthermore,
$\operatorname{mub}(\hat{S}\cap{\downarrow}c)\leq c$, so we also have
$s_{1},\ldots,s_{n}\leq c$. This implies that

$\operatorname{mub}(s_{1},\ldots,s_{n})\leq\operatorname{mub}(S\cap{\downarrow}c),$

which completes the proof. ∎

Encouraged by Proposition 2.3, we will just write $\alpha$ instead of
$\alpha_{S}$, if there is no risk of confusion. Before moving on to the main
theorem of this section, we require one more lemma.

######  Lemma 2.4.

Let $\mathcal{C}$ be strongly bounded from above, and let
$S\subseteq\mathcal{C}$ be a finite subset. Then
$\hat{S}\cap{\downarrow}\alpha(c)=\hat{S}\cap{\downarrow}c$ for all
$c\in\mathcal{C}$.

###### Proof.

Let $c\in\mathcal{C}$. We immediately see that
$\hat{S}\cap{\downarrow}\alpha(c)\subseteq\hat{S}\cap{\downarrow}c$, because
$\alpha(c)\leq c$. Suppose that $d\in\hat{S}\cap{\downarrow}c$. We need to
show that $d\leq\alpha(c)$. This follows from Proposition 2.3, because now

$\alpha(c)=\alpha_{\hat{S}}(c)=\operatorname{mub}(\hat{S}\cap{\downarrow}c).$

∎

Let $\mathcal{C}$ be strongly bounded from above, let $M$ be an
$R\mathcal{C}$-module, and let $S\subseteq\mathcal{C}$ be a finite subset. The
morphism $\alpha$ gives rise to a natural transformation

$T_{\alpha}\colon\operatorname{res}_{\alpha}\operatorname{res}_{\tilde{S}}M\rightarrow
M,$

where for any $c\in\mathcal{C}$, $T_{\alpha,c}$ is the morphism

$M(\alpha(c)\leq
c)\colon(\operatorname{res}_{\alpha}\operatorname{res}_{\tilde{S}}M)(c)=M(\alpha(c))\rightarrow
M(c).$

We are now able to prove our main result

######  Theorem 2.5.

Let $\mathcal{C}$ be strongly bounded from above, and let $M$ be an
$R\mathcal{C}$-module. Given a finite subset $S\subseteq\mathcal{C}$, the
following conditions are equivalent:

  * 1)

For all $c\leq d$ in $\mathcal{C}$,

$S\cap{\downarrow}c=S\cap{\downarrow}d\ \Rightarrow\ M(c\leq d)\text{ is an
isomorphism};$

  * 2)

$T_{\alpha}\colon\operatorname{res}_{\alpha}\operatorname{res}_{\tilde{S}}M\rightarrow
M$ is an isomorphism;

  * 3)

$\alpha$ is an encoding of $M$.

If $\min(\mathcal{C})\in S$, then condition 1) says that $M$ is
$S$-determined.

###### Proof.

Suppose first that 1) holds. We can safely assume that $S$ includes the
minimum element of $\mathcal{C}$, so that
$\operatorname{Supp}(M)\subseteq{\uparrow}S=\mathcal{C}$. This will not affect
the sets $\hat{S}$ or $\tilde{S}$, nor the functor $\alpha$. Therefore $M$ is
$S$-determined. We have proved in [Hyry]*p. 25, Cor. 4.13 that an
$S$-determined module is $\hat{\hat{S}}$-presented. Lemma 2.2 now tells us
that $M$ is $\hat{S}$-presented. So
$M\cong\operatorname{ind}_{\hat{S}}\operatorname{res}_{\hat{S}}M$. This
implies that for $c\in\mathcal{C}$,

$(\operatorname{res}_{\alpha}\operatorname{res}_{\tilde{S}}M)(c)=M(\alpha(c))\cong\underset{d\leq\alpha(c),\
d\in\hat{S}}{\operatorname{colim}}M(d).$

Furthermore, by Lemma 2.4, we get

$\underset{d\leq\alpha(c),\
d\in\hat{S}}{\operatorname{colim}}M(d)=\underset{d\leq c,\
d\in\hat{S}}{\operatorname{colim}}M(d)=(\operatorname{ind}_{\hat{S}}\operatorname{res}_{\hat{S}}M)(c)\cong
M(c).$

If 2) holds, we immediately see that the functor $\alpha$ with the
$R\tilde{S}$-module $\operatorname{res}_{\tilde{S}}M$ is an encoding of $M$.

Finally, suppose that 3) is true. Assume that $c\leq d$ and
$S\cap{\downarrow}c=S\cap{\downarrow}d$. We need to show that $M(c\leq d)$ is
an isomorphism. Since $\alpha$ is an encoding of $M$, there exists an
$R\tilde{S}$-module $N$ such that $\operatorname{res}_{\alpha}N\cong M$. Here
$\operatorname{res}_{\alpha}N(c\leq d)$ is the morphism
$N(\alpha(c)\leq\alpha(d))$. We note that

$\alpha(c)=\operatorname{mub}(S\cap{\downarrow}c)=\operatorname{mub}(S\cap{\downarrow}d)=\alpha(d),$

so the morphism $\operatorname{res}_{\alpha}N(c\leq d)$ is an isomorphism.
Thus $M(c\leq d)$ is an isomorphism. Therefore 1) holds true. ∎

##  3\. Adding infinitary points

One approach to understand $R\mathbb{Z}^{n}$-modules better is to expand the
set $\mathbb{Z}^{n}$ to include points at infinity. This idea has been
utilized by Perling in [Perling]. Set
$\overline{\mathbb{Z}}:=\mathbb{Z}\cup\\{-\infty\\}$. It is easy to see that
$\overline{\mathbb{Z}}^{n}$ inherits a poset structure from $\mathbb{Z}^{n}$.
Any $R\mathbb{Z}^{n}$-module $M$ can be naturally extended to an
$R\overline{\mathbb{Z}}^{n}$-module $\overline{M}$ by setting

$\overline{M}(c)=\lim_{d\geq c,\ d\in\mathbb{Z}^{n}}M(d)$

for all $c\in\overline{\mathbb{Z}}^{n}$. More formally, this is the
coinduction of $M$ with respect to the inclusion
$\mathbb{Z}^{n}\rightarrow\overline{\mathbb{Z}}^{n}$. The functor
$M\mapsto\overline{M}$ establishes an equivalence of categories between the
category $R\mathbb{Z}^{n}\text{-}\textbf{Mod}$ and its essential image in
$R\overline{\mathbb{Z}}^{n}\text{-}\textbf{Mod}$.

Let $S\subseteq\overline{\mathbb{Z}}^{n}$ be a finite non-empty subset. We
denote by $\operatorname{mlb}(S)$ the (unique) maximal lower bound of $S$. In
this section, we will define a morphism $\beta$ “dual” to $\alpha$. The idea
is to map an element to the maximal lower bound of the elements of $S$ above
it. The morphism $\beta$ will play a crucial role in the proof of our main
result, Theorem 4.7.

We restrict ourselves to _cartesian_ subsets of $\overline{\mathbb{Z}}^{n}$,
i.e. subsets of the form $S=S_{1}\times\cdots\times S_{n}$, where
$S_{1},\ldots,S_{n}$ are subsets of $\overline{\mathbb{Z}}$. In this
situation, we can calculate $\alpha$ and $\beta$ coordinatewise. We begin with
the following observation.

######  Proposition 3.1.

Let $p_{i}\colon\overline{\mathbb{Z}}^{n}\rightarrow\overline{\mathbb{Z}}$ be
the canonical projection for every $i\in\\{1,\ldots,n\\}$, and let
$S\subseteq\overline{\mathbb{Z}}^{n}$ be a finite non-empty subset. Then

  * 1)

$\operatorname{mub}(S)=(\max(p_{1}(S)),\ldots,\max(p_{n}(S)))$;

  * 2)

$\operatorname{mlb}(S)=(\min(p_{1}(S)),\ldots,\min(p_{n}(S)))$.

###### Proof.

Since both 1) and 2) are proved in the same way, we will only present the
proof of 1) here. Let $i\in\\{1,\ldots,n\\}$. The existence of
$\max(p_{i}(S))$ follows from the fact that $p_{i}(S)$ is non-empty, linearly
ordered and finite. Write

$d=(d_{1},\ldots,d_{n}):=\operatorname{mub}(S).$

We will show that $d_{i}=\max(p_{i}(S))$. First, since $d$ is an upper bound
of $S$ and the canonical projection $p_{i}$ preserves order, we see that
$d_{i}=p_{i}(d)\geq\max(p_{i}(S))$. Secondly, if $\max(p_{i}(S))<d_{i}$, then

$d^{\prime}:=(d_{1},\ldots,d_{i-1},\max(p_{i}(S)),d_{i+1},\ldots,d_{n})$

is an upper bound of $S$ such that $d^{\prime}<d$, contradicting the
minimality of $d$. Thus $d_{i}=\max(p_{i}(S))$. ∎

Let $S:=S_{1}\times\cdots\times S_{n}\subseteq\overline{\mathbb{Z}}^{n}$ be a
cartesian subset. We write

$\overline{S}=\tilde{S_{1}}\times\cdots\times\tilde{S_{n}},$

where $\widetilde{S_{i}}=S_{i}\cup\\{-\infty\\}\subseteq\overline{\mathbb{Z}}$
for all $i\in\\{1,\ldots,n\\}$. Note that if $S$ is finite, then so is
$\overline{S}$.

######  Example 3.2.

Let $a\leq b$ in $\mathbb{Z}^{n}$. We write $a=(a_{1},\ldots,a_{n})$ and
$b=(b_{1},\ldots,b_{n})$. For the closed interval

$\displaystyle[a,b]$ $\displaystyle=\\{c\in\mathbb{Z}^{n}\mid a\leq c\leq
b\\}=[a_{1},b_{1}]\times\cdots\times[a_{n},b_{n}],$

we have

$\displaystyle\overline{[a,b]}$
$\displaystyle=\widetilde{[a_{1},b_{1}]}\times\cdots\times\widetilde{[a_{n},b_{n}]}$
$\displaystyle=\\{(c_{1},\ldots,c_{n})\mid a_{i}\leq c_{i}\leq b_{i}\
\text{or}\ c_{i}=-\infty\,\ (i\in\\{1,\ldots,n\\})\\}.$

We now have

######  Lemma 3.3.

Let $S:=S_{1}\times\cdots\times S_{n}\subseteq\overline{\mathbb{Z}}^{n}$ be a
finite cartesian subset, and let $T\subseteq S$ be a finite non-empty subset.
Then

  * 1)

$\operatorname{mub}(T)\in S$;

  * 2)

$\operatorname{mlb}(T)\in S$;

  * 3)

$\tilde{\overline{S}}=\overline{S}$.

###### Proof.

To prove 1), let $p_{i}$ be the canonical projection
$\overline{\mathbb{Z}}^{n}\rightarrow\overline{\mathbb{Z}}$ for all
$i\in\\{1,\ldots,n\\}$. From Proposition 3.1 1), we get that

$\operatorname{mub}(T)=(\max(p_{1}(T)),\ldots,\max(p_{n}(T))).$

Thus $\operatorname{mub}(T)\in S$, because $p_{i}(T)\subseteq p_{i}(S)=S_{i}$
for all $i\in\\{1,\ldots,n\\}$.

Next, the proof for 2) is done in the same way as 1), this time using
Proposition 3.1 2).

Finally, for 3), we note that $\overline{S}$ is finite and cartesian, so 1)
implies $\hat{\overline{S}}=\overline{S}$. Since $\overline{S}$ already
contains the minimum element of $\overline{\mathbb{Z}}^{n}$, we get

$\tilde{\overline{S}}=\hat{\overline{S}}\cup\\{(-\infty,\ldots,-\infty)\\}=\overline{S}\cup\\{(-\infty,\ldots,-\infty)\\}=\overline{S}.$

∎

Let $S:=S_{1}\times\cdots\times S_{n}\subseteq\overline{\mathbb{Z}}^{n}$ be a
finite cartesian subset. Since $\tilde{\overline{S}}=\overline{S}$ by Lemma
3.3 3), we have a poset morphism
$\alpha:=\alpha_{\overline{S}}\colon\overline{\mathbb{Z}}^{n}\rightarrow\overline{S}$,
where

$\alpha(c)=\operatorname{mub}(\overline{S}\cap{\downarrow}c)$

for all $c\in\overline{\mathbb{Z}}^{n}$. By Lemma 3.3 2), we now can define a
“dual” poset morphism $\beta:=\beta_{S}\colon\overline{S}\rightarrow S$ by
setting

$\beta(c)=\operatorname{mlb}(S\cap{\uparrow}c)$

for all $c\in\overline{S}$. Here the set $S\cap{\uparrow}c$ is always non-
empty, because $S$ is final in $\overline{S}$.

We can now give coordinatewise formulas for $\alpha$ and $\beta$.

######  Proposition 3.4.

We write $\alpha_{i}:=\alpha_{\overline{S_{i}}}$ and
$\beta_{i}:=\beta_{S_{i}}$ for all $i\in\\{1,\ldots,n\\}$. For
$c:=(c_{1},\ldots,c_{n})\in\overline{\mathbb{Z}}^{n}$, we have

  * 1)

$\alpha(c)=(\alpha_{1}(c_{1}),\ldots,\alpha_{n}(c_{n}))$;

  * 2)

if $c\in\overline{S}$, then
$\beta(c)=(\beta_{1}(c_{1}),\ldots,\beta_{n}(c_{n}))$.

###### Proof.

To prove 1), we will first show that

$p_{i}(\overline{S}\cap{\downarrow}c)=\overline{S_{i}}\cap{\downarrow}c_{i},$

where $p_{i}\colon\overline{\mathbb{Z}}^{n}\rightarrow\overline{\mathbb{Z}}$
is the canonical projection for all $i\in\\{1,\ldots,n\\}$. Since
$p_{i}(\overline{S})=\overline{S_{i}}$ and
$p_{i}({\downarrow}c)={\downarrow}c_{i}$, we see that
$p_{i}(\overline{S}\cap{\downarrow}c)\subseteq\overline{S_{i}}\cap{\downarrow}c_{i}$.
For the other direction, suppose that
$d\in\overline{S_{i}}\cap{\downarrow}c_{i}$. Then $d\leq c_{i}$, so we have an
element

$d^{\prime}:=(-\infty,\ldots,-\infty,d,-\infty,\ldots,-\infty)\in\overline{S}\cap{\downarrow}c$

such that $p_{i}(d^{\prime})=d$. Hence
$p_{i}(\overline{S}\cap{\downarrow}c)=\overline{S_{i}}\cap{\downarrow}c_{i}$.
Now, using this result and Proposition 3.1 1), we get

$\displaystyle\alpha(c)$
$\displaystyle=\operatorname{mub}(\overline{S}\cap{\downarrow}c)$
$\displaystyle=(\max(\overline{S_{1}}\cap{\downarrow}c_{1}),\ldots,\max(\overline{S_{n}}\cap{\downarrow}c_{n}))$
$\displaystyle=(\alpha_{1}(c_{1}),\ldots,\alpha_{n}(c_{n})).$

For 2), the proof is similar. Let $c\in\overline{S}$. We will first show that

$p_{i}(S\cap{\uparrow}c)=S_{i}\cap{\uparrow}c.$

From $p_{i}(S)=S_{i}$ and $p_{i}({\uparrow}c)={\uparrow}c_{i}$, we see that
$p_{i}(S\cap{\uparrow}c)\subseteq S_{i}\cap{\uparrow}c_{i}$. Next, suppose
that $d\in S_{i}\cap{\uparrow}c_{i}$. Since $c\in\overline{S}$, there is an
element $s:=(s_{1},\ldots,s_{n})\in S$ such that $s\geq c$. Because $d\geq
c_{i}$ and $S$ is cartesian, we again have an element

$d^{\prime}:=(s_{1},\ldots,s_{i-1},d,s_{i+1},\ldots,s_{n})\in
S\cap{\uparrow}c$

such that $p_{i}(d^{\prime})=d$. Thus
$p_{i}(S\cap{\uparrow}c)=S_{i}\cap{\uparrow}c$. To finish the proof, we use
Proposition 3.1 2):

$\displaystyle\beta(c)$ $\displaystyle=\operatorname{mlb}(S\cap{\uparrow}c)$
$\displaystyle=(\min(S_{1}\cap{\uparrow}c_{1}),\ldots,\min(S_{n}\cap{\uparrow}c_{n}))$
$\displaystyle=(\beta_{1}(c_{1}),\ldots,\beta_{n}(c_{n})).$

∎

We note that $\alpha$ and $\beta\circ\alpha$ are “continuous” in the following
sense.

######  Proposition 3.5.

Let $c:=(c_{1},\ldots,c_{n})\in\overline{\mathbb{Z}}^{n}$.

  * 1)

If $N$ is an $R\overline{S}$-module, then

$\lim_{d\geq c,\ d\in\mathbb{Z}^{n}}N(\alpha(d))\cong N(\alpha(c)).$

  * 2)

If $Q$ is an $RS$-module, then

$\lim_{d\geq c,\ d\in\mathbb{Z}^{n}}Q((\beta\circ\alpha)(d))\cong
Q((\beta\circ\alpha)(c)).$

###### Proof.

For 1), suppose that $N$ is an $R\overline{S}$-module. Let
$c^{\prime}:=(c^{\prime}_{1},\ldots,c^{\prime}_{n})\in\overline{\mathbb{Z}}^{n}$
as follows: For any $i\in\\{1,\ldots,n\\}$, we set
$a_{i}=\min(S_{i}\cap\mathbb{Z})$, if it exists, and

$c^{\prime}_{i}:=\left\\{\begin{aligned} &\max(c_{i},0),\ \text{if}\
S_{i}\cap\mathbb{Z}=\emptyset;\\\ &\max(c_{i},a_{i}-1),\
\text{otherwise}.\end{aligned}\right.$

This guarantees that we always have $c\leq c^{\prime}$ and
$c^{\prime}\in\mathbb{Z}^{n}$. With the notation from Proposition 3.4, we may
write

$\alpha(c)=(\alpha_{1}(c_{1}),\ldots,\alpha_{n}(c_{n})).$

Let $i\in\\{1,\ldots,n\\}$. If $S_{i}\cap\mathbb{Z}=\emptyset$, then
$\alpha_{i}(c^{\prime}_{i})=-\infty=\alpha_{i}(c_{i})$. Similarly, if
$c_{i}^{\prime}=a_{i}-1$, then
$\alpha_{i}(c_{i}^{\prime})=-\infty=\alpha_{i}(c_{i})$. Thus
$\alpha(c)=\alpha(c^{\prime})$ in all cases. Since $\alpha$ is a poset
morphism, we see that for all $d\in\mathbb{Z}^{n}$ such that $c\leq d\leq
c^{\prime}$,

$\alpha(c)=\alpha(d)=\alpha(c^{\prime}),$

and therefore

$N(\alpha(c))=N(\alpha(d))=N(\alpha(c^{\prime})).$

Furthermore, because the set $\\{d\in\mathbb{Z}^{n}\mid c\leq d\leq
c^{\prime}\\}$ is an initial subset of the set $\\{d\in\mathbb{Z}^{n}\mid
c\leq d\\}$, we have

$\lim_{d\geq c,\ d\in\mathbb{Z}^{n}}N(\alpha(d))\cong\lim_{c\leq d\leq
c^{\prime},\ d\in\mathbb{Z}^{n}}N(\alpha(d))\cong N(\alpha(c)).$

Next, for 2), let $Q$ be an $RS$-module. Now $\operatorname{res}_{\beta}Q$ is
an $R\overline{S}$-module, so by 1), we have

$\lim_{d\geq c,\
d\in\mathbb{Z}^{n}}(\operatorname{res}_{\beta}Q)(\alpha(d))\cong(\operatorname{res}_{\beta}Q)(\alpha(c)).$

On the other hand, by definition, for all $e\in\overline{\mathbb{Z}}^{n}$,

$(\operatorname{res}_{\beta}Q)(\alpha(e))=Q(\beta(\alpha(e)))=Q((\beta\circ\alpha)(e)).$

This means that we may write the above isomorphism as

$\lim_{d\geq c,\ d\in\mathbb{Z}^{n}}Q((\beta\circ\alpha)(d))\cong
Q((\beta\circ\alpha)(c)).$

∎

######  Corollary 3.6.

Let $N$ be an $R\overline{\mathbb{Z}}^{n}$-module, and let
$c\in\overline{\mathbb{Z}}^{n}$. Then

  * 1)

$\displaystyle\lim_{d\geq c,\ d\in\mathbb{Z}^{n}}N(\alpha(d))\cong
N(\alpha(c))$;

  * 2)

$\displaystyle\lim_{d\geq c,\ d\in\mathbb{Z}^{n}}N((\beta\circ\alpha)(d))\cong
N((\beta\circ\alpha)(c))$.

###### Proof.

For 1), we note that $\operatorname{res}_{\overline{S}}N$ is an
$R\overline{S}$-module, where $(\operatorname{res}_{\overline{S}}N)(d)=N(d)$
for all $d\in\overline{S}$. We may then apply Proposition 3.5 1) to get the
result. For 2), we use Proposition 3.5 2) on the $RS$-module
$\operatorname{res}_{S}N$. ∎

##  4\. Finitely determined modules

Let $M$ be an $R\mathcal{C}$-module. We say that $M$ is _pointwise finitely
presented_ if $M(c)$ is finitely presented for all $c\in\mathcal{C}$. Slightly
generalizing the definition of Miller in [Miller]*p. 25, Ex. 4.5, where $R=k$
is a field, we say that an $R\mathbb{Z}^{n}$-module $M$ is _finitely
determined_ , if $M$ is pointwise finitely presented, and for some $a\leq b$
in $\mathbb{Z}^{n}$, the convex projection
$\pi\colon\mathbb{Z}^{n}\rightarrow[a,b]$ gives $M$ an encoding by the closed
interval $[a,b]\subseteq\mathbb{Z}^{n}$. Here the convex projection $\pi$
takes every point in $\mathbb{Z}^{n}$ to its closest point in the interval
$[a,b]$. If $a=(a_{1},\ldots,a_{n})$ and $b=(b_{1},\ldots,b_{n})$, we have for
any $c:=(c_{1},\ldots,c_{n})\in\mathbb{Z}^{n}$,

$\pi(c)=(\pi_{1}(c_{1}),\ldots,\pi_{n}(c_{n})),$

where

$\pi_{i}(c_{i})=\max(a_{i},\min(c_{i},b_{i}))$

for all $i\in\\{1,\ldots,n\\}$. Note that a pointwise finitely presented
$R\mathbb{Z}^{n}$-module $M$ is finitely determined if and only if there
exists a closed interval $[a,b]\subseteq\mathbb{Z}^{n}$ such that the
morphisms $M(c\leq c+e_{i})$ ($i=1,\ldots,n$) are isomorphisms whenever
$c_{i}$ lies outside $[a_{i},b_{i}]$.

######  Remark 4.1.

Let $M$ be an $R\mathbb{Z}^{n}$-module. Then $M$ is encoded by the closed
interval $[a,b]$ with the convex projection
$\pi\colon\mathbb{Z}^{n}\rightarrow[a,b]$ if and only if
$M\cong\operatorname{res}_{\pi}\operatorname{res}_{[a,b]}M$. Indeed, if
$M\cong\operatorname{res}_{\pi}N$ for some $R[a,b]$-module $N$, then for all
$c\in\mathbb{Z}^{n}$, we have

$M(c)\cong(\operatorname{res}_{\pi}N)(c)=N(\pi(c))=N(\pi(\pi(c)))\cong
M(\pi(c))$

because for all $c\in\mathbb{Z}^{n}$, $\pi(\pi(c))=\pi(c)$.

We would now like to investigate how the notion of finite determinacy relates
to our notion of $S$-determinacy, when $S$ is finite and $M$ is pointwise
finitely presented. While the requirement that
$\operatorname{Supp}(M)\subseteq{\uparrow}S$ does not necessarily hold for
finitely determined modules, we do have the following:

######  Proposition 4.2.

Let $M$ be an $R\mathbb{Z}^{n}$-module, and $a,b\in\mathbb{Z}^{n}$ such that
$a\leq b$. Set $u:=(1,1,\ldots,1)\in\mathbb{Z}^{n}$. If $M$ is
$[a+u,b]$-determined, then $M$ has an encoding by the closed interval $[a,b]$
with the convex projection $\pi\colon\mathbb{Z}^{n}\rightarrow[a,b]$. The
converse implication holds if $\operatorname{Supp}(M)\subseteq{\uparrow}a$.

###### Proof.

For the first implication, suppose that $M$ is $[a+u,b]$-determined. We write
$a=(a_{1},\ldots,a_{n})$ and $b=(b_{1},\ldots,b_{n})$. Let
$c:=(c_{1},\ldots,c_{n})\in\mathbb{Z}^{n}$. We note that if $c_{i}\leq a_{i}$
for some $i\in\\{1,\ldots,n\\}$, then also $\pi_{i}(c_{i})\leq a_{i}$, so that
$c,\pi(c)\notin\operatorname{Supp}(M)$. Otherwise $c>a$, in which case
$\pi(c)\leq c$ and $[a+u,b]\cap{\downarrow}\pi(c)=[a+u,b]\cap{\downarrow}c$.
Thus $M(\pi(c))\rightarrow M(c)$ is an isomorphism by the definition of
$[a+u,b]$-determined modules, and
$M\cong\operatorname{res}_{\pi}\operatorname{res}_{[a,b]}M$.

To prove the converse, assume that
$\operatorname{Supp}(M)\subseteq{\uparrow}a$ and $M$ has an encoding by the
closed interval $[a,b]$ with the encoding convex projection
$\pi\colon\mathbb{Z}^{n}\rightarrow[a,b]$. Let
$c:=(c_{1},\ldots,c_{n})\in\mathbb{Z}^{n}$. Suppose that $c_{i}<a_{i}$ for
some $i\in\\{1,\ldots,n\\}$. From the condition
$\operatorname{Supp}(M)\subseteq{\uparrow}a$, we see that $M(c)=0$. Since $M$
is finitely determined, we also have $M(\pi(c))=M(c)=0$. Thus $M(c)=0$ if
$c_{i}\leq a_{i}$ for some $i\in\\{1,\ldots,n\\}$. If this is not the case, we
have $c\geq a+u$. Let $c\leq d$ in $\mathcal{C}$ such that $a+u\leq c\leq d$
and $[a+u,b]\cap{\downarrow c}=[a+u,b]\cap{\downarrow d}$. This implies that
$\pi(c)=\pi(d)$, so $M(c\leq d)$ is an isomorphism. ∎

To proceed, we have to shift our focus to
$R\overline{\mathbb{Z}}^{n}$-modules. Let $a\leq b$ in $\mathbb{Z}^{n}$. With
the notation from section 3, we will view the case $S=[a,b]$. In particular,
we have $\alpha=\alpha_{\overline{[a,b]}}$ and $\beta=\beta_{[a,b]}$.
Proposition 3.4 gives us formulas for $\alpha$ and $\beta$. If
$c:=(c_{1},\ldots,c_{n})\in\overline{\mathbb{Z}}^{n}$ and
$d:=(d_{1},\ldots,d_{n})\in\overline{[a,b]}$, then

$\alpha(c)=(\alpha_{1}(c_{1}),\ldots,\alpha_{n}(c_{n}))\quad\text{and}\quad\beta(d)=(\beta_{1}(d_{1}),\ldots,\beta_{n}(d_{n})).$

Here $\alpha_{i}:=\alpha_{\overline{S_{i}}}$ and $\beta_{i}:=\beta_{S_{i}}$
for all $i\in\\{1,\ldots,n\\}$. Explicitly,

$\alpha_{i}(c_{i})=\left\\{\begin{aligned} -\infty,\ &\text{if}\
c_{i}<a_{i};\\\ c_{i},\ &\text{if}\ a_{i}\leq c_{i}\leq b_{i};\\\ b_{i},\
&\text{if}\
c_{i}>b_{i}\end{aligned}\right.\quad\text{and}\quad\beta_{i}(d_{i})=\left\\{\begin{aligned}
a_{i},\ &\text{if}\ d_{i}=-\infty;\\\ d_{i},\
&\text{otherwise}\end{aligned}\right.$

for every $i\in\\{1,\ldots,n\\}$. The next proposition shows us that the
composition $\beta\circ\alpha$ is an extension of the convex projection $\pi$
from $\mathbb{Z}^{n}$ to $\overline{\mathbb{Z}}^{n}$.

######  Proposition 4.3.

Let $\pi\colon\mathbb{Z}^{n}\rightarrow[a,b]$ be the convex projection. Then
for any $c:=(c_{1},\ldots,c_{n})\in\mathbb{Z}^{n}$,

$\pi(c)=(\beta\circ\alpha)(c).$

###### Proof.

Suppose first that $n=1$. Recall that $\pi(c)=\max(a,\min(c,b))$. Now there
are three cases:

  * •

If $c\in[a,b]$, then $(\beta\circ\alpha)(c)=\beta(c)=c=\pi(c)$;

  * •

If $c<a$, then $(\beta\circ\alpha)(c)=\beta(-\infty)=a=\pi(c)$;

  * •

If $c>b$, then $(\beta\circ\alpha)(c)=\beta(b)=b=\pi(c)$.

Suppose next that $n>1$. Using Proposition 3.4, we may write

$\alpha(c)=(\alpha_{1}(c_{1}),\ldots,\alpha_{n}(c_{n}))\quad\text{and}\quad\beta(d)=(\beta_{1}(d_{1}),\ldots,\beta_{n}(d_{n}))$

for all $d\in\overline{[a,b]}$. Similarly, recall that

$\pi(c)=(\pi_{1}(c_{1}),\ldots,\pi_{n}(c_{n})).$

It now follows from the case $n=1$ that

$\displaystyle(\beta\circ\alpha)(c)$
$\displaystyle=\beta(\alpha_{1}(c_{1}),\ldots,\alpha_{n}(c_{n}))$
$\displaystyle=((\beta_{1}\circ\alpha_{1})(c_{1}),\ldots,(\beta_{n}\circ\alpha_{n})(c_{n}))$
$\displaystyle=(\pi_{1}(c_{1}),\ldots,\pi_{n}(c_{n}))$ $\displaystyle=\pi(c).$

∎

######  Remark 4.4.

In an effort to keep the notation simpler, we only defined $\beta$ for the
elements in the image of $\alpha$. Of course, we could have defined $\beta$ in
a fully dual fashion to $\alpha$, starting from posets that are strongly
bounded from below, adding the point $\infty$ to $\mathbb{Z}$, and defining a
set $\underline{S}$ dually to $\overline{S}$. This would have resulted in the
situation where

$(\beta|_{\overline{S}}\circ\alpha)(c)=(\alpha|_{\underline{S}}\circ\beta)(c)=\pi(c)$

for all $c\in\mathbb{Z}^{n}$. In other words, the same result would have been
achieved.

We saw in Remark 4.1 that if $M$ is encoded by a closed interval $[a,b]$ with
the convex projection $\pi\colon\mathbb{Z}^{n}\rightarrow[a,b]$, we have
$M(c)\cong M(\pi(c))$ for all $c\in\mathbb{Z}^{n}$. On the other hand, by
Theorem 2.5, we have $\overline{M}(\alpha(c))\cong\overline{M}(c)$ for all
$c\in\overline{\mathbb{Z}}^{n}$ if $M$ is $S$-determined and
$S\subseteq\overline{\mathbb{Z}}^{n}$ is finite. In preparation for the proof
of Theorem 4.7, we will now show that a similar result applies to $\beta$ in
both cases.

######  Proposition 4.5.

Set $u:=(1,1,\ldots,1)\in\mathbb{Z}^{n}$. Let $M$ be an
$R\mathbb{Z}^{n}$-module, and let $c\in\overline{[a,b]}$.

  * 1)

If $M$ has an encoding by the closed interval $[a,b]$ with the convex
projection $\pi\colon\mathbb{Z}^{n}\rightarrow[a,b]$, then
$\overline{M}(c)\cong M(\beta(c))$.

  * 2)

If $\overline{M}$ is $\overline{[a+u,b]}$-determined, then
$\overline{M}(c)\cong M(\beta(c))$.

###### Proof.

To show 1), suppose that $M$ has an encoding by the closed interval $[a,b]$
with the convex projection $\pi\colon\mathbb{Z}^{n}\rightarrow[a,b]$. Then, by
the definition of $\overline{M}$,

$\overline{M}(c)=\lim_{d\geq c,\ d\in\mathbb{Z}^{n}}M(d).$

The encoding gives us $M(d)\cong M(\pi(d))$ for all $d\in\mathbb{Z}^{n}$. This
implies that

$\overline{M}(c)\cong\lim_{d\geq c,\ d\in\mathbb{Z}^{n}}M(\pi(d)).$

We may now apply Corollary 3.6 to see that $\overline{M}(c)\cong
M(\beta(\alpha(c)))$. Note that $c\in\overline{[a,b]}$ implies $\alpha(c)=c$.
Thus $\overline{M}(c)\cong M(\beta(c))$.

Next, to prove 2), let $\overline{M}$ be $\overline{[a+u,b]}$-determined.
Since $c\leq\beta(c)$, it is then enough to show that
$\overline{[a+u,b]}\cap{\downarrow}c=\overline{[a+u,b]}\cap{\downarrow}\beta(c)$.
We instantly have ${\downarrow}c\subseteq{\downarrow}\beta(c)$. For the other
direction, let
$d:=(d_{1},\ldots,d_{n})\in\overline{[a+u,b]}\cap{\downarrow}\beta(c)$. We
want to show that $d\leq c$. Recall that we may write
$\beta(c)=(\beta_{1}(c_{1}),\ldots,\beta_{n}(c_{n}))$, where

$\beta_{i}(c_{i})=\left\\{\begin{aligned} &a_{i},\ \text{if}\
c_{i}=-\infty;\\\ &c_{i},\ \text{otherwise}.\end{aligned}\right.$

for all $i\in\\{1,\ldots,n\\}$. Suppose that $i\in\\{1,\ldots,n\\}$. If
$\beta_{i}(c_{i})=c_{i}$, we have $d_{i}\leq\beta_{i}(c_{i})=c_{i}$.
Otherwise, if $\beta_{i}(c_{i})=a_{i}$, we must have $d_{i}=c_{i}=-\infty$,
because $d_{i},c_{i}\in\overline{[a_{i}+1,b_{i}]}$. We conclude that $d\leq
c$. ∎

######  Remark 4.6.

Let $M$ be a pointwise finitely presented $R\mathbb{Z}^{n}$-module and let
$c\in\overline{\mathbb{Z}}^{n}$. If $M$ is finitely determined with the convex
projection $\pi\colon\mathbb{Z}^{n}\rightarrow[a,b]$, then from the proof of
Proposition 4.5, we have $\overline{M}(c)\cong M((\beta\circ\alpha)(c))$, so
that $\overline{M}$ is pointwise finitely presented.

We are now ready to state

######  Theorem 4.7.

Let $M$ be a pointwise finitely presented $R\mathbb{Z}^{n}$-module. Then the
following are equivalent:

  * 1)

$M$ is finitely determined;

  * 2)

$\overline{M}$ is $S$-determined for some finite
$S\subseteq\overline{\mathbb{Z}}^{n}$;

  * 3)

$\overline{M}$ is finitely presented.

###### Proof.

We will first show the equivalence of 1) and 2). Note that for any finite
subset $S\subseteq\overline{\mathbb{Z}}^{n}$, we can always find
$a,b\in\mathbb{Z}^{n}$ such that $S\subseteq\overline{[a+u,b]}$. Consider the
functor

$\alpha^{\prime}=\alpha_{\overline{[a+u,b]}}\colon\overline{\mathbb{Z}}^{n}\rightarrow\overline{[a+u,b]},$

and denote its restriction to $\mathbb{Z}^{n}$ by $\overline{\alpha}$. By
Theorem 2.5, $\overline{M}$ is $\overline{[a+u,b]}$-determined if and only if
$\overline{M}$ is encoded by $\alpha^{\prime}$. That is,
$\overline{M}\cong\operatorname{res}_{\alpha^{\prime}}N$ for some
$R\overline{[a+u,b]}$-module $N$. By restricting to $\mathbb{Z}^{n}$, we see
that

$M\cong\operatorname{res}_{\mathbb{Z}^{n}}\operatorname{res}_{\alpha^{\prime}}N=\operatorname{res}_{\overline{\alpha}}N,$

so $\overline{\alpha}$ encodes $M$. Conversely, if $M$ is encoded by
$\overline{\alpha}$, then $\overline{M}$ has an obvious encoding by
$\alpha^{\prime}$, because $\overline{\alpha}$ is a surjection on objects.
Next, we note that the restriction of $\beta$ to $\overline{[a+u,b]}$,

$\overline{\beta}\colon\overline{[a+u,b]}\rightarrow[a,b].$

is an isomorphism of posets. Therefore
$\overline{\beta}\circ\overline{\alpha}$ is an encoding of $M$ if and only if
$\overline{\alpha}$ is an encoding of $M$. These conditions are equivalent to
$M$ being finitely determined, because
$\overline{\beta}\circ\overline{\alpha}=\pi$. Namely, for all
$c\in\mathbb{Z}^{n}$, we have
$(\overline{\beta}\circ\overline{\alpha})(c)=(\beta\circ\alpha^{\prime})(c)$,
where

$\displaystyle(\beta\circ\alpha^{\prime})(c)_{i}$
$\displaystyle=\left\\{\begin{aligned} &\beta_{i}(-\infty),\ \text{if}\
c_{i}=a_{i},\\\
&\beta_{i}(\alpha_{i}(c_{i})),\text{else.}\end{aligned}\right.$
$\displaystyle=\left\\{\begin{aligned} &a_{i},\ \text{if}\ c_{i}=a_{i},\\\
&\pi_{i}(c_{i}),\text{else.}\end{aligned}\right.$ $\displaystyle=\pi(c)_{i}$

for all $i\in\\{1,\ldots,n\\}$.

Finally, we observe that the equivalence of 2) and 3) follows from the main
result of our previous paper, [Hyry]*p. 25, Thm. 4.15. For
$\overline{\mathbb{Z}}^{n}$-modules, it states that being pointwise finitely
presented and $S$-determined for some finite
$S\subseteq\overline{\mathbb{Z}}^{n}$ is equivalent to being finitely
presented. Also note Remark 4.6, which shows us that $\overline{M}$ is
pointwise finitely presented. ∎

We are now able to give a “sharpened” version of Proposition 4.2.

######  Corollary 4.8.

If $M$ is an $R\mathbb{Z}^{n}$-module and $a,b\in\mathbb{Z}^{n}$ such that
$a\leq b$, then the following are equivalent:

  * 1)

$M$ is encoded by the convex projection
$\pi\colon\mathbb{Z}^{n}\rightarrow[a,b]$;

  * 2)

$\overline{M}$ is $\overline{[a+u,b]}$-determined, where
$u:=(1,\ldots,1)\in\mathbb{Z}^{n}$.

###### Proof.

We showed in the proof of Theorem 4.7 that 2) implies 1). Conversely, suppose
that 1) holds. Let $c\leq d$ in $\overline{\mathbb{Z}}^{n}$ such that
$\overline{[a+u,b]}\cap{\downarrow}c=\overline{[a+u,b]}\cap{\downarrow}d$.
Coordinatewise, for $i=\\{1,\ldots,n\\}$, this implies that either
$c_{i}=d_{i}$, $b_{i}\leq c_{i}<d_{i}$ or $c_{i}<d_{i}\leq a_{i}$. In any
case, $(\beta\circ\alpha)(c)=(\beta\circ\alpha)(d)$, so that

$\overline{M}(c)\cong
M((\beta\circ\alpha)(c))=M((\beta\circ\alpha)(d))\cong\overline{M}(d).$

Thus $M(c\leq d)$ is an isomorphism, and $M$ is
$\overline{[a+u,b]}$-determined. ∎

To demonstrate Theorem 4.7 and Corollary 4.8, it is convenient to take the
point of view of topological data analysis, and consider the births and deaths
of elements of a module. Given an $R\mathbb{Z}^{n}$ module $M$, one can track
how an element $x\in\overline{M}(c)$, where $c\in\overline{\mathbb{Z}}^{n}$,
evolves when mapped with the homomorphisms $M(c\leq c^{\prime})$,
($c,c^{\prime}\in\overline{\mathbb{Z}}^{n}$). We say that the element $x$ is
born at $c$ if it is not in the image of any morphism $M(c^{\prime}\leq c)$,
where $c^{\prime}<c$. On the other hand, the element $x$ dies at
$c^{\prime\prime}$ if $M(c\leq c^{\prime\prime})(x)=0$, but $M(c\leq
c^{\prime})(m)\neq 0$ for all $c\leq c^{\prime}<c^{\prime\prime}$.

Consider now an $R\mathbb{Z}^{2}$-module $M$ that is finitely determined, and
let $\pi\colon\mathbb{Z}^{2}\rightarrow[a,b]$ be the accompanying convex
projection. Note that no new elements are born or die in the leftmost edge or
the bottom edge of the box $[a,b]$. This follows from the fact that every
element on these two edges has already appeared infinite times before, and was
born at some infinitary point. Let us write $a=(a_{1},a_{2})$. For example, if
an element, say $x\in M((a_{1},c))$, maps to zero on the leftmost edge of
$[a,b]$, in $M((a_{1},c+1))$, then $x\in\overline{M}((-\infty,c))$ will also
map to zero in $\overline{M}((-\infty,c+1))$. Thus $x$ does not “die” at the
point $(a_{1},c+1)$, but rather at the infinitary point
$(-\infty,c+1)\in\overline{[a+u,b]}$.

######  Remark 4.9.

Let $M$ be an $R\mathbb{Z}^{n}$-module and $c\in\overline{\mathbb{Z}}^{n}$.
Consider the natural homomorphism

$\lambda_{\overline{M},c}\colon\underset{d\leq
c,d\in\overline{[a+u,b]}}{\operatorname{colim}}\overline{M}(d)\rightarrow\overline{M}(c).$

Following [Hyry]*p. 15, Def. 3.6, we say that $c$ is a _birth_ if
$\lambda_{\overline{M},c}$ is a non-epimorphism, and a _death_ if
$\lambda_{\overline{M},c}$ is a non-monomorpism. Furthermore, suppose that
$\overline{M}$ is $\overline{[a+u,b]}$-determined, and the births are “well-
behaved” enough. That is, for any birth $c$, the module
$\overline{M}(c)/\operatorname{Im}{\lambda_{\overline{M},c}}$ is projective.
The latter of course holds if $R$ is a field. Then, as we discussed in
[Hyry]*p. 21, Remark 3.27, births and deaths show the positions of the minimal
generators and relations of $M$.

In the next example, we will demonstrate how, for a finitely determined module
$M$, the extension $\overline{M}$ has births and deaths at infinitary points
that guarantee the existence of a finite presentation of $\overline{M}$.

######  Example 4.10.

Let $M$ be an $R\mathbb{Z}^{2}$-module that is defined on objects by

$M(c)=\left\\{\begin{aligned} &R,\ \text{if}\ c\leq(0,0);\\\ &0,\
\text{otherwise},\end{aligned}\right.$

for all $c\in\mathbb{Z}^{2}$, and where a morphism $R\rightarrow R$ is always
$\operatorname{id}_{R}$. Then $M$ is finitely determined with the convex
projection $\pi\colon\mathbb{Z}^{2}\rightarrow[(0,0),(1,1)]$. Now, by Remark
4.8, $\overline{M}$ is $\overline{[(1,1),(1,1)]}$-determined. Here
$\overline{[(1,1),(1,1)]}$ is the set

$\\{(-\infty,-\infty),(1,-\infty),(-\infty,1),(1,1)\\}.$

In particular, we have $\overline{M}((-\infty,-\infty))=R$, and

$\overline{M}((-\infty,1))=\overline{M}((1,-\infty))=\overline{M}((1,1))=0.$

Furthermore, by Theorem 4.7, $\overline{M}$ is now finitely presented. In more
concrete terms, we have an exact sequence of
$R\overline{\mathbb{Z}}^{2}$-modules

$K\rightarrow N\rightarrow\overline{M}\rightarrow 0,$

where

$N=R[\operatorname{Mor}_{\overline{\mathbb{Z}}^{2}}((-\infty,-\infty),-)]$

and

$K=R[\operatorname{Mor}_{\overline{\mathbb{Z}}^{2}}((1,-\infty),-)]\oplus
R[\operatorname{Mor}_{\overline{\mathbb{Z}}^{2}}((-\infty,1),-)].$

Here $(-\infty,-\infty)$ is the only birth of $M$, while $(1,-\infty)$ and
$(-\infty,1)$ are the deaths.

######  Example 4.11.

If $k$ is a field, then it is well known that finitely generated
$k\mathbb{Z}^{n}$-modules are finitely presented. This result, however, does
not apply to $k\overline{\mathbb{Z}}^{n}$-modules. For a counterexample,
consider a $k\mathbb{Z}^{2}$-module $M$, where

$M((x,y))=\left\\{\begin{aligned} &k,\text{ if }x+y<0;\\\ &0,\text{
otherwise}.\end{aligned}\right.$

Clearly $\overline{M}$ is finitely generated with its only birth in
$(-\infty,-\infty)$. It is not finitely presented, since the deaths happen at
points $(n,-n)$ for all $n\in\mathbb{Z}$.

Finally, we want to relate Theorem 4.7 to the work of Perling ([Perling]).
Recall that a subset $L\subseteq\overline{\mathbb{Z}}^{n}$ is a join-
sublattice if $\operatorname{mub}(S)\in L$ for every finite subset $S\subseteq
L$. Note that this is equivalent to the condition that $L=\hat{L}$. Given a
join-sublattice $L\subseteq\overline{\mathbb{Z}}^{n}$, following Perling in
[Perling]*pp. 16-19, Ch. 3.1, we define the zip-functor

$\operatorname{zip}_{L}\colon R\mathbb{Z}^{n}\text{-}\textbf{Mod}\rightarrow
RL\text{-}\textbf{Mod}$

and the unzip-functor

$\operatorname{unzip}_{L}\colon RL\text{-}\textbf{Mod}\rightarrow
R\overline{\mathbb{Z}}^{n}\text{-}\textbf{Mod}.$

Contrary to Perling, we do not assume that $R$ is a field. The zip-functor
maps an $R\mathbb{Z}^{n}$-module $M$ to the $RL$-module
$\operatorname{res}_{L}\overline{M}$, whereas he unzip-functor maps an
$RL$-module $N$ to an $R\overline{\mathbb{Z}}^{n}$-module
$\operatorname{unzip}_{L}N$ defined by

$(\operatorname{unzip}_{L}N)(c)=\left\\{\begin{aligned}
&N(\operatorname{mub}(L\cap{\downarrow}c)),\ \text{if}\
L\cap{\downarrow}c\neq\emptyset;\\\ &0,\ \text{otherwise}\end{aligned}\right.$

for all $c\in\overline{\mathbb{Z}}^{n}$. Note that
$\operatorname{Supp}(\operatorname{unzip}_{L}N)\subseteq{\uparrow}L$.

######  Remark 4.12.

It turns out that $\operatorname{unzip}_{L}$ is essentially the same thing as
$\operatorname{res}_{\alpha}$, when $L$ is finite and $\alpha:=\alpha_{L}$.
There is the slight complication that $\operatorname{unzip}_{L}$ is defined
for $RL$-modules, while $\operatorname{res}_{\alpha}$ is defined for
$R\tilde{L}$-modules. We may, however, extend an $RL$-module $N$ to an
$R{\tilde{L}}$-module $\tilde{N}$ by setting

$\tilde{N}((-\infty,\ldots,-\infty))=0,$

if $(-\infty,\ldots,-\infty)\notin L$, and $\tilde{N}(c)=N(c)$, otherwise.
Having defined the module $\tilde{N}$ in this way, we see that
$\operatorname{unzip}_{L}N\cong\operatorname{res}_{\alpha}\tilde{N}$.

Given an $R\mathbb{Z}^{n}$-module $M$, the join-sublattice $L$ is called _$M$
-admissible_ in [Perling]*p. 18, Def. 3.4 if the condition
$\overline{M}\cong\operatorname{unzip}_{L}\operatorname{zip}_{L}M$ is
satisfied. This leads us to the following proposition.

######  Proposition 4.13.

Let $M$ be an $R\mathbb{Z}^{n}$-module, and $L$ a finite join-sublattice. Then
$L$ is $M$-admissible if and only if $\overline{M}$ is $L$-determined.

###### Proof.

Let $c\in\overline{\mathbb{Z}}^{n}$. With the earlier notation, we see that

$\operatorname{unzip}_{L}\operatorname{zip}_{L}M=\operatorname{unzip}_{L}\operatorname{res}_{L}\overline{M}\cong\operatorname{res}_{\alpha}\widetilde{\operatorname{res}_{L}\overline{M}},$

where

$(\operatorname{res}_{\alpha}\widetilde{\operatorname{res}_{L}\overline{M}})(c)=\left\\{\begin{aligned}
&(\operatorname{res}_{\alpha}\operatorname{res}_{\tilde{L}}\overline{M})(c),\
\text{if}\ L\cap{\downarrow}c\neq\emptyset;\\\ &0,\
\text{otherwise.}\end{aligned}\right.$

Assume first that
$\overline{M}\cong\operatorname{unzip}_{L}\operatorname{zip}_{L}M$. If
$L\cap{\downarrow}c=\emptyset$, we have $\overline{M}(c)=0$ by the definition
of the functor $\operatorname{unzip}_{L}$. But in this case $\alpha(c)\leq c$,
so that $L\cap{\downarrow}\alpha(c)=\emptyset$. Using the definition of
$\operatorname{unzip}_{L}$ again, we get

$(\operatorname{res}_{\alpha}\operatorname{res}_{\tilde{L}}\overline{M})(c)=\overline{M}(\alpha(c))=0.$

On the other hand, if there is an element $d\in L\cap{\downarrow}c$, then, by
the above formula,
$\overline{M}(c)\cong(\operatorname{res}_{\alpha}\operatorname{res}_{\tilde{L}}\overline{M})(c)$.
Thus,

$\overline{M}\cong\operatorname{res}_{\alpha}\operatorname{res}_{\tilde{L}}\overline{M}$

and $\operatorname{Supp}(\overline{M})\subseteq{\uparrow}L$, so $\overline{M}$
is $L$-determined by Theorem 2.5.

Conversely, suppose that $\overline{M}$ is $L$-determined. By Theorem 2.5, we
have
$\overline{M}\cong\operatorname{res}_{\alpha}\operatorname{res}_{\tilde{L}}\overline{M}$
and $\operatorname{Supp}(\overline{M})\subseteq{\uparrow}L$. The above formula
shows us that

$(\operatorname{unzip}_{L}\operatorname{zip}_{L}M)(c)=(\operatorname{res}_{\alpha}\operatorname{res}_{\tilde{L}}\overline{M})(c)$

for all $c\in{\uparrow}L$. If $c\notin{\uparrow}L$, then
$c\notin\operatorname{Supp}(\overline{M})$, which means that
$\overline{M}(c)=0$. In this case, we also have
$(\operatorname{unzip}_{L}\operatorname{zip}_{L}M)(c)=0$ by the definition of
the functor $\operatorname{unzip}_{L}$. Thus we have an isomorphism

$\overline{M}\cong\operatorname{unzip}_{L}\operatorname{zip}_{L}M.$

∎

## References