# Topological incidence rings, their groups of units and incidence functors

João V. P. e Silva

###### Abstract

We focus on working on incidence rings and their group of units. Some general
properties about them are given, including how they are always the inverse
limit of finite matrix groups/rings, giving a natural way to define their
topology. Such construction can also be translated as a functor from the
category of locally finite, preordered sets to the category of topological
rings. It is shown that under some assumptions on the ordered sets, simple
matrix groups over a field can be built from the group of units of such rings.

##  1 Introduction

The study of infinite matrices traces back to Poincaré in the discussion of
the well known Hill’s equation. In [Ber68], Bernkopf gives an introduction to
the history of the study of infinite matrices and their importance for
operator theory.
At the present paper we focus on incidence rings, a class of rings described
using locally finite ordered sets, that is, the sets $[x,y]=\\{z;\ x\leqslant
z\leqslant y\\}$ are finite for every $x,y$ in the ordered set. Such class of
rings was first defined at [Sta70] restricted to partially ordered sets, and
it was used to give a counter-example for a non-published conjecture. Further
work on incidence rings was done at [Vos80, Haa84, Fro85, PSS90, DvW96,
AHdR99, AHdR02, Khr10] with results about Morita duality, the isomorphism
problem on incidence rings and other general properties of such rings.
In this article we focus on the theory of topological incidence rings and
their groups of units. Alternative proof for the isomorphism problem of
incidence rings of partially ordered sets [Vos80, Theorem 1] is given using
topological arguments. We also show connections between topological incidence
rings and the category of locally finite preordered sets by constructing
incidence functors over a ring $P$, a collection of contravariant functors
from the category of locally finite preordered sets to the category of
topological rings that map colimits to limits. Using results from the
literature, we also show other interesting properties of the incidence
functors.
We also study the groups of units of incidence rings. A generalization of the
construction of matrix groups and rings from [GRW20] is given for ordered sets
that are not $\mathbb{Z}$-like [GRW20, Definition 2.1]. We also generalize the
construction of non-compactly generated, non-discrete, simple groups [GRW20,
Theorem 5.2] for a bigger class of groups that are not necessarily totally
disconnected, nor locally compact, nor second countable.
Groups similar to the ones studied here have been described previously. The
groups denoted here as $\textup{aGL}_{\mathbf{N}}(\mathbb{F}_{q})$, for
$\mathbf{N}$ the set of naturals with usual order and $\mathbb{F}_{q}$ the
finite field with $q$ elements, was first built at [VK98] as the group GLB.
Some first study on the representation of such groups was done at [GKV14].
Study of the group GLB and similarly defined groups over more general rings
are also described in [HLY17, GH12, GH15, Hoł02b].
We also prove sufficient conditions for the group of units of an incidence
ring to be quasi-discrete or solvable. It then follows that if $G$ contains a
matrix group with such property as an open, compact subgroup, then $G$ is
always an elementary topological group in the Wesolek sense [Wes15].
All rings in this article are assumed to be associative with identity. Rings
are assumed to be non-trivial, unless explicitly stated. Ring homomorphisms
send the multiplicative identity to the multiplicative identity. Topological
rings and groups are always Hausdorff spaces.

###  1.1 Structure of the article

Section 2 of this article will focus on giving the necessary background for
the definitions and results from the article. In Subsection 2.1 some
definitions for ordered sets are given. These will be central for the
definition of the incidence rings, their groups of units, and to prove results
for such classes of objects. Subsection 2.2 gives some results from category
theory that will be used in a later section.
In Section 3 we give the definition of the incidence rings. It is also shown
that if the locally finite preordered set is infinite, then these rings are
inverse limits of finite matrix rings, allowing us to give a topological
structure to the incidence rings. An alternative proof of the isomorphism
problem of incidence rings of partially ordered sets using topological
arguments [Vos80, Theorem 1] is also given, using a similar argument to
[Khr10, Theorem 5].
Section 4 focuses on connecting some properties from the incidence rings with
their partially ordered sets. The initial focus is describing a category for
locally finite preordered sets so that there exists a functor from this
category to the category of topological rings. It is also shown that under
such properties the functor is contravariant and sends colimits to limits. A
corollary of such a result is that all matrix rings over $P$ indexed by
partially ordered sets can be built from a countable collection of finite
matrices over $P$ under the operations of direct products, inverse limits, and
pullbacks.
Section 5 focuses on the groups of units of an incidence ring. A description
of such groups is given, and a generalization of the construction of GLB from
[VK98] and $\textup{AU}_{\Lambda}(\mathbf{F}_{q})$ from [GRW20] is also given.
The main focus of the section is generalizing the construction of simple
groups as in [GRW20] and showing it can be done for an arbitrary topological
field. We also show properties of topological groups that are locally the
group of units of incidence rings of “not too big” preordered sets.
Section 6 gives a list of questions and alternative constructions that might
also lead to interesting results.

Acknowledgements: This article is part of my PhD thesis written under the
supervision of George Willis, Colin Reid, and Stephan Tornier. I would like to
thank them for advising me while writing the article and proofreading it. A
special thanks as well to Mykola Khrypchenko for giving me references on the
past work done on incidence rings, and for some nice discussion on some of the
properties of such rings.

##  2 Background

###  2.1 Ordered sets

For the context of our rings and groups, a definition that proves necessary is
the following:

######  Definition 2.1.

Let $(\Lambda,\ \preceq)$ be a proset (preordered set). Given $s_{1}\preceq
s_{2}\in\Lambda$ we define $[s_{1},\ s_{2}]:=\\{s\in\Lambda;\ s_{1}\preceq
s\preceq s_{2}\\}$. We will call these subsets of $\Lambda$ intervals. We say
that the proset is locally finite if for every $s_{1}\preceq s_{2}\in S$ the
set $[s_{1},\ s_{2}]_{\preceq}$ is finite.

This locally finite type of property allows us to define multiplication and
addition on these rings in a similar way to the finite-dimensional matrix
rings. It will also be essential to build a nice topology for such
groups/rings.
As the central object of this text is algebraic structures defined in relation
to an ordered set, defining some subsets and relations in the ordered sets
will be essential for simplifying the proofs. This subsection will focus on
giving these basic definitions and providing some results about the structure
of ordered sets and their subsets.

######  Notation 2.2.

Let $\Lambda$ be a proset and $s_{1}\preceq s_{2}\in\Lambda$. We denote:

  * •

$s_{1}\precnsim s_{2}$ if $s_{1}\preceq s_{2}$ and $s_{1}\nsim s_{2}$.

  * •

$s_{1}\precnapprox s_{2}$ if $s_{1}\preceq s_{2}$ and $s_{1}\neq s_{2}$.

  * •

$s_{1}\npreceq s_{2}$ if it is not the case $s_{1}\preceq s_{2}$.

Notice that for $\Lambda$ a proset the set $[s,\ s]$ is the set of all
elements equivalent to $s$ under the preorder. Hence it is not always the case
$[s,\ s]=\\{s\\}$. Notice that if for every $s\in\Lambda$, $[s,\ s]=\\{s\\}$
then $\Lambda$ is a poset (partially ordered set).

######  Definition 2.3.

Given a proset $\Lambda$, we define the following:

  * •

Given $s\in\Lambda$, we define the neighbourhoods of $s$ as:

$\mathcal{N}_{n}(s):=\left\\{\begin{array}[]{ll}[s,\ s]&\mbox{for }n=0\\\
\\{t\in\Lambda;\ t\preceq s\mbox{ or }s\preceq t\\}&\mbox{for }n=1\\\
\bigcup_{t\in\mathcal{N}_{n-1}(s)}\mathcal{N}_{1}(t)&\mbox{for
}n>1\end{array}\right.$

The full collection of neighbours $s$ is defined as
$\mathcal{N}_{\omega}(s)=\bigcup_{n\in\mathbb{N}}\mathcal{N}_{n}(s)$.

  * •

We say $s_{1},\ s_{2}\in\Lambda$ are independent if
$s_{1}\notin\mathcal{N}_{\omega}(s_{2})$.

  * •

A subset $\Lambda^{\prime}\subset\Lambda$ is said to be closed under intervals
if for every $s_{1}\preceq s_{2}\in\Lambda^{\prime}$ then $[s_{1},\
s_{2}]_{\preceq}\subset\Lambda^{\prime}$.

  * •

Given $\Lambda^{\prime}\subset\Lambda$ a subset, we say $\Lambda^{\prime}$ is
convex if it is closed under intervals and for every $s_{1},\ s_{2}\in\Lambda$
there are $t_{1},\ t_{2},\ \ldots,\ t_{n}\in\Lambda^{\prime}$ such that
$t_{1}\in\mathcal{N}_{1}(s_{1})$, $t_{i+1}\in\mathcal{N}_{1}(t_{i})$,
$s_{2}\in\mathcal{N}_{1}(t_{n})$ for $1\leqslant i\leqslant n-1$. In other
words, there is a path of intervals inside $\Lambda^{\prime}$ connecting
$s_{1}$ and $s_{2}$.

  * •

Given $\\{\Lambda_{i}\\}_{i\in I}$ a collection such that for all $i\in I$,
$\Lambda_{i}\subset\Lambda$ is convex in $\Lambda$ and for $i\neq j\in I$ then
$\Lambda_{i}\bigcap\Lambda_{j}=\emptyset$, then $\\{\Lambda_{i}\\}_{i\in I}$
is called a locally convex collection of $\Lambda$.

######  Definition 2.4.

Let $\Lambda$ be a proset and $s\in\Lambda$. We say that:

  * •

$s$ is a maximal element if given $t\in\Lambda$ is such that $s\preceq t$ then
$s=t$.

  * •

$s$ is a minimal element if given $t\in\Lambda$ is such that $t\preceq s$ then
$s=t$.

  * •

$s$ is the maximum element if for all $t\in\Lambda$ we have $t\preceq s$.

  * •

$s$ is the minimum element if for all $t\in\Lambda$ we have $s\preceq t$.

The following examples illustrate the definition of intervals, neighbourhoods,
and convex subsets as denoted above.

######  Example 2.5.

  * •

$\mathbf{Q}$: The rational numbers with usual order is a partially ordered
sets that is not locally finite.

  * •

$\mathbf{Zig}$: This is the poset with elements in the integers and order
given by $2k>2k+1$ and $2k>2k-1$, for $k$ an integer.

${-2}$${0}$${2}$${\ldots}$${\ldots}$${-1}$${1}$${3}$

The only intervals for Zig are of the form $[n,\ n]=\\{n\\}$, $[2n,\
2n-1]=\\{2n,\ 2n-1\\}$ and $[2n,\ 2n+1]=\\{2n,\ 2n+1\\}$ for some
$n\in\mathbb{Z}$. In this case, $\mathcal{N}_{0}(0)=\\{0\\}$,
$\mathcal{N}_{n}(0)=\\{i\\}_{-n\leqslant i\leqslant n}$ and, because no two
elements are independent, $\mathcal{N}_{\omega}(0)=\mathbf{Zig}$. A subset
$S\subset\mathbf{Zig}$ is convex if, and only if, there are $n_{1},\
n_{2}\in\mathbf{Z}\bigcup\\{\infty,\ -\infty\\}$ such that
$S=\\{i\\}_{n_{1}<i<n_{2}}$. This poset has no independent element. Every
element is either maximal or minimal, but it doesn’t have any maximum or
minimum element.

  * •

$\mathbf{N}^{*}_{d}$: The non-zero natural numbers with order given by
$a\leqslant b$ if $b=ak$ for some $k$ natural number.
Order on the first $6$ natural numbers.

${1}$${2}$${4}$${3}$${6}$${5}$

The intervals of this poset can be described using arithmetic properties. For
example $[\leqslant 30]=\\{1,\ 2,\ 3,\ 5,\ 6,\ 10,\ 15,\
30\\}_{\mathbf{N}_{d}^{*}}$ (the divisors of $30$) and
$[30\leqslant]_{\mathbf{N}_{d}^{*}}=\\{30n\\}_{n\in\mathbf{N}_{d}^{*}}$
(multiples of $30$). The interval $[3,\ 30]=\\{3,\ 6,\ 15,\ 30\\}$, which are
all multiples of $3$ that divide $30$. Observe that
$\mathcal{N}_{1}(1)=\mathbf{N}_{d}^{*}$ but if $n\neq 1$ then
$\mathcal{N}_{1}(n)$ is the set of all multiples and divisors of $n$, but not
all $\mathbf{N}^{*}$. But, as $1$ is a divisor of all natural numbers,
$\mathcal{N}_{2}(n)=\mathbf{N}_{d}^{*}$ for all $n\in\mathbf{N}_{d}^{*}$. The
element $1$ is the minimum, but there is no maximal element.

We will also denote some commonly used ordered sets as follows:

  * •

$\mathbf{n<}$: The poset with elements $\\{0,\ 1,\ \ldots,\ n-1\\}$ and order
given by $0<1<\ldots<n-1$.

  * •

$S$: the poset with elements in the set $S$ and order given by equality. This
is the poset with elements on $S$ such that all elements are independent.
Example: $S=\\{0,\ 1,\ 2\\}$.

${0}$${1}$${2}$

  * •

$\overline{S}$: the proset with elements in the set $S$ and order given by
$i\preceq j$ for all $i,j$. In other words, it is the proset where all
elements are equivalent. In the case the set is $\\{0,\ 1,\ 2,\ \ldots,\
n-1\\}$ we will just denote it as n.
Example: 3

${1}$${0}$${2}$

  * •

$\mathbf{N}$: The natural numbers with the usual order, that is,
$0<1<2<\ldots<n<\ldots$.

  * •

$\mathbf{Z}$: The integers with the usual order, that is,
$\ldots<-1<0<1<\ldots.$

  * •

$\mathbf{m}\leftarrow\mathbf{n}$: the set with elements $\\{0^{\prime},\
1^{\prime},\ \ldots,\ m-1^{\prime},\ 0,\ 1,\ \ldots,\ n-1\\}$ and order given
by for every $i,\ j\in\\{0^{\prime},\ 1^{\prime},\ \ldots,\ m-1^{\prime}\\}$,
for every $k,\ l\in\\{0,\ 1,\ \ldots,\ n-1\\}$ we have $i\sim j$,
$i\precnapprox k$ and $k\sim l$.
Example: $\mathbf{2}\leftarrow\mathbf{2}$

${0^{\prime}}$${1^{\prime}}$${0}$${1}$

Notice that the prosets of the form $\mathbf{n}$ can be described as
$\mathbf{0}\leftarrow\mathbf{n}$, and the proset $\mathbf{2}<$ can be
described as $\mathbf{1}\leftarrow\mathbf{1}$. The prosets of the form
$\textbf{m}\leftarrow\textbf{n}$ are all the finite, irreducible prosets with
at most two equivalence classes of elements.

######  Definition 2.6.

Let $(\Lambda,\ \preceq)$ a proset. If there is a collection
$\\{\Lambda_{i}\\}_{i\in I}$ of convex subsets of $\Lambda$ such that
$\Lambda=\bigcup_{i\in I}\Lambda_{i}$, for every $i\neq j$ we have
$\Lambda_{i}\bigcap\Lambda_{j}=\emptyset$ and for every $s_{i}\in\Lambda_{i}$,
$s_{j}\in\Lambda_{j}$ we have that if $i\neq j$ then $s_{i}$ and $s_{j}$ are
independent, we say $\Lambda$ is the internal disjoint union of
$\\{\Lambda_{i}\\}_{i\in I}$. We denote it by $\Lambda=\sqcup_{i\in
I}\Lambda_{i}$.
If $\Lambda$ cannot be written as disjoint union of two non-empty subsets we
say $\Lambda$ is irreducible.
Let $\\{(\Lambda_{i},\ \preceq_{i})\\}_{i\in I}$ a collection of prosets. We
define the external disjoint union of this collection as the proset
$(\sqcup_{i\in I}\Lambda_{i},\ \preceq_{\sqcup})$ with elements in
$\sqcup_{i\in I}\Lambda_{i}$ and order $\preceq_{\sqcup}$ given by
$s_{1}\preceq_{\sqcup}s_{2}\in\sqcup_{i\in I}\Lambda_{i}$ if, and only if,
there is $i\in I$ such that $s_{1}\preceq_{i}s_{2}\in\Lambda_{i}$.

For the ones familiar with category theory, the disjoint union is the
coproduct on the category of prosets. Given the posets $\mathbf{2}\\!<=\\{0,\
1\\}$ and $\mathbf{3}\\!<=\\{0^{\prime},\ 1^{\prime},\ 2^{\prime}\\}$ then
$\mathbf{2\\!<}\\!\sqcup\\!\mathbf{3\\!<}$ is the poset with elements $\\{0,\
1,\ 0^{\prime},\ 1^{\prime},\ 2^{\prime}\\}$ and Hasse diagram as follows:

${1}$${2^{\prime}}$${1^{\prime}}$${0}$${0^{\prime}}$

The main use of disjoint union is breaking down ordered sets into irreducible
parts, as shown below:

######  Proposition 2.7.

Let $\Lambda$ be a non-empty preordered set. Then there is a collection
$\\{\Lambda_{i}\\}_{i\in I}$ of irreducible subsets of $\Lambda$ such that
$\Lambda=\sqcup_{i\in I}\Lambda_{i}$.

###### Proof.

Notice that for every $s\in\Lambda$, every element $t$ on the set
$\mathcal{N}_{\omega}(s)$ is independent from every element $t^{\prime}$ on
the set $\Lambda\backslash\mathcal{N}_{\omega}(s)$. It is also the case that
for every $s\in\Lambda$, the set $\mathcal{N}_{\omega}(s)$ is irreducible.
Now let $I\subset\Lambda$ a maximal subset of pairwise independent elements of
$\Lambda$ (such set exists by Zorn’s Lemma). Define for each $i\in I$ the
irreducible subset $\Lambda_{i}:=\mathcal{N}_{\omega}(i)$. As the set $I$ is
maximal, the observation above implies

$\Lambda=\sqcup_{i\in I}\Lambda_{i}.$

∎

We call the irreducible subsets $\Lambda_{i}=\mathcal{N}_{\omega}(i)$ the
components of $\Lambda$. The irreducible sets also give us a finite convex
hull for finite subsets of prosets as seen below.

######  Proposition 2.8.

Let $\Lambda$ an irreducible locally finite proset and $S\subset\Lambda$ a
finite subset. There is a finite convex subset $\alpha\subset\Lambda$ such
that $S\subset\alpha$.

###### Proof.

The result is trivial for $S$ the empty set. If $|S|>0$, let $s\in S$. For
each $s^{\prime}\in S$ let $t_{1}^{s^{\prime}},\ t_{2}^{s^{\prime}},\ \ldots,\
t_{n}^{s^{\prime}}$ be elements such that
$t_{1}^{s^{\prime}}\in\mathcal{N}_{1}(s)$,
$t^{s^{\prime}}_{i+1}\in\mathcal{N}_{1}(t^{s^{\prime}}_{i})$,
$s^{\prime}\in\mathcal{N}_{1}(t^{s^{\prime}}_{n})$. As $\Lambda$ is
irreducible such elements exist. Let
$S^{\prime}=S\bigcup\left(\cup_{s^{\prime}\in
S\backslash\\{s\\}}\\{t_{1}^{s^{\prime}},\ t_{2}^{s^{\prime}},\ \ldots,\
t_{n}^{s^{\prime}}\\}\right)$. This set is a finite union of finite sets,
hence is finite. Let $\alpha=\bigcup_{s_{1}\leqslant s_{2},\ s_{1},s_{2}\in
S^{\prime}}[s_{1},\ s_{2}]$. This set is a finite union of finite intervals
hence it is also finite.
To see $\alpha$ is convex, let $t_{1},\ t_{2}\in\alpha$ be such that
$t_{1}\preceq t_{2}$ in $\Lambda$. By the definition of $\alpha$ there are
$s_{1},\ s_{2},\ s_{1}^{\prime},\ s_{2}^{\prime}\in S^{\prime}$ such that
$t_{1}\in[s_{1},\ s_{2}]$ and $t_{2}\in[s_{1}^{\prime},\ s_{2}^{\prime}]$.
Notice that $s_{1}\preceq t_{1}\preceq t_{2}\preceq s_{2}^{\prime}$, hence
$[t_{1},\ t_{2}]\subset[s_{1},\ s_{2}^{\prime}]$. That is, $\alpha$ is closed
under intervals. By definition it is also convex. ∎

By looking at the components that intersect a finite subset $S\subset\Lambda$
individually, we obtain the following corollary:

######  Corollary 2.9.

Given $\Lambda$ a locally finite proset and $S\subset\Lambda$ a finite subset,
there is a locally convex collection $\\{\Lambda_{i}\\}_{i\in I}$ such that
$S\subset\bigcup_{i\in I}\Lambda_{i}$ and $\bigcup_{i\in I}\Lambda_{i}$ is
finite.

######  Definition 2.10.

Let $f:\Lambda_{1}\rightarrow\Lambda_{2}$ a function between preordered sets.
We say $f$ is an bijective order preserving map if it is a set bijection, and
for every $s_{1},\ s_{2}\in\Lambda_{1}$ we have that $s_{1}\preceq s_{2}$ if,
and only if, $f(s_{1})\preceq f(s_{2})$.

######  Definition 2.11.

Let $\Lambda$ be a set and $\preceq_{1}$, $\preceq_{2}$ two preorders on the
set $\Lambda$. We say $\preceq_{1}$ is a suborder of $\preceq_{2}$ if
$s_{1}\preceq_{1}s_{2}$ implies $s_{1}\preceq_{2}s_{2}$. If $\preceq_{1}$ is a
partial order we say it is a partial suborder of $\preceq_{2}$. If
$\preceq_{1}$ gives an irreducible partial order we say it is an irreducible
partial suborder of $\preceq_{2}$.
Given a collection $\\{\preceq_{i}\\}_{i\in I}$ of preorders of $\Lambda$, we
define their transitive closure as the minimal preorder
$\preceq_{\cup}:=\bigcup_{i\in I}\preceq_{i}$ such that if
$s_{1}\preceq_{i}s_{2}$ for some $i\in I$ then $s_{1}\preceq_{\cup}s_{2}$.

The following are useful to describe a proset $\Lambda$ in relation to a
proset $\Lambda^{\prime}$, with same elements as $\Lambda$, and a collection
of finite disjoint subsets. These definitions and results are essential for
building the simple topological groups in Subsection 5.3.

######  Definition 2.12.

Given proset $\Lambda$ and, for some indexing set $I$, $\\{S_{i}\\}_{i\in I}$
a collection of disjoint subsets of $\Lambda$, we define $(\Lambda+\sum_{i\in
I}S_{i},\ \preceq_{+})$ the proset with order given by:

$s_{1}\preceq_{+}s_{2}\textit{ if }\left\\{\begin{array}[]{l}s_{1}\preceq
s_{2}\text{ in }\Lambda\text{,\ or}\\\ \exists i_{1},\ i_{2},\ldots,i_{n}\in
I,\ \exists t_{i_{k}},\ q_{i_{k}}\in S_{i_{k}}\textit{ such that for
}1\leqslant k\leqslant n,\ s_{1}\preceq t_{i_{1}},\\\ t_{i_{k}}\preceq
q_{i_{k}}\textit{ and }q_{i_{n}}\preceq s_{2}.\end{array}\right.$

Notice that under such an order, for each $i\in I$ all elements in $S_{i}$ are
equivalent. This is the minimal order containing $\preceq$ as a suborder and
with such property.

######  Example 2.13.

Let $\Lambda=\mathbf{Zig}$ as in Example 2.5 and let $S_{1}=\\{a_{0},\
a_{3}\\}$, $S_{2}=\\{a_{2},\ a_{4}\\}$. Then the proset
$(\Lambda+(S_{1}+S_{2}),\ \preceq_{+})$ has the following order:

${a_{-2}}$${a_{0}}$${a_{2}}$${a_{4}}$${\cdots}$${\cdots}$${a_{-1}}$${a_{1}}$${a_{3}}$${a_{5}}$${0}$

where the dashed arrows are the new relations.

######  Proposition 2.14.

Let $(\Lambda,\ \preceq)$ be a proset. There are $\\{S_{i}\\}_{i\in I}$
disjoint subsets of $\Lambda$ and $\leqslant$ a partial suborder of $\preceq$
such that $(\Lambda,\ \preceq)\cong(\Lambda+\sum_{i\in I}S_{i},\
\leqslant_{+})$.

###### Proof.

First we need to get a partial order $\leqslant$ on $\Lambda$ in such a way we
can add equivalence relations and get our proset structure. Let $s_{1},\
s_{2}\in\Lambda$. If $s_{1}=s_{2}$ or $s_{1}\precnsim s_{2}$, define
$s_{1}\leqslant s_{2}$. For each $s\in\Lambda$, define $\leqslant$ restricted
to $\mathcal{N}_{0}(s)$ as a total order, that is, for every $s_{1},\
s_{2}\in\mathcal{N}_{0}(s)$ either $s_{1}\leqslant s_{2}$ or $s_{2}\leqslant
s_{1}$. By definition of $(\Lambda,\ \leqslant)$, it is then the case that
$(\Lambda,\ \preceq)\cong(\Lambda+\sum_{s\in\Lambda}\mathcal{N}_{0}(s),\
\leqslant_{+})$ ∎

######  Corollary 2.15.

If $\Lambda$ is a locally proset, then for every finite subset
$S^{\prime}\subset\Lambda$ the proset $(\Lambda+S^{\prime},\ \preceq_{+})$ is
locally finite.

###  2.2 The units group of a matrix ring

Given $P$ a ring, we denote the group of units of $P$ as $P^{*}$. We will
denote the ring of $n\times n$ matrices over $P$ by $\textup{M}_{n}(P)$, the
group of $n\times n$ invertible matrices by $\textup{GL}_{n}(P)$, and the
group of $n\times n$ invertible matrices with determinant $1$ as
$\textup{SL}_{n}(P)$.

######  Theorem 2.16 (Theorem 2.6 [Igu]).

The functor that maps a ring $P$ to its group of units is right adjoint.

######  Definition 2.17 (Definition 2.1 [Tar23]).

An absolute topological ring is a topological ring such that its group of
units with the subspace topology is a topological groups.

As topological fields are defined so the inverse map is continuous, every
topological field is an absolute topological ring.

######  Lemma 2.18 (§§103–105 [Dic01]).

Let $F$ be a field and let $n\in\mathbb{N}$ such that $n\geqslant 2$ ; in the
case $n=2$, assume $|F|>3$. Then every proper normal subgroup
$\textup{SL}_{n}(F)$ is central, and every noncentral normal subgroup of
$\textup{GL}_{n}(F)$ contains the groups $\textup{SL}_{n}(F)$.

######  Theorem 2.19 (Theorem 4.19 [How]).

Let $P$ be a commutative ring. Then for every $A\in\textup{M}_{n}(P)$ we have

$A\textup{adj}(A)=\textup{adj}(A)A=\det(A)I_{n}.$

Hence, $A\in\textup{M}_{n}(P)$ is invertible if, and only if, $\det(A)$ is
invertible in $P$.

######  Theorem 2.20 (Theorem 4.21 [How]).

Let $P$ be a commutative ring. If $A\in\textup{GL}_{n}(R)$ then

$A^{-1}=\frac{1}{\det(A)}\textup{adj}(A).$

######  Theorem 2.21 (Theorem 4.27 [How]).

[Cayley-Hamilton Theorem] Let $P$ be a commutative ring. If
$A\in\textup{GL}_{n}(R)$ then

$\textup{adj}(A)=-\sum_{i=1}^{n}c_{i}A^{i-1}$

where $c_{i}$ are the coefficients of the characteristic polynomial of $A$.

###  2.3 Category theory

######  Proposition 2.22 (Proposition 9.14 [Awo10]).

Right adjoint functors preserve limits and left adjoint functors preserve
colimits.

Let $\mathbf{U}:\mathbf{Rings}\rightarrow\mathbf{Grps}$ be the functor that
sends a ring $R$ to its group of units. This functor is right adjoint, hence
it preserve limits. Such property allow us to translate some results from
Section 3 and Section 4 to the group of units of matrices in Section 5.

######  Proposition 2.23 (Proposition 5.14 [Awo10]).

A category $\mathcal{C}$ has finite products (coproducts) and equalizers
(coequalizers) if, and only if, it has pullbacks (pushouts) and a terminal
object (initial object).

######  Proposition 2.24 (Corollary 5.22, Theorem 5.23 [Awo10]).

A category $\mathcal{C}$ has all its limits (colimits) of some cardinality
$\kappa$ if it is has all equalizers (coequalizers) and products (coproducts)
of cardinality $\kappa$.

######  Proposition 2.25 (Exercise 2 [Lan71]).

If $\mathcal{C}$ has all limits (colimits) and
$F:\mathcal{C}\rightarrow\mathcal{D}$ is a functor that preserves all products
(coproducts) of some cardinality $\kappa$ and all equalizers (coequalizers),
then $F$ maps limits (colimits) to limits (colimits).

Propositions 2.24 and 2.25 gives us tools to prove, under some conditions, the
existence of a faithful functor from the category or partially ordered sets to
the category of topological rings, sending the partially ordered sets to the
rings of matrices indexed by the partially ordered set.

###  2.4 Topological groups

######  Definition 2.26.

Let $G$ be a totally disconnected, locally compact, second countable group
(t.d.l.c.s.c.). We say $G$ is regionally elliptic if every compact subset of
$G$ is contained in a compact subgroup.

######  Definition 2.27 (Introduction [BM00]).

Let $G$ be a topological group. We define the quasicentre of $G$ as:

$QZ(G)=\\{g\in G;\ C_{G}(g)\text{ is open in }G\\}.$

When studying totally disconnected, locally compact groups, a class of groups
that play an important role are the elementary groups.

######  Definition 2.28 (Definition 1.1[Wes15]).

The class of elementary groups is the smallest class $\mathcal{E}$ of
t.d.l.c.s.c. groups such that:

  1. (i)

$\mathcal{E}$ contains all second countable profinite groups and countable
discrete groups;

  2. (ii)

$\mathcal{E}$ is closed under taking group extensions, that is, if there is
$N\trianglelefteq H$ closed subgroup such that $N\in\mathcal{E}$ and
$H/N\in\mathcal{E}$ then $H\in\mathcal{E}$;

  3. (iii)

$\mathcal{E}$ is closed under taking closed subgroups;

  4. (iv)

$\mathcal{E}$ is closed under taking quotients by closed normal subgroups;

  5. (v)

If $G$ is a t.d.l.c.s.c. group and $\bigcup_{i\in\mathbb{N}}O_{i}=G$, where
$\\{O_{i}\\}_{i\in\mathbb{N}}$ is an $\subset$-increasing sequence of open
subgroups of $G$ with $O_{i}\in\mathcal{E}$ for each $i$, then
$G\in\mathcal{E}$. We say that $\mathcal{E}$ is closed under countable
increasing union.

The class $\mathcal{E}$ of elementary groups admits a well-behaved rank $\xi$
[Wes15, Lemma 4.12]. This rank roughly measures how many steps it is necessary
to build an elementary group $G$ from the discrete and profinite groups under
operations $(ii)$ until $(v)$.

######  Proposition 2.29 (Proposition 6.1 [Wes15]).

Let $G$ be a t.d.l.c.s.c. group. Assume $G$ contains an open compact subgroup
$U\leqslant G$ such that $\textup{QZ}(U)$ is dense in $U$. Then $G$ is
elementary and $\xi(G)\leqslant 3$.

######  Definition 2.30.

The derived series of a topological group $G$ is the sequence of closed normal
groups $\\{G^{(n)}\\}_{n\in\mathbb{N}}$ defined by $G^{(0)}=G$ and
$G^{(n+1)}=\overline{[G^{(n)},G^{(n)}]}$. A group $G$ is solvable if this
sequence stabilizes at $\\{1\\}$ after finitely many steps. A group $G$ is
residually solvable if $\bigcap_{n\in\mathbb{N}}G^{(n)}=\\{1\\}$.

######  Theorem 2.31 (Theorem 8.1, Proposition 4.19 [Wes15]).

Let $G$ be a t.d.l.c.s.c. group. If there is $U$ an open, compact subgroup of
$G$ such that $U$ is solvable, then $G$ is elementary with $\xi(G)<\omega$.

##  3 Topological incidence rings

###  3.1 First definitions

######  Definition 3.1.

[Bel73, Definition 1.1] Let $P$ be a ring and $(\Lambda,\ \preceq)$ a locally
finite proset. We define the incidence ring of $\Lambda$ over $P$ as

$\textup{M}_{\Lambda}(P):=\left\\{(a_{s_{1},s_{2}})\in
P^{\Lambda\times\Lambda};\textit{ if }a_{s_{1},s_{2}}\neq 0\textit{ then
}s_{1}\preceq s_{2}\right\\}$

with coordinatewise sum and multiplication defined as follows: given two
element $(a_{s_{1},s_{2}}),(b_{s_{1},s_{2}})\in\textup{M}_{\Lambda}(P)$ then
$(a_{s_{1},s_{2}}).(b_{s_{1},s_{2}})=(c_{s_{1},s_{2}})$, where
$c_{s_{1},s_{2}}=\sum_{t\in[s_{1},s_{2}]}a_{s_{1},t}b_{t,s_{2}}$.

Note that this multiplication is defined in such a way that for any finite
$n\times n$ block of the matrix centered on the main diagonal, the
multiplication follows the same rule as $\textup{M}_{\mathbf{n}}(P)$. We will
denote the elements of $\textup{M}_{\Lambda}(P)$ as $A,B,C$. We denote
$A_{s_{1},s_{2}}$ as the coordinate $(s_{1},\ s_{2})$ in the matrix $A$. When
talking about the coordinate of the product $A_{1}A_{2}\ldots A_{n}$ we denote
it as $(A_{1}A_{2}\ldots A_{n})_{s_{1},s_{2}}$. Following are some examples of
incidence rings.

######  Example 3.2.

  1. 1.

Given $n\in\mathbb{N}$, the ring $\textup{M}_{\mathbf{n}}(P)$ is the ring of
$n\times n$ matrices over the ring $P$ with the usual multiplication.

  2. 2.

Given $n\in\mathbb{N}$, the ring $\textup{M}_{\mathbf{n<}}(P)$ is the ring of
$n\times n$ upper triangular matrices over $P$ with the usual multiplication.

  3. 3.

Given $S$ a set, the ring $\textup{M}_{S}(P)$ is isomorphic to $\prod_{s\in
S}P$ with the coordinate-wise multiplication.

  4. 4.

The ring $\textup{M}_{\mathbf{N}^{*}_{d}}(P)$ has elements of the form:

$A:=\left(\begin{array}[]{cccccc}A_{1,1}&A_{1,2}&A_{1,3}&A_{1,4}&A_{1,5}&\cdots\\\
0&A_{2,2}&0&A_{2,4}&0&\cdots\\\ 0&0&A_{3,3}&0&0&\cdots\\\
0&0&0&A_{4,4}&0&\cdots\\\ 0&0&0&0&A_{5,5}&\cdots\\\ 0&0&0&0&0&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right)$

and the multiplication of two matrices
$A,B\in\textup{M}_{\mathbf{N}^{*}_{d}}(P)$ is:

$AB:=\left(\begin{array}[]{ccccccc}A_{1,1}B_{1,1}&\sum_{d|2}A_{1,d}B_{d,2}&\sum_{d|3}A_{1,d}B_{d,3}&\sum_{d|4}A_{1,d}B_{d,4}&\sum_{d|5}A_{1,d}B_{d,5}&\cdots\\\
0&A_{2,2}B_{2,2}&0&\sum_{d|4}A_{2,d}B_{d,4}&0&\cdots\\\
0&0&A_{3,3}B_{3,3}&0&0&\cdots\\\ 0&0&0&A_{4,4}B_{4,4}&0&\cdots\\\
0&0&0&0&A_{5,5}B_{5,5}&\cdots\\\ 0&0&0&0&0&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right)$

Given $S\subset\textup{M}_{\Lambda}(P)$ a subset, we define
$S_{s_{1},s_{2}}=\\{A_{s_{1},s_{2}};\ A\in S\\}\subset P$. Note though that
$R\subset\textup{M}_{\Lambda}(P)$ a subset is a subring if $0,1\in R$ and for
every $s_{1}\preceq s_{2}$ in $\Lambda$ we have that

$R_{(s_{1},s_{2})}+R_{(s_{1},s_{2})}\subset R_{(s_{1},s_{2})},$
$\sum_{s\in[s_{1},s_{2}]}R_{s_{1},s}R_{s,s_{2}}\subset R_{s_{1},s_{2}}.$

And $I\subset\textup{M}_{\Lambda}(P)$ a subset is a two-sided ideal if $0\in
I$ and for every $s_{1}\preceq s_{2}$ in $\Lambda$ we have that

$R_{(s_{1},s_{2})}+R_{(s_{1},s_{2})}\subset R_{(s_{1},s_{2})},$
$\sum_{s\in[s_{1},s_{2}]}I_{s_{1},s}R_{s,s_{2}}\subset I_{s_{1},s_{2}},$
$\sum_{s\in[s_{1},s_{2}]}R_{s_{1},s}I_{s,s_{2}}\subset I_{s_{1},s_{2}}.$

This allows us to easily describe some ideals and subrings of
$\textup{M}_{\Lambda}(P)$ central to our work. These also allow us to prove
the existence of a topology on $\textup{M}_{\Lambda}(P)$ in a natural way,
preserving properties of the finite blocks of matrices.

######  Notation 3.3 (Ideals of $\textup{M}_{\Lambda}(P)$).

Let $\Lambda$ a locally finite proset and $P$ a ring. We denote the following
two-sided ideals:

  * •

Given $s_{1}\preceq s_{2}$ in $\Lambda$ we define

$I^{\Lambda}_{[s_{1},s_{2}]}(P)=\\{A\in\textup{M}_{\Lambda}(P);\ \text{ if
}t_{1},\ t_{2}\in[s_{1},\ s_{2}]\text{ then }A_{t_{1},t_{2}}=0\\}.$

This ideal has quotient

$\textup{M}_{\Lambda}(P)/I^{\Lambda}_{[s_{1},s_{2}]}(P)\cong\textup{M}_{[s_{1},s_{2}]}(P).$

  * •

For a convex set $\Lambda^{\prime}\subset\Lambda$, the ideal

$I^{\Lambda}_{\Lambda^{\prime}}=\bigcap_{s_{1}\preceq
s_{2}\in\Lambda^{\prime}}I^{\Lambda}_{[s_{1},s_{2}]}.$

This ideal has quotient

$\textup{M}_{\Lambda}(P)/I^{(\Lambda,\
\preceq)}_{\Lambda^{\prime}}\cong\textup{M}_{\Lambda^{\prime}}(P).$

  * •

For $\\{\Lambda_{i}\\}_{i\in I}$ a locally convex set of $\Lambda$ we define
the ideal

$I^{\Lambda}_{\\{\Lambda_{i}\\}_{i\in I}}=\bigcap_{i\in
I}I^{\Lambda}_{\Lambda_{i}}.$

This ideal has quotient

$\textup{M}_{\Lambda}(P)/I^{\Lambda}_{\\{\Lambda_{i}\\}_{i\in
I}}\cong\prod_{i\in I}\textup{M}_{\Lambda_{i}}(P).$

  * •

Given $J\leqslant P$ a two-sided ideal we define

$\textup{M}_{\Lambda}(J):=\\{A\in\textup{M}_{\Lambda}(P);\text{ for every
}t_{1},t_{2}\in\Lambda\text{ then }A_{t_{1},t_{2}}\in
J\\}\leqslant\textup{M}_{\Lambda}(P).$

This ideal has quotient

$\textup{M}_{\Lambda}(P)/\textup{M}_{\Lambda}(J)\cong\textup{M}_{\Lambda}(P/J).$

  * •

Given $\\{\Lambda_{i}\\}_{i\in I}$ a locally convex set of $\Lambda$ and
$J\leqslant P$ a two-sided ideal of $P$, then the ideal

$I_{\\{\Lambda_{i}\\}_{i\in I}}+\textup{M}_{\Lambda}(J)$

has quotient

$\textup{M}_{\Lambda}(P)/(I_{\\{\Lambda_{i}\\}_{i\in
I}}+\textup{M}_{\Lambda}(J))\cong\prod_{i\in I}\textup{M}_{\Lambda_{i}}(P/J).$

Some of the ideals above were initially defined at [Bel73].

######  Example 3.4.

In the case of the matrix ring $\textup{M}_{\mathbf{N}}(\mathbb{Z}_{p})$ and
interval be $[0,\ 2]=\\{0,\ 1,\ 2\\}$, we have that $I_{[0,2]}$ has elements
of the form

$\left(\begin{array}[]{cccccc}0&0&0&A_{0,3}&A_{0,4}&\cdots\\\
0&0&0&A_{1,3}&A_{1,4}&\cdots\\\ 0&0&0&A_{2,3}&A_{2,4}&\cdots\\\
0&0&0&A_{3,3}&A_{3,4}&\cdots\\\ 0&0&0&0&A_{4,4}&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right)$

and the quotient $\textup{M}_{\mathbf{N}}(\mathbb{Z}_{p})/I_{[0,2]}$ is
isomorphic to $\textup{M}_{\mathbf{3<}}(\mathbb{Z}_{p})$, the ring of upper
$3\times 3$ triangular matrices over $\mathbb{Z}_{p}$. Given the ideal
$p\mathbb{Z}_{p}$ of $\mathbb{Z}_{p}$, the quotient
$\textup{M}_{\mathbf{N}}(\mathbb{Z}_{p})/\textup{M}_{\mathbf{N}}(p\mathbb{Z}_{p})$
is isomorphic to $\textup{M}_{\mathbf{N}}(\mathbb{Z}/p\mathbb{Z})$, and
$\textup{M}_{\mathbf{N}}(\mathbb{Z}_{p})/(I_{[0,2]}+\textup{M}_{\mathbf{N}}(p\mathbb{Z}_{p}))$
is isomorphic to $\textup{M}_{\mathbf{3<}}(\mathbb{Z}/p\mathbb{Z}).$

###  3.2 Infinite matrices as limits of finite matrices

To prove the results in this section we will look at $\textup{M}_{\Lambda}(P)$
as objects in the category of associative topological rings with identity and
the morphisms are the continuous homomorphism of rings. This category will be
denoted as TopRings. For the next result, notice that if $P$ is a topological
ring and $\Lambda$ is a finite proset, then the ring $\textup{M}_{\Lambda}(P)$
is a topological ring with subspace topology in relation to the product space
$P^{\Lambda\times\Lambda}$. In a similar way, for $\Lambda$ an infinite,
locally finite proset we define the topology on $\textup{M}_{\Lambda}(P)$ as
the subspace topology on $P^{\Lambda\times\Lambda}$. Our main goal in this
subsection is proving that with this topology, $\textup{M}_{\Lambda}(P)$ is a
topological ring.
Notice that if $\Lambda^{\prime}\subset\Lambda$ is a finite convex subset, say
$\Lambda^{\prime}=\\{s_{1},\ s_{2},\ \ldots,\ s_{n}\\}$, then
$\textup{M}_{\Lambda^{\prime}}(P)$ can be seen as a subring of
$\textup{M}_{\mathbf{n}}(P)$ with embedding determined by the enumeration of
$\Lambda^{\prime}$. The ideals of the form $I^{\Lambda}_{\Lambda^{\prime}}$,
for $\Lambda^{\prime}$ a finite convex subset, will then be essential to build
our inverse limit and the topology, as the quotient under such ideals are
topological rings.
Given $\Lambda$ a proset, define $\Gamma(\Lambda)=\\{\alpha\subset\Lambda;\
\alpha\text{ is finite and convex}\\}$ the set of all finite convex subsets of
$\Lambda$. The next result follows directly from Proposition 2.8.

######  Corollary 3.5.

If $\Lambda$ is an irreducible poset then for convex subsets $\alpha,\
\beta\in\Gamma(\Lambda)$ there is $\gamma\in\Gamma(\Lambda)$ such that
$\alpha\subset\gamma$ and $\beta\subset\gamma$, that is, under the $\subset$
order the set $\Gamma(\Lambda)$ is a directed set.

We can then prove the following Theorem:

######  Theorem 3.6.

Let $P$ a ring and $(\Lambda,\ \preceq)$ an infinite irreducible locally
proset. Then
$\textup{M}_{\Lambda}(P)\cong\varprojlim_{\alpha\in\Gamma(\Lambda)}\textup{M}_{\alpha}(P)$.
If $P$ is a topological ring then $\textup{M}_{\Lambda}(P)$ is a topological
ring with topology given by the inverse limit.

###### Proof.

By Corollary 3.5 we have that $\Gamma(\Lambda)$ is a directed set. Given
$\alpha\subset\beta\in\Gamma(\Lambda)$ define
$\pi_{\beta\rightarrow\alpha}:\textup{M}_{\beta}(P)\longrightarrow\textup{M}_{\alpha}(P)$
as the natural projection of $\textup{M}_{\beta}(P)$ with kernel
$I^{\beta}_{\alpha}$. It is easy to see that given
$\alpha\subset\beta\subset\gamma$ then
$\pi_{\gamma\rightarrow\alpha}=\pi_{\beta\rightarrow\alpha}\circ\pi_{\gamma\rightarrow\beta}$,
that is, the following diagram commutes.

${\textup{M}_{\gamma}(P)}$${\textup{M}_{\beta}(P)}$${\textup{M}_{\alpha}(P)}$$\scriptstyle{\pi_{\gamma\rightarrow\alpha}}$$\scriptstyle{\pi_{\gamma\rightarrow\beta}}$$\scriptstyle{\pi_{\beta\rightarrow\alpha}}$

The collection
$((\textup{M}_{\alpha})_{\alpha\in\Gamma(\Lambda)},(\pi_{\beta\rightarrow\alpha})_{\alpha\subset\beta})$
is then an inverse system. As the category of associative rings with identity
is closed under inverse limits, the inverse limit of this system exists and is
unique. Denote $\varprojlim_{\alpha\in\Gamma(\Lambda)}\textup{M}_{\alpha}(P)$
as the inverse limit of this system. Our objective now is proving
$\textup{M}_{\Lambda}\cong\varprojlim_{\alpha\in\Gamma(\Lambda)}\textup{M}_{\alpha}(P)$.
Observe that for every $\alpha\in\Gamma(\Lambda)$ there is
$\psi_{\alpha}:\textup{M}_{\Lambda}(P)\longrightarrow\textup{M}_{\alpha}(P)$
given by the natural projection of $\textup{M}_{\Lambda}$ under
$I^{\Lambda}_{\alpha}$, and these projections are such that the outer triangle
of the following diagram commutes.

${\textup{M}_{\Lambda}(P)}$${/}$${\varprojlim_{\alpha\in\Gamma(\Lambda)}\textup{M}_{\alpha}(P)}$${\textup{M}_{\alpha_{2}}(P)}$${\textup{M}_{\alpha_{1}}(P)}$$\scriptstyle{f}$$\scriptstyle{\psi_{\alpha_{1}}}$$\scriptstyle{\psi_{\alpha_{2}}}$$\scriptstyle{\pi_{\alpha_{1}}}$$\scriptstyle{\pi_{\alpha_{2}}}$$\scriptstyle{\pi_{\alpha_{2}\rightarrow\alpha_{1}}}$

By the universal property of inverse limits, there exists an unique ring
homomorphism
$f:\textup{M}_{\Lambda}(P)\longrightarrow\varprojlim_{\alpha\in\Gamma(\Lambda)}\textup{M}_{\alpha}(P)$
such that the whole diagram commutes. It remains to show the function $f$ is a
bijection.
Let $A\in\textup{M}_{\Lambda}(P)$ be such that $f(A)=0$. Because the diagram
commutes we have that $\psi_{\alpha}(A)=0$ for all $\alpha\in\Gamma(\Lambda)$.
Because these projections are such that $\psi_{\alpha}(A)=0$ if, and only if,
$A_{t_{1},t_{2}}=0$ for all $t_{1},\ t_{2}\in\alpha$ we have that $A=0$. Hence
$f$ is injective.
It remains to prove that $f$ is surjective. For that, given
$\overline{A}\in\varprojlim_{\alpha\in\Gamma(\Lambda)}\textup{M}_{\alpha}(P)$
we need to find $A\in\textup{M}_{\Lambda}(P)$ such that $f(A)=\overline{A}$.
We will describe such $A$ coordinate-wise. Notice that as $f$ is a ring
homomorphism, it will an isomorphism if, and only if, $f$ is an isomorphism of
abelian groups (when we forget the multiplication structure on our ring).
Hence we can treat these as abelian groups to simplify our proof.
Observe that by looking at these as abelian groups, for every proset $\Lambda$
we have that $\textup{M}_{\Lambda}(P)$ is a subgroup of
$P^{\Lambda\times\Lambda}$. Hence for every $s_{0},\ t_{0}\in\Lambda$ there is
a projection

$\begin{array}[]{c}\pi^{\Lambda}_{s_{0},t_{0}}:\textup{M}_{\Lambda}(P)\rightarrow
P\\\ A\mapsto A_{s_{0},t_{0}}.\end{array}$

Similarly, for $\alpha_{1},\alpha_{2}\in\Gamma(\Lambda)$, with
$\alpha_{1}\subset\alpha_{2}$, and $s_{0},\ t_{0}\in\alpha_{1}$ there are
projection
$\pi^{\alpha_{1}}_{s_{0},t_{0}}:\textup{M}_{\alpha_{1}}(P)\rightarrow P$ and
$\pi^{\alpha_{2}}_{s_{0},t_{0}}:\textup{M}_{\alpha_{2}}(P)\rightarrow P$ such
that
$\pi_{s_{0},t_{0}}^{\alpha_{1}}\circ\psi_{\alpha_{1}}=\pi_{s_{0},t_{0}}^{\alpha_{2}}\circ\psi_{\alpha_{2}}$,
and the following diagram commutes:

${\textup{M}_{\Lambda}(P)}$${/}$${\varprojlim_{\alpha\in\Gamma(\Lambda)}\textup{M}_{\alpha}(P)}$${\textup{M}_{\alpha_{2}}(P)}$${\textup{M}_{\alpha_{1}}(P)}$${P}$$\scriptstyle{f}$$\scriptstyle{\psi_{\alpha_{1}}}$$\scriptstyle{\psi_{\alpha_{2}}}$$\scriptstyle{\pi_{\alpha_{1}}}$$\scriptstyle{\pi_{\alpha_{2}}}$$\scriptstyle{\pi_{\alpha_{2}\rightarrow\alpha_{1}}}$$\scriptstyle{\pi^{\alpha_{2}}_{s_{0},t_{0}}}$$\scriptstyle{\pi^{\alpha_{1}}_{s_{0},t_{0}}}$

That is, the coordinate $A_{s_{0},t_{0}}$ is the same for all
$\alpha\in\Gamma(\Lambda)$ such that $s_{0},\ t_{0}\in\alpha$.
Notice that
$\pi_{s_{0},t_{0}}^{\beta_{1}}\circ\pi_{\alpha}(\overline{A})=\pi_{s_{0},t_{0}}^{\beta_{2}}\circ\pi_{\alpha}(\overline{A})$
for every $\beta_{1},\ \beta_{2}\in\Gamma(\Lambda)$ such that $[s_{0},\
t_{0}]\subset\beta_{i}$, $i=1,\ 2$. We define $A$ coordinatewise as
$A_{s_{0},t_{0}}=\pi^{\alpha}_{s_{0},t_{0}}\circ\pi_{\alpha}(\overline{A})$,
for $s_{0},\ t_{0}\in\alpha$ and $\alpha\in\Gamma(\Lambda)$. As the diagram
commutes, $\pi^{\alpha}_{s_{0},t_{0}}\circ\pi_{\alpha}\circ
f=\pi^{\alpha}_{s_{0},t_{0}}\circ\psi_{\alpha}$ for all $s_{0}\preceq t_{0}$
such that $s_{0},\ t_{0}\in\alpha$, and $\alpha\in\Gamma(\Lambda)$. Hence $A$
is so that $f(A)=\overline{A}$. That is, $f$ is an isomorphism of rings.
By the inverse limit property, $\textup{M}_{\Lambda}(P)$ is a closed subset of
$\prod_{\alpha\in\Gamma(\Lambda)}\textup{M}_{\alpha}(P)$ under the product
topology, hence it can be seen as a topological ring.
Notice that, as observed earlier for finite prosets, it then follows that
$\textup{M}_{\Lambda}(P)$ is a closed subset of
$\prod_{\alpha\in\Gamma(\Lambda)}P^{\alpha\times\alpha}$. As $\Lambda$ is
infinite, we have that $|\Gamma(\Lambda)|=|\Lambda|$. Hence there exists a
homeomorphism between $\prod_{\alpha\in\Gamma(\Lambda)}P^{\alpha\times\alpha}$
and $P^{\Lambda\times\Lambda}$ given by a bijection from
$\bigsqcup_{\alpha\in\Gamma{(\Lambda)}}\alpha\\!\times\\!\alpha$ to
$\Lambda\\!\times\\!\Lambda$. That is, the topological ring structure on
$\textup{M}_{\Lambda}(P)$ can be seen as the subspace topology in relation to
$P^{\Lambda\times\Lambda}$, as in the finite proset case. ∎

######  Proposition 3.7.

[Fro85, Lemma 2.2] Let $P$ be a ring and $\Lambda=\sqcup_{i\in I}\Lambda_{i}$
a preordered set. Then $\textup{M}_{\Lambda}(P)\cong\prod_{i\in
I}\textup{M}_{\Lambda_{i}}(P)$. If $P$ is a topological ring then
$\textup{M}_{\Lambda}(P)$ is a topological ring with topology given by the
product topology on the $\textup{M}_{\Lambda_{i}}(P)$.

Notice that Theorem 3.6 and Proposition 3.7 implies that all matrix rings
indexed by prosets are limits of finite matrix rings. Both results then imply
that the topology given to $\textup{M}_{\Lambda}(P)$ in relation to
$\textup{P}^{\Lambda\times\Lambda}$ makes $\textup{M}_{\Lambda}(P)$ into a
topological ring. More than that, this is the topology given by the inverse
limit.
For an example, let $\mathbf{2}\\!<$ be our poset and $P$ be any ring. We can
define an isomorphism of rings from
$f:\textup{M}_{\mathbf{2}<\sqcup\mathbf{2}<}(P)\rightarrow\textup{M}_{\mathbf{2}<}(P)\times\textup{M}_{\mathbf{2}<}(P)$
as follows:

$\left(\begin{array}[]{cccc}A_{0,0}&A_{0,1}&0&0\\\ 0&A_{1,1}&0&0\\\
0&0&A_{3,3}&A_{3,4}\\\
0&0&0&A_{4,4}\end{array}\right)\mapsto\left(\begin{array}[]{cc}A_{0,0}&A_{0,1}\\\
0&A_{1,1}\end{array}\right)\times\left(\begin{array}[]{cc}A_{0^{\prime},0^{\prime}}&A_{0^{\prime},1^{\prime}}\\\
0&A_{1^{\prime},1^{\prime}}\end{array}\right).$

###  3.3 Topological structure

We now use Theorem 3.6 and Proposition 3.7 to prove properties of the rings as
topological spaces. Given $P$ a topological ring, $\Lambda$ a proset and
$S\subset P$ a non-empty subset such that $0\in S$, we define

$\textup{M}_{\Lambda}(S):=\left\\{A\in\textup{M}_{\Lambda}(P);\
A_{s_{1},s_{2}}\in S\text{ for all }s_{1},\
s_{2}\in\Lambda\right\\}\subset\textup{M}_{\Lambda}(P).$

######  Remark 3.8.

Let $\\{X_{i}\\}_{i\in I}$ be a collection of topological spaces. Define
$X=\prod_{i\in I}X_{i}$ a topological space given by the product topology on
the sets $X_{i}$. For each $i\in I$, let $\pi_{i}:X\rightarrow X_{i}$ be the
projection on the $i$-th coordinate. Then:

  * •

The subsets $O\subset X$ such that for all $i\in I$ the set $\pi_{i}(O)$ is an
open subset of $X_{i}$, and there is a finite subset $J\subset I$ such that if
$i\in I\backslash J$ then $\pi_{i}(O)=X_{i}$ form a basis of open subsets of
$\prod_{i\in I}X_{i}$.

  * •

Given $\\{x_{a}\\}_{a\in A}$ a net in $X$ converges, it converges to $x$ if,
and only if, for every $i\in I$ the net $\\{\pi_{i}(x_{a})\\}_{a\in A}$
converges to $\pi_{i}(x)$ in $X_{i}$.

The following results follow directly from Remark 3.8 and the topology given
on $\textup{M}_{\Lambda}(P)$ as a topological subspace of
$P^{\Lambda\times\Lambda}$.

######  Corollary 3.9.

Let $P$ be a topological ring, $\\{S_{\sigma}\\}_{\sigma\in\Sigma}$ a basis of
the neighbourhood of $0$ and $\Lambda$ a locally proset. Then
$\\{I^{\Lambda}_{\alpha}(P)+\textup{M}_{\Lambda}(S_{\sigma})\\}_{\alpha\in\Gamma(\Lambda),\
\sigma\in\Sigma}$ is a basis of the neighbourhood of $0$ in
$\textup{M}_{\Lambda}(P)$.

######  Corollary 3.10.

Let $P$ be a topological ring and $\Lambda$ a locally finite proset. Given $J$
a two-sided open ideal of $P$ and $\\{\Lambda_{i}\\}_{i\in I}\subset\Lambda$
locally convex, then the following are true:

  1. 1.

The ideals $I^{\Lambda}_{\\{\Lambda_{i}\\}_{i\in I}}$ and
$\textup{M}_{\Lambda}(J)$ are closed in $\textup{M}_{\Lambda}(P)$.

  2. 2.

The ideal $I^{\Lambda}_{\\{\Lambda_{i}\\}_{i\in I}}$ is open if, and only if,
$\bigsqcup_{i\in I}\Lambda_{i}$ is finite and $P$ has the discrete topology.

  3. 3.

The ideal $\textup{M}_{\Lambda}(J)$ is open if, and only if, $\Lambda$ is
finite.

######  Corollary 3.11.

Let $P$ be a topological ring, $\Lambda$ an irreducible proset and
$\\{A^{(i)}\\}_{i\in I}\subset\textup{M}_{\Lambda}(P)$ a net. The net is
convergent if, and only if, for every $\alpha\in\Gamma(\Lambda)$ we have that
$\\{\pi_{\alpha}(A^{(i)})\\}_{i\in I}$ is convergent, and the sequence
congerges to $A$ defined coordinate-wise as follows:

$A_{s_{0},t_{0}}:=\left\\{\begin{array}[]{ll}\lim_{i\in
I}A^{(i)}_{s_{0},t_{0}},&\mbox{if }s_{0}\preceq t_{0}\in\Lambda\\\
0,&\mbox{otherwise.}\end{array}\right.$

We can also easily describe a dense subset:

######  Notation 3.12.

Given $P$ a ring and $\Lambda$ a proset, we denote the following elements from
$\textup{M}_{\Lambda}(P)$ as:

  * •

For $p\in P$ we define the element $p^{\Lambda}$ coordinatewise as:

$p^{\Lambda}_{s_{1},s_{2}}:=\left\\{\begin{array}[]{ll}p,&\mbox{if
}s_{1}=s_{2}\in\Lambda\\\ 0,&\mbox{otherwise.}\end{array}\right.$

  * •

For $S\subset\Lambda$ a subset we define the element $1^{S}$ coordinatewise
as:

$1^{S}_{s_{1},s_{2}}:=\left\\{\begin{array}[]{ll}1,&\mbox{if }s_{1}=s_{2}\in
S\\\ 0,&\mbox{otherwise.}\end{array}\right.$

  * •

For $t_{1}\preceq t_{2}\in\Lambda$ we define the element $e^{(t_{1},t_{2})}$
coordinatewise as:

$e^{(t_{1},t_{2})}_{s_{1},s_{2}}:=\left\\{\begin{array}[]{ll}1,&\mbox{if
}s_{1}=t_{1}\text{ and }s_{2}=t_{2}\\\ 0,&\mbox{otherwise.}\end{array}\right.$

Notice that for every $s\in\Lambda$, by definition, $e^{(s,s)}=1^{\\{s\\}}$.
Given $P$ a ring and $S\subset P$ a subset we will define $\langle S\rangle$
the minimal subring of $P$ containing $S$.

######  Proposition 3.13.

Let $P$ be a topological ring, $S\subset P$ a dense subset of $P$ and
$\Lambda$ a proset. Then

$\textup{M}_{\Lambda}(P)=\overline{\langle\\{p^{\Lambda}\\}_{p\in S},\
\\{1^{\\{t\\}}\\}_{t\in\Lambda},\
\\{e^{(s_{1},s_{2})}\\}_{s_{1}\precnapprox_{2}\in\Lambda}\rangle}=\overline{\langle\\{p^{\Lambda}\\}_{p\in
S},\ \\{e^{(s_{1},s_{2})}\\}_{s_{1}\preceq s_{2}\in\Lambda}\rangle}.$

###### Proof.

Define $R:=\langle\\{p^{\Lambda}\\}_{p\in S},\
\\{1^{\\{t\\}}\\}_{t\in\Lambda},\ \\{e^{(s_{1},s_{2})}\\}_{s_{1}\preceq
s_{2}\in\Lambda}\rangle$ and $\overline{R}$ the closure of $R$ in
$\textup{M}_{\Lambda}(P)$. Because $S\subset P$ is dense it is clear that
$\\{p^{\Lambda}\\}_{p\in P}\subset\overline{R}$, hence we can use these
elements to write our convergent nets.
This result is clearly true for $\Lambda$ finite, as the subring $R$ contains
all the elements with finitely many non-zero coordinates. Assume $\Lambda$ is
infinite and let $\Gamma(\Lambda)$ be as in Theorem 3.6. For each
$\alpha\in\Gamma(\Lambda)$ define

$A^{(\alpha)}_{s_{1},s_{2}}:=\left\\{\begin{array}[]{ll}A_{s_{1},s_{2}},&\mbox{if
}s_{1},\ s_{2}\in\alpha\\\ 0,&\mbox{otherwise.}\end{array}\right.$

It is clear that these are all in $R$. By Corollary 3.11 it follows that this
net is convergent and it converges to $A$. Hence
$\overline{R}=\textup{M}_{\Lambda}(P)$. ∎

######  Corollary 3.14.

Let $\Lambda$ be a proset and $P$ a topological ring. Then
$\textup{M}_{\Lambda}(P)$ is second countable if, and only if, $P$ is second
countable and $\Lambda$ is countable.

###### Proof.

If $P$ is second countable then there is $S\subset P$ a countable dense
subset. By Proposition 3.13 it follows that the set
$\langle\\{p^{\Lambda}\\}_{p\in S},\ \\{1^{\\{t\\}}\\}_{t\in\Lambda},\
\\{e^{(s_{1},s_{2})}\\}_{s_{1}\preceq s_{2}\in\Lambda}\rangle$ is dense in
$\textup{M}_{\Lambda}(P)$. Because $S$ and $\Lambda$ are countable it follows
that the dense subring $\langle\\{p^{\Lambda}\\}_{p\in S},\
\\{1^{\\{t\\}}\\}_{t\in\Lambda},\ \\{e^{(s_{1},s_{2})}\\}_{s_{1}\preceq
s_{2}\in\Lambda}\rangle$ is also countable, hence $\textup{M}_{\Lambda}(P)$ is
also second countable.
For the other side, notice that $\\{A\in\textup{M}_{\Lambda}(P);\ \text{if
}s_{1}\neq s_{2}\text{ then }A_{s_{1},s_{2}}=0\\}\cong\prod_{s\in\Lambda}P$ is
a closed subring of $\textup{M}_{\Lambda}(P)$, and this subring is second
countable if, and only if, $P$ is second countable and $\Lambda$ is countable.
∎

###  3.4 Isomorphism problems for matrix rings over posets

In this section we assume that our ordered sets are partially ordered sets and
that $P$ has only $0$ and $1$ as its idempotents. The topology given to $P$ is
the discrete topology.
In this section we give conditions on the ring $P$ so two incidence rings are
isomorphic if, and only if, their partially ordered sets are isomorphic. For
our proofs, we will use the idempotents of our topological incidence rings to
recreate the structure of our poset, proving that if there is an isomorphism
of matrix rings, their posets are isomorphic as well. For that, first we will
write down some relations between the generators of $\textup{M}_{\Lambda}$
that follow easily from the definitions, and use such properties to defined
our equivalence classes of idempotents. An alternative proof of this result
using Jacobison radical is given at [Vos80, Theorem 1].

######  Remark 3.15.

Given $\Lambda$ a proset and $P$ a topological ring note that:

  * •

For every $s_{1}\leqslant s_{2}\in\Lambda$, $t_{1}\leqslant t_{2}\in\Lambda$

$e^{(s_{1},s_{2})}e^{(t_{1},t_{2})}=\left\\{\begin{array}[]{ll}e^{(s_{1},t_{2})}&\mbox{if
}s_{2}=t_{1}\\\ 0&\mbox{otherwise.}\end{array}\right.$

  * •

For every $s\in\Lambda$,

$e^{(s,s)}=1^{\\{s\\}}.$

  * •

For every $S_{1},\ S_{2}$ subsets of $\Lambda$

$1^{S_{1}}1^{S_{2}}=\left\\{\begin{array}[]{ll}1^{S_{1}\cap S_{2}}&\mbox{if
}S_{1}\cap S_{2}\neq\emptyset\\\ 0&\mbox{otherwise.}\end{array}\right.$

.

  * •

For every $S\subset\Lambda$, $s_{1}\leqslant s_{2}\in\Lambda$

$1^{S}e^{(s_{1},s_{2})}=\left\\{\begin{array}[]{ll}e^{(s_{1},s_{2})}&\mbox{if
}s_{1}\in S\\\ 0&\mbox{otherwise.}\end{array}\right.$
$e^{(s_{1},s_{2})}1^{S}=\left\\{\begin{array}[]{ll}e^{(s_{1},s_{2})}&\mbox{if
}s_{2}\in S\\\ 0&\mbox{otherwise.}\end{array}\right.$

  * •

For every $S\subset\Lambda$, $p\in P$

$p^{\Lambda}1^{S}=1^{S}p^{\Lambda}.$

  * •

For every $s_{1}\leqslant s_{2}\in\Lambda$, $p\in P$

$p^{\Lambda}e^{(s_{1},s_{2})}=e^{(s_{1},s_{2})}p^{\Lambda}.$

  * •

Given $s,s_{1},s_{2}\in\Lambda$ and $A\in\textup{M}_{\Lambda}(P)$, then

$(1^{\\{s\\}}A)_{t_{1},t_{2}}=\left\\{\begin{array}[]{ll}A_{t_{1},t_{2}},&\mbox{if
}t_{1}=s\\\ 0,&\mbox{otherwise.}\end{array}\right.$
$(A1^{\\{s\\}})_{t_{1},t_{2}}=\left\\{\begin{array}[]{ll}A_{t_{1},t_{2}},&\mbox{if
}t_{2}=s\\\ 0,&\mbox{otherwise.}\end{array}\right.$
$(1^{\\{s_{1}\\}}A1^{\\{s_{2}\\}})_{t_{1},t_{2}}=\left\\{\begin{array}[]{ll}A_{t_{1},t_{2}},&\mbox{if
}t_{1}=s_{1}\text{ and }t_{2}=s_{2}\\\ 0,&\mbox{otherwise.}\end{array}\right.$
$(A1^{\\{s\\}}A)_{t_{1},t_{2}}=\left\\{\begin{array}[]{ll}A_{t_{1},s}A_{s,t_{2}},&\mbox{if
}s\in[t_{1},\ t_{2}]\\\ 0,&\mbox{otherwise.}\end{array}\right.$

These all can be proven by expanding the sums in each coordinate $(t_{1},\
t_{2})$.

We will denote $\mathcal{B}(P):=\\{a\in P;\ a^{2}=a\\}$ the set of idempotents
of $P$, also known as the Boolean ring of $P$. This set will be given a poset
structure by defining $a\leqslant b\in\mathcal{B}(P)$ if, and only if,
$ab=ba=a$. Note that under this order $1$ is the maximum element and $0$ is
the minimum element.
Given $\Lambda$ a poset and $P$ a ring and, given $S\subset\Lambda$, the
elements $1^{S}$ as defined in Notation 3.12 are all idempotents. The
following will be used throughout the whole section:

######  Lemma 3.16.

Let $P$ be a ring such that $\mathcal{B}(P)=\\{0,\ 1\\}$, $\Lambda$ a poset
and $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))$. Then for every $s\in\Lambda$
we have $A_{s,s}\in\\{0,\ 1\\}$. If $A_{s,s}=0$ for every $s\in\Lambda$ then
$A=0$.

###### Proof.

If $A$ is idempotent then $(A^{2})_{s,s}=(A_{s,s})^{2}=A_{s,s}$ for every
$s\in\Lambda$, hence $A_{s,s}\in\mathcal{B}(P)=\\{0,1\\}$.
Assume $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))$ is such that $A_{s,s}=0$ for
all $s\in\Lambda$. Let $s_{1}\leqslant s_{2}$. As $A^{2}=A$ it follows that
$\sum_{t\in[s_{1},s_{2}]}A_{s_{1},t}A_{t,s_{2}}=A_{s_{1},s_{2}}$. If
$|[s_{1},s_{2}]|=2$ this implies that $A_{s_{1},s_{2}}=0$. An induction under
$|[s_{1},s_{2}]|$ with the fact all intervals are finite implies
$A_{s_{1},s_{2}}=0$, for every $s_{1}\leqslant s_{2}$. Hence $A=0$. ∎

######  Lemma 3.17.

Let $P$ be a ring such that $\mathcal{B}(P)=\\{0,\ 1\\}$, $\Lambda$ a poset
and $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))\backslash\\{0\\}$. If
$s\in\\{s\in\Lambda;\ \pi_{[s,s]}(A)=1\\}$ then
$A-A1^{\\{s\\}}A\in\mathcal{B}(\textup{M}_{\Lambda}(P))$ and
$A-A1^{\\{s\\}}A<A$.

###### Proof.

To show $A-A1^{s}A$ is an idempotent, notice that

$(A-A1^{\\{s\\}}A)^{2}=A^{2}-A^{2}1^{\\{s\\}}A-A1^{\\{s\\}}A^{2}-A1^{\\{s\\}}A^{2}1^{\\{s\\}}A=A-2A1^{\\{s\\}}A+A1^{\\{s\\}}A1^{\\{s\\}}A.$

By Remark 3.15, $1^{\\{s\\}}A1^{\\{s\\}}=1^{\\{s\\}}$ hence
$(A-A1^{\\{s\\}}A)^{2}=A-A1^{\\{s\\}}A.$
We have that

$A(A-A1^{\\{s\\}}A)=A^{2}-A^{2}1^{\\{s\\}}A=A-A1^{\\{s\\}}A=A^{2}-A1^{\\{s\\}}A^{2}=(A-A1^{\\{s\\}}A)A.$

Then $A-A1^{\\{s\\}}A\leqslant A$.
To show it is not an equality, assume that $A=A-A1^{\\{s\\}}A$. It follows
that $A1^{\\{s\\}}A=0$. By Remark 3.15 this is true if, and only if,
$A_{s,s}=0$, a contradiction to our assumption. ∎

######  Notation 3.18.

We now define, for $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))$, the set
$b(A)=\\{s\in\Lambda;\ \pi_{[s,s]}(A)=1\\}$. Lemma 3.17 shows that if $A$ is a
non-zero idempotent, then this set is non-empty.

Given a non-zero element $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))$ this
result gives us a way to erase one of the diagonal elements in $A$ and get an
element strictly less than it under the idempotent order. We want to show that
the partial order on the idempotents of $\textup{M}_{\Lambda}(P)$ is similar
to the partial order on the power set of $\Lambda$ with subset order.

######  Notation 3.19.

Let $P$ be a ring and $\Lambda$ a locally finite poset. Let
$A\in\textup{M}_{\Lambda}(P)$ and $S\subset b(A)$ be a finite subset. We
define the following:

  * •

If $S=\\{s_{1}\\}$ we define

$A^{S}=A-A1^{\\{s_{1}\\}}A.$

  * •

If $A^{S}$ was defined for $S=\\{s_{1},\ \ldots,\ s_{n-1}\\}$, and $s_{n}\in
b(A)\backslash S$, we define

$A^{S\cup\\{s_{n}\\}}=A^{S}-A^{S}1^{\\{s_{n}\\}}A^{S}.$

######  Lemma 3.20.

Let $\Lambda$ be a poset and $P$ be a ring such that $\mathcal{B}(P)=\\{0,\
1\\}$. Let $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))$ such that $b(A)$ has at
least two elements. Let $s_{1}\neq s_{2}\in b(A)$. Then
$A^{\\{s_{1},s_{2}\\}}=A^{\\{s_{2},s_{1}\\}}$.

###### Proof.

We have that

$A^{\\{s_{1},s_{2}\\}}=A-A1^{\\{s_{1}\\}}A-A1^{\\{s_{2}\\}}A+A1^{\\{s_{1}\\}}A1^{\\{s_{2}\\}}A+A1^{\\{s_{2}\\}}A1^{\\{s_{1}\\}}A-A1^{\\{s_{1}\\}}A1^{\\{s_{2}\\}}A1^{\\{s_{1}\\}}A$
$A^{\\{s_{2},s_{1}\\}}=A-A1^{\\{s_{2}\\}}A-A1^{\\{s_{1}\\}}A+A1^{\\{s_{2}\\}}A1^{\\{s_{1}\\}}A+A1^{\\{s_{1}\\}}A1^{\\{s_{2}\\}}A-A1^{\\{s_{2}\\}}A1^{\\{s_{1}\\}}A1^{\\{s_{2}\\}}A$

By Remark 3.15, we get that
$1^{\\{s_{1}\\}}A1^{\\{s_{2}\\}}A1^{\\{s_{1}\\}}=1^{\\{s_{2}\\}}A1^{\\{s_{1}\\}}A1^{\\{s_{2}\\}}=0$.
Hence $A^{\\{s_{1},s_{2}\\}}=A^{\\{s_{2},s_{1}\\}}$. ∎

It follows that for every finite set $S$ the construction of $A^{S}$ does not
depend on the order of $S$. We can then define $A^{S}$ for the case $S$ is
infinite as follows:
If $S$ is infinite, let $I=\\{S^{\prime}\subset S;\ S^{\prime}\text{ is
finite}\\}$ with order given by containment. Define $A^{S}$ as the limit of
the net $\\{A^{S^{\prime}}\\}_{S^{\prime}\in I}$, that is,

$A^{S}=\lim_{S^{\prime}\in I}A^{S^{\prime}}.$

This limit is well-defined as $P$ has the discrete topology. By Corollary 3.11
this net is convergent. By Lemma 3.20, the definition of $A^{S}$ also doesn’t
depend on the ordering of $S$ even in the case $S$ is infinite, hence $A^{S}$
is uniquely defined. Proposition 3.17 implies the element $A^{S}$ is an
idempotent.

######  Theorem 3.21.

Let $P$ be a ring such that $\mathcal{B}(P)=\\{0,\ 1\\}$, $\Lambda$ a poset
and $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))\backslash\\{0\\}$. Given
$S\subset b(A)$ a non-empty subset, then $A^{S}<A$ and $A^{S}_{s,s}=1$ if, and
only if, $s\in b(A)\backslash S$. More than that, if $S_{1}\subset
S_{2}\subset b(A)$ then $A^{S_{2}}\leqslant A^{S_{1}}$ with equality if, and
only if, $S_{1}=S_{2}$.

###### Proof.

The proof for the finite case follows directly from the definition and Lemma
3.17.
By the definition of the net and the idempotent order it clear that, for all
$S^{\prime}\subset S$ finite subsets,
$AA^{S^{\prime}}=A^{S^{\prime}}A=A^{S^{\prime}}$. Hence $A^{S}A=\lim
AA^{S^{\prime}}=\lim A^{S^{\prime}}A=\lim A^{S^{\prime}}=A^{S}$. We then
proved the first part of the statement.
For the second part, let $S_{1}\subset S_{2}\subset b(A)$. The result is
trivial for the case $S_{1}$ is finite, hence we assume $S_{1}$ is infinite.
Then

$A^{S_{1}}A^{S_{2}}=\lim_{S^{\prime}_{1}\subset_{\text{finite}}S_{1}}A^{S^{\prime}_{1}}A^{S_{2}}=\lim_{S^{\prime}_{1}\subset_{\text{finite}}S_{1}}A^{S_{2}}A^{S^{\prime}_{1}}=A^{S_{2}},$

hence $A^{S_{2}}\leqslant A^{S_{1}}$. If $S_{1}\neq S_{2}$, then given $s\in
S_{2}\backslash S_{1}$, by Lemma 3.17,we have that $A^{S_{2}}\leqslant
A^{S_{1}\cup\\{s\\}}<A^{S_{1}}$, that is, $A^{S_{1}}\neq A^{S_{2}}$.

∎

######  Corollary 3.22.

Let $\Lambda$ be a poset and $P$ a ring such that $\mathcal{B}(P)=\\{0,\
1\\}$. If $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))\backslash\\{0\\}$ is such
that $A$ has only one non-zero diagonal coordinate, say $A_{s,s}=1$, then
$A-A1^{(s)}A=0$.

The initial idea from this section is using idempotents of the form
$1^{\\{s\\}}$ to recover our poset structure. The problem is that isomorphirms
don’t necessarily send elements with only one non-zero coordinate to other
elements with only one non-zero coordinate. Following is an example for
$2\times 2$ triangular matrices:

$\left(\begin{array}[]{ll}1&a\\\
0&1\end{array}\right)\cdot\left(\begin{array}[]{ll}1&0\\\
0&0\end{array}\right)\cdot\left(\begin{array}[]{ll}1&a\\\
0&1\end{array}\right)^{-1}=\left(\begin{array}[]{cc}1&-a\\\
0&0\end{array}\right)$

The next few definitions will be central to tackle such problem:

######  Definition 3.23 (Definition 86.4.8 [Sta23]).

Let $P$ be a topological ring. We say an element $r\in P$ is topologically
nilpotent if $\lim_{n\in\mathbb{N}}r^{n}=0$.

In [Sta23] the notion of topologically nilpotent is given for topological
rings with a neighbourhood of $0$ consisting of ideals, known as linearly
topologized [Sta23, Definition 15.36.1]. As our rings might not be linearly
topologized, we define topologically nilpotent in a more general way.

######  Example 3.24.

  * •

Given $\mathbb{R}$ the ring of real numbers with the usual topology, for all
$r\in(-1,\ 1)$ we have that $r$ is topologically nilpotent.

  * •

Given $\mathbb{Z}_{p}$ the p-adic integers, then for all $r\in
p\mathbb{Z}_{p}$ we have that $r$ is topologically nilpotent.

  * •

Let $P$ a topological ring and $\Lambda$ a poset. If
$A\in\textup{M}_{\Lambda}(P)$ is such that for all $s\in\Lambda$, $A_{s,s}=0$,
then $A$ is topologically nilpotent. Notice that if $|[s_{1},\ s_{2}]|=2$ then

$(A^{2})_{s_{1},s_{2}}=A_{s_{1},s_{1}}A_{s_{1},s_{2}}+A_{s_{1},s_{2}}A_{s_{2},s_{2}}=0.$

In general, the poset structure on $\Lambda$ implies that if $|[s_{1},\
s_{2}]|=n$ then $(A^{n})_{s_{1},s_{2}}=0$. This fact can easily be proved by
induction. Corollary 3.11 then implies such $A$ is topologically nilpotent.

The following two definitions will allow us to order the idempotents of
$\textup{M}_{\Lambda}(P)$ in such a way we can recover the poset structure of
$\Lambda$ from the ring of matrices.

######  Definition 3.25.

Let $P$ be a topological ring with only $0$ and $1$ as its idempotents and
$\Lambda$ a poset. We define an relation $\sim$ in
$\mathcal{B}(\textup{M}_{\Lambda}(P))$ as $A\sim B$ if, and only if, $A-B$ is
topologically nilpotent. The equivalence class of an idempotent
$A\in\mathcal{B}(P)$ is denoted as $[A]$.

######  Definition 3.26.

Let $\Lambda$ be a poset. Given that $\Lambda$ has $0$ a minimum element, we
say $s\in\Lambda$ is minimal non-zero if $[0,\ s]=\\{0,\ s\\}$, that is, there
is no element $t\in\Lambda$ such that $0<t$ and $t<s$.

######  Lemma 3.27.

Let $\Lambda$ be a poset and $P$ a ring such that $\mathcal{B}(P)=\\{0,\
1\\}$. The following are true for the relation $\sim$ in
$\textup{M}_{\Lambda}(P)$:

  1. 1.

$A\sim B$ in $\mathcal{B}(\textup{M}_{\Lambda(P)})$ if, and only if,
$b(A)=b(B)$.

  2. 2.

It is an equivalence relation.

  3. 3.

For every $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))$ there exists an unique
$S\subset\Lambda$ such that $A\sim 1^{S}$.

  4. 4.

Given $[A],\ [B]$ equivalence classes, the order given by $[A]\leqslant[B]$ if
there are $A\in[A]$, $B\in[B]$ such that $A\leqslant B$ in relation to the
idempotent order is a partial order.

  5. 5.

There is an order preserving bijection from the power set of $\Lambda$ and the
equivalence classes of idempotents $\mathcal{B}(\textup{M}_{\Lambda}(P))$. The
elements of $\Lambda$ can be identified with the minimal non-zero elements of
$\mathcal{B}(\textup{M}_{\Lambda}(P))$.

###### Proof.

$\mathit{1.}$ Let $A,\ B\in\mathcal{B}(\textup{M}_{\Lambda}(P))$. Notice that
if $b(A)\neq b(B)$, there is $t\in\Lambda$ such that $(A-B)_{s,s}\in\\{1,\
-1\\}$. Theorem 3.21 then implies that $A-B$ is not topologically nilpotent.
On the other hand, if $b(A)=b(B)$ by Theorem 3.21 and Lemma 3.16 we have that
$A-B$ is topologically nilpotent, proving the argument.
Statement $\mathit{2.}$ and $\mathit{3.}$ follows directly from $\mathit{1.}$,
and statements $\mathit{4.}$ and $\mathit{5.}$ follows directly from
$\mathit{3}$. ∎

######  Example 3.28.

Notice that for the case $\Lambda$ is a proset this is not necessarily an
equivalence relation on $\textup{M}_{\Lambda}(P)$. For example, on the case
$\Lambda=\mathbf{2}$ and $P=\mathbb{Z}_{p}$ we have that

$\left(\begin{array}[]{cc}1&p\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}1&0\\\
p&0\end{array}\right)=\left(\begin{array}[]{cc}0&p\\\ -p&0\end{array}\right)$

and

$\left(\begin{array}[]{cc}0&p\\\
-p&0\end{array}\right)^{2n}=\left(\begin{array}[]{cc}(-1)^{n}p^{2n}&0\\\
0&(-1)^{n}p^{2n}\end{array}\right)\xrightarrow{n\rightarrow\infty}\left(\begin{array}[]{cc}0&0\\\
0&0\end{array}\right)$ $\left(\begin{array}[]{cc}0&p\\\
-p&0\end{array}\right)^{2n+1}=\left(\begin{array}[]{cc}0&(-1)^{n}p^{2n+1}\\\
-(-1)^{n}p^{2n+1}&0\end{array}\right)\xrightarrow{n\rightarrow\infty}\left(\begin{array}[]{cc}0&0\\\
0&0\end{array}\right)$

hence it is topologically nilpotent. It is also the case that

$\left(\begin{array}[]{cc}1&p\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}1&1\\\
0&0\end{array}\right)=\left(\begin{array}[]{cc}0&p-1\\\ 0&0\end{array}\right)$
$\left(\begin{array}[]{cc}1&0\\\
p&0\end{array}\right)-\left(\begin{array}[]{cc}1&0\\\
1&0\end{array}\right)=\left(\begin{array}[]{cc}0&0\\\ p-1&0\end{array}\right)$

are nilpotent, hence topologically nilpotent. But

$\left(\left(\begin{array}[]{cc}1&1\\\
0&0\end{array}\right)-\left(\begin{array}[]{cc}1&0\\\
1&0\end{array}\right)\right)^{2n}=\left(\begin{array}[]{cc}(-1)^{n}&0\\\
0&(-1)^{n}\end{array}\right)$

is not topologically nilpotent.

######  Lemma 3.29.

Let $P$ be a ring such that $\mathcal{B}(P)=\\{0,\ 1\\}$ and $\Lambda$ a
poset. Given $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))\backslash\\{0\\}$
idempotent, then $A=1^{S}+A^{\prime}$ for some $S\subset\Lambda$, $A^{\prime}$
topologically nilpotent with $A^{\prime}_{s,s}=0$ for all $s\in\Lambda$. If
$[A]$ is minimal non-zero then
$A^{\prime}=1^{\\{s\\}}A^{\prime}+A^{\prime}1^{\\{s\\}}+A^{\prime}1^{\\{s\\}}A^{\prime}$.

###### Proof.

Because $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))\backslash\\{0\\}$ then, by
Lemma 3.27, there is $S\subset\Lambda$ such that $A\sim 1^{S}$. Hence, by
definition of the equivalence relation, $A-1^{S}=A^{\prime}$ for $A^{\prime}$
topologically nilpotent. Then $A=1^{S}+A^{\prime}$, where $A^{\prime}_{s,s}=0$
for all $s\in\Lambda$.
If $[A]$ is minimal non-zero then there is $s\in\Lambda$ such that
$[A]=[1^{\\{s\\}}]$. By Lemmas 3.17 and 3.16, $A-A1^{\\{s\\}}A=0$. We can
write $A$ as $1^{\\{s\\}}+A^{\prime}$, as shown above. Using the property $A$
is an idempotent we get

$(1^{\\{s\\}}+A^{\prime})^{2}=1^{\\{s\\}}+1^{\\{s\\}}A^{\prime}+A^{\prime}1^{\\{s\\}}+A^{\prime}1^{\\{s\\}}A^{\prime}=1^{\\{s\\}}+A^{\prime}.$

Rearranging the equation then gives us
$A^{\prime}=1^{\\{s\\}}A^{\prime}+A^{\prime}1^{\\{s\\}}+A^{\prime}1^{\\{s\\}}A^{\prime}$.
∎

######  Corollary 3.30.

Let $A\in\mathcal{B}(\textup{M}_{\Lambda}(P))$ be minimal non-zero and
$s\in\Lambda$ be such that $A\sim 1^{\\{s\\}}$. Then $A_{s_{1},s_{2}}$ can be
non-zero if, and only if, $s\in[s_{1},\ s_{2}]$.

###### Proof.

A direct corollary from Lemma 3.29 and from Remark 3.15. ∎

######  Lemma 3.31.

Let $\Lambda_{1},\Lambda_{2}$ be two posets and $P$ a topological ring such
that $\mathcal{B}(P)=\\{0,\ 1\\}$. If
$f:\textup{M}_{\Lambda_{1}}(P)\rightarrow\textup{M}_{\Lambda_{2}}(P)$ is a
continuous homomorphism of rings then, for every $S_{1}\subset\Lambda_{1}$
there is $S_{2}\subset\Lambda_{2}$ such that
$f([1^{S_{1}}])\subset[1^{S_{2}}]$. It is also the case that if $S\subset
S^{\prime}\subset\Lambda$ then $[f(1^{S})]\leqslant[f(1^{S^{\prime}})]$.

###### Proof.

Let $A\in[1^{S_{1}}]$. As homomorphisms send idempotents to idempotents, Lemma
3.29 implies $f(A)=1^{S_{2}}+f(A)^{\prime}$ for some $S_{2}\subset\Lambda_{2}$
and $f(A)^{\prime}\in\textup{M}_{\Lambda_{2}}(P)$ topologically nilpotent.
Hence $f(A)\in[1^{S_{2}}]$.
Suppose now $B\in[1^{S_{1}}]$. Then $B-A$ is topologically nilpotent and,
because $f$ is continuous, it follows that $f(B-A)$ is topologically
nilpotent. Then $f(B)\in[f(A)]=[1^{S_{2}}]$ for every $B\in[1^{S_{1}}]$. Hence
$f[1^{S_{1}}]\subset[f(1^{S_{1}})]=[1^{S_{2}}]$, for some
$S_{2}\subset\Lambda_{2}$.
For the last statement, assume $[1^{S}]<[1^{S^{\prime}}]$. Hence
$f(1^{S^{\prime}})f(1^{S})=f(1^{S})f(1^{S^{\prime}})=f((1^{S}+1^{S^{\prime}\backslash
S^{\prime}})(1^{S}))=f(1^{S})$, that is,
$[f(1^{S})]\leqslant[f(1^{S^{\prime}})]$. ∎

The next result is essential for the proof of the isomorphism problem on the
rings of matrices indexed by ordered sets. It connects properties of the
minimal non-zero elements and the ordering of our ordered set.

######  Lemma 3.32.

Let $[1^{\\{s_{1}\\}}],[1^{\\{s_{2}\\}}]$ be minimal non-zero classes of
$\textup{M}_{\Lambda}(P)$, for a ring $P$ such that $\mathcal{B}(P)=\\{0,\
1\\}$ and $\Lambda$ a poset. Then there is $A\in[1^{\\{s_{1}\\}}]$,
$B\in[1^{\\{s_{2}\\}}]$ such that $AB\neq 0$ if, and only if, $s_{1}\leqslant
s_{2}$. It is the case $s_{1}=s_{2}$ if, and only if, $AB\neq 0$ for every
$A\in[1^{\\{s_{1}\\}}]$ and $B\in[1^{\\{s_{2}\\}}]$.

###### Proof.

Let $[1^{\\{s_{1}\\}}],\ [1^{\\{s_{2}\\}}]$ be two minimal non-zero classes
and let $A\in[1^{\\{s_{1}\\}}],\ B\in[1^{\\{s_{2}\\}}]$. Write
$A=1^{\\{s_{1}\\}}+A^{\prime}$ and $B=1^{\\{s_{2}\\}}+B^{\prime}$, as in Lemma
3.29. We first show that if $AB\neq 0$ then $s_{1}\leqslant s_{2}$. Notice
that $AB$ can be written as follows

$AB=1^{\\{s_{1}\\}}1^{\\{s_{2}\\}}+1^{\\{s_{1}\\}}B^{\prime}+A^{\prime}1^{\\{s_{2}\\}}+A^{\prime}B^{\prime}.$

Hence, if $AB\neq 0$, at least one of the terms above is non-zero. We now work
on necessary conditions for each term to be non-zero:

  * •

Remark 3.15 implies $1^{\\{s_{1}\\}}1^{\\{s_{2}\\}}\neq 0$ if, and only if
$s_{1}=s_{2}$.

  * •

By Remark 3.15 and Corollary 3.30 we have that $1^{\\{s_{1}\\}}B^{\prime}$ is
non-zero if, and only if, $s_{1}\leqslant s_{2}$ and
$B^{\prime}_{s_{1},s_{2}}\neq 0$.

  * •

Similar to the case $1^{\\{s_{1}\\}}B^{\prime}$, $A^{\prime}1^{\\{s_{2}\\}}$
is non-zero if, and only if, $s_{1}\leqslant s_{2}$ and
$A^{\prime}_{s_{1},s_{2}}\neq 0$.

  * •

For $A^{\prime}B^{\prime}$, notice that by Lemma 3.29 we can rewrite it as:

$(1^{\\{s_{1}\\}}A^{\prime}+A^{\prime}1^{\\{s_{1}\\}}+A^{\prime}1^{\\{s_{1}\\}}A^{\prime})(1^{\\{s_{2}\\}}B^{\prime}+B^{\prime}1^{\\{s_{2}\\}}+B^{\prime}1^{\\{s_{2}\\}}B^{\prime}).$

After expanding, the terms are of the form $X1^{\\{s_{1}\\}}Y1^{\\{s_{2}\\}}Z$
for $X,Y,Z\in\\{1^{\Lambda},A^{\prime},B^{\prime},A^{\prime}B^{\prime}\\}$. By
Remark 3.15 if $s_{1}\nleqslant s_{2}$ then
$1^{\\{s_{1}\\}}Y1^{\\{s_{2}\\}}=0$. Hence if $A^{\prime}B^{\prime}\neq 0$ we
have $s_{1}\leqslant s_{2}$.

That is, if $AB\neq 0$ then $s_{1}\leqslant s_{2}$. We now need to proof the
other side.
Let $s_{1}\leqslant s_{2}$. Define the matrix $A$ coordinate-wise as follows:

$A_{t_{1},t_{2}}=\begin{cases}1&\mbox{if }t_{1}=t_{2}=s_{1}\mbox{ or
}t_{1}=s_{1},\ t_{2}=s_{2}\\\ 0&\mbox{otherwise.}\end{cases}$

By Remark 3.15, $A1^{s_{2}}\neq 0$. Hence, we get the only if side.
Now assume $s_{1}=s_{2}=s$. By Lemma 3.29 it follows that for all
$A,B\in[1^{\\{s\\}}]$ we have
$AB=1^{\\{s\\}}+1^{\\{s\\}}B^{\prime}+A^{\prime}1^{\\{s\\}}+A^{\prime}B^{\prime}$.
As $1^{\\{s\\}}B^{\prime}+A^{\prime}1^{\\{s\\}}+A^{\prime}B^{\prime}$ is
topologically nilpotent and $1^{\\{s\\}}$ is not, we have that $AB\neq 0$.
In the case $s_{1}<s_{2}$, $1^{\\{s_{1}\\}}1^{\\{s_{2}\\}}=0$. Hence there are
$A\in[1^{\\{s_{1}\\}}]$, $B\in[1^{\\{s_{2}\\}}]$ such that $AB=0$. ∎

Now the process of rebuilding our poset using our matrices ring is as follows:

  * •

The elements of our poset will be the minimal non-zero classes of idempotents,
which can be written as $[1^{\\{s\\}}]$ for $s\in\Lambda$.

  * •

If there are $A\in[1^{\\{s_{1}\\}}]$ and $B\in[1^{\\{s_{2}\\}}]$ such that
$AB\neq 0$ we get $s_{1}\leqslant s_{2}$. If it is also the case that there
are idempotents $A\in[1^{\\{s_{1}\\}}]$ and $B\in[1^{\\{s_{1}\\}}]$ such that
$AB=0$, then $s_{1}=s_{2}$.

By Lemma 3.32 it follows that the set of minimal non-zero idempotents with
order as given above is the same as $\Lambda$. With Lemma 3.31 we get that
isomorphisms send minimal non-zero to minimal non-zero. We can then prove the
following:

######  Theorem 3.33.

[Vos80, Theorem 1] Let $\Lambda_{1},\ \Lambda_{2}$ be posets and $P$ be a
topological ring such that $\mathcal{B}(P)=\\{0,\ 1\\}$. Then
$\textup{M}_{\Lambda_{1}}(P)\cong\textup{M}_{\Lambda_{2}}(P)$ if, and only if,
$\Lambda_{1}\cong\Lambda_{2}$, that is, there is a bijective order preserving
map from $\Lambda_{1}$ to $\Lambda_{2}$.

###### Proof.

Assume $f:\textup{M}_{\Lambda_{1}}(P)\rightarrow\textup{M}_{\Lambda_{2}}(P)$
is a continuous isomorphism. Because $f$ is a ring isomorphism, Lemma 3.31
implies that for all $s_{1}\in\Lambda_{1}$ there exists an unique
$s_{2}\in\Lambda_{2}$ such that $f([1^{\\{s_{1}\\}}])=[1^{\\{s_{2}\\}}]$, as
these are minimal non-zero. Hence $f$ gives a bijection from
$\\{[1^{\\{s_{1}\\}}];\ s_{1}\in\Lambda_{1}\\}$ and $\\{[1^{\\{s_{2}\\}}];\
s_{2}\in\Lambda_{2}\\}$. Notice that the continuous isomorphism of rings also
preserve all the properties from Lemma 3.32. Hence the map
$\tilde{f}:\Lambda_{1}\rightarrow\Lambda_{2}$ defined by $s_{1}\mapsto s_{2}$
if, and only if, $f([1^{\\{s_{1}\\}}])=[1^{\\{s_{2}\\}}]$ is an order
preserving bijective map. Hence $\Lambda_{1}\cong\Lambda_{2}$.
On the other hand, let $f:\Lambda_{1}\rightarrow\Lambda_{2}$ an order
preserving bijective map. It is easy to see that the map

$\begin{array}[]{llll}\tilde{f}:&\textup{M}_{\Lambda_{2}}(P)&\longrightarrow&\textup{M}_{\Lambda_{1}}(P)\\\
&A&\longmapsto&\tilde{f}(A)\end{array}$

where $\tilde{f}(A)_{s_{1},s_{2}}=A_{f(s_{1}),f(s_{2})}$, is a continuous
isomorphism of rings. ∎

Theorem 3.33 is not always true for topological rings. In fact, [Vos80,
Proposition 1] proves that if $\Lambda$ is a locally finite pre-ordered set
then there exists $P$ a ring such that $P\cong\textup{M}_{\Lambda}(P)$. In the
same article, it is also shown that if $P$ is a ring that is simple Artinian
modulo the Jacobson radical, then
$\textup{M}_{\Lambda_{1}}(P)\cong\textup{M}_{\Lambda_{2}}(P)$ if, and only if,
$\Lambda_{1}\cong\Lambda_{2}$. A list of answers for the isomorphism problem
in different classes of incidence rings can be seen at [AHdR].

##  4 Categorical properties

###  4.1 The functor $\mathbf{M}_{\\_}(P)$

Theorem 3.33 closely relates the structure of matrix rings of the form
$\textup{M}_{\Lambda}(P)$ with their respective partially ordered sets. In
this section we will show that this can be extended to defining a
contravariant functor from a category of partially ordered sets to the
category of rings, showing that even products and inverse limits of these
rings can be understood just by looking at limits and coproducts on the
category or partially ordered sets. This question arise naturally from
Proposition 3.7 and Theorem 3.6, which points to the direction that coproducts
are mapped to products and colimits to limits under such map.
For this section we will need to restrict the maps we accept for the category
of ordered sets, as not all maps would work. We then give arguments for why
each condition is added.

######  Definition 4.1.

Given $\Lambda,\ \Lambda^{\prime}$ two prosets and
$f:\Lambda\rightarrow\Lambda^{\prime}$ a set function. We say that:

  * •

$f$ is order preserving if for $a\preceq b\in\Lambda$ then $f(a)\preceq
f(b)\in\Lambda^{\prime}$.

  * •

$f$ is a convex function if it is an order preserving function and whenever
$\alpha\subset\Lambda$ is a convex subset then $f(\alpha)$ is a convex subset
of $\Lambda^{\prime}$.

  * •

$f$ is a convex embedding if it is a convex function and injective.

  * •

$f$ is constant if $f(\Lambda)=\\{t\\}$, for some $t\in\Lambda^{\prime}$

  * •

Given that $\sqcup_{i\in I}\Lambda_{i}$ is the decomposition of $\Lambda$ into
its components. we say $f$ factors out into constant functions and convex
embeddings (FCC) if for each $i\in I$ either the function
$f|_{\Lambda_{i}}:\Lambda_{i}\rightarrow\Lambda^{\prime}$ is a constant
function or a convex embedding.

Let $f:\Lambda_{1}\rightarrow\Lambda_{2}$ be an order preserving morphism. A
first idea for a map
$\tilde{f}:\textup{M}_{\Lambda_{1}}(P)\rightarrow\textup{M}_{\Lambda_{2}}(P)$
is as in Theorem 3.33. We then have $\tilde{f}(A)$ is defined coordinatewise
as

$(\tilde{f}(A))_{s_{1},s_{2}}=\sum A_{t_{1},t_{2}}$

for $t_{1},t_{2}\in\Lambda_{1}$, $f(t_{1})=s_{1}$, $f(t_{2})=s_{2}$.

  1. 1.

If $f$ is injective, but not surjective, then $\tilde{f}(1^{\Lambda_{1}})\neq
1^{\Lambda_{2}}$, hence it is not a ring homomorphism. For example:

$f:2<\rightarrow 3<$ $f(0)=0,\ \ f(1)=2$

then

$\tilde{f}:\textup{M}_{2<}(P)\rightarrow\textup{M}_{3<}(P)$
$\left(\begin{array}[]{cc}a_{0,0}&a_{0,1}\\\
0&a_{1,1}\end{array}\right)\mapsto\left(\begin{array}[]{ccc}a_{0,0}&0&a_{0,1}\\\
0&0&0\\\ 0&0&a_{1,1}\end{array}\right).$

  2. 2.

If $f$ is surjective, but not injective, it does not preserve multiplication.
For example:

$f:2<\rightarrow 1$

then

$\tilde{f}:\textup{M}_{2<}(P)\rightarrow\textup{M}_{1}(P)$
$\left(\begin{array}[]{cc}a_{0,0}&a_{0,1}\\\
0&a_{1,1}\end{array}\right)\mapsto a_{0,0}+a_{0,1}+a_{1,1}.$

Hence it is necessary to make some tweaks when defining a morphism between
matrix rings in relation to a morphism between prosets.
A second idea is reversing the direction of
$f:\Lambda_{1}\rightarrow\Lambda_{2}$. We can then define $\mathbf{M}[f]$ on
the set of generators (Proposition 3.13) of $\textup{M}_{\Lambda_{2}}(P)$ as
follows:

$\begin{array}[]{llll}\mathbf{M}[f]:&\textup{M}_{\Lambda_{2}}(P)&\longrightarrow&\textup{M}_{\Lambda_{1}}(P)\\\
&p^{\Lambda_{2}}&\longmapsto&p^{\Lambda_{1}}\\\
&e^{(s_{1},s_{2})}&\longmapsto&\left\\{\begin{array}[]{ll}\sum_{t_{1},t_{2}}e^{(t_{1},t_{2})}&\mbox{for
}t_{1}\preceq t_{2}\mbox{ and }f(t_{1})=s_{1},\ f(t_{2})=s_{2}\\\
0&\mbox{otherwise}\end{array}\right.\end{array}$

In this case we have the following:

######  Proposition 4.2.

Assume $\Lambda_{1},\ \Lambda_{2}$ are prosets and $P$ is a ring. Let
$f:\Lambda_{1}\rightarrow\Lambda_{2}$ be an order preserving morphism that is
not convex. Then $\mathbf{M}[f]$ is not well-defined.

###### Proof.

Let $f$ be an order preserving that is not convex. If $f$ is not convex then
there is $\alpha\in\Lambda_{1}$ a convex set such that $f(\alpha)$ is not
convex in $\Lambda_{2}$. Hence either $f(\alpha)$ is not closed under
intervals or it is not connected.
If $f(\alpha)$ is not connected, there are $s_{1},\ s_{2}\in\alpha$ such that
$[s_{1},\ s_{2}]\neq\emptyset$, but $[f(s_{1}),\ f(s_{2})]=\emptyset$. That
is, $s_{1}\preceq s_{2}$ but $f(s_{1})\npreceq f(s_{2})$, a contradiction to
the assumption $f$ is order preserving.
If $f$ is not closed under intervals, there are $s_{1}\preceq
s_{2}\in\Lambda_{1}$ such that $[f(s_{1}),\ f(s_{2})]$ is not in $f(\alpha)$.
Then there is $f(s_{1})\precnapprox t\precnapprox f(s_{2})$ such that
$f^{-1}(t)=\emptyset$. In this case we then have

$\mathbf{M}[f](e^{(f(s_{1}),t)})\mathbf{M}[f](e^{(t,f(s_{2}))})=0$

but

$\mathbf{M}[f](e^{(f(s_{1}),f(s_{2}))})$

contains $e^{(s_{1},s_{2})}$ as a term, hence is non-zero. That is,

$\mathbf{M}[f](e^{(f(s_{1}),t)})\mathbf{M}[f](e^{(t,f(s_{2}))})\neq\mathbf{M}[f](e^{(f(s_{1}),f(s_{2}))}).$

∎

An example of the result above is as follows: let
$f:\mathbf{2}<\longrightarrow\mathbf{3}<$ defined by $f(0)=0$, $f(1)=2$. This
map is order preserving but not convex. The
$\mathbf{M}[f]:{M}_{\mathbf{3}<}(P)\rightarrow{M}_{\mathbf{2}<}(P)$ would be
defined as follows:

$\left(\begin{array}[]{lll}a_{0,0}&a_{0,1}&a_{0,2}\\\ 0&a_{1,1}&a_{1,2}\\\
0&0&a_{2,2}\end{array}\right)\mapsto\left(\begin{array}[]{ll}a_{0,0}&a_{0,2}\\\
0&a_{2,2}\end{array}\right)$

which is easily seen not to preserve multiplication.
It is still not known if it the morphism being FCC is a necessary condition
for $\mathbf{M}[f]$ to be well-defined, but there are some examples showing
that being convex is not sufficient. For instance, let
$f:\mathbf{4<}\rightarrow\mathbf{2<}$ defined as $f(0)=f(1)=0$, $f(2)=f(3)=1$.
The map $\mathbf{M}[f]:{M}_{\mathbf{2}<}(P)\rightarrow{M}_{\mathbf{4}<}(P)$
would be defined as follows:

$\left(\begin{array}[]{ll}a_{0,0}&a_{0,1}\\\
0&a_{1,1}\end{array}\right)\mapsto\left(\begin{array}[]{llll}a_{0,0}&0&a_{0,1}&a_{0,1}\\\
0&a_{0,0}&a_{0,1}&a_{0,1}\\\ 0&0&a_{1,1}&0\\\ 0&0&0&a_{1,1}\end{array}\right)$

which is clearly not a ring homomorphism.

######  Remark 4.3.

Let $\Lambda^{1}$ be a proset and let $\sqcup_{i\in I}\Lambda_{i}$ the
decomposition of $\Lambda^{1}$ into its components. Notice that if
$f:\Lambda^{1}\rightarrow\Lambda^{2}$ is FCC then for every $s\in\Lambda^{2}$,
$i\in I$ we have that either $f^{-1}(s)\cap\Lambda_{i}=\emptyset$,
$f^{-1}(s)\cap\Lambda_{i}=\\{t_{i}\\}$ or
$f^{-1}(s)\cap\Lambda_{i}=\Lambda_{i}$.

######  Definition 4.4.

We define the category category of locally finite prosets
$\textbf{{prosets}}_{\textup{fin}}$ as:

  * •

Objects: all locally finite prosets,

  * •

Arrows: all the FCC morphisms between locally finite prosets.

Notice that the category $\textbf{{prosets}}_{\textup{fin}}$ has $0=\emptyset$
as its initial object and $1$ as its terminal object.
Our goal now is to show that if $f$ is FCC then $\textbf{M}[f]$ can be
extended to a ring homomorphism. Not only that, we can define a contravariant
functor from the category $\textbf{prosets}_{\text{fin}}$ to TopRings.

######  Theorem 4.5.

Let $P$ be a topological ring. The map

$\begin{array}[]{cccc}\textbf{{M}}_{\\_}(P):&\textbf{{prosets}}_{\textup{fin}}&\Longrightarrow&\textbf{{TopRings}}\\\
&\Lambda&\longmapsto&\textup{M}_{\Lambda}(P)\\\
&f&\longmapsto&\textbf{{M}}[f]\end{array}$

is a contravariant functor.

###### Proof.

Given $P$ a topological ring, the functor $\mathbf{M}_{\\_}(P)$ maps the
object $\Lambda$ from $posets$ to
$\mathbf{M}_{\\_}(P)(\Lambda):=\textup{M}_{\Lambda}(P)$, where
$\textup{M}_{\emptyset}(P):=\\{0\\}$ is defined to be the trivial ring. We now
need to show the functor map FCC maps to continuous ring homomorphims. To
simplify the notation we will denote $\mathbf{M}_{\\_}(P)(f):=\mathbf{M}[f]$,
for $f$ an FCC map. The proof will follow by defining a map from the
generators of the domain and showing it extends to a continuous ring
homomorphism. Remark 3.15 and Proposition 3.13 will be central for this part
of the proof. It will then remain to show that composition of functions is
mapped to composition of functions.
If $\Lambda_{1}$ is the empty poset and $f:\emptyset\rightarrow\Lambda_{2}$ is
the unique morphism from the empty set to $\Lambda_{2}$, define
$\mathbf{M}[f]$ as the unique morphism from $\textup{M}_{\Lambda_{2}}(P)$ to
$\textup{M}_{\emptyset}(P)$.
For $\Lambda_{2}$ a non-empty set, we extend $\mathbf{M}[f]$ for some finite
sums as follows:

  * •

If $A,B\in\\{p^{\Lambda}\\}_{p\in P}\cup\\{e^{(s_{1},s_{2})}\\}_{s_{1}\preceq
s_{2}\in\Lambda_{2}}$, the generating set, we define

$\mathbf{M}[f](A+B):=\mathbf{M}[f](A)+\mathbf{M}[f](B).$

  * •

If $A\in\\{e^{(s_{1},s_{2})}\\}_{s_{1}\preceq s_{2}\in\Lambda_{2}}$ and $p\in
P$, we define

$p^{\Lambda_{1}}\mathbf{M}[f](A)=\mathbf{M}[f](p^{\Lambda_{2}}A)=\mathbf{M}[f](Ap^{\Lambda_{2}})=\mathbf{M}[f](A)p^{\Lambda_{1}}.$

We want now to show that such a map can be extended to a continuous ring
homomorphism. First, we will show that such a map can be extended to a ring
homomorphism on the dense subset from Proposition 3.13 and then extend it
continuously to the whole ring. For that let $\Lambda_{1}=\sqcup_{i\in
I}\Lambda_{i}$ the decomposition of $\Lambda_{1}$ into its irreducible
components.

Claim 1: Let $t_{1}\precnapprox t_{2}\in\Lambda_{1}$ be such that
$f(t_{1})=s_{1}$ and $f(t_{2})=s_{2}$ for some $s_{1}\precnapprox
s_{2}\in\Lambda_{2}$. If $s_{1}\precnapprox s_{2}\precnapprox s_{3}$, for some
$s_{3}\in\Lambda_{2}$ then one of the following happen:

  * •

$e^{(t_{1},t_{2})}\mathbf{M}[f](e^{(s_{2},s_{3})})=0$, or

  * •

$e^{(t_{1},t_{2})}\mathbf{M}[f](e^{(s_{2},s_{3})})=e^{(t_{1},t_{3})}$ for a
unique $t_{3}\in\Lambda_{1}$ such that $t_{1}\precnapprox t_{2}\precnapprox
t_{3}$ and $f(t_{3})=s_{3}$.

Assume $e^{(t_{1},t_{2})}\mathbf{M}[f](e^{(s_{2},s_{3})})\neq 0$. Then
expanding it, we have the right-hand side of

$e^{(t_{1},t_{2})}\mathbf{M}[f](e^{(s_{2},s_{3})})=e^{(t_{1},t_{2})}\left(\sum
e^{(t_{2}^{\prime},t_{3}^{\prime})}\right)$

has at least one pair $(t_{2}^{\prime},\ t_{3}^{\prime})$ such that
$f(t_{2}^{\prime})=s_{2}$, $t_{2}^{\prime}=t_{2}$, and
$f(t_{3}^{\prime})=s_{3}$ (Remark 3.15). Hence, restricted to the component
containing $t_{1},\ t_{2}$ the function $f$ is a convex embedding. By Remark
4.3, such the pair $(t_{2}^{\prime},\ t_{3}^{\prime})$ is unique, as it has to
be on the same component as $t_{1},\ t_{2}$. That is, if
$e^{(t_{1},t_{2})}\mathbf{M}[f](e^{(s_{2},s_{3})})\neq 0$ then
$e^{(t_{1},t_{2})}\mathbf{M}[f](e^{(s_{2},s_{3})})=e^{(t_{1},t_{3}^{\prime})}$.
Claim 2: Let $s_{1}\precnapprox s_{2}$ be elements in $\Lambda_{2}$. Then
$\mathbf{M}[f](e^{(s_{1},s_{2})})\mathbf{M}[f](e^{(t_{1},t_{2})})=\mathbf{M}[f](e^{(s_{1},s_{2})}e^{(t_{1},t_{2})})$
for every $t_{1},\ t_{2}\in\Lambda_{2}$.
Assume $s_{1},\ s_{2},\ t_{1},\ t_{2}\in\Lambda_{2}$ are such that
$s_{1}\precnapprox s_{2}$, $s_{2}=t_{1}$ and $t_{1}\precnapprox t_{2}$, as
otherwise the result would be trivially $0$ on both sides. Then
$e^{(s_{1},s_{2})}e^{(t_{1},t_{2})}=e^{(s_{1},t_{2})}$ and, as $f$ is FCC, by
Remark 4.3 the right-hand side is

$\mathbf{M}[f](e^{(s_{1},t_{2})})=\sum_{j\in J}e^{(q_{j_{1}},q_{j_{2}})}$ (1)

for $J$ a subset of $I$, $q_{j_{1}}\precnapprox q_{j_{2}}\in\Lambda_{j}$ and
$f^{-1}(s_{1})\cap\Lambda_{j}=\\{q_{j_{1}}\\}$,
$f^{-1}(t_{2})\cap\Lambda_{j}=\\{q_{j_{2}}\\}$. For the left-hand side,

$\mathbf{M}[f](e^{(s_{1},s_{2})})\mathbf{M}[f](e^{(t_{1},t_{2})})=\left(\sum_{k\in
K}e^{(s_{k_{1}},s_{k_{2}})}\right)\left(\sum_{l\in
L}e^{(t_{l_{1}},t_{l_{2}})}\right)$ (2)

where $K$ and $L$ are subsets of $I$, for $k\in K$ we have
$s_{k_{1}}\precnapprox s_{k_{2}}\in\Lambda_{k}$,
$f^{-1}(s_{1})\cap\Lambda_{k}=\\{s_{k_{1}}\\}$,
$f^{-1}(s_{2})\cap\Lambda_{k}=\\{s_{k_{2}}\\}$ and for $l\in L$ we have
$t_{l_{1}}\precnapprox t_{l_{2}}\in\Lambda_{l}$,
$f^{-1}(t_{1})\cap\Lambda_{l}=\\{t_{l_{1}}\\}$,
$f^{-1}(t_{2})\cap\Lambda_{l}=\\{t_{l_{2}}\\}$. As we assumed
$s_{1}\precnapprox s_{2}=t_{1}\precnapprox t_{2}$ we have that

$\mathbf{M}[f](e^{(s_{1},s_{2})})\mathbf{M}[f](e^{(t_{1},t_{2})})=\left(\sum_{p\in
K\cap L}e^{(s_{p_{1}},t_{p_{2}})}\right).$

We need to show $K\cap L=J$, and the result will follow from Remark 4.3. If
$j\in J$ then there are $q_{j_{1}}\precnapprox q_{j_{2}}$ such that
$f(q_{j_{1}})=s_{1}$, $f(q_{j_{2}})=t_{2}$, hence $f|_{\Lambda_{j}}$ is convex
embedding. Then there is $q\in[q_{j_{1}},q_{j_{2}}]$ such that
$f(q)=s_{2}=t_{1}$, that is, $e^{(q_{j_{1}},q)}$ and $e^{(q,q_{j_{2}})}$ show
up as terms in the right hand side of 2. Hence $j\in K\cap L$.
On the other hand, if $p\in K\cap L$, then there are $s_{p_{1}}\precnapprox
t\precnapprox t_{p_{2}}\in\Lambda_{p}$ such that $f(s_{p_{1}})=s_{1}$,
$f(t)=s_{2}=t_{1}$, $f(t_{p_{2}})=t_{2}$. Hence the term $e^{(s_{1},t_{2})}$
appears on the right-hand side of Equation 1. That is, $p\in J$.
Claim 3: $\mathbf{M}[f]$ is well-defined and continuous.
Claim $1.$ and $2.$ implies it is well-defined for all elements that are
finite sums and finite products of the generators. Hence it is well-defined
for a dense subset of $\textup{M}_{\Lambda_{2}}(P)$ (Corollary 3.13). Given
$\\{A_{\alpha}\\}_{\alpha\in\Gamma(\Lambda_{2})}\subset\langle\\{p^{\Lambda_{2}}\\}_{p\in
S},\ \\{e^{(s_{1},s_{2})}\\}_{s_{1}\preceq s_{2}\in\Lambda_{2}}\rangle$ a
convergent net, Corollary 3.11 then implies

$\mathbf{M}[f]\left(\lim_{\alpha\in\Gamma(\Lambda_{2})}(A_{\alpha})\right)=\lim_{\alpha\in\Gamma(\Lambda_{2})}\mathbf{M}[f](A_{\alpha}).$

Hence the morphism can be extended to a continuous ring homomorphism.
Claim 4: Given $f_{1}:\Lambda_{1}\rightarrow\Lambda_{2}$ and
$f_{2}:\Lambda_{2}\rightarrow\Lambda_{3}$ then $\mathbf{M}[f_{2}\circ
f_{1}]=\mathbf{M}[f_{1}]\circ\mathbf{M}[f_{2}]$.
The result is trivial when $\Lambda_{1}=\emptyset$. Assume
$\Lambda_{1}\neq\emptyset$. The proof will be given by looking at where the
generators of $\textup{M}_{\Lambda_{3}}$ are mapped.
It is easy to see that for $p\in P$

$\textbf{M}[f_{2}\circ
f_{1}](p^{(\Lambda_{3}})=p^{\Lambda_{1}}=\textbf{M}[f_{1}]\circ\textbf{M}[f_{2}](p^{\Lambda_{3}}).$

For the $e^{(s_{1},s_{2})}$ case, notice that if $\mathbf{M}[f_{2}\circ
f_{1}](e^{(s_{1},s_{2})})\neq 0$ then:

$\begin{array}[]{lll}\mathbf{M}[f_{2}\circ
f_{1}](e^{(s_{1},s_{2})})&=\sum_{t_{1},t_{2}}e^{(t_{1},t_{2})}&\mbox{for
}t_{1}\preceq t_{2}\in\Lambda_{1},\ \mbox{and }f_{2}\circ
f_{1}(t_{1})=s_{1},\\\ &&f_{2}\circ f_{1}(t_{2})=s_{2}\end{array}$ (3)

and

$\begin{array}[]{lll}\mathbf{M}[f_{1}]\circ\mathbf{M}[f_{2}](e^{(s_{1},s_{2})})&=\mathbf{M}[f_{1}](\sum_{q_{1},q_{2}}e^{(q_{1},q_{2})})&\mbox{where
}q_{1}\preceq q_{2}\in\Lambda_{2}\mbox{ and }f_{2}(q_{1})=s_{1},\\\
&&f_{2}(q_{2})=s_{2}\\\
&=\sum_{q_{1},q_{2}}(\sum_{t_{1},t_{2}}e^{(t_{1},t_{2})})&\mbox{for
}t_{1}\preceq t_{2}\in\Lambda_{1},\ \mbox{ and }f_{2}(q_{1})=s_{1},\\\
&&f_{2}(q_{2})=s_{2},\ f_{1}(t_{1})=q_{1},\ f_{1}(t_{2})=q_{2}.\end{array}$
(4)

Claim 1 implies that the right-hand side of Equation (4) has to be the same as
the right-hand side of Equation (3). Then
$\mathbf{M}[f_{1}]\circ\mathbf{M}[f_{2}](e^{(s_{1},s_{2})})=\mathbf{M}[f_{2}\circ
f_{1}](e^{(s_{1},s_{2})})$.
It then follows that $\mathbf{M}_{\\_}(P)$ is a contravariant functor.

∎

We call the functor $\mathbf{M}_{\\_}(P)$ the incidence functor.

######  Remark 4.6.

Notice that given $f:\Lambda^{1}\rightarrow\Lambda^{2}$ an FCC map and two
elements $e^{(s_{1},s_{2})},\ e^{(t_{1},t_{2})}$ on the generator set of
$M_{\Lambda^{2}}(P)$, if they are distinct and have no trivial image then the
image under $\mathbf{M}[f]$ of both elements are disjoint. Hence if $f$ is a
surjective FCC map we have that $\mathbf{M}[f]$ is injective.

###  4.2 Sending colimits to limits

In this section we will work with properties of the category $posets$ that
later can be used to show our contravariant functor maps colimits to limits.

######  Lemma 4.7.

Given $\\{\Lambda_{i}\\}_{i\in I}$ be an arbitrary collection of objects in
$\textbf{{prosets}}_{\textup{fin}}$, their coproduct exist and it is
$\sqcup_{i\in I}\Lambda_{i}$, as defined in Definition 2.6.

###### Proof.

Let $\\{\Lambda_{i}\\}_{i\in I}$ as in the statement and define
$\Lambda:=\sqcup_{i\in I}\Lambda_{i}$. For each $i\in I$ let
$i_{i}:\Lambda_{i}\rightarrow\Lambda$ be the embeddings sending $\Lambda_{i}$
to its copy in $\Lambda$.
Claim: $\Lambda$ is the coproduct of $\\{\Lambda_{i}\\}_{i\in I}$.
Let $\Lambda^{\prime}$ be another locally finite proset and, for each $i\in
I$, let $f_{i}:\Lambda_{i}\rightarrow\Lambda^{\prime}$ a FCC map. Define
$(f_{i})_{i\in I}:\Lambda\rightarrow\Lambda^{\prime}$ by $(f_{i})_{i\in
I}(s_{i})=f_{i}(s_{i})$ if $s_{i}\in\Lambda_{i}$. It is clear that
$(f_{i})_{i\in I}\circ i_{i}=f_{i}$ and $(f_{i})_{i\in I}$ is the unique FCC
morphism satisfying this property. Hence $\Lambda$ is the coproduct of the
collection $\\{\Lambda_{i}\\}_{i\in I}$.

${\Lambda}$${\Lambda_{i}}$${\Lambda^{\prime}}$$\scriptstyle{(f_{i})_{i\in
I}}$$\scriptstyle{i_{i}}$$\scriptstyle{f_{i}}$

∎

The following follows directly from Lemma 4.7 and Proposition 3.7.

######  Corollary 4.8.

The incidence functor $\mathbf{M}_{\\_}(P)$ maps coproducts to products.

######  Definition 4.9 (Quotient order).

Let $\Lambda=\sqcup_{i\in I}\Lambda_{i}$ be a poset with order given by the
disjoint union. We will say $\equiv\subset\Lambda\times\Lambda$ is a FCC
equivalence relation if there is a FCC surjective map
$f:\Lambda\rightarrow\Lambda^{\prime}$ such that given $s_{1},s_{2}\in\Lambda$
then $s_{1}\equiv s_{2}$ if, and only if, $f(s_{1})=f(s_{2})$. We write
$\Lambda^{\prime}:=\Lambda/\equiv$ and call this set the FCC quotient under
$\equiv$.

For the next result we will prove the existence of pushouts in the category
$\textbf{{prosets}}_{\textup{fin}}$. We will first give a remark on how such
construction works in the case the proset $\Lambda$ is irreducible, and the
proof will then be given for the general case.

######  Remark 4.10.

Let $\Lambda$, $\Lambda_{1}$ and $\Lambda_{2}$ be prosets and
$f:\Lambda\rightarrow\Lambda_{1}$, $g:\Lambda\rightarrow\Lambda_{2}$ be FCC
maps. We want to build the pushout diagram:

${\Lambda}$${\Lambda_{1}}$${\Lambda_{2}}$${{\Lambda_{f,g}}}$$\scriptstyle{g}$$\scriptstyle{f}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$

For the case $\Lambda=\emptyset$, as the empty set is the initial object of
$\textbf{{prosets}}_{\textup{fin}}$,
$\Lambda_{f,g}=\Lambda_{1}\sqcup\Lambda_{2}$ and $p_{1}$, $p_{2}$ are the
embeddings of $\Lambda_{1},\ \Lambda_{2}$ into $\Lambda_{1}\sqcup\Lambda_{2}$.
Assume $\Lambda$ is irreducible and non-empty. As $\Lambda$ is irreducible,
there exists an unique $j_{0}\in J$ and $k_{0}\in K$ such that
$f(\Lambda)\subset\Lambda_{j}$, $g(\Lambda)\subset\Lambda_{k}$. The desired
proset $\Lambda_{f,g}$ will be as follows:

$\Lambda_{f,g}=\left(\bigsqcup_{j\in
J\backslash\\{j_{0}\\}}\Lambda_{j}\right)\sqcup\left(\bigsqcup_{k\in
K\backslash\\{k_{0}\\}}\Lambda_{k}\right)\sqcup\Lambda_{\\{j_{0},k_{0}\\}}.$

The proset $\Lambda_{\\{j_{0},k_{0}\\}}$ is the FCC quootient
$\Lambda_{j_{0}}\sqcup\Lambda_{k_{0}}$ with equivalence relation given by:

  * •

for every $t\in\Lambda$, $f(t)\equiv g(t)$,

  * •

if $\Lambda\neq 1$, $f$ is a convex embedding and $g$ is a constant function
then for every $s_{1},\ s_{2}\in\Lambda_{j_{0}}$ we have $s_{1}\equiv s_{2}$,

  * •

if $\Lambda\neq 1$, $g$ is a convex embedding and $f$ is a constant function
then for every $s_{1},\ s_{2}\in\Lambda_{k_{0}}$ we have $s_{1}\equiv s_{2}$.

This is the maximal FCC quotient of $\Lambda_{j_{0}}\sqcup\Lambda_{k_{0}}$
such that $\pi\circ i_{j_{0}}\circ f=\pi\circ i_{k_{0}}\circ g$. Here the maps
$\pi:\Lambda_{j_{0}}\sqcup\Lambda_{k_{0}}\rightarrow\Lambda_{\\{j_{0},k_{0}\\}}$
is given by the FCC quotient under $\equiv$, and $i_{j_{0}}$, $i_{k_{0}}$ are
the embeddings from $\Lambda_{j_{0}}$, $\Lambda_{k_{0}}$ into their respective
components in the disjoint union $\Lambda_{j_{0}}\sqcup\Lambda_{k_{0}}$. It is
then the case

  * •

$p_{1}|_{\Lambda_{j_{0}}}=\pi\circ i_{j_{0}}$,

  * •

for $j\neq j_{0}$, $p_{1}|_{\Lambda_{j}}=i_{j}$,

  * •

$p_{2}|_{\Lambda_{k_{0}}}=\pi\circ i_{k_{0}}$,

  * •

and for $k\neq k_{0}$, $p_{2}|_{\Lambda_{k}}=i_{k}$.

As $p_{1}$, $p_{2}$ is defined on all the components of $\Lambda_{\\{f,g\\}}$,
it is uniquely defined for the whole proset. Notice that $\Lambda_{f,g}$ is an
FCC quotient of $\Lambda_{1}\sqcup\Lambda_{2}$.

######  Lemma 4.11.

Let $\Lambda,\Lambda_{1},\Lambda_{2}$ be locally finite prosets and
$f:\Lambda\rightarrow\Lambda_{1}$, $g:\Lambda\rightarrow\Lambda_{2}$ FCC maps.
Then there exists $\Lambda_{f,g}$ a locally finite proset and
$p_{1}:\Lambda_{1}\rightarrow\Lambda_{f,g}$,
$p_{2}:\Lambda_{2}\rightarrow\Lambda_{f,g}$ FCC maps such that the diagram

${\Lambda}$${\Lambda_{1}}$${\Lambda_{2}}$${{\Lambda_{f,g}}}$$\scriptstyle{g}$$\scriptstyle{f}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$

is a pushout diagram.

###### Proof.

For the proof we will write $\Lambda=\sqcup_{i\in I}\Lambda_{i}$,
$\Lambda_{1}=\sqcup_{j\in J}\Lambda_{j}$ and $\Lambda_{2}=\sqcup_{k\in
K}\Lambda_{k}$ the decomposition of these sets into their components.
Define the sets:

$\begin{array}[]{l}J_{0}=\\{j\in J;\ \text{for all }i\in I,\
f(\Lambda_{i})\cap\Lambda_{j}=\emptyset\\}\subset J\\\ J_{1}=\\{j\in
J\backslash J_{0};\ \forall i\in I,\ \text{ if
}f(\Lambda_{i})\subset\Lambda_{j}\text{ then }f|_{\Lambda_{i}}\text{ is a
convex embedding}\\}\subset J\\\ J_{2}=\\{j\in J\backslash J_{0};\ \forall
i\in I,\ \text{ if }\Lambda_{i}\neq 1,\ f(\Lambda_{i})\subset\Lambda_{j}\text{
then }f|_{\Lambda_{i}}\text{ is a constant function}\\}\subset J\\\
J_{3}=J\backslash(J_{0}\cup J_{1}\cup J_{2})\\\ K_{0}=\\{k\in K;\ \text{for
all }i\in I,\ g(\Lambda_{i})\cap\Lambda_{k}=\emptyset\\}\subset K\\\
K_{1}=\\{k\in K\backslash K_{0};\ \forall i\in I,\ \text{ if
}g(\Lambda_{i})\subset\Lambda_{k}\text{ then }g|_{\Lambda_{i}}\text{ is a
convex embedding}\\}\subset K\\\ K_{2}=\\{k\in K\backslash K_{0};\ \forall
i\in I,\ \text{ if }\Lambda_{i}\neq 1,\ g(\Lambda_{i})\subset\Lambda_{k}\text{
then }g|_{\Lambda_{i}}\text{ is a constant function}\\}\subset K\\\
K_{3}=K\backslash(K_{0}\cup K_{1}\cup K_{2}).\end{array}$

For each $l\in J\sqcup L$ also define the set $\langle l\rangle$ as the
minimal set such that:

  * •

$l\in\langle l\rangle$.

  * •

Let $k\in K$ be such that $k\in\langle l\rangle$. If $i\in I$ is such that
$g(\Lambda_{i})\subset\Lambda_{k}$ and $f(\Lambda_{i})\subset\Lambda_{j}$,
then $j\in\langle l\rangle$.

  * •

Let $j\in J$ be such that $j\in\langle l\rangle$. If $i\in I$ is such
$f(\Lambda_{i})\subset\Lambda_{j}$ and $g(\Lambda_{i})\subset\Lambda_{k}$,
then $k\in\langle l\rangle$.

In other words, $\langle l\rangle$ is the set of components that contains
$\Lambda_{l}$ and that will be “glued together” under the pushout. Notice that
if $l\in J_{0}\sqcup K_{0}$ then $\langle l\rangle=\\{l\\}$.
Let $L=\\{\langle l\rangle;\ l\in J\sqcup K\\}$ the partition of $J\sqcup K$
defined as above. Define $\Lambda_{f,g}$ as follows:

$\Lambda_{f,g}:=\bigsqcup_{\langle l\rangle\in L}\Lambda_{\langle l\rangle}$

where $\Lambda_{\langle l\rangle}$ is defined as

$\Lambda_{\langle l\rangle}=\left.\left(\bigsqcup_{l\in\langle
l\rangle}\Lambda_{l}\right)\right/\equiv_{l}$

where

  * •

for every $s\in\Lambda$, if there is $l\in\langle l\rangle$ such
$f(s)\in\Lambda_{l}$ then $f(s)\equiv_{l}g(s)$.

  * •

if $l\in\langle l\rangle$ is such that $l\in J_{3}\cup K_{3}$ then for every
$s_{1},\ s_{2}\in\Lambda_{l}$ we have $s_{1}\equiv_{l}s_{2}$.

  * •

if $j\in J_{1}$ and $k\in K_{2}$ are such that there is $i\in I$ with
$f(\Lambda_{i})\subset\Lambda_{j}$, $g(\Lambda_{i})\subset\Lambda_{k}$ then
for every $s_{1},\ s_{2}\in\Lambda_{j}$ we have $s_{1}\equiv_{l}s_{2}$.

  * •

if $j\in J_{2}$ and $k\in K_{1}$ are such that there is $i\in I$ with
$f(\Lambda_{i})\subset\Lambda_{j}$, $g(\Lambda_{i})\subset\Lambda_{k}$ then
for every $s_{1},\ s_{2}\in\Lambda_{k}$ we have $s_{1}\equiv_{l}s_{2}$.

with the order given by the FCC quotient. Notice that the construction is
similar to the irreducible case, but here we have more components being “glued
together”. Notice that $\Lambda_{f,g}=(\Lambda_{1}\sqcup\Lambda_{2})/\equiv$,
for the $\equiv$ FCC equivalence relation defined above on each component
$\Lambda_{\langle l\rangle}$. Let
$i_{1}:\Lambda_{1}\rightarrow\Lambda_{1}\sqcup\Lambda_{2}$ and
$i_{2}:\Lambda_{2}\rightarrow\Lambda_{1}\sqcup\Lambda_{2}$ the embeddings from
the coproduct and $\pi:\Lambda_{1}\sqcup\Lambda_{2}\rightarrow\Lambda_{f,g}$
given by the FCC quotient. We then define $p_{1}:=\pi\circ i_{1}$ and
$p_{2}:=\pi\circ i_{2}$.
Notice that $\Lambda_{f,g}$ is defined so that it is the maximal FCC quotient
of $\Lambda_{1}\sqcup\Lambda_{2}$ such that there exists an unique FCC maps
$p_{1}$, $p_{2}$ with $p_{1}\circ f=p_{2}\circ g$. The construction implies
that if there are FCC maps $f^{\prime}:\Lambda_{1}\rightarrow\Lambda^{\prime}$
and $g^{\prime}:\Lambda_{2}\rightarrow\Lambda^{\prime}$, then there exists a
unique FCC map $p^{\prime}:\Lambda_{f,g}\rightarrow\Lambda^{\prime}$ such that
$f^{\prime}=p^{\prime}\circ p_{1}$ and $g^{\prime}=p^{\prime}\circ p_{2}$.

${\Lambda}$${\Lambda_{1}}$${\Lambda_{2}}$${{\Lambda_{f,g}}}$${\Lambda^{\prime}}$$\scriptstyle{g}$$\scriptstyle{f}$$\scriptstyle{p_{1}}$$\scriptstyle{f^{\prime}}$$\scriptstyle{p_{2}}$$\scriptstyle{g^{\prime}}$$\scriptstyle{p^{\prime}}$

∎

Note that the intuitive idea of pushouts of prosets is getting two prosets,
$\Lambda_{1}$ and $\Lambda_{2}$, with some common convex subsets, given by the
maps $f$ and $g$ on each component of $\Lambda$, and making these subsets an
intersection of them. Below we have an example of that:

######  Example 4.12.

Let $\alpha=\mathbf{2}<=\\{0,1\\}$, $\Lambda_{1}=\mathbf{Z}$ and
$\Lambda_{2}=\mathbf{Zig}$ and define

$\begin{array}[]{llll}i_{1}:&\mathbf{2}<&\longrightarrow&\mathbf{Z}\\\
&0&\longmapsto&0\\\ &1&\longmapsto&-1,\end{array}$

and

$\begin{array}[]{llll}i_{2}:&\mathbf{2}<&\longrightarrow&\mathbf{Zig}\\\
&0&\longmapsto&0\\\ &1&\longmapsto&-1.\end{array}$

The pushout of this diagram is the following poset:

${\iddots}$${1}$${-2}$${0}$${2}$${\ldots}$${-1}$${1}$${\ldots}$${-2}$${\iddots}$

Where the normal arrow represents the element of $\mathbf{Zig}$, the dashed
arrows are the elements from $\mathbf{Z}$ and the double arrow represent the
elements “glued together” under the pushout.

Notice that as $\emptyset$ is an initial object of
$\textbf{{prosets}}_{\textup{fin}}$ then this category has coequalizers
(Proposition 2.23). We then get the following:

######  Corollary 4.13.

The category $\textbf{{prosets}}_{\textup{fin}}$ has all colimits.

###### Proof.

By Lemmas 4.7 and 4.11 and Proposition 2.24 our results follows. ∎

And as the category of $\textbf{{prosets}}_{\textup{fin}}$ has only FCC maps
and locally finite ordered sets, we get the following:

######  Corollary 4.14.

Colimits of FCC maps are FCC maps and they preserve the property of being
locally finite.

In other words, direct limits, pushouts and disjoint unions of locally posets
under FCC maps are locally finite posets. We now will use these results to
show that coequalizers are mapped to equalizers via $\mathbf{M}_{\\_}(P)$.

######  Remark 4.15.

Let $f,\ g:\Lambda^{1}\rightarrow\Lambda^{2}$ be two maps. Let
$\\{\Lambda_{i}\\}_{i\in I}$ the set of components of $\Lambda^{1}$ and
$\\{\Lambda_{j}\\}_{j\in J}$ the set of components of $\Lambda^{2}$. We define
$\textup{Coeq}(f,\ g):=\Lambda^{2}/\equiv_{\textup{Coeq}}$ as follows:

  1. 1.

For all $t\in\Lambda^{1}$, $f(t)\equiv_{\textup{Coeq}}g(t)$.

  2. 2.

If $j\in J$ is such that there is $i\in I$, $t\in\Lambda_{i}$ with $f(t),\
g(t)\in\Lambda_{j}$ and $f(t)\neq g(t)$, then for every $s_{1},\
s_{2}\in\Lambda_{j}$, $s_{1}\equiv_{\textup{Coeq}}s_{2}$.

and the map $p:\Lambda^{2}\rightarrow\textup{Coeq}(f,\ g)$ as the quotient
map. Notice that the first condition makes $p\circ f=p\circ g$, and the second
condition is a minimal condition for the map $p$ to be FCC.

######  Lemma 4.16.

Let $f_{1},\ f_{2}:\Lambda^{1}\rightarrow\Lambda^{2}$ and
$p:\Lambda^{2}\rightarrow\textup{Coeq}(f_{1},\ f_{2})$ be the coequalizer of
$f_{1}$ and $f_{2}$. Then $\mathbf{M}[p]$ is the equalizer of
$\mathbf{M}[f_{1}],\ \mathbf{M}[f_{2}]$.

###### Proof.

First, let $f_{1},f_{2},\ p$ as in the statement. We first show that
$\mathbf{M}[f_{1}]\circ\mathbf{M}[p]=\mathbf{M}[f_{2}]\circ\mathbf{M}[p]$. We
then prove $\mathbf{M}[p]$ is the equalizer of
$\mathbf{M}[f_{1}],\mathbf{M}[f_{2}]$.
Notice that for $n=1,\ 2$:

$\begin{array}[]{llll}\mathbf{M}[f_{n}]\circ\mathbf{M}[p]:&\textup{M}_{\textup{Coeq}(f_{1},\
f_{2})}(P)&\longrightarrow&\textup{M}_{\Lambda^{1}}(P)\\\
&r^{\textup{Coeq}(f_{1},\ f_{2})}&\longmapsto&r^{\Lambda^{1}}\\\
&1^{\\{s\\}}&\longmapsto&\left\\{\begin{array}[]{ll}1^{(p\circ
f_{n})^{-1}(s)}&\mbox{if }(p\circ f_{n})^{-1}\neq\emptyset\\\
0&\mbox{otherwise}\end{array}\right.\\\
&e^{(s_{1},s_{2})}&\longmapsto&\left\\{\begin{array}[]{ll}\sum_{t_{1},t_{2}}e^{(t_{1},t_{2})}&\mbox{for
}t_{1}\precnapprox t_{2}\mbox{ and }p\circ f_{n}(t_{1})=s_{1},\\\ &p\circ
f_{n}(t_{2})=s_{2}\\\ 0&\mbox{otherwise}\end{array}\right.\end{array}$

Because $p$ is the coequalizer of $f_{1}$ and $f_{2}$ it follows that
$\mathbf{M}[f_{1}]\circ\mathbf{M}[p]=\mathbf{M}[f_{2}]\circ\mathbf{M}[p]$.
Notice that as $p$ is surjective, Remark 4.6 implies $\mathbf{M}[p]$ is
injective.
Now let $\Lambda^{\prime}$ be a poset and
$h^{\prime}:{M}_{\Lambda^{\prime}}(P)\rightarrow\textup{M}_{\Lambda^{2}}(P)$
be such that $\mathbf{M}[f_{1}]\circ h^{\prime}=\mathbf{M}[f_{2}]\circ
h^{\prime}$ and $h^{\prime}(r^{\Lambda^{\prime}})=r^{\Lambda^{2}}$, for every
$r\in P$. Define
$h:\textup{M}_{\Lambda^{\prime}}(P)\rightarrow\textup{M}_{\textup{Coeq}(f_{1},\
f_{2})}(P)$ as:

$h(A)=\left\\{\begin{array}[]{ll}(\mathbf{M}[p])^{-1}(h^{\prime}(A))&\mbox{if
}(\mathbf{M}[p])^{-1}(h^{\prime}(A))\neq\emptyset\\\
0&\mbox{otherwise.}\end{array}\right.$

This function is well-defined because $\mathbf{M}[p]$ is injective. Hence the
following diagram commute:

${\textup{M}_{\textup{Coeq}(f_{1},\
f_{2})}(P)}$${\textup{M}_{\Lambda^{2}}(P)}$${\textup{M}_{\Lambda^{1}}(P)}$${\textup{M}_{\Lambda^{\prime}}(P)}$$\scriptstyle{\mathbf{M}[p]}$$\scriptstyle{\mathbf{M}[f_{1}]}$$\scriptstyle{\mathbf{M}[f_{2}]}$$\scriptstyle{h}$$\scriptstyle{h^{\prime}}$

That is, $\mathbf{M}[p]$ is the equalizer of $\mathbf{M}[f_{1}],\
\mathbf{M}[f_{2}]$. ∎

######  Corollary 4.17.

The incidence functor $\mathbf{M}_{\\_}$ sends colimits to limits.

###### Proof.

By Lemma 4.16 and Corollary 4.8 we have that the functor
${}^{op}\circ\mathbf{M}_{\\_}(P):\textbf{{prosets}}_{\textup{fin}}\Rightarrow\textbf{TopRings}^{op}$
maps coequalizers to coequalizers and coproducts to coproducts. The result
then follows from Proposition 2.25. ∎

###  4.3 Generating the category $\textbf{{prosets}}_{\textup{fin}}$

In this section, we focus on showing that the prosets of the form
$\mathbf{m}\leftarrow\mathbf{n}$, for $n,\ m\in\mathbb{N}$ can generate all
locally finite prosets under direct limits, disjoint unions, and pushouts of
FCC maps.
The next result is the version of Theorem 3.6 in the category
$\textbf{{prosets}}_{\textup{fin}}$.

######  Theorem 4.18.

Let $\Lambda$ be an irreducible, locally finite proset. Then
$\Lambda\cong\varinjlim_{\alpha\in\Gamma(\Lambda)}\alpha$, where
$\Gamma(\Lambda)=\\{\alpha\subset\Lambda;$ is finite and convex$\\}$ with
subset order.

###### Proof.

For each $\alpha_{1}\subset\alpha_{2}\in\Gamma(\Lambda)$ let
$j_{\alpha_{2}\rightarrow\alpha_{1}}:\alpha_{1}\rightarrow\alpha_{2}$ be the
convex embedding given by the subset structure. It is clear that if
$\alpha_{1}\subset\alpha_{2}\subset\alpha_{3}$ then
$j_{\alpha_{1}\rightarrow\alpha_{2}}\circ
j_{\alpha_{2}\rightarrow\alpha_{3}}=j_{\alpha_{1}\rightarrow\alpha_{3}}$,
hence the direct limit $\varinjlim_{\alpha\in\Gamma(\Lambda)}\alpha$ exists.
For each $\alpha_{i}\in\Gamma(\Lambda)$ let
$i_{\alpha_{i}}:\alpha_{i}\rightarrow\Lambda$ be the embedding of $\alpha_{i}$
into $\Lambda$. Because the outer triangle of the following diagram commutes,
there exists an unique
$f:\varinjlim_{\alpha\in\Gamma(\Lambda)}\alpha\rightarrow\Lambda$ such that
the whole diagram commutes.

${\Lambda}$${\varinjlim_{\alpha\in\Gamma(\Lambda)}\alpha}$${\alpha_{2}}$${\alpha_{1}}$$\scriptstyle{f}$$\scriptstyle{i_{\alpha_{2}}}$$\scriptstyle{j_{\alpha_{2}}}$$\scriptstyle{i_{\alpha_{1}}}$$\scriptstyle{j_{\alpha_{1}}}$$\scriptstyle{j_{\alpha_{2}\rightarrow\alpha_{1}}}$

By the direct limit property, $f$ is an FCC map and a direct limit of
embeddings. Hence it is a convex embedding. To complete our proof, we will
show that $f$ is surjective.
Let $a\in\Lambda$. The set $\mathcal{N}_{0}(a)$ is a finite convex subset of
$\Lambda$ containing $a$. It is clear that $f\circ
j_{\mathcal{N}_{0}(a)}(a)=a$, hence $f$ is surjective. ∎

Lastly we can show a property which can be taken from the category
$\textbf{prosets}_{\text{fin}}$ and translated to the context of matrix rings
over $P$ indexed by a poset.

######  Theorem 4.19.

All objects in $\textbf{{prosets}}_{\textup{fin}}$ are generated by
$\mathbf{m}\leftarrow\mathbf{n}$, for $m,\ n\in\mathbb{N}$, and under disjoint
unions, direct limits and pushouts of convex embeddings.

###### Proof.

We first prove by induction that we can generate all the irreducible, finite
prosets. from The proof for this case will be by induction on the cardinality
of $\Lambda$.
Our generating set generates contains all irreducible posets of cardinalities
$0$, $1$ and $2$, proving the base cases, as noted in Example 2.5. Assume the
result is true for every irreducible $\Lambda$ with cardinality $k$, for some
$k>2$.
Let $\Lambda\notin\\{\mathbf{i}\leftarrow\mathbf{j}\\}_{i+j=k+1}$ be an
irreducible proset with cardinality $k+1$. Let $a\neq b\in\Lambda$ such that
$\Lambda\backslash\mathcal{N}_{0}(a)$, $\Lambda\backslash\mathcal{N}_{0}(b)$,
and $\Lambda\backslash(\mathcal{N}_{0}(a)\cup\mathcal{N}_{0}(b))$ are
irreducible. Let
$\alpha=\Lambda\backslash(\mathcal{N}_{0}(a)\cup\mathcal{N}_{0}(b))$,
$\Lambda_{1}=\Lambda\backslash\mathcal{N}_{0}(a)$ and
$\Lambda_{2}=\Lambda\backslash\mathcal{N}_{0}(b)$. Such elements $a,b$ exists
as every finite connected graph with at least $3$ vertices has at least two
non-cut vertices (a proof can be seen in [Bas]). The conditions on $\Lambda$
imply $\Lambda\backslash(\mathcal{N}_{0}(a)\cup\mathcal{N}_{0}(b))$ is a non-
empty set.
Define $i_{1}:\alpha\rightarrow\Lambda_{1}$,
$i_{2}:\alpha\rightarrow\Lambda_{2}$ the embeddings given by the subset
structure. It follows that $|\Lambda_{1}|,|\Lambda_{2}|,|\alpha|<k+1$ and
$\Lambda$ is the pushout of these prosets, proving our induction.
Theorem 4.18 implies that all infinite irreducible prosets are generated by
finite, irreducible prosets under direct limits. Proposition 2.6 implies that
if all the irreducible posets are generated, so are the reducible ones under
disjoint unions. ∎

######  Corollary 4.20.

The subclass of all rings of the form $\textup{M}_{\Lambda}(P)$, for $\Lambda$
a poset, is generated by $\textup{M}_{\mathbf{m}\leftarrow\mathbf{n}}(P)$, for
$m,n\in\mathbb{N}$, under products, inverse limits and pullbacks.

###### Proof.

A direct corollary from Corollary 4.17 and Theorem 4.19. ∎

This corollary can be useful for proving general properties of these rings.
Instead of proving for all of them individually, you can prove for the
generators and show that these properties are preserved by inverse limits and
pullbacks.

##  5 Group of invertible matrices

###  5.1 $\textup{GL}_{\Lambda}(P)$ and its properties

In this section we will work with the group of units of the triangular matrix
rings and show how to use them to build simple topological groups. In this
section we assume $P$ is a commutative ring.

######  Definition 5.1.

Given $P$ a ring and $\Lambda$ a proset. We define the general linear group of
$P$ in relation to $\Lambda$ as:

$\textup{GL}_{\Lambda}(P):=\left\\{A\in\textup{M}_{\Lambda}(P);\
A^{-1}\in\textup{M}_{\Lambda}(P)\right\\}.$

That is, the group of units of $\textup{M}_{\Lambda}(P)$.

######  Definition 5.2.

Let $\Lambda$ be a locally finite proset and $P$ be a ring. Let
$A\in\textup{M}_{\Lambda}(P)$. We define the set $\textup{DET}(A)$ as:

$\textup{DET}(A)=\\{\det(\pi_{\alpha}(A));\ \alpha\in\Gamma(\Lambda)\\}\subset
P^{\Gamma(\Lambda)}.$

The following is a directly corollary of Theorem 2.20 and the fact the groups
of the form $\textup{GL}_{\Lambda}(P)$ are inverse limits of finite matrix
rings.

######  Proposition 5.3.

Let $P$ be a commutative ring and $\Lambda$ a preordered set, and
$A\in\textup{M}_{\Lambda}(P)$. Then $A\in\textup{GL}_{\Lambda}(P)$ if, and
only if, $\pi_{\alpha}(A)$ is invertible for all $\alpha\in\Gamma(\Lambda)$.
That is, $A\in\textup{GL}_{\Lambda}(P)$ if, and only if,
$\textup{DET}(A)\subset\prod_{\alpha\in\Gamma(\Lambda)}P^{*}$.

A stronger version of Proposition 5.3 is given at [Bel73, Theorem 1.16].

######  Corollary 5.4.

Let $P$ be a commutative, absolute topological ring. Then for every finite
preordered set $\Lambda$ we have that $\textup{M}_{\Lambda}(P)$ is an absolute
topological ring.

###### Proof.

Let $A$ be an element of $\textup{GL}_{\Lambda}(P)$. By Theorem 2.21 we have
that for every $\alpha\in\Gamma(\Lambda)$, the map
$\pi_{\alpha}(A)\mapsto\textup{adj}(\pi_{\alpha}(A))$ is a polynomial, hence
it is continuous. As $P$ is an absolute topological ring and
$\det(\pi_{\alpha}(A))\in P^{*}$, we have by Theorem 2.20 and Proposition 5.3
that if $\pi_{\alpha}(A)\in\textup{GL}_{\alpha}(P)$, the map
$\pi_{\alpha}(A)\mapsto\pi_{\alpha}(A^{-1})$ is continuous. As the inverse map
is continous for all $\textup{GL}_{\alpha}(P)$, $\alpha\in\Gamma(\Lambda)$,
the inverse limit property implies the map $A\mapsto A^{-1}$ is continuous for
$\textup{GL}_{\Lambda}(P)$. ∎

As $P$ is commutative, when $\Lambda$ is a poset we can define
$\textup{GL}_{\Lambda}(P)$ as all the infinite matrices such that the diagonal
elements belongs to $P^{*}$. In the case $P$ is not commutative this is not
always the case, as shown in articles [Hoł02a] and [Asp59].
Given that the functor that sends a ring to its units group is left adjoint
(Proposition 2.22), Corollaries 5.4, 4.17, and Theorem 3.6 allow us to prove
the following for the case $\textup{GL}_{\Lambda}(P)$:

######  Proposition 5.5.

Let $P$ be an absolute topological ring and $\Lambda$ a locally finite proset.
The following are true:

  1. 1.

If $\Lambda$ is infinite, then
$\textup{GL}_{\Lambda}(P)\cong\varprojlim_{\alpha\in\Gamma(\Lambda)}\textup{GL}_{\alpha}(P)$
is a topological group.

  2. 2.

Given $\Lambda$ is a proset, then $\textup{GL}_{\Lambda}(P)$ can be generated
by $\textup{GL}_{\mathbf{m}\leftarrow\mathbf{n}}(P)$ under products, inverse
limits and pullbacks.

We can also describe some normal subgroups of $\textup{GL}_{\Lambda}(P)$ in a
similar way to the ideals of $\textup{M}_{\Lambda}(P)$, as follows:

######  Notation 5.6 (Normal subgroups).

Let $\Lambda$ a locally finite proset and $P$ a commutative ring. We denote
the following normal subgroups:

  * •

Given $s_{1}\preceq s_{2}$ in $\Lambda$ we define

$N^{\Lambda}_{[s_{1},s_{2}]}(P)=\\{A\in\textup{GL}_{\Lambda}(P);\ \text{if
}s\in[s_{1},\ s_{2}],\ A_{s,s}=1\text{,\ and if }t_{1}\neq t_{2}\in[s_{1},\
s_{2}],\ \ A_{t_{1},t_{2}}=0\\}.$

This normal subgroup has quotient

$\textup{GL}_{\Lambda}(P)/N^{\Lambda}_{[s_{1},\
s_{2}]}(P)\cong\textup{GL}_{[s_{1},\ s_{2}]}(P).$

  * •

For a convex set $\Lambda^{\prime}\subset\Lambda$, the normal subgroup

$N^{\Lambda}_{\Lambda^{\prime}}=\bigcap_{s_{1}\preceq
s_{2}\in\Lambda^{\prime}}N^{\Lambda}_{[s_{1},\ s_{2}]}.$

This normal subgroup has quotient

$\textup{GL}_{\Lambda}(P)/N^{(\Lambda,\
\preceq)}_{\Lambda^{\prime}}\cong\textup{GL}_{\Lambda^{\prime}}(P).$

  * •

For $\\{\Lambda_{i}\\}_{i\in I}$ a locally convex set of $\Lambda$ we define
the normal subgroup

$N^{\Lambda}_{\\{\Lambda_{i}\\}_{i\in I}}=\bigcap_{i\in
I}N^{\Lambda}_{\Lambda_{i}}.$

This normal subgroup has quotient

$\textup{GL}_{\Lambda}(P)/N^{\Lambda}_{\\{\Lambda_{i}\\}_{i\in
I}}\cong\prod_{i\in I}\textup{GL}_{\Lambda_{i}}(P).$

######  Example 5.7.

Let $\Lambda=\mathbf{3}<$, $P=\mathbb{Z}$ and $\alpha=\\{0,1\\}$. Then
$N^{\mathbf{3}<}_{\alpha}$ is the normal subgroup with elements of the form:

$\left(\begin{array}[]{lll}1&0&a_{0,2}\\\ 0&1&a_{1,2}\\\
0&0&a_{2,2}\end{array}\right)$

where $a_{2,2}\in\\{-1,1\\}$.

The following is the equivalent for Corollary 3.10 for the groups
$\textup{GL}_{\Lambda}(P)$.

######  Corollary 5.8.

Let $P$ a commutative, absolute topological ring, $\Lambda$ a locally finite
proset and $\\{\Lambda_{i}\\}_{i\in I}\subset\Lambda$ locally convex. Then the
following are true:

  1. 1.

The normal subgroup $N^{\Lambda}_{\\{\Lambda_{i}\\}_{i\in I}}$ is closed in
$\textup{GL}_{\Lambda}(P)$.

  2. 2.

The normal subgroup $N^{\Lambda}_{\\{\Lambda_{i}\\}_{i\in I}}$ is open if, and
only if, $\bigsqcup_{i\in I}\Lambda_{i}$ is a finite set and $P$ has the
discrete topology.

One can also get an equivalent to Theorem 4.5 for the groups of units. For
that we denote $\mathbf{U}:\mathbf{Rings}\rightarrow\mathbf{Grps}$ the functor
that maps a ring to its group of units. The following is a direct consequence
from Proposition 2.22, Theorem 4.5, Corollary 4.17 and Corollary 5.4:

######  Theorem 5.9.

Let $P$ be a commutative, absolute topological ring. The functor
$\mathbf{GL}_{\\_}(P):=\mathbf{U}\circ\mathbf{M}_{\\_}(P)$ is a contravariant
functor from the category $\mathbf{prosets_{fin}}$ to the category of
topological groups, $\mathbf{TopGrps}$. This functor maps colimits to limits.

###  5.2 The center

For the next result, given $A\in\textup{M}_{\Lambda}(P)$, we say that $A$ is
scalar if $A_{s,s}=A_{t,t}$ for every $t,s\in\Lambda$, and if $t\neq s$ then
$A_{t,s}=0$. We will denote $Z(\textup{M}_{\Lambda}(P))$ as the center of the
ring of matrices and $Z(\textup{GL}_{\Lambda}(P))$ the center of the group of
units of matrices.

######  Proposition 5.10.

[Bel73, Theorem 1.23] Let $\Lambda$ be a finite irreducible proset and $P$ a
commutative ring. Then $A\in Z(\textup{M}_{\Lambda}(P))$ if, and only if, $A$
is scalar.

######  Proposition 5.11.

Let $\Lambda$ be a finite irreducible proset and $P$ a commutative ring such
that $\exists p_{1},\ p_{2}\in P^{*}$ with $p_{1}-p_{2}\in P^{*}$. Then $A\in
Z(\textup{GL}_{\Lambda}(P))$ if, and only if, $A$ is scalar.

###### Proof.

First we will use induction to prove for the case $\Lambda$ is a poset. We
then use it to generalize for the prosets case.
Given that $|\Lambda|=1$, both results follow trivially, as every element is
scalar. If $|\Lambda|=2$, as $\Lambda$ is irreducible then
$\Lambda=\mathbf{2}<$. Let $A\in\textup{M}_{\Lambda}(P)$. Then $A$ can be
written as follows.

$A:=\begin{pmatrix}A_{0,0}&A_{0,1}\\\ 0&A_{1,1}\end{pmatrix}.$

Assume now that $A\in Z(\textup{GL}_{\Lambda}(P))$. Hence, given
$B\in\textup{GL}_{\Lambda}(P)$ any element, the following equations are true
for every $B_{0,0},B_{0,1},B_{1,1}\in P$

$\left\\{\begin{array}[]{l}A_{0,0}B_{0,0}=B_{0,0}A_{0,0}\\\
A_{1,1}B_{1,1}=B_{1,1}A_{1,1}\\\
B_{0,0}A_{0,1}+B_{0,1}A_{1,1}=A_{0,0}B_{0,1}+A_{0,1}B_{1,1}.\end{array}\right.$

By making $B_{0,0}=B_{1,1}=B_{0,1}=1$ we get $A_{0,0}=A_{1,1}$. On the other
hand, by fixing $B_{0,0}=B_{1,1}=1$ and letting $B_{0,1}\in P$ any element, it
follows that $A_{0,0},\ A_{1,1}\in P^{*}$. The third equation also tell us
that for every $p_{1},\ p_{2}\in P^{*}$ we have that
$A_{0,1}p_{1}=A_{0,1}p_{2}$. Hence $A_{0,1}(p_{1}-p_{2})=0$. Using $p_{1},\
p_{2}$ from the assumption and using the fact that units cannot be zero
divisors, it follows that $A_{0,1}=0$.
Assume that for every $k\leqslant n$ it is true that if $|\Lambda|=k$ and
$\Lambda$ is an irreducible poset than the statements are true.
Let $|\Lambda|=n+1$. Let $s_{1}\leqslant s_{2}$. If $|[s_{1},\ s_{2}]|\leq n$,
then the result follows on the submatrix with coordinates on $[s_{1},\ s_{2}]$
by induction. Hence the result is always true when $|\Lambda|\leqslant n+1$
and $\Lambda\neq\mathbf{n+1}<$.
If $\Lambda:=\mathbf{n+1}<$, by looking at the submatrices indexed by
$\\{0,1,\ldots,n-1\\}$ and $\\{1,2,\ldots,n\\}$ it only remains to show
$A_{0,\ n}=0$. Given $A\in\textup{M}_{\Lambda}(P)$ then, if $AB=BA$ the
coordinate $A_{0,\ n}$ implies that

$\sum_{0\leq i\leq n}A_{0,i}B_{i,n}=\sum_{0\leq i\leq n}B_{0,i}A_{i,n}$

By the induction argument, as $A\in Z(\textup{M}_{\Lambda}(P))$ we have that
$A_{0,0}=A_{n,n}\in Z(P)$ and for $i\neq j$ and $(i,\ j)\neq(0,\ n)$ we have
that $A_{i,j}=0$. Hence the equation is reduced to

$A_{0,0}B_{0,\ n}+A_{0,n}B_{n,n}=B_{0,0}A_{0,n}+B_{0,n}A_{n,n}$ (5)

By letting $B_{0,0}=B_{n,n}=B_{0,n}=1$ it follows that $A_{0,0}=A_{n,n}$. On
the other hand, by fixing $B_{0,0}=B_{1,1}=1$ and letting $B_{0,n}\in P$ any
element, it follows that $A_{0,0},\ A_{n,n}\in P^{*}$. The third equation also
tell us that for every $p_{1},\ p_{2}\in P^{*}$ we have that
$A_{0,n}p_{1}=A_{0,n}p_{2}$. Hence $A_{0,n}(p_{1}-p_{2})=0$. Using $p_{1},\
p_{2}$ from the assumption and using the fact that units cannot be zero
divisors, it follows that $A_{0,n}=0$.
For the proset case, denote the proset as $(\Lambda,\ \preceq)$. Notice that
every scalar element $A$ with coordinates in $P^{*}$ belongs to
$\textup{GL}_{(\Lambda,\ \preceq)}(P)$. It is also the case that for any
irreducible partial suborder $\leqslant$ of $\preceq$ (Definition 2.11), if
$A\in Z(\textup{GL}_{(\Lambda,\leqslant)})$ then $A$ is scalar. Hence, $A$ can
only belong to the center of $\textup{GL}_{(\Lambda,\ \preceq)}$ if it is
scalar. ∎

Note that the group part of the result is not true for the case
$\Lambda=\mathbf{2}<$ and $\mathbb{F}_{2}$, as
$Z(\textup{M}_{\mathbf{2}}(\mathbb{F}_{2}))$ has only scalar matrices but
$\textup{GL}_{\mathbf{2}}(\mathbb{F}_{2})\cong\mathbb{F}_{2}$ is an abelian
group. This also gives us an example that given $P$ a non-commutative ring, it
is not necessarily true that $Z(P)\cap P^{*}=Z(P^{*})$.

######  Corollary 5.12.

Let $\Lambda$ be an irreducible proset and $P$ a commutative, absolute
topological ring. Then $Z(\textup{M}_{\Lambda}(P))\cong P$. If there are
$p_{1},\ p_{2}\in P^{*}$ such that $p_{1}-p_{2}\in P^{*}$ then
$Z(\textup{GL}_{\Lambda}(P))\cong P^{*}$.

###  5.3 Building simple topological groups

######  Definition 5.13.

Given $\Lambda$ a poset and $P$ a ring, define the almost general linear group
of $P$ in relation to $\Lambda$ as

$\textup{aGL}_{\Lambda}(P):=\left\\{A\in{\textup{GL}_{\Lambda+S}}\text{ for
some }S\subset\Lambda\text{ finite}\right\\}.$

In other words, the set of all matrices that accept finitely many non-zero
entries without relation in $\Lambda$.

######  Example 5.14.

Let $\Lambda=\mathbf{N}$. Then $A\in\textup{aGL}_{\mathbf{N}}(P)$ if, and only
if, $A\in\textup{GL}_{\mathbf{N}+\\{0,\ 1,\ldots,\ n\\}}(P)$ for some
$n\in\mathbb{N}$.

######  Proposition 5.15.

Let $\Lambda$ be a poset and $P$ a ring. The group $\textup{aGL}_{\Lambda}(P)$
is equal to the direct limit
$\varinjlim_{S\subset\Lambda\text{finite}}\textup{GL}_{\Lambda+S}(P)$.

###### Proof.

For each $S$ finite subsets of $\Lambda$, define
$j_{S}:\textup{GL}_{\Lambda}\rightarrow\textup{GL}_{\Lambda+S}$ the embedding
such that $A_{s_{1},s_{2}}=(j_{S}(A))_{s_{1},s_{2}}$. If $S_{1}\subset S_{2}$
are subsets of $\Lambda$, define $j_{S_{1}\rightarrow
S_{2}}:\textup{GL}_{\Lambda+S_{1}}\rightarrow\textup{GL}_{\Lambda+S_{2}}$ the
embedding such that $A_{s_{1},s_{2}}=(j_{S_{1}\rightarrow
S_{2}}(A))_{s_{1},s_{2}}$. These morphisms are such that if $S_{1}\subset
S_{2}\subset S_{3}$ are finite subsets, then $j_{S_{1}\rightarrow
S_{3}}=j_{S_{2}\rightarrow S_{3}}\circ j_{S_{1}\rightarrow S_{2}}$ and
$j_{S_{1}}=j_{S_{2}}\circ j_{S_{1}\rightarrow S_{2}}$. Hence the direct limit
of this system, $\varinjlim\textup{GL}_{\Lambda+S}(P)$ exist. We now need to
prove $\varinjlim\textup{GL}_{\Lambda+S}(P)=\textup{aGL}_{\Lambda}(P)$.
For each $S$ finite subset of $\Lambda$, define
$i_{S}:\textup{GL}_{\Lambda+S}\rightarrow\textup{aGL}_{\Lambda}(P)$ the
embedding such that $A_{s_{1},s_{2}}=(i_{S}(A))_{s_{1},s_{2}}$. Notice that
these morphisms are such that the outer triangle of the following diagram
commutes:

${\textup{aGL}_{\Lambda}(P)}$${\varinjlim\textup{GL}_{\Lambda+S}(P)}$${\textup{GL}_{\Lambda+S_{2}}(P)}$${\textup{GL}_{\Lambda+S_{1}}(P)}$$\scriptstyle{i}$$\scriptstyle{i_{S_{2}}}$$\scriptstyle{j_{S_{2}}}$$\scriptstyle{i_{S_{1}}}$$\scriptstyle{j_{S_{1}}}$$\scriptstyle{j_{S_{1}\rightarrow
S_{2}}}$

Hence, there exists an unique
$i:\varinjlim\textup{GL}_{\Lambda+S}(P)\rightarrow\textup{aGL}_{\Lambda}(P)$.
By definition of all the morphisms and of $\textup{aGL}_{\Lambda}(P)$ it
follows that $i$ is an isomorphism. ∎

The following corollary follows directly from the last proposition.

######  Corollary 5.16.

Let $\Lambda$ be a poset and $P$ a ring. If $\Lambda$ is countable,
$\Lambda=\\{s_{n}\\}_{n\in\mathbb{N}}$ is an enumeration then
$\textup{aGL}_{\Lambda}(P)=\bigcup_{n\in\mathbb{N}}\textup{GL}_{\Lambda+\\{s_{1},s_{2},\ldots,s_{n}\\}}(P)$.
If $P$ is an absolute topological ring, $\textup{aGL}_{\Lambda}(P)$ is a
topological group under the final topology.

For the next result, we use the following notation:

  * •

Let $S$ be a finite set with cardinality $n$ and $F$ a field. We denote
$SL_{\overline{S}}(F)\leqslant\textup{GL}_{\overline{S}}(F)$ the square
matrices indexed by $S$ with determinant $1$

  * •

Given $\Lambda$ an irreducible proset and $S\subset\Lambda$ a finite set,
define $(\overline{S},\ \preceq_{+})\subset\Lambda$ the minimal convex subset
containing $S$ with order given by all elements being equivalent.

######  Theorem 5.17.

Let $F$ be a topological field and $\Lambda$ an infinite, locally finite
proset. Then $Z_{\Lambda}(F):=Z(\textup{aGL}_{\Lambda}(F))$ is the unique
largest proper closed normal subgroup of $\textup{aGL}_{\Lambda}(F)$. In
particular,
$\textup{aPGL}_{\Lambda}(P):=\textup{aGL}_{\Lambda}(F)/Z_{\Lambda}(F)$ is
topologically simple.

###### Proof.

The group $Z_{\Lambda}(F)$ is clearly closed and normal in
$\textup{aGL}_{\Lambda}(F)$. To complete the proof we will show that if $M$ is
a normal subgroup of $\textup{aGL}_{\Lambda}(F)$ is not contained in
$Z_{\Lambda}(F)$, then $M=\textup{aGL}_{\Lambda}(F)$. For the proof we will
look at the intersection of $M$ with the subgroups of
$\textup{aGL}_{\Lambda}(F)$ of the form $\textup{GL}_{\Lambda+S}(F)$, for $S$
a finite subset of $\Lambda$, and show all the elements of
$\textup{GL}_{\Lambda+S}(F)$ are in $M$.
Fix such $M$ and let $A\in M$ which is not scalar. Let $S$ be a finite set
containing $\\{s,t\in\Lambda;A_{s,t}\neq 0\textit{ and }s\npreceq t\\}$. Then
$A$ can be seen as an element in $\textup{GL}_{\Lambda+S}(F)$.
Now, let $N^{\Lambda+S}_{S}(F)$ the normal subgroup of
$\textup{GL}_{\Lambda+S}(F)$ as in Notation 5.6. For each $S\subset
S^{\prime}\subset\Lambda$ finite convex subset of $\Lambda+S$, define:

$\textup{M}_{S^{\prime}}:=(\textup{GL}_{\Lambda+S}(F)\cap
M)/N^{\Lambda+S}_{S^{\prime}}.$

That can be seen as a normal subgroup of
$\textup{GL}_{\overline{S^{\prime}}}(F)$. Notice that as
$AN_{S^{\prime}}^{\Lambda+S}$ is a non-central element in
$\textup{GL}_{\overline{S^{\prime}}}(F)$, hence $\textup{M}_{S^{\prime}}$ is
non-central in $\textup{GL}_{\overline{S^{\prime}}}(F)$. By Lemma 2.18 we have
that $SL_{\overline{S^{\prime}}}(F)\leqslant\textup{M}_{S^{\prime}}$. We now
want to show $\textup{M}_{S^{\prime}}=\textup{GL}_{\overline{S^{\prime}}}(F)$.
Given $A^{\prime}\in\textup{GL}_{\overline{S^{\prime}}}(F)$, there is $B$ an
element of $\textup{GL}_{\Lambda+S^{\prime}}(F)$ such that for

$B_{i,j}=\left\\{\begin{array}[]{ll}A^{\prime}_{i,j}&\mbox{if }i,j\in
S^{\prime},\\\ 0&\mbox{if }i,j\notin S^{\prime}\mbox{ and }i\neq j,\\\
\mbox{is an element of }P^{*}&\mbox{if }i,j\notin S^{\prime}\mbox{ and
}i=j.\end{array}\right.$

Moreover, by changing the coordinates $(i,i)$ for $i\notin S^{\prime}$, $B$
may be chosen with $BN^{\Lambda+S^{\prime}}_{S^{\prime\prime}}$ is an element
$SL_{S^{\prime\prime}}(F)$ for some finite convex subset $S^{\prime\prime}$ of
$\Lambda+S^{\prime}$ that strictly contains $\overline{S^{\prime}}$.
Let $A$ the fixed non-scalar element in $M$. Let
$A^{\prime}=AN^{\Lambda+S}_{S^{\prime}}\in\textup{GL}_{\overline{S^{\prime}}}(F)$.
Let $B$ be the element in $SL_{S^{\prime\prime}}(F)$ generated in relation to
$A^{\prime}$, as seen above. Since $N_{S^{\prime\prime}}\geqslant
SL_{S^{\prime\prime}}(F)$ and $\textup{GL}_{\overline{S^{\prime}}}(F)$ can be
seen as a subgroup of $SL_{S^{\prime\prime}}(F)$, the previous argument and
the fact $A=BN^{\Lambda+S}_{S^{\prime}}$ implies
$\textup{M}_{S^{\prime}}=\textup{GL}_{\overline{S^{\prime}}}(F)$. Hence, for
every $S\subset\Lambda$ finite subset, $\textup{GL}_{\Lambda+S}(F)\leqslant
M$. Proposition 5.15 then implies $M=\textup{aGL}_{\Lambda}(F)$.
In particular, any nontrivial closed normal subgroup of
$\textup{aGL}_{\Lambda}(F)/Z_{\Lambda}(F)$ has preimage equal to
$\textup{aGL}_{\Lambda}(F)$ and $\textup{aGL}_{\Lambda}(F)/Z_{\Lambda}(F)$ is
topologically simple. ∎

Notice that if $\Lambda$ is a finite proset, then:

$\textup{aGL}_{\Lambda}(G)/Z_{\Lambda}(F)=\textup{PGL}_{|\Lambda|}(F);$

the latter group fails to be topologically simple if $|F|>2$, because the
determinant homomorphism.
One can see that this construction can make simple topological groups of large
cardinalities over different fields. Under some conditions the group created
can have an interesting topology, for example, if your field is finite with
discrete topology, the group generated will be a totally disconnected locally