# Limits and colimits, generators and relations of partial groups

Edoardo Salati<EMAIL_ADDRESS>
###### Abstract.

We analyse limits and colimits in the category $Part$ of partial groups,
algebraic structures introduced by A. Chermak. We will prove that $Part$ is
both complete and cocomplete and, in addition, that the full subcategory of
finite partial groups is both finitely complete and finitely cocomplete.
Cocompleteness is then used in order to define quotients of partial groups. We
will also identify a category richer than $Set$ (the category of sets and set-
maps) and build the free partial groups over objects is such category; this
yields a larger class of free partial groups, eventually allowing to prove
that every partial group is the quotient of a free partial group.

## Introduction

The notion of a partial group was developed in connection with fusion systems;
roughly speaking, a fusion system is a category whose objects are the
subgroups of a finite $p$-group $S$ and whose morphisms are injective group
homomorphisms, including at least all the conjugations by elements of $S$.
Research on fusion systems is currently very active in connection to Local
Finite Group Theory, Modular Representation Theory and Algebraic Topology. In
this last field it has been a tool employed in the proof of the Martino-Priddy
conjecture, asserting that, if $p$ is a prime, the $p$-completed classifying
space associated to a finite group $G$ is homotopy equivalent to the
classifying space associated to the fusion system $\mathcal{F}_{S}(G)$
(described, for example, in [AKO11, Part I, Section 1]), where
$S\in\mathrm{Syl}_{p}(G)$. With regard to this, partial groups played a
prominent role in the definition of the classifying space associated to an
abstract saturated fusion system $\mathcal{F}$; this was achieved by replacing
$\mathcal{F}$ with a different, slightly richer, category, the so-called
_centric linking system_ , and then defining the classifying space of
$\mathcal{F}$ as the $p$-completed geometric realization of such associated
centric linking system, whose existence and uniqueness are therefore necessary
conditions to the definition. Partial groups provide a translation of the
categorical language into one more similar to that of groups, which in turn
allowed A. Chermak in [Che13] to tackle and solve the problem by means of the
well established group-theoretic machinery. The interested reader may find a
beautiful survey in [AO16], with Sections 3 and 4 covering the interplay
between Algebra and Homotopy Theory.

There is a natural notion of morphisms of partial groups, as defined in
[Che22], and in the same paper Chermak deals with further properties of the
category $Part$ of partial groups, including completeness and cocompleteness.
In particular, in [Che22] the existence of colimits is proved exclusively over
diagrams with specific properties; the main goal of this paper is to correct a
mistake appeared in the proof of the cocompleteness of $Part$ in an
unpublished early work of Chermak and, at the same time, prove the following
result.

######  Theorem A.

The category $Part$ is complete and cocomplete. In addition, if $FinPart$ is
the full subcategory of $Part$ with objects the finite partial groups, then
$FinPart$ is finitely complete and finitely cocomplete.

Early examples of how colimits in $Part$ may be exploited and used as a tool
are the _elementary expansions_ defined by Chermak in [Che16], which consist
in the main construction appearing in [Che13, Section 5] and building the
centric linking system over a fusion system, as well as the construction
realized by A. Díaz Ramos, R. Molinier and A. Viruel in [RMV21] for proving
that the category $Part$ is universal.
We will also take advantage of the categorical setting in order to define
quotients for all partial groups and to study free partial groups: this will
set the necessary background for the following result.

######  Proposition B.

Every partial group is a quotient of a free partial group.

The notion of free partial groups over sets is analyzed very early, in Section
1, as it is used to build a counterexample to Chermak’s construction of
colimits of partial groups. Such counterexample is presented in Section 2,
together with a proof of the cocompleteness of $Part$. We will also report a
proof of the completeness of $Part$ in Section 3, although this can also be
found in [Che22], with the purpose of collecting all coherent results in one
single paper. Combining results of Sections 2 and 3 yields a proof of Theorem
A.
Existence of colimits is exploited in Section 4 in order to define quotients
of partial groups. Quotients were previously defined only for very specific
subclasses, namely for groups and for the so-called localities, as defined in
[Che22, Sections 2 to 4] and it will be proved that the definition given in
this paper specializes to the already existing notions. Section 5 is dedicated
to the proof of Proposition B; as a consequence of the analysis undertaken in
Section 1 it turns out that the notion of free partial groups over sets is too
weak for such a result. After replacing the category $Set$ of sets and set-
maps with a better suited category, denoted by $Set^{s}$, the notions of
quotients and of free partial groups over $Set^{s}$ allow a proof in terms of
generators and relations, in analogy with what happens for groups. The section
is ended with an example of how these tools may be used: in detail, it is
shown that Chermak’s elementary expansions construction may be realized in
terms of generators and relations.

## Acknowledgements

First of all, to my supervisor Ellen Henke, who has introduced me to the world
of fusion systems, partial groups and localities, many thanks for reading the
paper and for the several pieces of advice she provided me with.
A special thank to Andrew Chermak, for reading the results and advancing the
idea of publishing them in a paper. He was the first to deal with the
categorical properties of partial groups and this paper, actually, stems from
previous work of his.
Thanks also to A. González and to A. Díaz Ramos, R. Molinier and A. Viruel for
their interest, the emails we exchanged and the suggestions I received from
them.

##  1\. Partial groups and free partial groups

Roughly speaking, a _partial group_ is a set $\mathcal{L}$ equipped with a
_partial_ operation, defined on a subset of the set of words on the alphabet
$\mathcal{L}$; in other words, it may not be possible to multiply all elements
as it happens with groups.
A little remark about the notation adopted throughout the paper: we will use
right-hand notation for functions and functors, i.e. writing $(x)f$, or simply
$xf$, whenever $f$ is a function or a functor and $xf$ is the object (or
morphism) associated to $x$ by $f$. Composition of functions or functors will
then be denoted by simple juxtaposition or by the symbol $\cdot$.

######  Definition 1.1.

Let $\mathcal{L}$ be a non-empty set and $W(\mathcal{L})=W$ be the free monoid
on $\mathcal{L}$; denote concatenation of words in $W$ by the symbol $\circ$.
Consider a subset $D=D(\mathcal{L})\subseteq W$, a map
$\Pi:D\rightarrow\mathcal{L}$, which will be referred to as the _multivariable
product_ of $\mathcal{L}$ (and sometimes just as the _product_ of
$\mathcal{L}$), and an involutory bijection
$i:\mathcal{L}\rightarrow\mathcal{L}$. Then the quadruple
$(\mathcal{L},D,\Pi,i)$ (often denoted just by $\mathcal{L}$ when the context
leaves no ambiguity) is a partial group provided that the following hold:

  1. (1)

$\mathcal{L}\subseteq D\qquad$ and moreover $\qquad u\circ v\in
D\quad\implies\quad u,v\in D$;

  2. (2)

$\Pi|_{\mathcal{L}}=id_{\mathcal{L}}$;

  3. (3)

if $u\circ v\circ w\in D$, then $\quad u\circ(v)\Pi\circ w\in D\quad$ and
$\quad(u\circ v\circ w)\Pi=(u\circ(v)\Pi\circ w)\Pi$;

  4. (4)

by extending $i$ to $W(\mathcal{L})$ defining
$(x_{1},\dots,x_{n})i=((x_{n})i,\dots,(x_{1})i)$, if $w\in D$ then we have
$\quad(w)i\circ w\in D\quad$ and $\quad((w)i\circ
w)\Pi=1_{\mathcal{L}}=(\emptyset)\Pi$, where $\emptyset$ is the empty word.

In analogy with terminology for groups, we call the element $1_{\mathcal{L}}$
the _unit_ or _neutral element_ of the partial group $\mathcal{L}$ and for
every $x\in\mathcal{L}$, $(x)i$ is the _inverse_ of $x$.
The partial group $(\mathcal{L},D,\Pi,i)$ is _finite_ if the set $\mathcal{L}$
is finite.

For a survey of some elementary properties of partial groups we refer to
[Che22, Section 1]. Clearly, $\mathcal{L}$ is a group if and only if
$D(\mathcal{L})=W(\mathcal{L})$ and, in such a case, $\Pi$ gives the group
multiplication thanks to associativity.
A. Chermak also defines a notion of morphisms of partial groups.

######  Definition 1.2.

Given partial groups $(\mathcal{L},D,\Pi,i)$ and
$(\mathcal{L}^{\prime},D^{\prime},\Pi^{\prime},i^{\prime})$, a morphism
$\beta$ of partial groups from $\mathcal{L}$ to $\mathcal{L}^{\prime}$ is a
set-wise map $\beta:\mathcal{L}\rightarrow\mathcal{L}^{\prime}$ such that, if
$\beta^{*}:W(\mathcal{L})\rightarrow W(\mathcal{L}^{\prime})$ is the
componentwise extension, i.e.
$(x_{1},\dots,x_{n})\beta^{*}:=((x_{1})\beta,\dots,(x_{n})\beta)$, then:

  * (a)

$(D)\beta^{*}\subseteq D^{\prime}$;

  * (b)

we have a commutative diagram

${D}$${\mathcal{L}}$${D^{\prime}}$${\mathcal{L}^{\prime}.}$$\scriptstyle{\Pi}$$\scriptstyle{\beta^{*}}$$\scriptstyle{\beta}$$\scriptstyle{\Pi^{\prime}}$

We therefore obtain a category $Part$ whose objects are partial groups and
whose morphisms are morphisms of partial groups according to the above
definition, clearly with the usual composition of maps. It is trivial to check
that $\beta$ is an isomorphism (in the categorical sense) if and only if
$\beta$ is a bijective morphism of partial groups satisfying
$(D)\beta^{*}=D^{\prime}$. Moreover, morphisms of partial groups behave well
with respect to units, that is if
$\beta:(\mathcal{L},D,\Pi,i)\longrightarrow(\mathcal{L}^{\prime},D^{\prime},\Pi^{\prime},i^{\prime})$
is a morphism between partial groups, then
$(1_{\mathcal{L}})\beta=1_{\mathcal{L}^{\prime}}$, as shown in [Che22, Lemma
1.13]. More generally, there is a fully-faithful inclusion functor
$Grp\hookrightarrow Part$.

######  Notation.

For any map $f:X\longrightarrow Y$ between sets $X,Y$, the notation $f^{*}$
will always denote the induced map $f^{*}:W(X)\longrightarrow W(Y)$ (or
restrictions of it) as defined in Definition 1.2. This applies also for maps
between partial groups or other structures with a forgetful functor in $Set$,
up to applying such forgetful functor. On the other hand, given a partial
group $(\mathcal{L},D,\Pi,i)$, in accordance to the notation in Definition 1.1
the symbol $i$ will denote also the extension of $i$ to $W(\mathcal{L})$ taken
by reversing order. We will also equivalently write $D(\mathcal{L})$ or
$D_{\mathcal{L}}$ for the domain of a partial group
$\mathcal{L}=(\mathcal{L},D_{\mathcal{L}},\Pi,i)$.

Define the category of pointed sets, $Set^{*}$, as the category whose objects
are pairs $(X,x)$, where $X$ is a non-empty set and $x\in X$, and whose
morphisms $f:(X,x)\longrightarrow(Y,y)$ are set-maps $f:X\longrightarrow Y$
with the property that $(x)f=y$. Composition is given by the usual composition
of maps. Since partial groups are naturally pointed at the unit, we will work
with pointed sets instead of sets; this will simplify the construction of free
partial groups. In better words, let’s consider first the forgetful functor
$U:Grp\longrightarrow Set$; we can actually define a slightly different
forgetful functor (still denoted by $U$) which remembers a little more, namely

$U:Grp\longrightarrow Set^{*},\quad(G)U=(G,1_{G})$

where $1_{G}$ is the unit of the group. This is possible since
$(1_{G})f=1_{H}$ whenever $f:G\longrightarrow H$ is a homomorphism between
groups. Since morphisms of partial group have the same property, we can
similarly define a forgetful functor

$U:Part\longrightarrow
Set^{*},\quad(\mathcal{L},D,\Pi,i)U=(\mathcal{L},1_{\mathcal{L}}).$

It is a classical situation that in which $U$ is a right adjoint to the free-
construction functor; indeed, we will prove that $U:Part\longrightarrow
Set^{*}$ is a right adjoint by building free partial groups over pointed sets.
It will become clear that the construction may be carried in a very similar
way also when dealing with the category $Set$ in place of $Set^{*}$ and yields
the same partial group.

###  1.1. Free partial groups over pointed sets

Suppose we have a pointed set $(X,1)$ (the base-point will become the unit of
the partial group). Consider the set $Y:=\\{1\\}\sqcup
X^{*}\sqcup\hat{X}^{*}$, where $X^{*}=X\setminus\\{1\\}$, $\sqcup$ denotes the
disjoint union and we use $\hat{X}^{*}$ to distinguish the two copies of
$X^{*}$ (the same notation will also be adopted for the respective elements).
Define an involutory bijection $i$ on $Y$ by $(1)i=1$ and $(x)i=\hat{x}$ and
set $D\subseteq W(\mathcal{L})$ to be the set of words obtained from
alternating finite strings of the form $(\dots,x,\hat{x},x,\hat{x},\dots)$,
for $x\in X^{*}$, by adding in any position any finite number of copies of the
element $1$. We say that such strings are built on the element $x$.
For $w\in D$ built on the element $x\in X^{*}$, define

$(w)\Pi=\begin{cases}x\quad\text{ if the number of $x$ is greater than that of
$\hat{x}$},\\\ 1\quad\text{ if the number of $x$ equals that of $\hat{x}$},\\\
\hat{x}\quad\text{ if the number of $x$ is lower than that of
$\hat{x}$}.\end{cases}$

We now prove that $(Y,D,\Pi,i)$ is a partial group. Clearly $Y\subseteq D$ and
$\Pi|_{Y}=\mathrm{id}_{Y}$; if $u,v$ are words in $W(Y)$ such that $u\circ
v\in D$, then it is a string built on some element $x\in X^{*}$ and,
therefore, also $u$ and $v$ are strings built on the same element $x$,
yielding $u,v\in D$. These are axioms (1) and (2) of Definition 1.1. To prove
axiom (3), first of all note that we may identify the $\hat{-}$ notation with
application of $i$. Suppose that $u\circ v\circ w\in D$; if $v$ is the empty
word or a word made only of the element $1$, there is nothing to prove,
otherwise, supposing that $v$ has the form $v=(a_{0},\dots,b_{0})$ with
$a_{0},b_{0}\in\\{1,x,\hat{x}\\}$, we may find elements
$a,b\in\\{x,\hat{x}\\}$ so that $v=(1,\dots,1,a,\dots,b,1,\dots,1)$. Then the
last occurrence of an element which is not $1$ in $u$ must be $\hat{a}$ and
the first occurrence of an element which is not $1$ in $w$ must be $\hat{b}$.
If $a=x$, then $(v)\Pi\in\\{1,x\\}$; when $(v)\Pi=1$, $b=\hat{x}$, so
$\hat{a}=\hat{x}$ and $\hat{b}=x$; this implies $u\circ(v)\Pi\circ w\in D$.
When $(v)\Pi=x$, then $b=x$ and in this case we have $\hat{a}=\hat{x}$,
$(v)\Pi=x$ and $\hat{b}=\hat{x}$, showing again that $u\circ(v)\Pi\circ w\in
D$. The case $a=\hat{x}$ is analogous; that $(u\circ v\circ
w)\Pi=(u\circ(v)\Pi\circ w)\Pi$ is trivial, so axiom (3) holds. Let now
$u=(u_{0},\dots,u_{n})\in D$ and suppose it is built on an element $x\in
X^{*}$; then $(u)i\circ u=((u_{n})i,\dots,(u_{0})i,u_{0},\dots,u_{n})$ and, by
definition of $i$, also $(u)i$ is built on $x$. Let $u_{k}$ be the first
occurrence of an element different from $1$ in $u$, then $(u_{k})i$ is the
last occurrence of element different from $1$ in $(u)i$, showing that
$(u)i\circ u$ is built on $x$. Thus $(u)i\circ u\in D$. Finally, let $d$ be
the difference between the number of occurrences of $x$ and the number of
occurrences of $\hat{x}$ in $u$. Then $d\in\\{-1,0,1\\}$ and the same
difference computed on $(u)i$ gives $-d$; this shows that $((u)i\circ
u)\Pi=1$, so axiom (4) also holds.

Define an association $F$ by setting $(X,1)F:=(Y,D,\Pi,i)$. In order to obtain
a functor it remains to define $F$ over morphisms; let
$f:(X,1_{X})\rightarrow(Z,1_{Z})$ be a morphism of pointed sets, then we
associate to $f$ the map $fF:(X,1_{X})F\rightarrow(Z,1_{Z})F$ defined by
setting $(1_{X})fF=1_{Y}$, $(x)fF=(x)f$ and $(\hat{x})fF=\widehat{(x)f}$. It
is straightforward to check that $fF$ is a partial group morphism and that $F$
is indeed a functor. A similar construction can be carried over sets and in
this case one ends with the free partial group over the set $X$ being exactly
$(X\sqcup\\{1\\},1)F$.
Let $U:Part\longrightarrow Set^{*}$ be the forgetful functor defined by
$(\mathcal{L},D,\Pi,i)U=(\mathcal{L},1_{\mathcal{L}})$; for every pointed set
$(X,1)$ there exists a natural inclusion $j:(X,1)\hookrightarrow(X,1)FU$
simply defined by $(x)j=x$ for every $x\in X$. Then $(X,1)F$ is the free
partial group over the pair $((X,1),j)$; indeed for any pointed-set map
$f:(X,1)\rightarrow(\mathcal{L},1_{\mathcal{L}})$, where $\mathcal{L}$ is a
partial group, there exists a unique morphism of partial groups
$\tilde{f}:(X,1)F\longrightarrow\mathcal{L}$ making the following diagram
commute.

${(X,1)}$${(X,1)FU}$${(\mathcal{L})U}$$\scriptstyle{j}$$\scriptstyle{f}$$\scriptstyle{(\tilde{f})U}$

$\tilde{f}$ is defined by $(1)\tilde{f}=1_{\mathcal{L}}$, $(x)\tilde{f}=(x)f$
and $(\hat{x})\tilde{f}\mapsto((x)f)^{-1}$, where $x\in X$ and the exponent
$-1$ denotes the inverse in $\mathcal{L}$. Note that the existence of the
morphism $\tilde{f}$ is guaranteed by the fact that the domain $(X,1)F$ is the
_smallest_ possible satisfying axioms (1) to (4) of Definition 1.1 (this will
become formally clear in Lemma 5.3), therefore universal initial with respect
to morphisms of partial groups. As an immediate consequence we have bijections

$Hom_{Part}((X,1)F,\mathcal{L})\cong Hom_{Set^{*}}((X,1),(\mathcal{L})U)$
($\prime$)

yielding the following lemma.

######  Lemma 1.1.

With the notation above, $F\dashv U:Part\longrightarrow Set^{*}$.

###### Proof.

We explicitly define a natural isomorphism between the bifunctors
$Hom_{Part}((-)F,-)$ and $Hom_{Set^{*}}(-,(-)U)$. Let $(X,1)\in Ob(Set^{*})$
and $\mathcal{L}=(\mathcal{L},D,\Pi,i)\in Ob(Part)$ and define

${\psi_{((X,1),\mathcal{L})}:}$${Hom_{Set^{*}}((X,1),(\mathcal{L})U)}$${Hom_{Part}((X,1)F,\mathcal{L})}$${f}$${\tilde{f}}$

where $\tilde{f}$ is the map defined above; then ($\prime$ ‣ 1.1) shows that
$\psi_{((X,1),\mathcal{L})}$ is a bijection.
We need to show naturality; let $g:(Y,1)\longrightarrow(X,1)$ be a morphism of
pointed sets,
$h:(\mathcal{L},D(\mathcal{L}),\Pi_{\mathcal{L}},i_{\mathcal{L}})\longrightarrow(\mathcal{M},D(\mathcal{M}),\Pi_{\mathcal{M}},i_{\mathcal{M}})$
be a morphism of partial groups and consider the following diagram

${Hom_{Set^{*}}((X,1),(\mathcal{L})U)}$${Hom_{Part}((X,1)F,\mathcal{L})}$${Hom_{Set^{*}}((Y,1),(\mathcal{M})U)}$${Hom_{Part}((Y,1)F,\mathcal{M})}$$\scriptstyle{\psi_{((X,1),\mathcal{L})}}$$\scriptstyle{\psi_{((Y,1),\mathcal{M})}}$

where the vertical arrows are pre- and post-composition with the morphisms $g$
and $h$ on the left side, $gF$ and $hF$ on the right side. Let $f\in
Hom_{Set^{*}}((X,1),(\mathcal{L})U)$; following the upper-right path we get

$f\mapsto\tilde{f}\mapsto gF\cdot\tilde{f}\cdot h,$

whereas the bottom-left path yields

$f\mapsto g\cdot f\cdot hU\mapsto\widetilde{g\cdot f\cdot hU}.$

By $\prime$ ‣ 1.1 these morphisms are determined by their image on $1$ and
$Y$, and for $y\in Y\cup\\{1\\}$ we get

$(y)gF\cdot\tilde{f}\cdot h=(y)g\tilde{f}\cdot h=((y)gf)h,$

and

$(y)\widetilde{g\cdot f\cdot hU}=(y)(g\cdot f\cdot hU)=((y)gf)h.$

This completes the proof. ∎

Note that neither $UF$ nor $FU$ are the identity functor; as a consequence
even though $F$ preserves colimits (so we can push colimits in $Set^{*}$ to
colimits in $Part$), we cannot use this fact to prove that $Part$ is
cocomplete as there is a change of objects.

##  2\. Colimits in $Part$

###  2.1. Colimits in $Set^{*}$

Before discussing about colimits in $Part$, we shall study in detail colimits
in the category $Set^{*}$. Recall that given a relation $R$ on a set $X$ there
exists a unique smallest equivalence relation on $X$ containing $R$, exactly
the intersection of all equivalence relations on $X$ containing $R$. In
addition, the functions with domain $X$ that factor through the relation $R$
are exactly the functions factoring through the smallest equivalence relation
containing $R$. In detail,

######  Lemma 2.1.

Let $f:X\rightarrow Y$ be a function between sets and $\sim^{\prime}$ a binary
relation on $X$, i.e. $\sim^{\prime}\subseteq X\times X$. Let $\sim$ be the
smallest equivalence relation on $X$ containing $\sim^{\prime}$. Then $f$
factors through $X/\sim$, as in the commutative diagram below, if and only if
$(u)f=(v)f$ for every $u\sim^{\prime}v$.

${f:}$${X}$${Y}$${\frac{X}{\sim}}$$\scriptstyle{f}$$\scriptstyle{\pi}$$\scriptstyle{\exists!\hat{f}}$

###### Proof.

Since $\sim^{\prime}\subseteq\sim$, the _only if_ part is trivial. As for the
_if_ part, sharing the same image by $f$ is an equivalence relation, that is
we may define

$\forall u,v\in X,\;u\sim^{\prime\prime}v\iff(u)f=(v)f.$

Then we have $\sim^{\prime}\subseteq\sim^{\prime\prime}$ and, as $\sim$ is the
smallest equivalence relation containing $\sim^{\prime}$, also
$\sim\subseteq\sim^{\prime\prime}$; clearly $f$ factors through
$\sim^{\prime\prime}$, thus it also factors through $\sim$. ∎

Turning our attention to colimits, let $\mathcal{C}$ be a small category and
$F:\mathcal{C}\rightarrow Set^{*}$ a functor; we call such a functor a
_diagram_ of type $\mathcal{C}$ in $Set^{*}$. For every $X\in Ob(\mathcal{C})$
denote by $(X,x)$ its image under $F$; similarly for every $j\in
Mor_{\mathcal{C}}(X,Y)$ denote by $jF$ the image of the morphism $j$ via $F$.
In order to avoid an excessively elaborate notation, we will write disjoint
unions of pointed sets, meaning by it the disjoint union of the sets and
forgetting about the base-points, the outcome being a set (not pointed). Then
one has

$colim(F)=\left(\dfrac{\displaystyle\bigsqcup_{X\in
Ob(\mathcal{C})}(X)F}{\sim},[x]\right)$ ($\star$)

where $\sim$ is the smallest equivalence relation on $\bigsqcup_{X\in
Ob(\mathcal{C})}(X)F$ containing all the pairs $(x,y)$ and $(z,(z)jF)$ for all
images $XF=(X,x),\,YF=(Y,y)$ and all $z\in(X,x)$ and $j\in
Mor_{\mathcal{C}}(X,Y)$; here $[x]$ denotes the equivalence class of a (and
therefore any) base-point $x$. The maps building a cocone to $colim(F)$ are
defined as $i_{X}:(X,x)\longrightarrow colim(F),\,z\mapsto[z]$. Indeed
consider a cocone $((R,r),t_{X})$ from $F$ to $(R,r)$

${(X,x)}$${colim(F)}$${(R,r)}$${(Y,y)}$$\scriptstyle{i_{X}}$$\scriptstyle{t_{X}}$$\scriptstyle{\exists!\psi}$$\scriptstyle{i_{Y}}$$\scriptstyle{t_{Y}}$

and define the morphism $\psi$ by $[z]\overset{\psi}{\rightarrow}(z)t_{X}$ for
$z\in(X,x)$. In order to prove that $\psi$ is well-defined we need to show
that it factors through $\sim$; in virtue of Lemma 2.1 we only need to prove
$(z)t_{X}=((z)jF)t_{Y}$ for any $z\in(X,x)$ and $j\in Mor_{\mathcal{C}}(X,Y)$,
but this is trivially true by the definition of a cocone. As for the
uniqueness, any other morphism $\varphi$ such that $i_{X}\cdot\varphi=t_{X}$
for every $X\in Ob(\mathcal{C})$ satisfies
$([z])\varphi=(z)i_{X}\cdot\varphi=(z)t_{X}$, thus $\varphi=\psi$.

###  2.2. Factorization of the multivariable product

We describe a possible attempt to prove cocompleteness of $Part$ through
colimits of $Set^{*}$. If $\mathcal{C}$ is a small category, $a\in
Ob(\mathcal{C})$ and $T:\mathcal{C}\rightarrow Part$ a functor, denote images
via $T$ by $aT=(aT,D(aT),\Pi_{a},i_{a})$ and, for each morphism
$x:a\rightarrow b$ in $\mathcal{C}$, its image under $T$ by
$\lambda_{x}:aT\rightarrow bT$. Now consider the forgetful functor
$U:Part\rightarrow Set^{*}$ and set $(\mathcal{L},1_{\mathcal{L}}):=colim(TU)$
to be the colimit in the category of pointed sets. We wish to define a partial
group structure over $\mathcal{L}$ which makes it the colimit over $T$ in
$Part$.
Since we will often refer to the following way of defining a domain, we
provide a numbering for referencing reasons.

######  2.3.

A natural way to define the domain $D\subseteq W(\mathcal{L})$ consists in
taking $D$ as the set of words $[w]=([w_{1}],\dots,[w_{n}])$ for which there
exist some $a\in Ob(\mathcal{C})$ such that $w_{i}\in aT$ for all
$i\in\\{1,\dots,n\\}$ and $(w_{1},\dots,w_{n})\in D(aT)$, where square
brackets $[-]$ are used to denote the equivalence classes of the relation
defining $\mathcal{L}$,. Equivalently, if $\rho$ is the projection over the
quotient defining $\mathcal{L}$, as in ($\star$ ‣ 2.1), we may restrict
$\rho^{*}$ to $\bigsqcup_{a\in Ob(\mathcal{C})}D(aT)$, that is
$(u_{1},\dots,u_{n})\rho^{*}=((u_{1})\rho,\dots,(u_{n})\rho)$ for any word
$(u_{1},\dots,u_{n})\in D(aT)$ and any $a\in Ob(\mathcal{C})$: then
$\displaystyle D=\left(\bigsqcup_{a\in Ob(\mathcal{C})}D(aT)\right)\rho^{*}$.

Clearly $D$ satisfies condition (1) of Definition 1. One would now hopefully
define a product $\Pi$ by lifting to products $\Pi_{a}$ of the partial groups
$aT$ for suitable objects $a\in Ob(\mathcal{C})$; for this purpose consider
the following diagram

${\displaystyle\bigsqcup_{a\in
Ob(\mathcal{C})}D(aT)}$${\displaystyle\bigsqcup_{a\in
Ob(\mathcal{C})}aTU}$${(\mathcal{L},1_{\mathcal{L}})}$${D}$$\scriptstyle{\rho^{*}}$$\scriptstyle{\sqcup\Pi_{a}}$$\scriptstyle{\rho}$$\scriptstyle{\Pi}$
($\dagger$)

One might now hope to factor $\rho^{*}$ via $\sqcup\Pi_{a}\cdot\rho$. This
argument would at least be complicated to prove and we will actually show it
wrong. Indeed suppose we have $u=(u_{1},\dots,u_{k})\in D(aT)$ and
$v=(v_{1},\dots,v_{h})\in D(bT)$ such that $u\rho^{*}=v\rho^{*}$, then we get
that $u$ and $v$ have equal length and for all the components we have
$[u_{i}]=[v_{i}]$. Now we need $[(u)\Pi_{a}]=[(v)\Pi_{b}]$, however since
there is no guarantee that the components $u_{i}$ and $v_{i}$ are identified
through a unique morphism of partial groups $aT\rightarrow bT$, we cannot use
property (b) of Definition 1.2 to prove our factorization.
We will provide a counterexample showing how a construction carried as above
may, in general, fail; in order to build a not too complicated counterexample
we will need to split the formation of colimits according to the following
result about categories.

######  Proposition 2.2.

Consider a locally small category $\mathcal{C}$; then every (co)limit is the
(co)equalizer of a (co)product. In particular $\mathcal{C}$ is (finitely)
(co)complete if and only if $\mathcal{C}$ admits all (finite) (co)products and
all (co)equalizers.

###### Proof.

See [Mac71, Chapter V, section 2], where the author proves the statement for
limits; the statement for colimits is obtained by duality. ∎

We start by showing that $Part$ has coproducts, since the factorization
discussed above works in this case and fails when dealing with coequalizers.

######  Lemma 2.3.

The category $Part$ has all coproducts.

###### Proof.

Let $\mathcal{C}$ be a small, discrete category, $T$ a diagram of type
$\mathcal{C}$ in $Part$, $U$ the forgetful functor $Part\longrightarrow
Set^{*}$. Our aim is to show that there exists a partial group structure over
the coproduct taken in $Set^{*}$, that is $colim(TU)$, which makes it the
coproduct in $Part$. We will extensively refer to the construction described
in Section 2.2 and adopt the same notation. For every $a\in Ob(\mathcal{C})$
set $\mathcal{L}_{a}=(\mathcal{L}_{a},D(\mathcal{L}_{a}),\Pi_{a},i_{a}):=aT$
and write $1_{a}$ for its unit. From Section 2.1 we obtain

$colim(TU)=\displaystyle\frac{\bigsqcup_{a\in
Ob(\mathcal{C})}(\mathcal{L}_{a})U}{\sim},$

where $\sim$ is the equivalence relation identifying exactly all the units
$1_{a}$. Set also $\mathcal{L}:=colim(TU)$ and define the domain
$D(\mathcal{L})$ of $\mathcal{L}$, as well as the maps $\rho$ and $\rho^{*}$,
as in 2.3. Let $u\in D(\mathcal{L}_{a})$ and $v\in D(\mathcal{L}_{b})$ be such
that $(u)\rho^{*}=(v)\rho^{*}$, then either all the components of $u$ and $v$
are the unit element or $\mathcal{L}_{a}=\mathcal{L}_{b}$ and $u=v$. Hence, in
this situation $\bigsqcup\Pi_{a}\cdot\rho$ factors through $\rho^{*}$; we take
the resulting map $\Pi:D(\mathcal{L})\rightarrow\mathcal{L}$ as the product on
$\mathcal{L}$. The inversion map $i$ on $\mathcal{L}$ is defined by
$([z])i=[(z)i_{a}]$ when $z\in\mathcal{L}_{a}$; note that for $[z]=[1_{a}]$,
$([1_{a}])i=[(1_{a})i_{a}]=[1_{a}]=[1_{b}]=([1_{b}])i$ for any $a,b\in
Ob(\mathcal{C})$, so $i$ is well-define.
We have already seen in Section 2.2 that axiom (1) of Definition 1.1 holds;
axiom (2) is, also, trivial. Suppose now that $u\circ v\circ w\in
D(\mathcal{L})$, then we can lift via $\rho^{*}$ to a word in
$D(\mathcal{L}_{a})$ for some $a\in\mathcal{C}$ and remap via
$\sqcup\Pi_{a}\cdot\rho$ to $\mathcal{L}$; by the commutativity of $\dagger$ ‣
2.2 we obtain $(u\circ v\circ w)\Pi$. Since $\mathcal{L}_{a}$ is a partial
group, axiom (3) holds for the lifting of $u\circ v\circ w$ and, therefore,
holds also in $\mathcal{L}$. Similarly also axiom (4) holds.
We have natural inclusion morphisms
$j_{a}:\mathcal{L}_{a}\rightarrow\mathcal{L}$, given by $(x)j_{a}=[x]$, which
are morphisms of partial groups and build a cocone $(\mathcal{L},j_{a})$ from
$T$. Then $(\mathcal{L},j_{a})$ is the coproduct; indeed consider another
cocone $(\mathcal{M},f_{a})$ from $T$, with
$\mathcal{M}=(\mathcal{M},D(\mathcal{M}),\Pi_{\mathcal{M}},i_{\mathcal{M}})$,

${\mathcal{L}_{a}}$${\mathcal{L}}$${\mathcal{M}}$${\mathcal{L}_{b}}$$\scriptstyle{j_{a}}$$\scriptstyle{f_{a}}$$\scriptstyle{\exists!\psi}$$\scriptstyle{j_{b}}$$\scriptstyle{f_{b}}$

Then all we need to show is that the pointed-set morphism $\psi$ defined in
Section 2.1, acting as $([x])\psi=(x)f_{a}$ for $x\in\mathcal{L}_{a}$, is a
morphism of partial groups. Note that $D(\mathcal{L}_{a})f_{a}\subseteq
D(\mathcal{M})$, so that also $D(\mathcal{L})\subseteq D(\mathcal{M})$; if
$u=([u_{1}],\dots,[u_{k}])\in D(\mathcal{L})$, then every class $[u_{k}]$,
with $u_{k}$ not a unit of some $\mathcal{L}_{a}$, is a singleton, whereas the
units form all together a single class. Thus all the $u_{k}$ appearing in $u$
which are not a unit must all be elements of the same $\mathcal{L}_{a}$ for
some $a\in Ob(\mathcal{C})$. By picking the units to be $1_{a}$, we can lift
$u$ to a word $(u_{1},\dots,u_{n})\in D(\mathcal{L}_{a})$. Consider now the
following diagram

${D(\mathcal{L})}$${\mathcal{L}}$${D(\mathcal{M})}$${\mathcal{M}}$${D(\mathcal{L}_{a})}$${\mathcal{L}_{a}}$$\scriptstyle{\psi^{*}}$$\scriptstyle{\Pi}$$\scriptstyle{\psi}$$\scriptstyle{\Pi_{\mathcal{M}}}$$\scriptstyle{f_{a}^{*}}$$\scriptstyle{\Pi_{a}}$$\scriptstyle{f_{a}}$

where the bottom square is commutative since $f_{a}$ is a morphism of partial
groups. Now
$(u)\Pi\cdot\psi=(u_{1},\dots,u_{n})\Pi_{a}\cdot\rho\cdot\psi=([(u_{1},\dots,u_{n})\Pi_{a}])\psi$.
On the other hand

$(u)\psi^{*}\cdot\Pi_{\mathcal{M}}=(([u_{1}])\psi,\dots,([u_{n}])\psi)\Pi_{\mathcal{M}}=((u_{1})f_{a},\dots,(u_{n})f_{a})\Pi_{\mathcal{M}}=(u_{1},\dots,u_{n})f_{a}^{*}\cdot\Pi_{\mathcal{M}}.$

Since $f_{a}=j_{a}\cdot\psi$, we have

$(u)\psi^{*}\cdot\Pi_{\mathcal{M}}=(u_{1},\dots,u_{n})\Pi_{a}\cdot
f_{a}=(u_{1},\dots,u_{n})\Pi_{a}\cdot
j_{a}\cdot\psi=(u_{1},\dots,u_{n})j_{a}^{*}\cdot\Pi\cdot\psi=([(u_{1},\dots,u_{n})\Pi_{a}])\psi,$

so also the upper square is commutative. This completes the proof. ∎

######  Remark.

Note that coproducts of a finite number of finite partial groups are again
finite; this will translate, after proving existence of coequalizers, in the
fact that the full subcategory of $Part$ of finite partial groups is finitely
cocomplete. This is probably among the major differences between the
categories $Part$ and $Grp$.

###  2.4. Coequalizers of partial groups

We now begin our analysis of coequalizers; in order to keep notation simple,
we omit writing the application of the forgetful functor
$U:Part\longrightarrow Set$ whenever it is clear by the context whether we are
considering a partial group or just its underlying set.
Consider two partial groups
$\mathcal{L}=(\mathcal{L},D(\mathcal{L}),\Pi_{\mathcal{L}},j)$ and
$\mathcal{L}^{\prime}=(\mathcal{L}^{\prime},D(\mathcal{L}^{\prime}),\Pi,i)$
and two parallel morphisms of partial groups
$f,g:\mathcal{L}\rightarrow\mathcal{L}^{\prime}$. The setwise coequalizer is
given by

$C:=coeq_{Set}(f,g)=\frac{\mathcal{L}^{\prime}}{\sim}$

paired with the quotient map
$q:\mathcal{L}^{\prime}\rightarrow(\mathcal{L}^{\prime}/\sim)$, where $\sim$
is the smallest equivalence relation on $\mathcal{L}^{\prime}$ containing the
pairs $(xf,xg)$ for $x\in\mathcal{L}$. This is a simplified realization of the
same object constructed in $\eqref{colimit}$; according to this we are
supposed to consider $(\mathcal{L}\sqcup\mathcal{L}^{\prime})/\mathcal{R}$,
with $\mathcal{R}$ the smallest equivalence relation containing all pairs
$(x,xf)$ and $(x,xg)$ for $x\in\mathcal{L}$. In particular $\mathcal{R}$
identifies all $xf$ with $xg$; it is a simple exercise to prove that there is
a bijection

$\frac{\mathcal{L}^{\prime}}{\sim}\cong\frac{\mathcal{L}\sqcup\mathcal{L}^{\prime}}{\mathcal{R}}$

which is the identity over representatives in $\mathcal{L}^{\prime}$.
We try to define a partial group structure on $C$; for the purpose, let
$q:\mathcal{L}^{\prime}\longrightarrow C$ be the canonical projection and
$D=D(C)$ be defined as in Section 2.3, so in this case as the set of words
$\overline{w}$ which are the pointwise projection via $q$ of a word $w\in
D(\mathcal{L}^{\prime})$. A word of caution: the element $\overline{w}$ is not
an equivalence class with representative $w$, but a word whose single entries
are elements of $C$, so equivalence classes with representatives in
$\mathcal{L}^{\prime}$. We can picture the situation through the following
diagram

${D(\mathcal{L}^{\prime})}$${D(C)}$${\mathcal{L}^{\prime}}$${C}$$\scriptstyle{\Pi}$$\scriptstyle{q^{*}}$$\scriptstyle{\pi}$$\scriptstyle{q}$

Commutativity of this diagram is a necessary condition since $q$ has to be a
morphism of partial groups. Thus the multivariable product
$\pi:D(C)\rightarrow C$ of $C$ has to be defined as $[w]\pi=(w)\Pi q$;
similarly, the inversion $l$ on $C$ must be given by $[w]l=(w)iq^{*}$.
That $l$ is well-defined is an immediate consequence of the fact that $f$ and
$g$ are partial group morphisms, so they send inverses to inverses; the
situation with $\pi$ is instead much more complex. The possibility of
factorizing $\Pi q$ via the map $q^{*}$ is equivalent to proving the
implication $(u)q^{*}=(v)q^{*}\implies u\Pi\sim v\Pi$, where $u,v\in
D(\mathcal{L}^{\prime})$. The relation $\sim$ is generated by the weaker
relation $\sim^{\prime}$ made exactly of the pairs $(xf,xg)$ for
$x\in\mathcal{L}$; being $fq=gq$ we get

$(xf)\Pi q=(x\Pi_{\mathcal{L}})fq=(x\Pi_{\mathcal{L}})gq=(xg)\Pi q,$

that is $(xf)\Pi\sim(xg)\Pi$. Combining with Proposition 2.1 we obtain that
$\pi$ is well defined on words of length 1.
Problems arise when considering words of length greater than 1; if we can
write $u=xf^{*}=(x_{1}f,\dots,x_{n}f)$ and $v=xg^{*}=(x_{1}g,\dots,x_{n}g)$
for $x=(x_{1},\dots,x_{n})\in D(\mathcal{L})$, then we clearly get

$u\Pi=(x)f^{*}\Pi=x\Pi_{\mathcal{L}}f\sim
x\Pi_{\mathcal{L}}g=(x)g^{*}\Pi=v\Pi.$

However if $u=(u_{1},\dots,u_{n})$ and $v=(v_{1},\dots,v_{n})$ are words in
$D(\mathcal{L}^{\prime})$ such that $uq^{*}=vq^{*}$, we only know that
$u_{i}\sim v_{i}$ for each $1\leq i\leq n$. So even if we are able to
represent $uq^{*}$ through an element of the form $\tilde{u}=((x_{i})h_{i})$,
where $h_{i}=f$ for some $i$ and $h_{j}=g$ for $j\neq i$, there is no
guarantee that $u\Pi\sim\tilde{u}\Pi$. To better understand the obstruction to
our factorization suppose $n=2$, so $u=(u_{1},u_{2})$ and $v=(v_{1},v_{2})$,
with $u_{1}=xf$, $v_{1}=xg$, $u_{2}=yf$ and $v_{2}=yg$. Clearly the pairs
$(xf,yf),(xf,yg),(xg,yg)$ and $(xg,yf)$ are all representatives of the same
word in $D(C)$; we have already observed that $(xf,yf)\Pi\sim(xg,yg)\Pi$,
however there is no evident reason, for example, for
$(xf,yf)\Pi\sim(xf,yg)\Pi$. The counterexample that follows is constructed by
following this exact line of reasoning.

######  Counterexample 2.4.

Consider the pointed set $(A,1)$ with $A=\\{1,a,b\\}$ and let $\mathcal{L}$ be
the free partial group on $(A,1)$; recall that
$\mathcal{L}=\\{1,a,\hat{a},b,\hat{b}\\}$. Consider the Klein group
$\displaystyle
M=\frac{\mathbb{Z}}{2\mathbb{Z}}\times\frac{\mathbb{Z}}{2\mathbb{Z}}$ ($(M,1)$
as a pointed-set) with generators $(x,1)=x$ and $(1,y)=y$ and define the
pointed-set functions

${\tilde{f}:}$${(A,1)}$${M}$${a}$${x}$${b}$${y}$${\tilde{g}:}$${(A,1)}$${M}$${a}$${xy}$${b}$${x}$

Then we get induced morphisms of partial groups
$f,g:\mathcal{L}\longrightarrow M$ making the diagram below commutative

${(A,1)}$${\mathcal{L}}$${M}$$\scriptstyle{\tilde{g}}$$\scriptstyle{\tilde{f}}$$\scriptstyle{f}$$\scriptstyle{g}$

We can therefore consider the coequalizer (taken in $Set^{*}$, i.e. after
application of the forgetful functor) of morphisms $f$ and $g$

${\mathcal{L}}$${M}$${\frac{M}{\sim}}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{q}$

with $(M/\sim\;,q)=coeq_{Set^{*}}(f,g)$. If $\Pi$ is the multivariable product
on $M$, we want to show that $\Pi q$ does not factor via $q^{*}$ to a product
of $M/\sim$. First of all note that

$\displaystyle x=af\sim ag=xy;$ $\displaystyle x=bg\sim bf=y;$

Since $af=\hat{a}f$, $bf=\hat{b}f$ and similarly for $g$, we get that $\sim$
identifies exactly the non-identity elements of $M$, leaving us with
$\displaystyle\frac{M}{\sim}\cong\frac{\mathbb{Z}}{2\mathbb{Z}}$, which has a
unique structure of partial group coinciding with that of the group. As words
$(af,bf)$ and $(af,bg)$ have equal image under $q^{*}$, they represent the
same word on $M/\sim$. However

$(af,bf)\Pi=(x,y)\Pi=xy,$ $(af,bg)\Pi=(x,x)\Pi=x^{2}=1,$

but $1\not\sim xy$ in $M/\sim$.

###  2.5. The category $Part$ is cocomplete

The situation depicted by the above counterexample is analogous to that
happening in the category of groups and group homomorphisms. When building
coequalizers in $Grp$, we need to factor out not only those relations afforded
by the morphisms in the diagram (namely $f$ and $g$), but the smallest _normal
subgroup_ containing those relations. Similarly, when considering a
coequalizer in $Part$, we need to quotient over a relation which allows the
multivariable product to factor through the quotient itself. The difficulties
of dealing with partial groups come from the lack of a substructure which
controls this factorization process in the same way as normal subgroups
control factor groups; therefore we need to identify the correct equivalence
relation in a different way.
Let’s consider any equivalence relation $\sim$ defined on the underlying set
of a partial group $\mathcal{L}^{\prime}$; let
$q:\mathcal{L}^{\prime}\longrightarrow\frac{\mathcal{L}^{\prime}}{\sim}$ be
the canonical projection and define
$D\left(\frac{\mathcal{L}^{\prime}}{\sim}\right)$ as in Section 2.4. Then the
fibers of the map $q^{*}:D(\mathcal{L}^{\prime})\longrightarrow
D\left(\frac{\mathcal{L}^{\prime}}{\sim}\right)$ induce an equivalence
relation $\equiv$ on $D(\mathcal{L}^{\prime})$ and clearly $q^{*}$ is
surjective being $q$ surjective. For $u=(u_{i})_{i},\,v=(v_{i})_{i}\in
D(\mathcal{L}^{\prime})$, the relation $\equiv$ is given by

$u\equiv v\iff u,v\in(u)(q^{*})^{-1}\iff u_{i}\sim v_{i}\;\forall i,$ (1)

in particular $u$ and $v$ have equal length. Hence we get

${D(\mathcal{L}^{\prime})}$${D\left(\frac{\mathcal{L}^{\prime}}{\sim}\right)=\frac{D(\mathcal{L}^{\prime})}{\equiv}}$${\mathcal{L}^{\prime}}$${\frac{\mathcal{L}^{\prime}}{\sim}}$$\scriptstyle{q^{*}}$$\scriptstyle{\Pi}$$\scriptstyle{q}$

The map $\Pi q$ is also surjective, so we can consider the equivalence
relation $\approx$ on $D(\mathcal{L}^{\prime})$ afforded by fibers of $\Pi q$
and obtain
$\frac{\mathcal{L}^{\prime}}{\sim}\cong\frac{D(\mathcal{L}^{\prime})}{\approx}$.
The square diagram above then becomes

${D(\mathcal{L}^{\prime})}$${D\left(\frac{\mathcal{L}^{\prime}}{\sim}\right)=\frac{D(\mathcal{L}^{\prime})}{\equiv}}$${\frac{\mathcal{L}^{\prime}}{\sim}=\frac{D(\mathcal{L}^{\prime})}{\approx}}$$\scriptstyle{q^{*}}$$\scriptstyle{\Pi
q}$

so that $\Pi q$ factors through $q^{*}$ if and only if
$\,\equiv\,\subseteq\,\approx$.
We are then led to search for an equivalence relation $\sim$ on
$\mathcal{L}^{\prime}$ such that the induced relations $\equiv$ and $\approx$
satisfy $\,\equiv\,\subseteq\,\approx$; we would then consider such a smallest
equivalence relation as a candidate for our colimit. Let’s also describe
$\approx$ in terms of $\sim$;

$u\approx v\iff(u)\Pi q=(v)\Pi q\iff u\Pi\sim v\Pi.$ (2)

Combining (1) and (2), we need to find $\sim$ such that, for
$u=(u_{i})_{i},v=(v_{i})_{i}\in D(\mathcal{L}^{\prime})$:

$u_{i}\sim v_{i}\,\forall i\implies u\Pi\sim v\Pi.$ (*)

Let now $\sim_{0}$ be the equivalence relation on $\mathcal{L}^{\prime}$
generated by the pairs of the form $(xf,xg)$ for some $x\in\mathcal{L}$ (this
is the equivalence relation affording the coequalizer in $Set$); clearly the
set

$E:=\\{\Re\subseteq\mathcal{L}^{\prime}\times\mathcal{L}^{\prime}\mid\Re\text{
is an eq. rel. containing $\sim_{0}$ and satisfying \eqref{condition of
sim}}\\}$

is not empty, since it contains
$\mathcal{L}^{\prime}\times\mathcal{L}^{\prime}$. Since inclusion of
$\sim_{0}$ and property $\eqref{condition of sim}$ are preserved by
intersections, there exists the smallest equivalence relation we are searching
for, namely

$\mathcal{R}=\bigcap_{\Re\in E}\Re.$

######  Lemma 2.5.

The category $Part$ has all coequalizers.

###### Proof.

Consider morphisms between partial groups
$f,g:\mathcal{L}\longrightarrow\mathcal{L}^{\prime}$, where
$\mathcal{L}=(\mathcal{L},D(\mathcal{L}),\Pi_{\mathcal{L}},i_{\mathcal{L}})$
and
$\mathcal{L}^{\prime}=(\mathcal{L}^{\prime},D(\mathcal{L}^{\prime}),\Pi,i)$;
build the setwise coequalizer $\frac{\mathcal{L}^{\prime}}{\sim}$ and the
relation $\mathcal{R}$ (which depends on $\sim$) on $\mathcal{L}^{\prime}$ as
above. Consider the projection
$t:\mathcal{L}^{\prime}\longrightarrow\frac{\mathcal{L}^{\prime}}{\mathcal{R}}$
and the diagram

${\mathcal{L}}$${\mathcal{L}^{\prime}}$${\frac{\mathcal{L}^{\prime}}{\mathcal{R}}}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{t}$

Then we prove that:

  1. (1)

$\mathcal{L}^{\prime}$ induces a structure of partial group on
$\frac{\mathcal{L}^{\prime}}{\mathcal{R}}$;

  2. (2)

$t$ is a morphism of partial groups and $ft=gt$;

  3. (3)

$\left(\frac{\mathcal{L}^{\prime}}{\mathcal{R}},t\right)=coeq_{Part}(f,g)$.

  1. (1)

Define $D:=D(\mathcal{L}^{\prime}/\mathcal{R})$ as in 2.3, that is as the set
of images of elements of $D(\mathcal{L}^{\prime})$ via the map
$t^{*}:W(\mathcal{L}^{\prime})\longrightarrow
W(\mathcal{L}^{\prime}/\mathcal{R})$. By construction $\mathcal{R}$ is the
smallest equivalence relation on $\mathcal{L}^{\prime}$ such that $\Pi t$
factors as in the following commutative diagram:

${D(\mathcal{L}^{\prime})}$${D\left(\frac{\mathcal{L}^{\prime}}{\mathcal{R}}\right)}$${\mathcal{L}^{\prime}}$${\frac{\mathcal{L}^{\prime}}{\mathcal{R}}}$$\scriptstyle{t^{*}}$$\scriptstyle{\Pi}$$\scriptstyle{\pi}$$\scriptstyle{t}$
($\top$)

For any $x\in\mathcal{L}^{\prime}$ set
$[x]\in\mathcal{L}^{\prime}/\mathcal{R}$ to be the class with representative
$x$ and let $\pi$ be the (unique) map that realizes such factorization, that
is $\Pi t=t^{*}\pi$. This means that for any word
$\overline{u}=([u_{1}],\dots,[u_{n}])\in D$ the representatives
$u_{1},\dots,u_{n}$ may be chosen so that $(u_{1},\dots,u_{n})\in
D(\mathcal{L})$ and then we have $(\overline{u})\pi=(u_{1},\dots,u_{n})\Pi t$.
If $i$ is the inversion map on $\mathcal{L}^{\prime}$, define an inversion map
$j$ on $\mathcal{L}^{\prime}/\mathcal{R}$ by $([x])j:=[(x)i]$. Since
$\mathcal{R}$ is by definition the smallest equivalence relation containing
$\sim$ and such that (* ‣ 2.5) holds, then it is generated by the pairs
$(x,y)\in\mathcal{L}^{\prime}\times\mathcal{L}^{\prime}$ with either $x=zf$
and $y=zg$ for some $z\in\mathcal{L}$ or $x=u\Pi$ and $y=v\Pi$ for some words
$u=(u_{k})_{k},\,v=(v_{k})_{k}\in D(\mathcal{L}^{\prime})$ of equal length
with $u_{k}\sim v_{k}$ for every $k$. By Lemma 2.1 we only need to check that
$j$ is constant over such pairs. In the first case, since $f$ and $g$ are
morphisms of partial groups we get $(x)i=((z)i_{\mathcal{L}})f$ and
$(y)i=((z)i_{\mathcal{L}})g$, so $((x)i,(y)i)\in\mathcal{R}$; in the other
case, since inverting $(u_{k})$ and $(v_{k})$ consists in taking inverses in
opposite order, we have
$(u\Pi)i=((u)i)\Pi\enspace\mathcal{R}\enspace((v)i)\Pi=(v\Pi)i$. Thus $j$ is
well defined.
We need to prove that $(\mathcal{L}^{\prime}/\mathcal{R},D,\pi,j)$ is a
partial group. Axiom (1) of Definition 1.1 is an immediate consequence of the
definition of $D$. For any $x\in\mathcal{L}^{\prime}$ we have
$([x])\pi=[(x)\Pi]=[x]$, which is axiom (2). Suppose now to have a word
$\overline{u}\circ\overline{v}\circ\overline{w}\in D$; then there exists a
lift of it, say $u\circ v\circ w\in D(\mathcal{L}^{\prime})$, via $t^{*}$. As
axiom (3) holds for $\mathcal{L}^{\prime}$, $u\circ v\Pi\circ w\in
D(\mathcal{L}^{\prime})$ and $(u\circ v\Pi\circ w)\Pi=(u\circ v\circ w)\Pi$;
but we also have $(u\circ v\Pi\circ
w)t^{*}=\overline{u}\circ\overline{v}\pi\circ\overline{w}$, so this is an
element of $D$ and
$(\overline{u}\circ\overline{v}\pi\circ\overline{w})\pi=(u\circ v\Pi\circ
w)t^{*}\pi=(u\circ v\Pi\circ w)\Pi t=(u\circ v\circ w)\Pi t=(u\circ v\circ
w)t^{*}\pi=(\overline{u}\circ\overline{v}\circ\overline{w})\pi$. This is axiom
(3) for $\mathcal{L}^{\prime}/\mathcal{R}$. Let now $\overline{u}\in D$ and
$u\in D(\mathcal{L}^{\prime})$ be a lift of $\overline{u}$ via $t^{*}$, then
$(u)i\circ u\in D(\mathcal{L}^{\prime})$ and $((u)i\circ
u)\Pi=1_{\mathcal{L}^{\prime}}$, yielding
$(\overline{u})j\circ\overline{u}=(u)it^{*}\circ(u)t^{*}=((u)i\circ u)t^{*}\in
D$ and $((\overline{u})j\circ\overline{u})\pi=(((u)i\circ
u)t^{*})\pi=((u)i\circ u)\Pi
t=(1_{\mathcal{L}^{\prime}})t=1_{\mathcal{L}^{\prime}/\mathcal{R}}$,
therefore, showing that also axiom (4) holds.

  2. (2)

By definition of $D$ and $t$, we have $(D(\mathcal{L}^{\prime}))t^{*}=D$;
moreover the product $\pi$ is defined so that the square ($\top$ ‣ 1) is
commutative. This shows that $t$ is a morphism of partial groups. As
$\sim\,\subseteq\,\mathcal{R}$, we also have $ft=gt$.

  3. (3)

If
$\tau:\mathcal{L}^{\prime}\longrightarrow\mathcal{M}=(\mathcal{M},D(\mathcal{M}),\Pi_{\mathcal{M}},i_{\mathcal{M}})$
is a partial group homomorphism such that $f\tau=g\tau$, define $\psi$ as in
the diagram below by:

$\psi:[x]\mapsto(x)\tau$
${\mathcal{L}}$${\mathcal{L}^{\prime}}$${\frac{\mathcal{L}^{\prime}}{\mathcal{R}}}$${\mathcal{M}}$$\scriptstyle{f}$$\scriptstyle{g}$$\scriptstyle{t}$$\scriptstyle{\tau}$$\scriptstyle{\psi}$

If $\psi$ is well defined, then it is clearly unique such that the diagram
above commutes. Consider now the following diagram, which is commutative since
$\tau$ is a morphism of partial groups.

${D(\mathcal{L}^{\prime})}$${D(\mathcal{M})}$${\mathcal{L}^{\prime}}$${\mathcal{M}}$$\scriptstyle{\tau^{*}}$$\scriptstyle{\Pi}$$\scriptstyle{\Pi_{\mathcal{M}}}$$\scriptstyle{\tau}$

Let $\mathcal{R}_{\tau}$ be the equivalence relation induced by the fibers of
$\tau$ on $\mathcal{L}^{\prime}$; it induces equivalence relations
$\equiv_{\tau}$ and $\approx_{\tau}$ on $D(\mathcal{L}^{\prime})$ via fibers
of, respectively, $\tau^{*}$ and $\Pi\tau$. The commutativity of the diagram
above yields the fact that $\mathcal{R}_{\tau}$ fulfills (* ‣ 2.5). Since
$f\tau=g\tau$ we also have that $\mathcal{R}_{\tau}$ contains all the pairs
$(xf,xg)$ with $x\in\mathcal{L}$. $\mathcal{R}$ being the smallest equivalence
relation on $\mathcal{L}^{\prime}$ satisfying these two properties, we have
$\mathcal{R}\subseteq\mathcal{R}_{\tau}$, which means exactly that $\psi$ is
well defined.

∎

We finally obtain

######  Theorem 2.6.

The category $Part$ is cocomplete.

###### Proof.

Combine Theorems 2.3, 2.5 and Proposition 2.2. ∎

######  Example 2.7.

We now compute the coequalizer $coeq(f,g)$ for the partial groups and maps
defined in Counterexample 2.4; in this specific case this computation is
particularly easy.
Recall that $\sim$ is the equivalence relation on $M$ such that $M/\sim$,
paired with the projection map, is the coequalizer of $f$ and $g$ in $Set$; we
have already seen that $|M/\sim|=2$ and also that $M/\sim$ cannot be the
coequalizer in $Part$. By Lemma 2.5 there exists an equivalence relation
$\mathcal{R}$ on $M$ such that $M/\mathcal{R}$ is the coequalizer of $f$ and
$g$ in the category of partial groups; in particular, such $\mathcal{R}$
satisfies $\sim\subseteq\mathcal{R}$. Then we get
$\sim\subsetneqq\mathcal{R}$, which implies $|M/\mathcal{R}|=1$; therefore
$coeq_{Part}(f,g)=M/\mathcal{R}=\\{1\\}$, the trivial (partial) group, paired
as usual with the projection over the quotient.

##  3\. Limits in $Part$

The situation regarding limits of partial groups is definitely simpler and A.
Chermak has already proven completeness of the category $Part$ in [Che22,
Appendix A]. For the purpose of collecting all results in one single paper, we
nonetheless provide a proof here. The reader who is already familiar with
Chermak’s proof may skip this section.
Let’s consider for a moment groups; there are a free-construction functor
$F:Set\longrightarrow Grp$ and a forgetful functor $U:Grp\longrightarrow Set$
such that $F\dashv U$, so in particular $U$ preserves limits, as it is a right
adjoint, . The situation with partial groups is the same: we have a free-
construction functor $F:Set^{*}\longrightarrow Part$ and a forgetful functor
$U:Part\longrightarrow Set^{*}$ such that $F\dashv U$. Hence a limit in $Part$
must have as underlying set the pointed setwise limit.

We now sketch a proof of existence of limits in $Part$.

######  Theorem 3.1.

The category Part is complete.

###### Proof.

Again we prove that $Part$ has all products and equalizers.

_Step 1._ If $\\{\mathcal{L}_{a}\\}_{a\in I}$ is a family of partial groups,
consider the setwise cartesian product $\mathcal{L}=\bigtimes\mathcal{L}_{a}$;
the projections $p_{a}:\mathcal{L}\rightarrow\mathcal{L}_{a}$ extend in the
usual way to $p_{a}^{*}:W(\mathcal{L})\rightarrow W(\mathcal{L}_{a})$. Define
$D(\mathcal{L})=D$ as

$D:=\\{w=((w_{a}^{1}),\dots,(w_{a}^{n}))_{a}\in W(\mathcal{L})\mid\forall a\in
I,wp_{a}^{*}=(w_{a}^{1},\dots,w_{a}^{n})\in D(\mathcal{L}_{a})\\}.$

and a product $\Pi$ on $\mathcal{L}$ by $(w)\Pi:=((wp_{a}^{*})\Pi_{a})_{a}$.
The inversion map $i$ on $\mathcal{L}$ is trivially defined componentwise; it
is a straightforward exercise to check that $(\mathcal{L},D,\Pi,i)$ is a
partial group. Considering the diagram below

${\mathcal{L}_{a}}$${\mathcal{L}}$${\mathcal{M}}$${\mathcal{L}_{b}}$$\scriptstyle{p_{a}}$$\scriptstyle{p_{b}}$$\scriptstyle{t}$$\scriptstyle{t_{a}}$$\scriptstyle{t_{b}}$

$t$ is defined by $t:m\mapsto(mt_{a})_{a}$. Notice that, for
$w=(m_{1},\dots,m_{n})\in D(\mathcal{M})$, we have
$wt_{a}^{*}=(m_{1}t_{a},\dots,m_{n}t_{a})\in D(\mathcal{L}_{a})$ for all $a\in
I$, so $wt\in D$. Clearly $t$ is a morphism of partial groups, unique
satisfying the universal property of products.

_Step 2._ Consider morphisms of partial groups
$f,g:\mathcal{L}\rightarrow\mathcal{L}^{\prime}$; define

$\mathcal{E}:=\\{x\in\mathcal{L}\mid xf=xg\\}.$

We have

${\mathcal{E}}$${\mathcal{L}}$${\mathcal{L}^{\prime};}$$\scriptstyle{j}$$\scriptstyle{f}$$\scriptstyle{g}$

it is a simple exercise to prove that $\mathcal{E}$ is a partial group, $j$ a
partial groups morphism and that the pair $(E,j)=eq(f,g)$.

As a consequence of Proposition 2.2 we obtain that $Part$ is complete. In
addition, it is now a trivial observation the fact that the full subcategory
of $Part$ whose objects are the finite partial groups is finitely complete. ∎

###### Proof of Theorem A.

This is obtained by combining Theorems 2.6 and 3.1 and the remark after Lemma
2.3. ∎

###  3.1. Partial groups as simplicial sets

It has been initially conjectured by C. Broto and then formalized in joint
work with A. González that partial groups carry a natural structure of
simplicial sets. Indeed it turns out that $Part$ is equivalent to a full
subcategory of the category $sSet$ of simplicial sets, as described in
[Gon15]. We wish to show that such subcategory is not closed under formation
of colimits in $sSet$, so the well-understood structure of colimits of
simplicial sets cannot be used for computing colimits of partial groups. We
begin with recalling some definitions and elementary results; more details may
be found in [GJ99].
Let $\Delta$ be the category of finite partially ordered sets with monotone,
non-decreasing functions as morphisms. It is possible to show that $\Delta$ is
equivalent to the category whose objects are the finite ordered sequences
$[n]=\\{0<1<\dots<n\\}$ and whose morphisms are all non-decreasing functions.
This is known as the skeletal subcategory; in the following we will restrict
ourselves to this subcategory and denote it as well by $\Delta$.

######  Definition 3.1.

A simplicial set is a functor

$X:\Delta^{op}\longrightarrow Set;$

hence we may think at a simplicial set as a sequence of sets $X_{n}:=[n]X$
with morphisms induced by the monotone functions in $\Delta$ taken with
opposite direction.
The elements of $X_{0}$ are called _vertices_ of the simplicial set $X$; if
$X_{0}=\\{v\\}$ (i.e. $X$ has a single vertex), then $X$ is said to be
_reduced_.

Simplicial sets together with natural transformations form a category, usually
denoted by $sSet$. The structure of colimits of simplicial sets is well known;
if $T:\mathcal{C}\longrightarrow sSet$ is a diagram over the small category
$\mathcal{C}$, then

$colim(T):\Delta^{op}\longrightarrow Set$

is the simplicial set with
$colim(T)_{n}=\underset{c\in\mathcal{C}}{colim}\;[n](cT)$ and suitable
morphism induced by the universal property of colimits (we will not need to
study in detail the morphisms).
In [Gon15, Theorem 4.8] González proves that partial groups, when regarded as
simplicial sets, are characterized by being reduced, among other properties.
Therefore, if $\mathcal{L}$ and $\mathcal{M}$ are partial groups, viewed as
simplicial sets they are reduced, i.e. $\mathcal{L}_{0}=\\{\star\\}$ and
$\mathcal{M}_{0}=\\{*\\}$; however the coproduct computed in $sSet$ is not
reduced, since
$(\mathcal{L}\sqcup\mathcal{M})_{0}=\mathcal{L}_{0}\sqcup\mathcal{M}_{0}=\\{\star,*\\}$,
thus it cannot be a partial group.

##  4\. Quotients of partial groups

It is common to control algebraic structures by means of a generating set and
a set of relations. The generating process is normally related to the
existence of free objects, whereas considering relations means looking at
colimits. We now want to investigate this situation for partial groups. First
of all we need to define quotients of partial groups; such a notion was
previously defined only for specific subclasses, namely for groups and for the
so-called _localities_ (see [Che22, Definition 2.7 and Sections 3 and 4]); in
order to define quotients for all partial groups we will take advantage of the
categorical point of view, generalizing what happens with groups and group
homomorphisms. We will then show that our definition coincides with that given
by A. Chermak for localities.
We start with defining substructures; the names assigned to them follow the
notation used by Chermak in [Che22] and [Che17].

######  Definition 4.1.

Let $\mathcal{L}=(\mathcal{L},D,\Pi,i)$ be a partial group and
$\mathcal{H}\subseteq\mathcal{L}$ a non-empty subset. For any
$g\in\mathcal{L}$ write $D(g):=\\{x\in\mathcal{L}\mid(g^{-1},x,g)\in D)\\}$.

  * (a)

The quadruple
$(\mathcal{H},E_{\mathcal{H}},\Pi_{E_{\mathcal{H}}},i_{\mathcal{H}})$ is an
_impartial subgroup_ of $\mathcal{L}$ if $E_{\mathcal{H}}\subseteq D\cap
W(\mathcal{H})$, $\Pi_{E_{\mathcal{H}}}$ and $i_{\mathcal{H}}$ are the
restrictions of, respectively, $\Pi$ to $E_{\mathcal{H}}$ and $i$ to
$\mathcal{H}$ and if
$(\mathcal{H},E_{\mathcal{H}},\Pi_{E_{\mathcal{H}}},i_{\mathcal{H}})$ is a
partial group.

  * (b)

An impartial subgroup
$\mathcal{H}=(\mathcal{H},E_{\mathcal{H}},\Pi_{E_{\mathcal{H}}},i_{\mathcal{H}})$
of $\mathcal{L}$ is a _partial subgroup_ if $E_{\mathcal{H}}=D\cap
W(\mathcal{H})$. In this case $E_{\mathcal{H}}$ is completely determined by
the subset $\mathcal{H}$ and the partial group $\mathcal{L}$.

  * (c)

$\mathcal{H}$ is a _partial normal subgroup_ of $\mathcal{L}$ if it is a
partial subgroup of $\mathcal{L}$ and, for every $x\in\mathcal{H}$ and
$g\in\mathcal{L}$ such that $x\in D(g)$, $(g^{-1},x,g)\Pi\in\mathcal{H}$
holds.

Note that
$(\mathcal{H},E_{\mathcal{H}},\Pi_{E_{\mathcal{H}}},i_{\mathcal{H}})$ being a
partial group in (a) implies that $E_{\mathcal{H}}$ is closed under
application of $\Pi_{E_{\mathcal{H}}}$ and also $\mathcal{H}$ is closed under
$i_{\mathcal{H}}$. The next results will make clear that impartial subgroups
are the correct substructure to consider from the categorical point view;
however it is often better to work with partial subgroups since they have
better properties. According to Chermak’s definition in [Che17]

  * (a’)

$\mathcal{H}$ is an impartial subgroup of $(\mathcal{L},D,\Pi,i)$ if
$\mathcal{H}$ is the homomorphic image of a partial group homomorphism
$f:\mathcal{M}\longrightarrow\mathcal{L}$ for some partial group
$(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$.

With regard to (a’), the homomorphic image $\mathcal{I}:=im(f)$ becomes a
partial group
$(\mathcal{I},D_{\mathcal{I}},\Pi_{\mathcal{I}},i_{\mathcal{I}})$ by setting
$D_{\mathcal{I}}:=(D_{\mathcal{M}})f^{*}$ and defining
$\Pi_{\mathcal{I}}:=\Pi|_{\mathcal{I}}$ and
$i_{\mathcal{I}}:=i|_{\mathcal{I}}$, respectively the restrictions of $\Pi$ to
$D_{\mathcal{I}}$ and of $i$ to $\mathcal{I}$. Details may be found in the
proof of the following Lemma.

######  Lemma 4.1.

The definitions (a) and (a’) of an impartial subgroup are equivalent.

###### Proof.

Note that, if $\mathcal{H}$ is an impartial subgroup of a partial group
$(\mathcal{L},D,\Pi,i)$ according to (a), then the inclusion map
$i:\mathcal{H}\longrightarrow\mathcal{L}$ is a morphism of partial groups;
thus (a) implies (a’).
Conversely, if $\mathcal{H}=im(f)$ for some partial group
$(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$ and
homomorphism $f:\mathcal{M}\longrightarrow\mathcal{L}$, then we have
$D_{\mathcal{H}}:=(D_{\mathcal{M}})f^{*}\subseteq W(\mathcal{H})\cap
D_{\mathcal{L}}$. Taking restrictions $\Pi_{\mathcal{H}}$ of $\Pi$ to
$D_{\mathcal{H}}$ and $i_{\mathcal{H}}$ of $i$ to $\mathcal{H}$, we need to
show that $(\mathcal{H},D_{\mathcal{H}},\Pi_{\mathcal{H}},i_{\mathcal{H}})$ is
a partial group. Since $\mathcal{M}\subseteq D_{\mathcal{M}}$, we have
$\mathcal{H}=(\mathcal{M})f\subseteq D_{\mathcal{H}}$; moreover suppose that
$u=(u_{1},\dots,u_{k}),\,v=(v_{1},\dots,v_{l})\in W(\mathcal{H})$ are such
that $u\circ v\in D_{\mathcal{H}}$, then there exists
$m=(m_{1},\dots,m_{k},n_{1},\dots,n_{l})\in D_{\mathcal{M}}$ such that $u\circ
v=(m)f^{*}$. In particular,
$m^{\prime}:=(m_{1},\dots,m_{k}),n^{\prime}:=(n_{1},\dots,n_{l})\in
D_{\mathcal{M}}$ and $u=(m^{\prime})f^{*},\,v=(n^{\prime})f^{*}$, so that
$u,v\in D_{\mathcal{H}}$. This proves axiom (1) of Definition 1.1. Axiom (2)
is trivial, since the product is defined by restricting that of $\mathcal{L}$.
Let now $u\circ v\circ w\in D_{\mathcal{H}}$ with
$u=(u_{1},\dots,u_{k}),\,v=(v_{1},\dots,v_{l})$ and $w=(w_{1},\dots,w_{r})$;
by axiom (1) $u,v,w\in D_{\mathcal{H}}$ and there exist
$m^{\prime}:=(m_{1},\dots,m_{k}),\,n^{\prime}:=(n_{1},\dots,n_{l}),\,t^{\prime}:=(t_{1},\dots,t_{r})\in
D_{\mathcal{M}}$ so that $m^{\prime}\circ n^{\prime}\circ t^{\prime}\in
D_{\mathcal{M}}$ and
$u=(m^{\prime})f^{*},\,v=(n^{\prime})f^{*},\,w=(t^{\prime})f^{*}$. Since
$m^{\prime}\circ(n^{\prime})\Pi_{\mathcal{M}}\circ t^{\prime}\in
D_{\mathcal{M}}$, we have $u\circ(v)\Pi_{\mathcal{H}}\circ
w=(m^{\prime})f^{*}\circ(n^{\prime})f^{*}\Pi\circ(t^{\prime})f^{*}=(m^{\prime}\circ(n^{\prime})\Pi_{\mathcal{M}}\circ
t^{\prime})f^{*}\in D_{\mathcal{H}}$. The equality
$(u\circ(v)\Pi_{\mathcal{H}}\circ w)\Pi_{\mathcal{H}}=(u\circ v\circ
w)\Pi_{\mathcal{H}}$ follows at once, recalling that these are, in particular,
elements of $D$ and $\Pi_{\mathcal{H}}$ is the restriction of $\Pi$. Let now
$u$ and $m^{\prime}$ be as above; then $(m^{\prime})i_{\mathcal{M}}\circ
m^{\prime}\in D_{\mathcal{M}}$ and it has product $1_{\mathcal{M}}$, thus
$((m^{\prime})i_{\mathcal{M}}\circ m^{\prime})f^{*}=(u)i_{\mathcal{H}}\circ
u\in D_{\mathcal{H}}$ and $((u)i_{\mathcal{H}}\circ
u)\Pi_{\mathcal{H}}=((m^{\prime})i_{\mathcal{M}}\circ
m^{\prime})f^{*}\Pi=((m^{\prime})i_{\mathcal{M}}\circ
m^{\prime})\Pi_{\mathcal{M}}f=(1_{\mathcal{M}})f=1_{\mathcal{L}}=1_{\mathcal{H}}$.
This completes the proof. ∎

In particular, the impartial subgroups of a partial group $\mathcal{L}$ are,
up to isomorphism, exactly the partial groups embedded in $\mathcal{L}$ and
play the same role of subgroups of a group; for example, they provide a
factorization of a morphism in an epimorphism followed by a monomorphism (the
inclusion).
With respect to quotients, we look at what happens with groups from the
categorical point of view. If $G$ is a group and $H$ a subgroup of $G$, it is
well known that $H$ affords a quotient structure if and only if it is normal.
Nevertheless, it is possible to _kill_ any subgroup $H$ through a quotient at
the price of replacing $H$ with its the normal closure, that is $N:=\langle
H^{G}\rangle$. Since the category $Grp$ is cocomplete and has a $0$-object,
namely the trivial group, one can build the trivial map $\hat{1}$ between any
two groups $G$ and $H$ by passing through the trivial group,

${H}$${G}$${1}$$\scriptstyle{\hat{1}}$$\scriptstyle{\hat{1}}$$\scriptstyle{\hat{1}}$

In particular, if $H\leq G$ and $i:H\longrightarrow G$ is the inclusion map,
then it is a straightforward computation checking that
$coker(i)=coeq(i,\hat{1})=G/N$, where $N:=\langle H^{G}\rangle$, so the
following is a coequalizer diagram

${H}$${G}$${\frac{G}{N}}$$\scriptstyle{i}$$\scriptstyle{\hat{1}}$$\scriptstyle{q}$

Thus it is reasonable to generalize the notion of a quotient to any subgroup
$H$ by setting $G/H:=coeq_{Grp}(i,\hat{1})=coker(i)=G/\langle H^{G}\rangle$,
$i:H\longrightarrow G$ being the inclusion map and $\hat{1}$ the trivial map.
This line of reasoning extends directly to partial groups since also $Part$
has all limits and colimits and is pointed, the trivial group $1$ being the
$0$-object. Denoting the trivial map always by
$\hat{1}:\mathcal{L}\longrightarrow\mathcal{M}$, for every morphism
$f:\mathcal{L}\longrightarrow\mathcal{M}$ one then has $ker(f)=eq(f,\hat{1})$
and $coker(f)=coeq(f,\hat{1})$. It is immediate to verify that our notion of
kernel of a morphism coincides with the one given by Chermak in [Che22,
Definition 1.11]; we are then led to the following definition.

######  Definition 4.2.

Let $\mathcal{L}$ be a partial group and $\mathcal{H}\subseteq\mathcal{L}$ be
an impartial subgroup. Let $j:\mathcal{H}\longrightarrow\mathcal{L}$ be the
inclusion morphism. Define the quotient partial group
$\mathcal{L}/\mathcal{H}$ by

$\mathcal{L}/\mathcal{H}:=coker(j)=coeq_{Part}(j,\hat{1}),$

where $\hat{1}$ is the trivial map from $\mathcal{H}$ to $\mathcal{L}$.

In particular we can factor any morphism of partial groups
$f:\mathcal{L}\longrightarrow\mathcal{M}$ as an epi followed by a mono, that
is

${f:}$${\mathcal{L}}$${\mathcal{M}}$${im(f)}$

and it is a simple exercise to see that $coker(f)=\mathcal{M}/im(f)$.

It is clear that, when $\mathcal{H}$ and $\mathcal{L}$ are groups, the
definition above coincides with the known quotient in $Grp$. We now show that
it also coincides with the definition of a _quotient locality_ given by
Chermak in [Che22]. Chermak’s construction applies over a finite locality
$\mathcal{L}$ (see [Che22, Definition 2.7]) and a partial normal subgroup
$\mathcal{N}\trianglelefteq\mathcal{L}$; in [Che22, Section 3] the author
defines an equivalence relation on $\mathcal{L}$ by identifying the maximal
classes (named the _cosets_ of $\mathcal{N}$) of a reflexive and transitive
relation. In particular, [Che22, Proposition 3.14] shows that cosets are all
of the form $\mathcal{N}f$ for some element $f\in\mathcal{L}$.

######  Notation.

We follow the notation adopted in [Che22, Definition 3.6 and Corollary 4.5].
Thus $\mathcal{L}=(\mathcal{L},\Delta,S)$ is a finite locality and
$\mathcal{N}\leq\mathcal{L}$ a partial normal subgroup,
$\overline{\mathcal{L}}:=\mathcal{L}/\mathcal{N}$ is the set of
$\mathcal{N}$-cosets of $\mathcal{L}$ and
$\rho:\mathcal{L}\longrightarrow\overline{\mathcal{L}}$ the projection map;
just for the following lemma the notation $\mathcal{L}/\mathcal{N}$ is used
exclusively for Chermak’s construction. Setting
$\overline{\Delta}:=\\{(P)\rho=:\overline{P}\mid P\in\Delta\\}$, [Che22, Lemma
3.16 and Theorem 4.3] show that
$(\overline{\mathcal{L}},\overline{\Delta},\overline{S})$ is a locality.

######  Lemma 4.2.

Following the notation above, if $\mathcal{L}=(\mathcal{L},\Delta,S)$ is a
finite locality, $\mathcal{N}\leq\mathcal{L}$ a partial normal subgroup of
$\mathcal{L}$ and $\rho:\mathcal{L}\longrightarrow\mathcal{L}/\mathcal{N}$ the
projection sending each element $g\in\mathcal{L}$ to the unique maximal coset
of $\mathcal{N}$ containing $g$, then the locality
$(\overline{\mathcal{L}}=\mathcal{L}/\mathcal{N},\overline{\Delta},\overline{S})$
is the quotient $coker(\mathcal{N}\hookrightarrow\mathcal{L})$ as in
Definition 4.2.

###### Proof.

Recall that also $coker(\mathcal{N}\hookrightarrow\mathcal{L})$ is defined by
setting an equivalence relation on $\mathcal{L}$. Since the domains, product
maps and inversion maps of $\mathcal{L}/\mathcal{N}$ and
$coker(\mathcal{N}\hookrightarrow\mathcal{L})$ are in both cases induced by
those of $\mathcal{L}$ ([Che22, Lemma 3.16]), all we need to show is that
$\mathcal{L}/\mathcal{N}=coker(\mathcal{N}\hookrightarrow\mathcal{L})$ as sets
(that is, the equivalence relations are the same) and $\rho$ is the projection
map also in the coequalizer diagram defining
$coker(\mathcal{N}\hookrightarrow\mathcal{L})$. Let $\approx$ be the
equivalence relation on $\mathcal{L}$ with classes $\mathcal{L}/\mathcal{N}$
and $\mathcal{R}$ be the one defining
$coker(\mathcal{N}\hookrightarrow\mathcal{L})$; since the product of
$\mathcal{L}$ factors over $\mathcal{L}/\mathcal{N}$, by definition of
$\mathcal{R}$ we have $\mathcal{R}\subseteq\approx$.
Suppose now that $\mathcal{E}$ is an equivalence class of $\approx$; by
[Che22, Proposition 3.14] there exists an element $f\in\mathcal{L}$ such that
$\mathcal{E}=\mathcal{N}f$. Let $g\in\mathcal{N}f$, so $g=nf=(n,f)\Pi$ for
some $n\in\mathcal{N}$; since the words $(n,f)$ and $(1,f)$ are clearly
componentwise $\mathcal{R}$-related, by the definition of $\mathcal{R}$ we
also have $g=(n,f)\Pi\;\mathcal{R}\;(1,f)\Pi=f$. This shows that
$\mathcal{E}=\mathcal{N}f$ is contained in an equivalence class of
$\mathcal{R}$. Thus $\approx\subseteq\mathcal{R}$, so equality holds.
That $\rho$ is also the quotient map over
$coker(\mathcal{N}\hookrightarrow\mathcal{L})$ is now trivial. ∎

We end this section showing that, even though impartial subgroups are the
natural categorical subojects in $Part$, they are after all redundant for the
purpose of computing quotients.

######  Lemma 4.3.

Let $\mathcal{L}=(\mathcal{L},D,\Pi,i)$ be a partial group,
$\mathcal{N}=(\mathcal{N},E,\Pi_{\mathcal{N}},i_{\mathcal{N}})$ be an
impartial subgroup of $\mathcal{L}$ and
$\mathcal{M}=(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$
be the partial subgroup of $\mathcal{L}$ generated by the subset $\mathcal{N}$
in the sense of [Che22, Lemma 1.8(e) and 1.9]. Then
$\mathcal{L}/\mathcal{N}\cong\mathcal{L}/\mathcal{M}$.

###### Proof.

We have an obvious inclusion morphism
$i:\mathcal{N}\hookrightarrow\mathcal{M}$, so we may consider the following
diagram

${\mathcal{N}}$${\mathcal{M}}$${\mathcal{L}}$${\mathcal{L}/\mathcal{M}}$${\mathcal{L}/\mathcal{N}}$$\scriptstyle{i}$$\scriptstyle{i_{\mathcal{M}}}$$\scriptstyle{q_{\mathcal{N}}}$$\scriptstyle{q_{\mathcal{M}}}$$\scriptstyle{\psi}$$\scriptstyle{\varphi}$

with $ii_{\mathcal{M}}=i_{\mathcal{N}}$, the inclusion morphism of
$\mathcal{N}$ in $\mathcal{L}$. Since
$i_{\mathcal{N}}q_{\mathcal{M}}=ii_{\mathcal{M}}q_{\mathcal{M}}=\hat{1}$, by
the universal property of the cokernel
$(\mathcal{L}/\mathcal{N},q_{\mathcal{N}})$ there exists a unique map
$\varphi:\mathcal{L}/\mathcal{N}\longrightarrow\mathcal{L}/\mathcal{M}$ such
that $q_{\mathcal{M}}=q_{\mathcal{N}}\varphi$. Now [Che22, Lemma 1.9] shows
that $\mathcal{M}=\bigcup_{i\geq 0}\mathcal{N}_{i}$, with
$\mathcal{N}_{0}:=\mathcal{N}$ and $\mathcal{N}_{i}:=\\{(n)\Pi\mid n\in
W(\mathcal{N}_{i-1})\cap D\\}$; if $m\in\mathcal{N}_{0}$, then
$(m)q_{\mathcal{N}}=1_{\mathcal{L}/\mathcal{N}}$. By induction, let $i\geq 1$
and $m\in\mathcal{N}_{i}$, then $m=(m^{\prime})\Pi$ for some $m^{\prime}\in
W(\mathcal{N}_{i-1})\cap D$ and
$(m)q_{\mathcal{N}}=((m^{\prime})\Pi)q_{\mathcal{N}}=((m^{\prime})q_{\mathcal{N}}^{*})\Pi_{\mathcal{L}/\mathcal{N}}=(m^{\prime})q_{\mathcal{N}}=1_{\mathcal{L}/\mathcal{N}}$.
This shows that $i_{\mathcal{M}}q_{\mathcal{N}}=\hat{1}$, so the universal
property of $\mathcal{L}/\mathcal{M}$ yields the existence of a unique
morphism $\psi:\mathcal{L}/\mathcal{M}\longrightarrow\mathcal{L}/\mathcal{N}$
such that $q_{\mathcal{M}}\psi=q_{\mathcal{N}}$.
Now $q_{\mathcal{N}}\varphi\psi=q_{\mathcal{M}}\psi=q_{\mathcal{N}}$ and
$q_{\mathcal{M}}\psi\varphi=q_{\mathcal{N}}\varphi=q_{\mathcal{M}}$, so the
uniqueness property yields $\varphi\psi=id_{\mathcal{L}/\mathcal{N}}$ and
$\psi\varphi=id_{\mathcal{L}/\mathcal{M}}$, hence
$\mathcal{L}/\mathcal{N}\cong\mathcal{L}/\mathcal{M}$. ∎

##  5\. Generators and relations for partial groups

This section is mostly dedicated to the proof of Proposition B. First of all,
we shall address the fact that the behavior of partial groups can be
counterintuitive; for example, in [Che13, Section 5] and [Che16, Chapter 3]
Chermak describes a way of expanding a given locality $\mathcal{L}$, the
result being a new locality $\mathcal{L}^{+}$ having the following properties:
$\mathcal{L}$ is an impartial subgroup of $\mathcal{L}^{+}$ and
$\mathcal{L}^{+}$ is generated by the set $\mathcal{L}$. In particular,
whenever $\mathcal{L}\subset\mathcal{L}^{+}$ we get a proper embedding between
two partial groups sharing an equal set of generators, namely $\mathcal{L}$.
This happens since the set of generated elements depends not only on the
generators, but also on the product or, better, on the domains
$D(\mathcal{L})$ and $D(\mathcal{L}^{+})$. Another important aspect to keep in
mind is the following: the free partial group over a set $X$ is very small if
compared to the free group over $X$; for example, if $X$ is finite, the free
partial group over $X$ is also finite, however the only finite free group is
the trivial group, free over the empty set. This is again a consequence of the
structure of the domain of a free partial group; being such domain very small,
the product is defined only over few words.
The idea now consists in providing additional information controlling the size
of the domain; this information will correspond to defining a forgetful
functor $U$ from $Part$ to a suitable category, which, for a given partial
group $(\mathcal{L},D(\mathcal{L}),\Pi,i)$, remembers the domain
$D(\mathcal{L})$ together with the set $\mathcal{L}$. It turns out that such a
functor must remember also the inversion map $i$ in order to obtain good
categorical properties. We start with the following definition.

######  Definition 5.1.

For any set $X$, denote as usual by $W(X)$ the free monoid over $X$. Define
the category $Set^{s}$ in the following way.

  * •

The objects of $Set^{s}$ are triples $(X,S_{X},i_{X})$, where $X$ is a set,
$S_{X}\subseteq W(X)$ and $i_{X}:X\longrightarrow X$ is an involutory
bijection.

  * •

A morphism $h:(X,S_{X},i_{X})\longrightarrow(Y,S_{Y},i_{Y})$ between two
objects is given by a set-wise map $h:X\longrightarrow Y$ that satisfies the
property $(S_{X})h^{*}\subseteq S_{Y}$ and makes the following diagram
commutative.

${X}$${Y}$${X}$${Y}$$\scriptstyle{h}$$\scriptstyle{i_{X}}$$\scriptstyle{i_{Y}}$$\scriptstyle{h}$
($\ddagger$)

  * •

Composition of morphisms is the usual composition of maps.

The set $S_{X}$ should be thought of as the set of words over which we wish to
have a product defined; we then define a forgetful functor with the following
shape:

$U:Part\longrightarrow Set^{s},\qquad(\mathcal{L},D,\Pi,i)U=(\mathcal{L},D,i)$

Indeed, whenever we have a morphism of partial groups
$f:\mathcal{L}\longrightarrow\mathcal{L}^{\prime}$, by definition we have
$(D)f^{*}\subseteq D^{\prime}$ where $D$ and $D^{\prime}$ are the domains of
$\mathcal{L}$, respectively $\mathcal{L}^{\prime}$, so that we can define
$fU=f$; since $f$ respects the inversion map, the diagram $\ddagger$ ‣ • ‣ 5.1
is commutative. The first question to answer is whether this newly defined
forgetful functor $U$ is a right adjoint; that is to say, whether it is
possible to build free partial groups over objects of $Set^{s}$.

###  5.1. Construction of free partial groups over $Set^{s}$

Let $(X,S_{X},i_{X})$ be an object of the category $Set^{s}$, then we identify
the free partial group over $(X,S_{X},i_{X})$ inside a suitable quotient of
the free group with set of generators $X$. For any object $(X,S_{X},i_{X})\in
Ob(Set^{s})$ let $W:=W(X)$ be the free monoid on $X$ and $G(X)$ the free group
on $X$. Set $G_{X}:=G(X)/R(X)$, where $R(X)$ is the normal closure of the
subgroup of $G(X)$ generated by the relations $(x)i_{X}\circ
x=\emptyset=x\circ(x)i_{X}$ for all $x\in X$; we will often denote an element
$g\in G_{X}$ simply by a representative $(x_{1},x_{2},\dots,x_{n})\in G(X)$.
If $x\in X$, denote by $\hat{x}$ its inverse in $G(X)$ (this notation will be
used throughout the rest of the paper) and note that we get
$R(X)\hat{x}=R(X)(x)i_{X}$ in $G_{X}$; then extend $i_{X}$ to an involutory
bijection of $G(X)$ by
$(y_{1},\dots,y_{n})i_{X}:=((y_{n})i_{X},\dots,(y_{1})i_{X})$ where
$y_{i}\in\\{x_{i},\hat{x}_{i}\\}$ for $x_{i}\in X$.

######  Lemma 5.1.

Let $X$ be a set, $i_{X}$ an involutory bijection on $X$ and $G(X)$ and
$G_{X}$ be as above. Define the monoid homomorphism $t:W(X)\longrightarrow
G(X)$ by $t:x\mapsto x$ for all $x\in X$. Then $i_{X}$ factors modulo $R(X)$
and affords the inversion map of the group $G_{X}$. Moreover

  * (a)

if $q:G(X)\longrightarrow G_{X}$ is the quotient map, then the composition
$\mu:=tq:W(X)\longrightarrow G_{X}$ is a homomorphism of monoids;

  * (b)

$\mu$ is surjective and $(1_{G_{X}})\mu^{-1}=R(X)\cap W(X)$; in particular, if
$u,v\in W(X)$ are such that $(u)\mu=(v)\mu$, then $u\circ(v)i_{X}\in R(X)\cap
W(X)$.

###### Proof.

The subgroup $R(X)$ is $i_{X}$-invariant by definition; in $G_{X}$ we have
$(R(X)(x_{1},\dots,x_{n}))i_{X}=R(X)((x_{1})i_{X},\dots,(x_{n})i_{X})=R(X)(\hat{x_{1}},\dots,\hat{x_{n}})$,
showing that $i_{X}$ is the inversion map. Now (a) is trivial, as $q$ is a
group homomorphism. Let $w=(w_{1},\dots,w_{n})\in G(X)$ be a representative of
$R(X)w\in G_{X}$; then each $w_{i}$ is either of the form $x_{i}$ or of the
form $\hat{x_{i}}$ for some $x_{i}\in X$. Define

$w^{\prime}_{i}:=\begin{cases*}w_{i}&\text{if $w_{i}\in X$}\\\
(w_{i})i_{X}&\text{if $w_{i}\not\in X$}\end{cases*}$

and set $w^{\prime}:=(w^{\prime}_{1},\dots,w^{\prime}_{n})$; in particular
$R(X)w_{i}=R(X)w^{\prime}_{i}$ for every $i\in\\{1,\dots,n\\}$. Then
$w^{\prime}\in W(X)$ and $R(X)w=R(X)w^{\prime}$, showing that $\mu$ is
surjective.
Note that $t$ is injective, as $W(X)$ contains no non-trivial inverses of
elements of $G(X)$, so $(1_{G_{X}})\mu^{-1}=R(X)\cap W(X)$ holds. Take now
$u=(u_{1},\dots,u_{n}),\,v=(v_{1},\dots,v_{m})\in W(X)$ with $(u)\mu=(v)\mu$,
then we get

$(u_{1},\dots,u_{n})\circ(v_{1},\dots,v_{m})i_{X}\in R(X)\cap W(X)$

which is (b). ∎

The morphism $\mu$ induces also a map

${\nu:S_{X}}$${W(G_{X})}$${(x_{1},\dots,x_{n})}$${((x_{1})\mu,\dots,(x_{n})\mu)=(R(X)x_{1},\dots,R(X)x_{n}),}$

which satisfies $\nu\Pi_{G_{X}}=\mu$; we will often write simply
$S_{X}\subseteq W(G_{X})$ in place of $(S_{X})\nu\subseteq W(G_{X})$.
We need the following lemma.

######  Lemma 5.2.

Let $(\mathcal{L},D,\Pi,i)$ be a partial group and
$(\mathcal{L}_{k},D_{k},\Pi_{k},i_{k})_{k\in K}$ a family of impartial
subgroups of $\mathcal{L}$ indexed over a set $K$. Set

$\mathcal{M}:=\bigcap_{k\in K}\mathcal{L}_{k},$

then there is a unique structure of partial group over $\mathcal{M}$, namely
$(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$, such that
the following are satisfied:

  * (i)

$(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$ is an
impartial subgroup of $\mathcal{L}$,

  * (ii)

if
$\mathcal{N}=(\mathcal{N},D_{\mathcal{N}},\Pi_{\mathcal{N}},i_{\mathcal{N}})$
is an impartial subgroup of each of the $\mathcal{L}_{k}$, then it is an
impartial subgroup of $\mathcal{M}$.

We define $\mathcal{M}$ as the _intersection_ of the impartial subgroups
$\mathcal{L}_{k}$ in $\mathcal{L}$.

###### Proof.

Let $D_{\mathcal{M}}:=\bigcap_{k\in K}D_{k}$,
$\Pi_{\mathcal{M}}:=\Pi|_{D_{\mathcal{M}}}$ and
$i_{\mathcal{M}}:=i|_{D_{\mathcal{M}}}$; we prove that
$(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$ has the
desired properties. Property (i) clearly follows once we show that
$(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$ is a partial
group. According to our definitions, $\mathcal{M}\subseteq D_{k}$ for every
$k\in K$, so $\mathcal{M}\subseteq D_{\mathcal{M}}$. Let $u\circ v\in
D_{\mathcal{M}}$, then $u\circ v\in D_{k}$ for every $k\in K$, so $u,v\in
D_{k}$ for all $k$; this gives $u,v\in D_{\mathcal{M}}$ and axiom (1) of the
definition of a partial group holds. Axiom (2) is trivial, since
$\Pi_{\mathcal{M}}$ is defined as the restriction of $\Pi$. Suppose now that
$u\circ v\circ w\in D_{\mathcal{M}}$, then $u\circ v\circ w\in D_{k}$ for all
$k\in K$, so

$u\circ(v)\Pi_{k}\circ w\in D_{k}\quad\text{and}\quad(u\circ(v)\Pi_{k}\circ
w)\Pi_{k}=(u\circ v\circ w)\Pi_{k}\qquad\forall k\in K$

Since $\Pi_{\mathcal{M}}=\Pi|_{D_{\mathcal{M}}}=\Pi_{k}|_{D_{\mathcal{M}}}$
for every $k\in K$, we have

$u\circ(v)\Pi_{\mathcal{M}}\circ w\in
D_{k}\quad\text{and}\quad(u\circ(v)\Pi_{\mathcal{M}}\circ
w)\Pi_{\mathcal{M}}=(u\circ v\circ w)\Pi_{\mathcal{M}}$

which is axiom (3). Let now $u\in D_{\mathcal{M}}$, then again $u\in D_{k}$
for every $k\in K$ and therefore

$(u)i_{k}\circ u\in D_{k}\quad\text{and}\quad((u)i_{k}\circ
u)\Pi_{k}=1_{\mathcal{L}}\qquad\forall k\in K.$

Thus, reasoning as for axiom (3) we get again

$(u)i_{\mathcal{M}}\circ u\in
D_{\mathcal{M}}\quad\text{and}\quad((u)i_{\mathcal{M}}\circ
u)\Pi_{\mathcal{M}}=1_{\mathcal{L}}=1_{\mathcal{M}}$

which is axiom (4). This proves (i).
Suppose now that $\mathcal{N}$ is as in (ii), then for every $k\in K$

$D_{\mathcal{N}}\subseteq
D_{k},\quad\Pi_{\mathcal{N}}=\Pi_{k}|_{\mathcal{N}}=\Pi_{\mathcal{M}}|_{D_{\mathcal{N}}}\quad\text{and}\quad
i_{\mathcal{N}}=i_{k}|_{D_{\mathcal{N}}}=i_{\mathcal{M}}|_{D_{\mathcal{N}}},$

which implies that the inclusion map $\mathcal{N}\hookrightarrow\mathcal{M}$
is a morphism of partial groups. This proves (ii). ∎

We follow the notation introduced in Section 5.1

######  Lemma 5.3.

Suppose that $(X,S_{X},i_{X})\in Set^{s}$, then there exist unique sets
$\mathcal{L}\subseteq G_{X}$ and $D\subseteq W(\mathcal{L})\subseteq W(G_{X})$
such that:

  1. (1)

$X\subseteq\mathcal{L}$, $S_{X}\subseteq D$ and
$(\mathcal{L})i_{G_{X}}\subseteq\mathcal{L}$;

  2. (2)

denoting $\Pi:=\Pi_{G_{X}}|_{D}$ and $j=i_{G_{X}}|_{\mathcal{L}}$, then
$(D)\Pi\subseteq\mathcal{L}$ and $(\mathcal{L},D,\Pi,j)$ is a partial group;
in particular, for any word $s=(x_{1},\dots,x_{n})\in S_{X}$ we have
$(s)\Pi\in\mathcal{L}$;

  3. (3)

any other pair of sets $\mathcal{L}^{\prime}\subseteq G_{X}$,
$D^{\prime}\subseteq W(\mathcal{L}^{\prime})$ satisfying (1) and (2) is such
that $(\mathcal{L},D,\Pi,j)$ is an impartial subgroup of
$(\mathcal{L}^{\prime},D^{\prime},\Pi^{\prime},j^{\prime})$, where
$\Pi^{\prime}$ and $j^{\prime}$ are the restrictions of, respectively,
$\Pi_{G_{X}}$ to $D^{\prime}$ and $i_{G_{X}}$ to $\mathcal{L}^{\prime}$.

###### Proof.

We show that the group $G_{X}$ (so with $W(G_{X})$ as domain) satisfies (1)
and (2) and, afterwards, that properties (1) and (2) are stable under taking
intersections of impartial subgroups.
For $G_{X}$ we have $\Pi=\Pi_{G_{X}}$, $j=i_{G_{X}}$ and
$(G_{X},W(G_{X}),\Pi,j)$ clearly satisfies (1) and (2). Suppose now to have a
family $\\{\mathcal{L}_{k}\\}_{k\in K}$ of subsets $\mathcal{L}_{k}\subseteq
G_{X}$ with associated subsets $D_{k}\subseteq W(\mathcal{L}_{k})$, indexed
over an index-set $K$, such that all pairs $(\mathcal{L}_{k},D_{k})_{k\in K}$
satisfy (1) and (2) and set $\Pi_{k}:=\Pi_{G_{X}}|_{D_{k}}$ and
$j_{k}:=i_{G_{X}}|_{\mathcal{L}_{k}}$. Then by (2) each
$(\mathcal{L}_{k},D_{k},\Pi_{k},j_{k})$ is an impartial subgroup of $G_{X}$.
Let
$\mathcal{M}=(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$
be the intersection impartial subgroup of $G_{X}$, taken over all the
$\\{\mathcal{L}_{k}\\}_{k\in K}$, as defined in Lemma 5.2. Clearly
$X\subseteq\mathcal{M}$, $S_{X}\subseteq D_{\mathcal{M}}$ and
$(\mathcal{M})i_{\mathcal{M}}\subseteq\mathcal{M}$; as
$(D_{k})\Pi_{k}\subseteq\mathcal{L}_{k}$ for every $k\in K$, we also have
$(D_{\mathcal{M}})\Pi_{\mathcal{M}}\subseteq\mathcal{M}$. Thus $\mathcal{M}$
satisfies properties (1) and (2).
Let now $\\{\mathcal{L}_{k}\\}_{k\in K}$ be the set of all impartial subgroups
$(\mathcal{L}_{k},D_{k},\Pi_{k},j_{k})$ of $G_{X}$ satisfying (1) and (2) and
$(\mathcal{L},D,\Pi,j)$ be the intersection impartial subgroup of $G_{X}$
taken over the family $\\{\mathcal{L}_{k}\\}$. Then it is a partial group
satisfying (1) and (2) and, by definition, it satisfies also property (3). ∎

What we have proven is, actually, that there exists a unique smallest
impartial subgroup of $G_{X}$ satisfying the properties (1) and (2).
Note that a morphism $f:(X,S_{X},i_{X})\longrightarrow(Y,S_{Y},i_{Y})$ in
$Set^{s}$ induces a map $\tilde{f}:G_{X}\longrightarrow G_{Y}$ in the
following way. First we build $f^{+}:G(X)\longrightarrow G(Y)$ by the
universal property of the free group $G(X)$, as in the following diagram

${X}$${G(X)}$${G(Y)}$$\scriptstyle{f}$$\scriptstyle{f^{+}}$

and note that $f^{+}$ extends the map $f^{*}:W(X)\longrightarrow W(Y)$. Then
$f^{+}$ is a group homomorphism and the properties of morphisms in $Set^{s}$
yield the commutativity of the diagram $\ddagger$ ‣ • ‣ 5.1, so that
$(R(X))f^{+}\subseteq R(Y)$ holds; thus we get a well-defined group
homomorphism $\tilde{f}$ induced by $f^{+}$.

######  Definition 5.2.

Let $(X,S_{X},i_{X}),(Y,S_{Y},i_{Y})$ be objects of $Set^{s}$ and
$f:(X,S_{X},i_{X})\longrightarrow(Y,S_{Y},i_{Y})$ be a morphism; let also
$\mathcal{L}=(\mathcal{L},D,\Pi,j)$ be the unique smallest impartial subgroup
of $G_{X}$ satisfying (1) and (2) of Lemma 5.3 with respect to
$(X,S_{X},i_{X})$. Define the functor $F:Set^{s}\longrightarrow Part$ by

$F:(X,S_{X},i_{X})\mapsto(\mathcal{L},D,\Pi,j)\quad\text{and}\quad
fF:=\tilde{f}|_{\mathcal{L}},$

where $\tilde{f}:G_{X}\longrightarrow G_{Y}$ is the map defined above.

It is a straightforward computation to verify that $F$ is a functor; as a
consequence of the next lemma, we consider $F$ the free-construction functor.
In order to simplify notation we simply write $X$ for an object
$(X,S_{X},i_{X})\in Set^{s}$ whenever the context leaves no ambiguity.

######  Lemma 5.4.

The following hold.

  * (a)

Let $(X,S_{X},i_{X})$ be an object in $Set^{s}$, then the partial group
$(X)F=:(\mathcal{L},D,\Pi,j)$ defined in Lemma 5.3 is the free partial group
over the object $(X,S_{X},i_{X})$.

  * (b)

The forgetful functor $U:Part\longrightarrow
Set^{s},\,(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})U=(\mathcal{M},D_{\mathcal{M}},i_{\mathcal{M}})$
is a right adjoint to the functor $F:Set^{s}\longrightarrow Part$.

###### Proof.

  * (a)

We follow the notation of Lemma 5.3. Consider the inclusion map
$\iota:(X,S_{X},i_{X})\hookrightarrow(X)FU=(\mathcal{L},D,j)$ naturally
induced by the inclusion $X\subseteq\mathcal{L}$; note here that $\iota^{*}$
over $S_{X}$ is the map $\nu$ defined after Lemma 5.1. Being the free partial
group over the object $(X,S_{X},i_{X})$ means that for any other partial group
$\mathcal{M}=(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$
and any morphism
$f:(X,S_{X},i_{X})\longrightarrow(\mathcal{M})U=(\mathcal{M},D_{\mathcal{M}},i_{\mathcal{M}})$
in the category $Set^{s}$ there exists a unique morphism of partial groups
$\tilde{f}:\mathcal{L}\longrightarrow\mathcal{M}$ such that
$f=\iota\cdot\tilde{f}U$, that is that makes the following diagram
commutative.

${(X,S_{X},i_{X})}$${(\mathcal{L},D,j)}$${(\mathcal{M},D_{\mathcal{M}},i_{\mathcal{M}})}$$\scriptstyle{\iota}$$\scriptstyle{\forall
f}$$\scriptstyle{\tilde{f}U}$ ($\Theta$)

Let $\mu:W(X)\longrightarrow G(X)$ be the monoid homomorphism defined in 5.1,
consider the following diagram

${W(X)}$${W(\mathcal{M})\supseteq
D_{\mathcal{M}}}$${G_{X}}$$\scriptstyle{f^{*}}$$\scriptstyle{\mu}$

and define $\mathcal{L}_{0}:=((D_{\mathcal{M}})(f^{*})^{-1})\mu\subseteq
G_{X}$. Since $\mu$ commutes with $i_{X}$ and $f$ is a morphism of $Set^{s}$,
the inversion map on $G_{X}$ restricts to an involutory bijection on
$\mathcal{L}_{0}$. Denote $G_{X}=(G_{X},W(G_{X}),\Pi_{G_{X}},i_{G_{X}})$ and
set

$D_{0}:=\\{w=(w_{1},\dots,w_{n})\in W(\mathcal{L}_{0})\mid\exists
w^{\prime}_{i}\in(w_{i})\mu^{-1}\text{ s.t. }w^{\prime}_{1}\circ\dots\circ
w^{\prime}_{n}\in(D_{\mathcal{M}})(f^{*})^{-1}\\};$

we show that $(\mathcal{L}_{0},D_{0},\Pi_{0},i_{0})$ is a partial group, where
$\Pi_{0}$ and $i_{0}$ are the restrictions to $D_{0}$, respectively
$\mathcal{L}_{0}$, of $\Pi_{G_{X}}$ and $i_{G_{X}}$. Note that, $\mu$ being a
monoid homomorphism, for any $w\in D_{0}$ there exists
$w^{\prime}\in(D_{\mathcal{M}})(f^{*})^{-1}$ such that
$(w)\Pi_{0}=(w)\Pi_{G_{X}}=(w^{\prime})\mu$; thus
$(D_{0})\Pi_{0}\subseteq\mathcal{L}_{0}$. Let now
$u=(u_{1},\dots,u_{k}),\,v=(v_{1},\dots,v_{r}),\,w=(w_{1},\dots,w_{s})\in
W(\mathcal{L}_{0})$; suppose that $u\circ v\in D_{0}$ and take preimages
$u^{\prime}_{i},v^{\prime}_{j}\in W(X)$ along $\mu$ with
$u^{\prime}_{1}\circ\dots\circ u^{\prime}_{k}\circ
v^{\prime}_{1}\circ\dots\circ v^{\prime}_{r}\in(D_{\mathcal{M}})(f^{*})^{-1}$,
then $D_{\mathcal{M}}\ni(u^{\prime}_{1}\circ\dots\circ u^{\prime}_{k}\circ
v^{\prime}_{1}\circ\dots\circ
v^{\prime}_{r})f^{*}=(u^{\prime}_{1}\circ\dots\circ
u^{\prime}_{k})f^{*}\circ(v^{\prime}_{1}\circ\dots\circ v^{\prime}_{r})f^{*}$.
Hence $(u^{\prime}_{1}\circ\dots\circ u^{\prime}_{k})f^{*}\in D_{\mathcal{M}}$
as well as $(v^{\prime}_{1}\circ\dots\circ v^{\prime}_{r})f^{*}\in
D_{\mathcal{M}}$, so $u^{\prime}_{1}\circ\dots\circ
u^{\prime}_{k},\,v^{\prime}_{1}\circ\dots\circ
v^{\prime}_{r}\in(D_{\mathcal{M}})(f^{*})^{-1}$, which implies $u,v\in D_{0}$.
Axioms (1) and (2) of Definition 1.1 now easily follow. In the same way one
proves axioms (3) and (4).
By restricting $f^{*}$ to the preimage $(D_{\mathcal{M}})(f^{*})^{-1}$ we may
define the composition $f^{*}\Pi_{\mathcal{M}}$ as in the diagram below

${(D_{\mathcal{M}})(f^{*})^{-1}}$${D_{\mathcal{M}}}$${\mathcal{M}}$${\mathcal{L}_{0}}$$\scriptstyle{f^{*}}$$\scriptstyle{\mu}$$\scriptstyle{\Pi_{\mathcal{M}}}$

Suppose there exist words
$(x_{1},\dots,x_{n}),\,(y_{1},\dots,y_{m})\in(D_{\mathcal{M}})(f^{*})^{-1}$
with $(x_{1},\dots,x_{n})\mu=(y_{1},\dots,y_{m})\mu$, then by Lemma 5.1(b)
$(x_{1},\dots,x_{n})\circ((y_{1},\dots,y_{m})i_{X})\in R(X)\cap W(X)$.
Moreover,
$(x_{1},\dots,x_{n})\mu\circ((y_{1},\dots,y_{m})\mu)i_{X}=(x_{1},\dots,x_{n},(y_{m})i_{X},\dots,(y_{1})i_{X})\mu\in
D_{0}$. Thus $(x_{1},\dots,x_{n})\mu\circ((y_{1},\dots,y_{m})i_{X})\mu$ may be
represented by an element in $R(X)\cap
W(X)\cap(D_{\mathcal{M}})(f^{*})^{-1})$. Since
$(R(X)\cap(D_{\mathcal{M}})(f^{*})^{-1})f^{*}\Pi_{\mathcal{M}}=\\{1_{\mathcal{M}}\\}$,
we get
$(x_{1},\dots,x_{n})f^{*}\Pi_{\mathcal{M}}=(y_{1},\dots,y_{m})f^{*}\Pi_{\mathcal{M}}$,
so $f^{*}\Pi_{\mathcal{M}}$ (restricted to $(D_{\mathcal{M}})(f^{*})^{-1}$)
factors via $\mu$ to a map
$\tilde{f}_{0}:\mathcal{L}_{0}\longrightarrow\mathcal{M}$. An immediate
consequence of the definition of $D_{0}$ is that
$(D_{0})\tilde{f}_{0}^{*}\subseteq D_{\mathcal{M}}$, whereas the commutativity
of the diagram

${D_{0}}$${\mathcal{L}_{0}}$${D_{\mathcal{M}}}$${\mathcal{M}}$$\scriptstyle{\tilde{f}_{0}^{*}}$$\scriptstyle{\Pi_{0}}$$\scriptstyle{\tilde{f}_{0}}$$\scriptstyle{\Pi_{\mathcal{M}}}$

follows from the definition of $\tilde{f}_{0}$, which is therefore a morphism
of partial groups.
Clearly $X\subseteq\mathcal{L}_{0}$ and $S_{X}\subseteq D_{0}$; then Lemma 5.3
shows that also $\mathcal{L}$ is embedded in $\mathcal{L}_{0}$ and $D\subseteq
D_{0}$, so we may restrict $\tilde{f}_{0}$ to a morphism

$\tilde{f}:\mathcal{L}\longrightarrow\mathcal{M},$

which makes ($\Theta$) commutative by construction. Uniqueness of $\tilde{f}$
follows since any other morphism of partial groups $g$ making ($\Theta$)
commutative has to agree with $\tilde{f}$ over every element of $X$ in
$\mathcal{L}$ and therefore induces a map $g^{*}:W(X)\longrightarrow
W(\mathcal{M})$ equal to $f^{*}$.

  * (b)

The way of proceeding is exactly as in Lemma 1.1, so we only sketch the proof.
Let $X=(X,S_{X},i_{X}),\,Y=(Y,S_{Y},i_{Y})\in Ob(Set^{s})$ and
$\mathcal{L}=(\mathcal{L},D_{\mathcal{L}},\Pi_{\mathcal{L}},i_{\mathcal{L}}),\mathcal{M}=(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$
be partial groups. The bijections

$\psi_{X,\mathcal{L}}:Hom_{Set^{s}}(X,(\mathcal{L})U)\longrightarrow
Hom_{Part}((X)F,\mathcal{L})$

are a consequence of (a). Let $h:Y\longrightarrow X$ be a morphism in
$Set^{S}$ and $g:\mathcal{L}\longrightarrow\mathcal{M}$ a morphism in $Part$,
then we must show that the following diagram is commutative

${Hom_{Set^{*}}(X,(\mathcal{L})U)}$${Hom_{Part}((X)F,\mathcal{L})}$${Hom_{Set^{*}}(Y,(\mathcal{M})U)}$${Hom_{Part}((Y)F,\mathcal{M})}$$\scriptstyle{\psi_{(X,\mathcal{L})}}$$\scriptstyle{\psi_{(Y,\mathcal{M})}}$

where the vertical morphisms are pre- and post-composition. Denoting
$(f)\psi_{(X,\mathcal{L})}:=\tilde{f}$ (and similarly for the entire
collection of maps), it means showing

$hF\cdot\tilde{f}\cdot g=\widetilde{h\cdot f\cdot gU}$

and the argument here relies again on looking at the action on elements of
$Y$.

∎

###  5.2. Relations in partial groups

If $G$ is a group and $x,y\in G$, it is always possible to rewrite a relation
$x=y$ in the form $xy^{-1}=1_{G}$; all relations are indeed identifications of
elements with the unit $1_{G}$. Given a set $R$ of relations on $G$, one can
identify the smallest normal subgroup $N$ of $G$ containing $R$ and form $G/N$
and one says that the relations in $R$ are satisfied by $G/N$. This line of
reasoning is, however, not so straightforward for partial groups. Let
$\mathcal{L}=(\mathcal{L},D,\Pi,i)$ be a partial group, $x,y\in\mathcal{L}$
and suppose that we would like to find a quotient of $\mathcal{L}$ fulfilling
the relation $x=y$: a priori there is no reason why the word $x\circ y^{-1}$
should be in $D$, however assuming that we can form such a quotient of
$\mathcal{L}$, axiom (4) of Definition 1.1 yields $x\circ x^{-1}\in D$ and
$(x\circ x^{-1})\Pi=1$, so the relation $x=y$ becomes equivalent to $x\circ
y^{-1}\in D$ and $(x\circ y^{-1})\Pi=1$. This shows that adding the element
$x\circ y^{-1}$ to the domain of $\mathcal{L}$ is a necessary condition for
the relation $x=y$ to make sense.
Observe that relations on a group $G$ can equivalently be defined as
equalities $w=u$ between words $w,u\in W(G)$. This suggests the following
definition.

######  Definition 5.3.

Let $\mathcal{L}=(\mathcal{L},D,\Pi,i)$ be a partial group, then a _relation_
on $\mathcal{L}$ is an element of $D$.

We now show that every partial group is a quotient of a free partial group.

###### Proof of Proposition B.

Let $\mathcal{L}=(\mathcal{L},D,\Pi,i)$ be a partial group and
$U:Part\longrightarrow
Set^{s},\,(\mathcal{L},D,\Pi,i)\mapsto(\mathcal{L},D,i)$ the forgetful
functor. Set
$\mathcal{M}=(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$
to be the free partial group over $(\mathcal{L},D,i)$, according to Lemmas 5.3
and 5.4, and define the set of relations of $\mathcal{L}$ as

$R_{0}:=\\{u\in D\mid(u)\Pi=1_{\mathcal{L}}\\}.$

Since $D\subseteq D_{\mathcal{M}}$ by the definition of $\mathcal{M}$, the
canonical inclusion $\mathcal{L}\hookrightarrow\mathcal{M},\,x\mapsto x$
extends to a map
$\phi:D\mapsto\mathcal{M},\,u\mapsto(u)\phi:=(u)\Pi_{\mathcal{M}}$; then set
$R:=(R_{0})\phi$ and let
$\mathcal{N}=(\mathcal{N},D_{\mathcal{N}},\Pi_{\mathcal{N}},i_{\mathcal{N}})$
be the partial subgroup of $\mathcal{M}$ generated by $R$ in the sense of
[Che22, Lemmas 1.8(e) and 1.9]. We wish to prove that $\mathcal{L}$ is
isomorphic to the quotient $\mathcal{M}/\mathcal{N}$.
The universal property of the free object $\mathcal{M}$ yields a map
$f:\mathcal{M}\longrightarrow\mathcal{L}$ extending the identity over
$\mathcal{L}$ (in particular, $f$ is surjective); let
$i:\mathcal{N}\longrightarrow\mathcal{M}$ be the inclusion map and consider
the pair $(\mathcal{M}/\mathcal{N},q)=coker(i)$, as in the following diagram

${\mathcal{N}}$${\mathcal{L}}$${\mathcal{M}}$${\mathcal{M}/\mathcal{N}}$${\mathcal{L}}$$\scriptstyle{i}$$\scriptstyle{id_{\mathcal{L}}}$$\scriptstyle{f}$$\scriptstyle{q}$$\scriptstyle{\hat{f}}$

The map $f$ acts in the following way: suppose that $u\in\mathcal{M}$ and
write it by representatives in $\mathcal{L}$, so $u=u_{1}\circ\dots\circ
u_{n}$ with $u_{i}\in\mathcal{L}$, then we may think at $u\in D_{\mathcal{M}}$
since $u=(u_{1},\dots,u_{n})\Pi_{\mathcal{M}}$, and $u=(u)f^{*}\in D$ and
$(u)f=(u)\Pi$, since $f$ extends the identity map. In particular, every
element of $\mathcal{M}$ can be identified with an element of $D$. The
morphism $\hat{f}$ is induced by the universal property of the cokernel
$\mathcal{M}/\mathcal{N}$, since for every element $v\in R$,
$(v)f=(v)\Pi=1_{\mathcal{L}}$ and $R$ generates $\mathcal{N}$. As $f$ is
surjective, $\hat{f}$ is so as well.
Clearly $(D(\mathcal{M}/\mathcal{N}))\hat{f}^{*}\subseteq D$ as $\hat{f}$ is a
morphism in $Part$; if $u\in D$, viewing $u=(u)f^{*}\in D$ together with the
commutativity of the diagram above (namely $f=q\hat{f}$) yields
$(D(\mathcal{M}/\mathcal{N}))\hat{f}^{*}=D$. Hence we are only left with
showing injectivity of $\hat{f}$; note that for any $u\in D$ axioms (3) and
(4) of the definition of a partial group yield $(u)i\circ(u)\Pi\in D$ and
$((u)i\circ(u)\Pi)\Pi=1_{\mathcal{L}}$, so that recalling $D\subseteq
D_{\mathcal{M}}$

$u\circ(u)i\Pi\circ(u)\Pi\in
D\quad\text{and}\quad(u\circ(u)i\Pi\circ(u)\Pi)\Pi_{\mathcal{M}}=u$

Moreover, $u\circ(u)i\Pi\in\mathcal{N}$. Suppose now that $u,v\in\mathcal{M}$
are such that $(u)f=(v)f$; we have already seen that this means
$(u)\Pi=(v)\Pi$ (viewing $u,v$ as elements of $D$), so that

$u\circ(u)i\Pi\circ(u)\Pi\quad\text{and}\quad v\circ(v)i\Pi\circ(v)\Pi$

can be regarded as words of $D_{\mathcal{M}}$ of length two, namely
$(u\circ(u)i\Pi,(u)\Pi)$ and $(v\circ(v)i\Pi,(v)\Pi)$. Let $\mathcal{R}$ be
the equivalence relation defining $\mathcal{M}/\mathcal{N}$ as in Section 2.5,
then $(u\circ(u)i\Pi)\;\mathcal{R}\;(v\circ(v)i\Pi)$, since both are in
$\mathcal{N}$, and $(u)\Pi=(v)\Pi$. As

$(u\circ(u)i\Pi\circ(u)\Pi)\Pi_{\mathcal{M}}=u\quad\text{and}\quad(v\circ(v)i\Pi\circ(v)\Pi)\Pi_{\mathcal{M}}=v$

by * ‣ 2.5 in Section 2.5 we get $u\;\mathcal{R}\;v$, that is $u=v$ in
$\mathcal{M}/\mathcal{N}$. This completes the proof. ∎

In general, given a partial group $\mathcal{L}=(\mathcal{L},D,\Pi,i)$, and a
subset $S\subseteq W(\mathcal{L})$, one can freely add the relations in $S$ in
the following way. Set $R:=\\{w\in D\mid(w)\Pi=1_{\mathcal{L}}\\}$ and build
the free partial group $\mathcal{M}^{+}=(\mathcal{M}^{+},D^{+},\Pi^{+},i^{+})$
over $(\mathcal{L},D\cup S,i)$; if $\mathcal{N}^{+}\subseteq\mathcal{M}^{+}$
is the partial subgroup of $\mathcal{M}^{+}$ generated by elements of $R$
(seen as elements of $\mathcal{M}^{+}$ as in the previous lemma), then the
domain of the quotient $\mathcal{M}^{+}/\mathcal{N}^{+}$ contains the words in
$S$. In particular, we have the following properties.

######  Lemma 5.5.

Let $\mathcal{L}$, $S$, $\mathcal{M}^{+}$, $R$ and $\mathcal{N}^{+}$ be as
above, then $\mathcal{M}^{+}/\mathcal{N}^{+}$ contains an isomorphic copy of
$\mathcal{L}$. If $\mathcal{N}^{\prime}$ is the partial subgroup of
$\mathcal{M}^{+}$ generated by the relations in $R\cup S$, then
$\mathcal{N}^{\prime}/\mathcal{N}^{+}$ is embedded in
$\mathcal{M}^{+}/\mathcal{N}^{+}$ and $\mathcal{M}^{+}/\mathcal{N}^{\prime}$
is isomorphic to
$(\mathcal{M}^{+}/\mathcal{N}^{+})/(\mathcal{N}^{\prime}/\mathcal{N}^{+})$.

###### Proof.

Let
$\mathcal{M}=(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$
be the free partial group over $(\mathcal{L},D,i)$ and $\mathcal{N}$ the
partial subgroup of $\mathcal{M}$ generated by $R$. Then Proposition B shows
that $\mathcal{M}/\mathcal{N}\cong\mathcal{L}$; consider the diagram below

${(\mathcal{L},D,i)}$${\mathcal{M}}$${\mathcal{M}/\mathcal{N}}$${(\mathcal{L},D\cup
S,i)}$${\mathcal{M}^{+}}$${\mathcal{M}^{+}/\mathcal{N}^{+}}$$\scriptstyle{q}$$\scriptstyle{j}$$\scriptstyle{\iota}$$\scriptstyle{q^{+}}$

where $j$ is induced by the universal property of the free group $\mathcal{M}$
and $\iota$ by the universal property of the cokernel
$\mathcal{M}/\mathcal{N}$, since $i_{\mathcal{N}}jq^{+}=\hat{1}$ (as
$(\mathcal{N})j\subseteq\mathcal{N}^{+}$) for $i_{\mathcal{N}}$ the inclusion
of $\mathcal{N}$ in $\mathcal{M}$. Since $\mathcal{N}$ and $\mathcal{N}^{+}$
are generated by the same elements, $\mathcal{N}$ is an impartial subgroup of
$\mathcal{M}^{+}$ and $\mathcal{N}^{+}$ the partial subgroup generated by
$\mathcal{N}$. Lemma 4.3 shows that
$\mathcal{M}^{+}/\mathcal{N}\cong\mathcal{M}^{+}/\mathcal{N}^{+}$, so $\iota$
is the identity morphism.
Consider now

${\mathcal{N}^{\prime}}$${\mathcal{N}^{\prime}/\mathcal{N}^{+}}$${\mathcal{M}^{+}}$${\mathcal{M}^{+}/\mathcal{N}^{+}}$${(\mathcal{M}^{+}/\mathcal{N}^{+})\diagup(\mathcal{N}^{\prime}/\mathcal{N}^{+})}$${\mathcal{M}^{+}/\mathcal{N}^{\prime}}$$\scriptstyle{h}$$\scriptstyle{q^{+}|_{\mathcal{N}^{\prime}}}$$\scriptstyle{h^{\prime}}$$\scriptstyle{q^{+}}$$\scriptstyle{g}$$\scriptstyle{t}$$\scriptstyle{\eta}$

where $t=coker(h^{\prime})$ and $g=coker(h)$, whereas the map $\eta$ is
induced by the universal property of the cokernel
$\mathcal{M}^{+}/\mathcal{N}^{+}$. Since $h^{\prime}\eta=\hat{1}$ and
$hq^{+}t=\hat{1}$, the universal properties of
$(\mathcal{M}^{+}/\mathcal{N}^{+})\diagup(\mathcal{N}^{\prime}/\mathcal{N}^{+})$
and $\mathcal{M}^{+}/\mathcal{N}^{\prime}$ yield maps
$\psi,\phi:(\mathcal{M}^{+}/\mathcal{N}^{+})\diagup(\mathcal{N}^{\prime}/\mathcal{N}^{+})\rightleftarrows\mathcal{M}^{+}/\mathcal{N}^{\prime}$
which are one inverse of the other, proving the wanted isomorphism. ∎

###  5.3. Chermak’s elementary expansions revisited

As an example, we show that Chermak’s construction in [Che13, Section 5],
renamed _elementary expansion_ in [Che16, Section 3], is a special case of a
construction made in terms of generators and relations; this was the
fundamental step which allowed Chermak to build the centric linking system
associated to a given saturated fusion system $\mathcal{F}$ and so eventually
defining the classifying space of $\mathcal{F}$. Some notions about fusion
systems together with a fairly broad background about localities are here
required; we will in particular use results from [Che13, Section 2, pages
57-64] or, equivalently, from [Che22, Sections 1 and 2]. As a note to the
reader, [Che16, Section 3] will be considered part of the background needed
for the example; we will indeed adopt the same notation.

For the entire section $\mathcal{L}=(\mathcal{L},\Delta,S)$ is a locality over
the fusion system $\mathcal{F}$ in the sense of [Che13, Remark 2.8 (2)], or
equivalently as in [Che16, Section 2], and write also
$\mathcal{L}=(\mathcal{L},D_{\mathcal{L}},\Pi,i)$. For every $g\in\mathcal{L}$
Chermak defines $S_{g}:=\\{x\in S\mid(g^{-1},x,g)\in
D_{\mathcal{L}},\,x^{g}\in S\\}$; in general, denote by $c_{g}$ the
conjugation map by the element $g\in\mathcal{L}$ and for every word
$w=(w_{1},\dots,w_{n})\in W(\mathcal{L})$ write

$S_{w}:=\\{x\in S\mid c_{w_{1}}\cdot\dots\cdot c_{w_{n}}\text{ is defined on
$x$ and maps $x$ into $S$}\\}.$

The goal consists in building a new locality
$\mathcal{L}^{+}=(\mathcal{L}^{+},D^{+},\Pi^{+},i^{+})$ which has the same
associated fusion system $\mathcal{F}$, but a larger set $\Delta^{+}$ of
objects. Therefore suppose we wish to add the subgroup $R\leq S$ to the set of
objects, then the definition of objectivity, as in [Che13, Definition 2.6],
implies that we need to add also the following:

  * (i)

all the elements in $R^{\mathcal{F}}$ to $\Delta$;

  * (ii)

new elements to $\mathcal{L}$, which afford conjugations among the newly added
objects.

This suggests to define $\Delta^{+}$ as the closure of $\Delta\cup
R^{\mathcal{F}}$ in the set of subsets of subgroups of $S$, with respect to
taking $\mathcal{F}$-conjugates and overgroups. We then build
$\mathcal{L}^{+}=(\mathcal{L}^{+},D^{+},\Pi^{+},i^{+})$ as a quotient of the
free partial group
$\mathcal{M}=(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$
over the object $(\mathcal{L},T,i)\in Set^{s}$, with $T$ being a suitable
subset $T\subset W(\mathcal{L})$ and $i$ the inversion map of $\mathcal{L}$.
Under few additional conditions (which were identified by Chermak) the partial
group $\mathcal{L}^{+}$ ends up possessing the structure of a locality of the
form $(\mathcal{L}^{+},\Delta^{+},S)$. The correct idea in order to identify
$T$ is due to Chermak and consists in applying [Che22, Corollary 2.7], so we
choose

$T:=\\{w\in W(\mathcal{L})\mid S_{w}\in\Delta^{+}\\}.$

Indeed consider the partial group
$(\mathcal{M},D_{\mathcal{M}},\Pi_{\mathcal{M}},i_{\mathcal{M}})$ just
defined. If $S_{w}\in\Delta$, then $w\in D_{\mathcal{L}}\subseteq
D_{\mathcal{M}}$ (as $\Delta\subseteq\Delta^{+}$); on the other hand, suppose
that $S_{w}\in\Delta^{+}\setminus\Delta$ for $w=(w_{1},\dots,w_{k})$, then
there exists an $\mathcal{F}$-conjugate $P_{0}$ of $R$ such that $P_{0}\leq
S_{w_{1}}$ and so $w\in D_{\mathcal{M}}$ via $P_{0}$. In particular, all the
elements $P_{i}:=P_{i-1}^{w_{i}}$ are $\mathcal{F}$ conjugates of $R$. Take
now an $\mathcal{F}$ conjugate of $R$, say $(R)f$, and suppose that $f$ is the
composition of restrictions of conjugacy maps induced by elements
$(x_{1},\dots,x_{n})\in W(\mathcal{L})$ (recall that $\mathcal{F}$ is
generated by such maps); then $\mathcal{M}$ has the property that
$(x_{1},\dots,x_{n})\in D_{\mathcal{M}}$ with
$(x_{1},\dots,x_{n})\Pi_{\mathcal{M}}=x_{1}\circ\dots\circ x_{n}$. In other
words, all the $\mathcal{F}$-conjugations between elements of
$R^{\mathcal{F}}$, and therefore also of $\Delta^{+}$, are realized as
conjugations by some element of $\mathcal{M}$; this will translate in
$(\mathcal{L}^{+},\Delta^{+})$ being objective.

Just as in the proof of Proposition B, define the set $I$ of inner relations
of $\mathcal{L}$ by $I:=\\{u\in D\mid(u)\Pi=1_{\mathcal{L}}\\}$ and set
$\mathcal{N}:=\langle I\rangle$, that is the partial subgroup of $\mathcal{L}$
generated by $I$. Then we define

$\displaystyle\mathcal{L}^{+}:=\frac{\mathcal{M}}{\mathcal{N}}$

By the definition of $T$ we have $D\subseteq T$, so Lemma 5.5 shows that
$\mathcal{L}$ is embedded in $\mathcal{L}^{+}$. The following set of
hypothesis was identified by Chermak in [Che16, Hypothesis 3.2] and guarantees
that $(\mathcal{L}^{+},\Delta^{+},S)$ is a locality.

######  Hypothesis 5.6.

The subgroup $R\leq S$ satisfies the following:

  * (i)

every strict overgroup of $R$ contained in $S$ is an element of $\Delta$;

  * (ii)

$R$ and $O_{p}(N_{\mathcal{F}}(R))$ are fully normalized in $\mathcal{F}$;

  * (iii)

$N_{\mathcal{L}}(R)$ is a subgroup of $\mathcal{L}$ such that
$N_{\mathcal{F}}(R)=\mathcal{F}_{N_{S}(R)}(N_{\mathcal{L}}(R))$.

In particular, (i) implies $\Delta^{+}=\Delta\cup R^{\mathcal{F}}$; moreover
the hypothesis of [Che16, Lemma 3.1] are fulfilled and points (b) and (c) of
the lemma imply that for any $V\in R^{\mathcal{F}}$ there exists an element
$y\in\mathcal{L}$ such that $V=R^{y}$ and $N_{S}(V)\leq S_{y^{-1}}$. Following
the notation in [Che16, Section 3], $\textbf{Y}_{V}$ denotes the set of such
elements $y$ and $\textbf{X}_{V}:=\\{y^{-1}\mid y\in\textbf{Y}_{V}\\}$. Now
for any element $g\in\mathcal{L}$ there are two possible cases:

  * (a)

$S_{g}$ does not contain an element of $R^{\mathcal{F}}$; in this case the
conjugation $c_{g}:S_{g}\longrightarrow S_{g}^{g}$ cannot be realized via an
element of $R^{\mathcal{F}}$;

  * (b)

$S_{g}$ contains an element $U\in R^{\mathcal{F}}$, then the word $g$ is in
$D_{\mathcal{M}}$ via $(U,U^{g})$.

The following lemma is the fundamental result connecting our construction with
that of Chermak; one sees that the results which allowed Chermak to build a
set and define a partial group structure are here used just to describe the
partial group $\mathcal{L}^{+}$. Most of the following proof consists indeed
in the first lines of the proof of [Che16, Lemma 3.8].

######  Lemma 5.7.

The following hold.

  * (i)

If $g\in\mathcal{L}$ is such that $S_{g}$ satisfies condition (b) above, then
$g$ is equivalent to $(x^{-1},h,y)$ in $\mathcal{L}^{+}$ for suitable
$x\in\textbf{X}_{U}$, $y\in\textbf{Y}_{V}$ and $h\in N_{\mathcal{L}}(R)$,
where $U,V\in R^{\mathcal{F}}$.

  * (ii)

If $w=(w_{1},\dots,w_{n})\in\mathcal{M}$, then either each of the $w_{i}$
satisfies (b) (and we say that $w$ is _of type (b)_), or none of them does; in
the first case there exist suitable $x_{i}\in\textbf{X}_{U_{i}}$,
$y_{i}\in\textbf{Y}_{V_{i}}$ and $h_{i}\in N_{\mathcal{L}}(R)$ such that the
word
$w^{\prime}:=(x_{1}^{-1},h_{1},y_{1},x_{2}^{-1},h_{2},y_{2},\dots,x_{n}^{-1},h_{n},y_{n})\in
D_{\mathcal{M}}$ and it is equivalent to $w$ in $\mathcal{L}^{+}$.

  * (iii)

We have that each $(y_{i},x_{i+1}^{-1})\in D_{\mathcal{L}}$ and also
$w_{0}^{\prime}:=(h_{1},y_{1}x_{2}^{-1},h_{2},y_{2}x_{3}^{-1},\dots,y_{n-1}x_{n}^{-1},h_{n})\in
D_{\mathcal{L}}$; in particular, $\mathcal{L}^{+}$ is finite.

###### Proof.

Suppose that $g\in D_{\mathcal{M}}$ via $U\in R^{\mathcal{F}}$ and set
$U^{g}=:V$; we have already seen that under the hypothesis 5.6 the sets
$\textbf{Y}_{V}$ and $\textbf{X}_{U}$ are not empty, thus pick
$y\in\textbf{Y}_{V}$ and $x\in\textbf{X}_{U}$ and set $u:=(x,g,y^{-1})$. Then

$u=(x,g,y^{-1})\in
D_{\mathcal{L}}\qquad\text{via}\qquad(N_{S_{g}}(U)^{x^{-1}},N_{S_{g}}(U),N_{S_{g^{-1}}}(V),N_{S_{g^{-1}}}(V)^{y^{-1}}),$

so define $h:=(u)\Pi$ and note that $h\in N_{\mathcal{L}}(R)$. Now the word
$u^{\prime}:=(x^{-1},x,g,y^{-1},y)\in D_{\mathcal{L}}$ via $N_{S_{u}}(U)$ and
we get $g=(u^{\prime})\Pi=(x^{-1},h,y)\Pi$, thus proving (i).
If any of the $w_{i}$ for $i\in\\{1,\dots,n\\}$ is such that $S_{w_{i}}$
contains an $\mathcal{F}$-conjugate of $R$, than all the preceding and all the
following $w_{j}$ share the same property, as the $S_{w_{j}}$ are, by
definition, obtained by $\mathcal{F}$-conjugation. Suppose now that (b) holds
for each $w_{i}$ and let $U_{i}$, $V_{i}$ be as $U$ and $V$ in (i), so that
$w_{i}\in D_{\mathcal{M}}$ via $(U_{i},V_{i})$; since $w\in D_{\mathcal{M}}$
we can choose the sequence of the $(U_{i},V_{i})$ such that
$U_{i}^{w_{i}}=V_{i}=U_{i+1}$. Since by (i) each $(x_{i}^{-1},h_{i},y_{i})$ is
equivalent to $w_{i}$ in $\mathcal{L}^{+}$, $w^{\prime}$ is equivalent to $w$;
this is (ii).
Note now that each $(y_{i},x_{i+1}^{-1})\in D_{\mathcal{L}}$ via
$(N_{S}(V_{i})^{y_{i}^{-1}},N_{S}(V_{i})=N_{S}(U_{i+1}),N_{S}(U_{i+1})^{x_{i}^{-1}})$;
in particular we get
$(y_{i},x_{i+1}^{-1})\Pi_{\mathcal{M}}=(y_{i},x_{i+1}^{-1})\Pi=y_{i}x_{i+1}^{-1}$
in $\mathcal{L}^{+}$ and, moreover, $y_{i}x_{i+1}^{-1}\in N_{\mathcal{L}}(R)$.
Then $w_{0}^{\prime}\in W(N_{\mathcal{L}}(R))$ and, $N_{\mathcal{L}}(R)$ being
a subgroup of $\mathcal{L}$, it is a word of $D_{\mathcal{L}}$. Moreover,
$w=w^{\prime}=(x_{1}^{-1},(w_{0}^{\prime})\Pi,y_{n})$ in $\mathcal{L}^{+}$.
Let $\textbf{Y}:=\bigcup_{V\in R^{\mathcal{F}}}\textbf{Y}_{V}$ and similarly
$\textbf{X}:=\bigcup_{V\in R^{\mathcal{F}}}\textbf{X}_{V}$. If $w$ is such
that no $w_{i}$ satisfies (b), then the word $(w_{1},\dots,w_{n})$ must be in
$D_{\mathcal{M}}$ via some element of $\Delta$; in other words,
$(w_{1},\dots,w_{n})\in D_{\mathcal{L}}$, so $w=(w_{1},\dots,w_{n})\Pi$ in
$\mathcal{L}^{+}$. On the other hand, we have shown that whenever each $w_{i}$
satisfies (b), then $w$ is equivalent to an element appearing in the set
$\textbf{X}\times N_{\mathcal{L}}(R)\times\textbf{Y}$, which is finite being
$\mathcal{L}$ finite. This shows that also $\mathcal{L}^{+}$ is finite,
proving (iii). ∎

The lemma just proven shows actually more; it relates our construction with
that of the set $\Phi$ and with the description given in [Che16, Section 3 and
description 3.9] (or equivalently the set $\Theta$ in [Che13, Section 5,
Lemmas 5.8 and 5.9]).
In [Che16, Proposition 3.15(c)] Chermak characterizes $\mathcal{L}^{+}$ via a
push-out diagram taken in $Part$. We show that our definition of
$\mathcal{L}^{+}$ coincides with that of Chermak by showing that also in our
case $\mathcal{L}^{+}$ is the push-out of the same diagram. Define
$\mathcal{L}_{0}:=\\{g\in\mathcal{L}\mid S_{g}\text{ contains an
$\mathcal{F}$-conjugate of $R$}\\}$ and $D_{0}:=\\{w\in D_{\mathcal{L}}\mid
S_{w}\text{ contains an $\mathcal{F}$-conjugate of $R$}\\}$; it is not
difficult to see that $\mathcal{L}_{0}$, paired with the domain $D_{0}$, forms
an impartial subgroup of $\mathcal{L}$. Define
$\mathcal{L}_{0}^{+}:=\\{u\in\mathcal{L}^{+}\mid\text{$u$ is the image of an
element of $\mathcal{M}$ of type (b)}\\}$; then $\mathcal{L}_{0}^{+}$ forms a
partial subgroup of $\mathcal{L}^{+}$. The embedding of $\mathcal{L}$ in
$\mathcal{L}^{+}$ induces an embedding of $\mathcal{L}_{0}$ in
$\mathcal{L}_{0}^{+}$, thus we get the following diagram

${\mathcal{L}_{0}^{+}}$${\mathcal{L}^{+}}$${\mathcal{L}_{0}}$${\mathcal{L}}$$\scriptstyle{\iota}$$\scriptstyle{\lambda_{0}}$$\scriptstyle{\iota_{0}}$$\scriptstyle{\lambda}$
($\circ$)

where every morphism is an embedding. Our definition of $\mathcal{L}_{0}$
clearly coincides with the same given in [Che16, Lemma 3.14]; as a consequence
of Lemma 5.7 and of [Che16, 3.9] also $\mathcal{L}_{0}^{+}$ coincides with
that defined by Chermak. Since the diagram ($\circ$ ‣ 5.3) is the same
appearing in [Che16, Proposition 3.15 (c)], we only need to show the
following.

######  Lemma 5.8.

The diagram ($\circ$ ‣ 5.3) is a push-out diagram in $Part$.

###### Proof.

Suppose to have partial group homomorphisms
$f:\mathcal{L}\longrightarrow\mathcal{H}$ and
$g:\mathcal{L}_{0}^{+}\longrightarrow\mathcal{H}$, for some partial group
$\mathcal{H}$, such that $\iota_{0}f=\lambda_{0}g$; by Lemma 5.7(ii) every
element in $\mathcal{L}^{+}$ is either an element of $\mathcal{L}$ or an
element of $\mathcal{L}_{0}^{+}$, thus we can define

$\psi:\mathcal{L}^{+}\longrightarrow\mathcal{H},\quad
u\mapsto\begin{cases*}(u)g\quad&\text{if $u\in\mathcal{L}_{0}^{+}$}\\\
(u)f\quad&\text{if $u\in\mathcal{L}$}\end{cases*}$

Note that $\mathcal{L}_{0}=\mathcal{L}_{0}^{+}\cap\mathcal{L}$ and that
($\circ$ ‣ 5.3) is clearly commutative, since all morphisms are inclusions;
thus the map $\psi$ is well-defined. Every word in the domain
$D_{\mathcal{L}^{+}}$ of $\mathcal{L}^{+}$ lifts to a representative in
$D_{\mathcal{M}}$; the proof of Lemma 5.7 shows that either all
representatives in $D_{\mathcal{M}}$ are of type (b), or none is. Then
$\psi^{*}$ coincides with $f^{*}$ over words which lift to a representative
not of type (b) and with $g^{*}$ over words lifting to representatives of type
(b). This implies that $\psi$ is a morphism of partial groups.
Finally, uniqueness of the map $\psi$ is trivial; any other candidate map
$\mathcal{L}^{+}\longrightarrow\mathcal{H}$ would restrict to $f$ over
$\mathcal{L}$ and to $g$ over $\mathcal{L}_{0}^{+}$, coinciding then with
$\psi$. ∎

The proof that $(\mathcal{L}^{+},D^{+},S)$ is a locality is now given in
[Che16, Proposition 3.17], whereas [Che16, Proposition 3.18] shows uniqueness
of such a locality.

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