# Duality in Non-Abelian Algebra V. Homological diagram lemmas

Kishan Kumar Dayaram  Department of Mathematics and Applied Mathematics,
University of Johannesburg, Auckland Park Kingsway Campus, P.O. Box 524,
Auckland Park 2006, South Africa<EMAIL_ADDRESS>,  Amartya
Goswami  Department of Mathematics and Applied Mathematics, University of
Johannesburg, Auckland Park Kingsway Campus, P.O. Box 524, Auckland Park 2006,
South Africa.  National Institute for Theoretical and Computational Sciences
(NITheCS), South Africa<EMAIL_ADDRESS>,  Zurab Janelidze  Department of
Mathematical Sciences, Stellenbosch University, Private Bag X1 Matieland,
7602, South Africa  National Institute for Theoretical and Computational
Sciences (NITheCS), South Africa<EMAIL_ADDRESS>and  Tim Van der Linden
Institut de Recherche en Mathématique et Physique, Université catholique de
Louvain, chemin du cyclotron 2 bte L7.01.02, B–1348 Louvain-la-Neuve, Belgium
<EMAIL_ADDRESS>
###### Abstract.

Through abelian categories, homological lemmas for modules admit a self-dual
treatment, where half of the proof of a lemma is sufficient to prove the full
lemma. As remarked in the prequel of the present paper (‘Duality in Non-
Abelian Algebra IV’), there is a self-dual treatment of these lemmas even in
the case of non-abelian categories of group-like structures. It is based on
the axiomatic context of a ‘noetherian form’ introduced in the same paper,
which is a generalization of an abelian category that includes among its
examples semi-abelian and Grandis exact categories, as well as many others.
The aim of the present article is to expose the proofs of homological diagram
lemmas in this general context.

###### Key words and phrases:

$3\times 3$ Lemma, diagram lemma, Dragon Lemma, Five Lemma, group-like
structure, Goursat’s Lemma, homomorphism induction, noetherian form, semi-
abelian category, Salamander Lemma, Słomiński algebra, Snail Lemma, Snake
Lemma, Spider Lemma, Weak Four Lemma

###### 2020 Mathematics Subject Classification:

18G50, 20J05, 18E13, 20J15, 18D30, 08B05

The fourth author is a Senior Research Associate of the Fonds de la Recherche
Scientifique–FNRS

## Introduction

In group theory, certain results may be formulated and proved using subgroups
in the place of elements. This observation goes back to the work of E. Noether
and her followers in the 1930’s. It was later taken further by S. Mac Lane,
who introduced a method of _chasing subobjects_ to prove diagram lemmas of
homological algebra in an abelian category, which mimics the method of chasing
elements (see [30, 31]).

Discovery and investigation of abelian categories was preceded by S. Mac
Lane’s observation of a number of duality phenomena in group theory (see
[29]). Duality here refers to change of orientation of morphisms in a
category. Axioms for an abelian category are invariant under duality, which
means that every result established in an abelian category has a ‘dual’
counterpart.

While for any ring $R$, the category of $R$-modules is an abelian category,
the same is not true for other group-like structures, which form for instance
the category of $R$-Lie algebras, or the category of non-abelian groups
themselves. Although S. Mac Lane anticipates in [29] that a self-dual
axiomatic framework for encompassing such categories should be possible, a
framework with the anticipated functionality was only arrived at much later,
in [12], building on the development of projective homological algebra by M.
Grandis [14], and homological [2] and semi-abelian [16] categories.

The framework described in [12] emerges from earlier investigations [21, 24,
25, 26] (see also [43, 33, 9]), which extend the elementary language of a
category with abstractly specified subobjects (rather than making use of
subobjects defined internally via monomorphisms). These abstract subobjects
form a category fibred over the original category. The functor that connects
the two categories becomes the key player here. Duality is now in reference to
this functor (i.e., it is based on considering the same functor between dual
categories). The switch from a category to a functor is what enables to reveal
the kind of duality phenomena for non-abelian group-like structures that are
necessary for borrowing the subobject chasing technique from the context of an
abelian category.

In [12], it was shown how one can establish the classical isomorphism theorems
in this new framework. It was also suggested in [12] that the same framework
should be suitable for proving homological lemmas. For various concrete
lemmas, this has already been checked in [38, 33, 20]. Many of these proofs
are in fact almost a copy of proofs that work in the particular case of
Grandis exact categories [14]. However, there is an obstacle in adapting the
proof of the Snake Lemma from [14]. This obstacle was first identified by the
third and fourth authors, while working on [38]. It had to do with the
construction of the connecting morphism. The missing ingredient, the
‘pyramid’, was developed in [12] and we recall it in the present paper. This
led to a proof of a more general version of the Snake Lemma than the one
formulated in [38] — see Remark 5.8 (note that there are also minor
differences in the axioms of a noetherian form and those used in [38]). The
proof was sketched in [20] and worked out in full detail in [10].

The present paper is mostly expository. There is a large overlap with the
unpublished thesis [10] of the first author (prepared under the supervision of
the second and the third authors) and [12]. Our goal in this paper is to
provide researchers working with group-like structures an efficient tool for
verifying old or new homological-style diagram lemmas in their specific
context. The efficiency comes from the use of duality, as in the abelian
context. For example, consider the Short Five Lemma for (non-abelian) groups,
which states that in a given commutative diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{s}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\scriptstyle{t}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{u}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x}$$\textstyle{B^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{y}$$\textstyle{C^{\prime}}$

where the rows are short exact sequences, we have: $t$ is an isomorphism if
both $s$ and $u$ are isomorphisms. In order to prove this, we would only need
to prove that $t$ is injective (or surjective) if both $s$ and $u$ are, and
the full result would then follow by applying duality.

The language we use in setting up the theory closely follows the language of
group theory, thus making it easily adaptable to group-like structures at
hand. We demonstrate such adaption on an example of a particular type of
group-like structures: Słomiński algebras [35].

All semi-abelian categories and Grandis exact categories fit the framework of
this paper. Among the concrete categories included in these two classes of
examples are the category of modules over a ring, the category of groups (as
well as the category of Słomiński algebras), the category of graded abelian
groups, categories of Lie algebras, the category of cocommutative Hopf
algebras over an algebraically closed field of characteristic zero, the
category of Heyting semilattices, the category of loops, the dual of the
category of pointed sets, the category of modular lattices and modular
connections, the category of sets and partial bijections, and many others.

In an attempt to make the present paper as self-contained and as accessible as
possible, we have included proofs of a number of results from [12] that
concern properties of noetherian forms useful for establishing the homological
diagram lemmas — even those proofs which have been omitted in [12] due to
their relative simplicity.

###### Contents

  1. 1 The Framework
  2. 2 An Example: Słomiński Algebras
  3. 3 Embedding and Projection Basics
  4. 4 The Pyramid
  5. 5 Classical Diagram Lemmas
  6. 6 Exercises

##  1\. The Framework

In this section, we describe an axiomatic framework for establishing
homomorphism theorems, introduced in [12] (see also the references there).
Some of the basic results given in [12] without proofs are completed here with
proofs. This framework is based on a language where primitives are given by

  * —

groups,

  * —

group (homo)morphisms (we will use the terms ‘morphism’ and ‘homomorphism’
interchangeably), as well as codomain/domain of a group morphism,

  * —

composition of group morphisms,

  * —

subgroup of a group,

  * —

subgroup inclusion, and

  * —

direct/inverse image of a subgroup along a morphism.

When applied to group theory, these terms obtain their usual meaning. The
framework can be applied to many other group-like structures as well, where
these terms are to be interpreted in a natural way. For instance, to apply the
framework to module theory, we must interpret

  * —

a group as a module,

  * —

(codomain/domain of) a group morphism as (codomain/domain of) a module
homomorphism,

  * —

composition of morphisms as the usual composition of module homomorphisms,

  * —

subgroup of a group as a submodule of a module,

  * —

subgroup inclusion as inclusion of submodules,

  * —

direct/inverse image of a subgroup along a morphism as direct/inverse image of
a submodule along a morphism.

The framework can also be applied to more general categories, such as semi-
abelian categories in the sense of [16]. In this case, we interpret

  * —

a group as an object in the category,

  * —

(codomain/domain of) a group morphism as (codomain/domain of) a morphism
between objects,

  * —

composition of morphisms as the usual composition of morphisms in the
category,

  * —

subgroup of a group as a subobject of an object,

  * —

subgroup inclusion as inclusion of subobjects.

  * —

direct/inverse image of a subgroup along a morphism as direct/inverse image of
a subobject along a morphism (which are given by regular image and pullback,
respectively).

Among examples of semi-abelian categories are categories of many different
kinds of non-abelian group-like structures: modules over a ring; all kinds of
algebras over a ring, including Lie algebras, associative algebras, non-
associative algebras, etc.; groups; rings without identity; Heyting meet semi-
lattices; loops; cocommutative Hopf algebras over an algebraically closed
field of characteristic zero (see [13]), and others. Thus, our framework is
applicable to all of these structures. It also includes the framework of exact
categories in projective homological algebra due to M. Grandis [14].
Furthermore, according to [41], any algebraic category (i.e., any variety of
universal algebras) is an example of our framework, although in some cases,
such as the category of sets itself, the notion of an exact sequence
trivializes (and so, homological lemmas there do not give anything
interesting).

This general framework is called a _noetherian form_ (see [42, 12]). As
explained in [12], it can be formalized as a pair of categories and a functor
between them, satisfying certain properties. Duality in the elementary theory
of a functor (see [31]) translates to duality of the framework as described in
Table 1. This table displays atomic expressions of the language of a
noetherian form in the first column and their dual counterparts in the second
column. The logical expressions formed using these atomic expressions are to
be dualized by dualizing each of the atomic expressions in it and keeping the
rest of the logical structure of an expression unchanged. Notice that double
dual will always give back the starting expression.

Statement | Dual Statement
---|---
$G$ is a group |  $G$ is a group
$S$ is a subgroup of $G$ (written as $S\in\mathsf{Sub}G$) |  $S$ is a subgroup of $G$ (written as $S\in\mathsf{Sub}G$)
$f$ is a morphism |  $f$ is a morphism
from the group $X$ (domain) to the group $Y$ (codomain) | from the group $Y$ (domain) to the group $X$ (codomain)
(written as $f\colon X\to Y$) | (written as $f\colon Y\to X$)
$S$ is contained in $T$ (written as $S\subseteq T$) |  $T$ is contained in $S$ (written as $T\subseteq S$)
(where $S$ and $T$ are subgroups of a group $G$) | (where $S$ and $T$ are subgroups of a group $G$)
$h=gf$ | $h=fg$
(where $f\colon X\rightarrow Y$, $g\colon Y\rightarrow Z$ and $h\colon X\rightarrow Z$) | (where $f\colon Y\rightarrow X$, $g\colon Z\rightarrow Y$ and $h\colon Z\rightarrow X$)
$T$ is a direct image of $S$ along $f$ (written as $T=fS$) |  $T$ is an inverse image of $S$ along $f$ (written as $T=f^{-1}S$)
(where $S\in\mathsf{Sub}X$ and $f\colon X\to Y$) | (where $S\in\mathsf{Sub}X$ and $f\colon Y\to X$)

Table 1. Atomic expressions and their duals

As hinted in the table, for a group $G$ in the abstract framework, we write
$\mathsf{Sub}G$ to denote the set of subgroups of $G$. We require that
inclusion of subgroups turns $\mathsf{Sub}G$ into a poset. This means that we
have the following properties:

  * (P1)

$A\subseteq A$ (reflexivity),

  * (P2)

$A\subseteq B$ and $B\subseteq C$ imply $A\subseteq C$ (transitivity),

  * (P3)

$A\subseteq B$ and $B\subseteq A$ imply $A=B$ (antisymmetry).

For a set $\mathcal{S}$ of subgroups of a group $G$, the _join_ of
$\mathcal{S}$ is an element $T$ of $\mathsf{Sub}G$, usually denoted
$T=\bigvee\mathcal{S}$, such that:

  1. (i)

$S\subseteq T$ for all $S\in\mathcal{S}$ and

  2. (ii)

if $S\subseteq U$ for all $S\in\mathcal{S}$ and for some $U\in\mathsf{Sub}G$,
then $T\subseteq U$.

Similarly, the _meet_ of $\mathcal{S}$ is a subgroup $M$, usually denoted
$M=\bigwedge\mathcal{S}$, such that:

  1. (i)

$M\subseteq S$ for all $S\in\mathcal{S}$ and

  2. (ii)

if $N\subseteq S$ for all $S\in\mathcal{S}$ and for some element
$N\in\mathsf{Sub}G$, then $N\subseteq M$.

Note that the notions of join and meet are dual to each other. We require that
each $\mathsf{Sub}G$ is a bounded lattice, i.e.,

  * (BL)

For each finite $\mathcal{S}\subseteq\mathsf{Sub}G$, the join
$\bigvee\mathcal{S}$ and meet $\bigwedge\mathcal{S}$ exist.

This includes the case when $\mathcal{S}=\varnothing$, in which case the join
is the smallest subgroup of $G$ (that which is contained in every other
subgroup of $G$), which we denote by $1$, and the meet is the largest subgroup
of $G$ (that which contains every other subgroup of $G$), which we denote by
the same $G$ (although formally, we distinguish them — otherwise, by duality,
we would have to identify $G$ with its smallest subgroup as well). When
$\mathcal{S}$ has non-zero but finitely many elements, e.g.,
$\mathcal{S}=\\{A_{1},\dots,A_{n}\\}$, we write $A_{1}\vee\dots\vee A_{n}$ for
$\bigvee\mathcal{S}$ and $A_{1}\wedge\dots\wedge A_{n}$ for
$\bigwedge\mathcal{S}$.

For any morphism $f\colon X\rightarrow Y$, the _image_ of $f$, written as
$\mathsf{Im}f$, is the direct image of the largest subgroup of $X$ under the
map $f$ and (dually) the _kernel_ of $f$, written as $\mathsf{Ker}f$, is the
inverse image of the smallest subgroup of $Y$ under the map $f$. We require
that for each $S\in\mathsf{Sub}X$, the direct image $fS$ of $S$ under $f$ is
an element of $\mathsf{Sub}Y$, while the inverse image of $T\in\mathsf{Sub}Y$
under $f$ is an element of $\mathsf{Sub}X$. Thus in particular,
$\mathsf{Im}f=fX\in\mathsf{Sub}Y$ and $\mathsf{Ker}f=f^{-1}1\in\mathsf{Sub}X$.
In fact, we require that the direct image and inverse image maps along a given
$f\colon X\rightarrow Y$ form a Galois connection between $\mathsf{Sub}X$ and
$\mathsf{Sub}Y$. This means that the following holds:

  * (G)

$fA\subseteq C\Leftrightarrow A\subseteq f^{-1}C$, when $A\in\mathsf{Sub}X$,
$C\in\mathsf{Sub}Y$ and $f\colon X\to Y$.

The well-known consequences of the definition of a Galois connection give us
the following properties:

  * (G1)

The direct and inverse image maps are monotone (i.e., they preserve subgroup
inclusions).

  * (G2)

For a morphism $f\colon X\rightarrow Y$ and subgroups $A$ of $X$ and $B$ of
$Y$, we have $ff^{-1}fA=fA$ and $f^{-1}ff^{-1}B=f^{-1}B$.

  * (G3)

The direct image map preserves joins of subgroups and the inverse image map
preserves meets of subgroups.

  * (G4)

Under a morphism, the direct image of the smallest subgroup is the smallest
subgroup and the inverse image of the largest subgroup. This is the same as
(G3) considered for empty joins and meets.

A morphism $f\colon X\to Y$ is said to be a zero morphism when
$X=\mathsf{Ker}f$. This turns out to be equivalent to requiring
$\mathsf{Im}f=1$, by direct-inverse image Galois connection, according to
which

$X\subseteq
f^{-1}1=\mathsf{Ker}f\quad\Leftrightarrow\quad\mathsf{Im}f=fX\subseteq 1;$

note that since $X$ is the largest subgroup of $X$ and $1$ is the smallest
subgroup of $Y$, by antisymmetry, $X\subseteq\mathsf{Ker}f$ is equivalent to
$X=\mathsf{Ker}f$ and $\mathsf{Im}f\subseteq 1$ is equivalent to
$\mathsf{Im}f=1$. The dual of a zero morphism is a zero morphism; in other
words, the notion of a zero morphism is _self-dual_.

A subgroup $S$ of a group $G$ is normal if it is the kernel of some morphism
$f\colon G\rightarrow H$. The dual of this notion is: $S$ is conormal if it is
the image of some morphism $g\colon H\rightarrow G$. In usual group theory,
every subgroup is of course conormal (as every subgroup is the image of its
associated embedding). This is not a requirement in the abstract context of a
noetherian form since if it were, by duality every subgroup would have to be
normal, which does not hold for (non-abelian) groups.

For a group $G$, an _identity morphism_ of $G$ is a morphism
$\mathsf{id}_{G}\colon G\rightarrow G$ such that $\mathsf{id}_{G}f=f$ and
$h\mathsf{id}_{G}=h$ for all morphisms $f\colon A\rightarrow G$ and $h\colon
G\rightarrow B$. There is at most one identity morphism for each $G$
($\mathsf{id}_{G}=\mathsf{id}_{G}\mathsf{id}^{\prime}_{G}=\mathsf{id}^{\prime}_{G}$),
and we require that there is at least one, and hence exactly one:

  * (I)

Every group $G$ admits an identity morphism $\mathsf{id}_{G}$.

A morphism $f\colon X\rightarrow Y$ is an isomorphism if there exists a
morphism $g\colon Y\rightarrow X$ such that $fg=\mathsf{id}_{Y}$ and
$gf=\mathsf{id}_{X}$. The notion of an isomorphism is also self-dual.

The following further properties that we require ensure that groups and
morphisms between them form a category and mapping each group to its subgroup
lattice yields a functor from the category of groups to the category of
bounded lattices, with Galois connections as morphisms:

  1. (A)

$f(gh)=(fg)h$ whenever $f\colon X\rightarrow Y$, $g\colon Y\rightarrow Z$ and
$h\colon Z\rightarrow G$ (i.e., composition of morphisms is associative),

  2. (F1)

$\mathsf{id}_{G}S=\mathsf{id}_{G}^{-1}S=S$, when $S\in\mathsf{Sub}G$,

  3. (F2)

$g(fS)=(gf)S$ and $f^{-1}(g^{-1}T)=(fg)^{-1}T$, when $f\colon X\rightarrow Y$,
$S\in\mathsf{Sub}X$ and $T\in\mathsf{Sub}Y$.

Note that each of the requirements (P1-P3), (I), (G), (A), (F1), (F2) are
self-dual. They constitute Axiom 1 in [12], which states that:

######  Axiom 1.

Groups and group morphisms form a category under composition of groups.
Moreover, for each group $G$, the subgroups of $G$ together with subgroup
inclusions form a poset. Furthermore, each morphism $f\colon X\rightarrow Y$
defines a monotone Galois connection

$\textstyle{\mathsf{SubX}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{\mathsf{SubY}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{-1}}$

given by direct and inverse image maps under $f$, and this defines a functor
between the category of groups and the category of posets with Galois
connections.

A useful consequence of Axiom 1 (more precisely, of (G) and (F1,2)) is the
following:

  * (GF)

The direct and inverse image maps corresponding to an isomorphism $f$ are the
same as the inverse image and direct image map, respectively, corresponding to
its inverse $f^{-1}$.

The requirement (BL) is part of Axiom 2 in [12], which states that:

######  Axiom 2.

The poset of subgroups of each group is a bounded lattice. Moreover, for a
morphism $f\colon X\rightarrow Y$ and subgroups $A\in\mathsf{Sub}X$ and
$B\in\mathsf{Sub}Y$, we have $ff^{-1}B=B\wedge\mathsf{Im}f$ and
$f^{-1}fA=A\vee\mathsf{Ker}f$.

The next lemma gives the ‘Restricted Modular Law’: a restricted version of the
standard modular law. Notice that this result follows from Axiom 1 and Axiom 2
only. The proof is reproduced from [12].

######  Lemma 1.1 (Restricted Modular Law, [12]).

For any three subgroups, $X$, $Y$, and $Z$ of a group $G$, if either $Y$ is
normal and $Z$ is conormal, or $Y$is conormal and $X$ is normal, then we have:

$X\subseteq Z\Rightarrow X\vee(Y\wedge Z)=(X\vee Y)\wedge Z.$

###### Proof.

Note the two cases are dual, so it suffices to prove the first one (where $Y$
is normal and $Z$ is conormal). Assume $Y=\mathsf{Ker}g$ and $Z=\mathsf{Im}f$
for some morphisms $f$ and $g$ respectively. Thus, if $X\subseteq
Z=\mathsf{Im}f$, we have $X=X\wedge\mathsf{Im}f$ and hence:

$\displaystyle X\vee(Y\wedge Z)$
$\displaystyle=X\vee(\mathsf{Ker}g\wedge\mathsf{Im}f)=(X\wedge\mathsf{Im}f)\vee(\mathsf{Ker}g\wedge\mathsf{Im}f)$
$\displaystyle=ff^{-1}X\vee ff^{-1}\mathsf{Ker}g=f(f^{-1}X\vee
f^{-1}\mathsf{Ker}g)$
$\displaystyle=f(f^{-1}X\vee\mathsf{Ker}gf)=f(gf)^{-1}(gf)f^{-1}X$
$\displaystyle=f(gf)^{-1}g(ff^{-1}X)=f(gf)^{-1}g(X\wedge\mathsf{Im}f)$
$\displaystyle=f(gf)^{-1}gX=ff^{-1}g^{-1}gX$
$\displaystyle=ff^{-1}(X\vee\mathsf{Ker}g)=(X\vee\mathsf{Ker}g)\wedge\mathsf{Im}f$
$\displaystyle=(X\vee Y)\wedge Z.\qed$

######  Example 1.2.

The join $X\vee Y$ of subgroups $X$ and $Y$ of an (ordinary) group $G$ is
given by the (set-theoretic) intersection of all subgroups which contain $X$
and $Y$. The join may be given by $X\vee Y=XY$ (the product of the subgroups
given by $XY=\\{xy\mid x\in X,y\in Y\\}$) in the special case when the
subgroups commute (i.e., $XY=YX$). The conditions in the statement of the
Restricted Modular Law guarantee that either $X$ or $Y$ is normal and hence
that $X$ and $Y$ commute. Therefore for ordinary groups, the Restricted
Modular Law follows from the modular property of groups which states that if
$X$, $Y$ and $Z$ are subgroups of a group $G$ with $X\subseteq Z$, then
$(XY)\cap Z=X(Y\cap Z)$ (note that in the case when $Y$ is normal, $Y\cap Z$
also commutes with $X$ since $X\subseteq Z$).

The remaining axioms, along with the Restricted Modular Law, allow us to
establish certain isomorphism theorems and diagram lemmas, such as the
Butterfly Lemma and the Middle $3\times 3$ Lemma, in this context.

######  Remark 1.3.

Note that it was missed in [12] that the Restricted Modular Law is needed for
the proof of the Butterfly Lemma: the Restricted Modular Law would be needed
to show that the dashed zigzag on the butterfly diagram in the proof does
induce a homomorphism.

In the context of ordinary group theory, for any subgroup $S$ of a group $G$,
if we consider $S$ as a group, the embedding morphism $\iota_{S}\colon
S\rightarrow G$ has the following universal property:
$\mathsf{Im}\iota_{S}\subseteq S$ and if $f\colon F\rightarrow G$ is any
morphism such that $\mathsf{Im}f\subseteq S$, then there exists a unique
morphism $u\colon F\rightarrow S$ such that $f=\iota_{S}u$, i.e., such that
the diagram

---
$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{S}}$$\textstyle{G}$$\textstyle{F\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{u}$$\scriptstyle{f}$

commutes. In the context of a noetherian form, we capture this phenomenon as
follows:

######  Definition 1.4.

Given a subgroup $S$ of a group $G$, a morphism $m:M\rightarrow G$ is said to
be an embedding associated to $S$ if $\mathsf{Im}m\subseteq S$ and for any
morphism $m^{\prime}\colon M^{\prime}\rightarrow G$ such that
$\mathsf{Im}m^{\prime}\subseteq S$, there exists a unique homomorphism
$u\colon M^{\prime}\rightarrow M$ such that $m^{\prime}=mu$.

The dual notion of this is:

######  Definition 1.5.

Given a subgroup $S$ of a group $G$, a morphism $e\colon G\rightarrow H$ is
said to be a projection associated to $S$ if $S\subseteq\mathsf{Ker}e$ and for
any morphism $e^{\prime}\colon G\rightarrow H^{\prime}$ such that
$S\subseteq\mathsf{Ker}e^{\prime}$, there exists a unique morphism $v\colon
H\rightarrow H^{\prime}$ such that $e^{\prime}=ve$.

When $S$ is a conormal subgroup of $G$, we write $\iota_{S}\colon
S/1\rightarrow G$ to denote an embedding associated to $S$, when it exists.
Dually, when $S$ is a normal subgroup of $G$, we write $\pi_{S}\colon
G\rightarrow G/S$ to denote a projection associated to $S$.

######  Axiom 3.

Every conormal subgroup has an embedding associated to it and every normal
subgroup has a projection associated to it (in other words, $\iota_{S}$ and
$\pi_{S}$ are defined whenever $S$ is conormal and normal, respectively).

Embeddings allow to define relative normality of subgroups.

######  Definition 1.6.

For two subgroups $A$ and $B$ of a group $G$, we say that $B$ is _normal to_
$A$, denoted $B\lhd A$, if:

  1. (RN1)

$B\subseteq A$,

  2. (RN2)

$A$ is a conormal subgroup of $G$, and,

  3. (RN3)

$\iota_{A}^{-1}B$ is normal in $A/1$.

To complete the description of the framework, the following axioms remain:

######  Axiom 4.

Any morphism $f\colon X\rightarrow Y$ factorizes as
$f=\iota_{\mathsf{Im}f}h\pi_{\mathsf{Ker}f}$, where $h$ is an isomorphism.

######  Axiom 5.

The join of any two normal subgroups is normal and the meet of any two
conormal subgroups is conormal.

##  2\. An Example: Słomiński Algebras

The axioms detailed in the previous section have been chosen in such a way
that they hold for the context of ordinary groups. Anyone sufficiently
familiar with group theory will immediately agree that all stated axioms hold
true for groups under the interpretation suggested at the start of the
previous section. To give the reader a taste for how little of the group
properties is actually required for these axioms to hold, we demonstrate their
validity in the wider class of Słomiński algebras [35]. We have already
mentioned in the previous section various examples that lie within the
framework of a noetherian form, along with references where the reader may
find out more about these examples. The example of Słomiński algebras falls
under the class of examples given by semi-abelian categories — the category of
Słomiński algebras is a semi-abelian category — so the fact that the axioms
presented in the previous section hold for Słomiński algebras is not new.
Moreover, a reader well-versed in universal algebra will find the exposition
below rather elementary: to such readers, we recommend interpreting and
verifying the axioms by themselves.

######  Definition 2.1.

A _Słomiński algebra_ is a set $X$ together with two binary operations, $p$
and $d$, and a constant (nullary operation) $0$ that satisfies the following
conditions:

  1. (i)

$d(x,x)=0$, for all $x\in X$,

  2. (ii)

$p(d(x,y),y)=x$, for all $x,y\in X$.

We now describe an interpretation of a noetherian form. We shall use some
basic notions of universal algebra for Słomiński algebras, such as
_subalgebra_ , _homomorphism_ , _isomorphism_ and _congruence_ , as well as
basic results from universal algebra relating to these notions (see e.g., [7]
for more details).

The subobject poset for a Słomiński algebra $X$ is the poset of its
subalgebras. Morphisms in this context will be the homomorphisms between
Słomiński algebras. For a homomorphism $f\colon X\rightarrow Y$ between
Słomiński algebras, and for subalgebras $A$ of $X$ and $B$ of $Y$, the direct
image of $A$ and the inverse image $B$ under $f$ are given as expected,
$fA=\\{f(x)\mid x\in A\\}$ and $f^{-1}B=\\{x\in X\mid f(x)\in B\\}$,
respectively. Hence the kernel and image of $f$ are given by
$\mathsf{Ker}f=f^{-1}0=\\{x\in X\mid f(x)=0\\}$ and
$\mathsf{Im}f=f(X)=\\{f(x)\mid x\in X\\}$, respectively.

Notice that, for an (ordinary) group $G$ and any two elements $a$ and $b$ of
$G$, we define the operation $p$ by $p(a,b)=a\cdot b$ and the operation $d$ by
$d(a,b)=a\cdot(b^{-1})$. This makes the group $G$ into a Słomiński algebra.
Moreover, Słomiński algebra homomorphisms between groups are precisely group
homomorphisms and the subalgebras of a group are subgroups. In this case,
kernels and images of homomorphisms obtain their usual meaning.

######  Lemma 2.2.

Axiom 1 holds for the Słomiński algebra interpretation.

###### Proof.

The class of Słomiński algebras forms a category since:

  * —

For morphisms $f\colon X\rightarrow Y$ and $g\colon Y\rightarrow Z$ between
Słomiński algebras, the composite ${gf\colon X\rightarrow Z}$ is given by
$gf(x)=g(f(x))$. This is indeed a morphism since

$gf(d(x,y))=g(f(d(x,y)))=g(d(f(x),f(y)))=d(gf(x),gf(y))$

and $gf(p(x,y))=p(gf(x),gf(y))$ by a similar calculation.

  * —

Composition is associative since

$(h(gf))(x)=h(gf(x))=h(g(f(x)))=hg(f(x))=((hg)f)(x).$

  * —

For a Słomiński algebra $X$, the identity mapping $\mathsf{id}_{X}\colon
X\rightarrow X$ (given by $\mathsf{id}_{X}(x)=x$) is a morphism since
$\mathsf{id}_{X}(d(x,y))=d(x,y)=d(\mathsf{id}_{X}(x),\mathsf{id}_{X}(y))$ and
similarly, $\mathsf{id}_{X}(p(x,y))=p(\mathsf{id}_{X}(x),\mathsf{id}_{X}(y))$.
Moreover, for any morphism $f\colon X\rightarrow Y$, we have
$f(x)=f(\mathsf{id}_{X}(x))=f\mathsf{id}_{X}(x)$ and therefore
$f=f\mathsf{id}_{X}$. We also obtain $g=\mathsf{id}_{X}g$, for any morphism
$g\colon Z\rightarrow X$, by a similar argument.

Moreover, the set of subalgebras, $\mathsf{Sub}X$ of a given algebra $X$,
under the operation of set inclusion ($\subseteq$), forms a poset since this
is true for any subset of the power set of a set along with set inclusion.

Notice that both the direct image of a subalgebra $A$ and the inverse image of
a subalgebra $B$ are subalgebras, since:

  * —

If $f(x),f(y)\in fA$, then

$d(f(x),f(y))=f(d(x,y))\in fA$

and similarly

$p(f(x),f(y))=f(p(x,y))\in fA.$

  * —

If $x,y\in f^{-1}B$, then

$f(d(x,y))=d(f(x),f(y))\in B$

and hence $d(x,y)\in f^{-1}B$. Similarly, $p(x,y)\in f^{-1}B$.

Therefore, the direct and the inverse images of a subalgebra under a morphism
$f$ give rise to maps $f_{*}\colon\mathsf{Sub}X\rightarrow\mathsf{Sub}Y$ and
$f^{*}\colon\mathsf{SubY}\rightarrow\mathsf{Sub}X$,

$f_{*}(A)=fA,\quad f^{*}(B)=f^{-1}B.$

Moreover, if $fA\subseteq B$, then $f(x)\in B$ for all $x\in A$ and hence
$A\subseteq f^{-1}B$. Similarly, if $A\subseteq f^{-1}B$, then $f(x)\in B$ for
all $x\in A$ and hence $fA\subseteq B$. Therefore, $fA\subseteq
B\Leftrightarrow A\subseteq f^{-1}B$, showing that the maps $f_{*}$ and
$f^{*}$ constitute a monotone Galois connection between $\mathsf{Sub}X$ and
$\mathsf{Sub}Y$. ∎

######  Lemma 2.3.

Axiom 2 holds for the Słomiński algebra interpretation.

###### Proof.

For a Słomiński algebra $X$, the subset $0$ (consisting of just the constant
$0$) and the subset $X$ are both subalgebras and therefore the poset
$\mathsf{Sub}X$ has bottom and top elements given by $0$ and $X$,
respectively. Given two subalgebras $C$ and $D$ of $X$, we may define the meet
of $C$ and $D$, denoted $C\wedge D$, to be the largest subalgebra of $X$
contained in both $C$ and $D$. It may be verified that this turns out to be
the set-theoretic intersection of $C$ and $D$ and thus the meet of two
subalgebras exists (since it may be verified that the intersection is closed
under the operations $p$ and $d$ and moreover, contains the constant $0$). In
a dual manner, we define the join of $C$ and $D$, denoted $C\vee D$, to be the
smallest subalgebra of $X$ containing both $C$ and $D$. It may be verified
that the join of $C$ and $D$ is given by the intersection of subalgebras of
$A$ containing both $C$ and $D$ (note that $X$ is a subalgebra of $X$ that
contains $C$ and $D$ and hence this intersection is not empty) and thus the
join of two subalgebras exists (since it may be verified that the intersection
is closed under the operations $p$ and $d$ and moreover, contains the constant
$0$).

Suppose we are given a morphism $f\colon X\rightarrow Y$ and subalgebras $A$
of $X$ and $B$ of $Y$. We will show that $ff^{-1}B=B\wedge\mathsf{Im}f$. Let
$x\in f^{-1}B$. Then $f(x)\in B$ and hence $ff^{-1}B\subseteq B$. Moreover,
$ff^{-1}B\subseteq\mathsf{Im}f$ and hence $ff^{-1}B\subseteq
B\wedge\mathsf{Im}f$. Now take $y\in B\wedge\mathsf{Im}f$. Then $y=f(x)$ for
some $x\in X$. Therefore $f(x)=y\in B$ and hence $x\in f^{-1}B$. Hence
$B\wedge\mathsf{Im}f\subseteq ff^{-1}B$ and thus
$ff^{-1}B=B\wedge\mathsf{Im}f$.

We will now show that $f^{-1}fA=A\vee\mathsf{Ker}f$. Let $x\in A$. Then
$f(x)\in fA$ and hence $x\in f^{-1}fA$. Therefore $A\subseteq f^{-1}fA$. Now
let $x\in\mathsf{Ker}f$. Then $f(x)=0\in fA$ and hence $x\in f^{-1}fA$.
Therefore $\mathsf{Ker}f\subseteq f^{-1}fA$ and hence
$A\vee\mathsf{Ker}f\subseteq f^{-1}fA$. Next, let $x\in f^{-1}fA$. Then
$f(x)\in fA$ and hence there exists $a\in A$ such that $f(x)=f(a)$. Therefore,
$0=d(f(x),f(a))=f(d(x,a))$ and hence $d(x,a)\in\mathsf{Ker}f$. Hence
$x=p(d(x,a),a)\in A\vee\mathsf{Ker}f$ and thus $f^{-1}fA\subseteq
A\vee\mathsf{Ker}f$. Therefore $f^{-1}fA=A\vee\mathsf{Ker}f$. ∎

######  Lemma 2.4.

In every Słomiński algebra, $d(x,y)=0\Leftrightarrow x=y$.

###### Proof.

By the definition of a Słomiński algebra, if $x=y$, then $d(x,y)=0$. It
remains to show that if $d(x,y)=0$, then $x=y$. Suppose $d(x,y)=0$, then
$x=p(d(x,y),y)=p(0,y)=p(d(y,y),y)=y$. ∎

######  Lemma 2.5.

Axiom 3 holds for the Słomiński algebra interpretation.

###### Proof.

Consider a subalgebra $A$ of a Słomiński algebra $X$. Define the map
$\iota_{A}\colon A\rightarrow X$ by $\iota_{A}(x)=x$ for all $x\in A$. Then
$\iota_{A}$ is a morphism since
$\iota_{A}(d(x,y))=d(x,y)=d(\iota_{A}(x),\iota_{A}(y))$ and
$\iota_{A}(p(x,y))=p(\iota_{A}(x),\iota_{A}(y))$ by a similar argument.
Moreover, given any morphism $f\colon B\rightarrow X$ such that
$\mathsf{Im}f\subseteq A$, we may define $f^{\prime}\colon B\rightarrow A$ by
$f^{\prime}(x)=f(x)$. Then clearly $f^{\prime}$ is a morphism since $f$ is a
morphism. Furthermore, $\iota_{A}f^{\prime}(x)=\iota_{A}f(x)=f(x)$ and hence
$f=\iota_{A}f^{\prime}$. If $f=\iota_{A}f^{\prime\prime}$, for a morphism
$f^{\prime\prime}\colon B\rightarrow A/1$, then
$\iota_{A}f^{\prime}=\iota_{A}f^{\prime\prime}$ and hence
$f^{\prime}(x)=f^{\prime\prime}(x)$ for all $x\in B$. Therefore
$f^{\prime}=f^{\prime\prime}$ and $f^{\prime}$ is the unique morphism such
that $f=\iota_{A}f^{\prime}$. Hence $\iota_{A}$ is the embedding associated to
$A$. Note that every subalgebra is conormal.

Now suppose $B$ is a normal subalgebra of $X$. Then there must exist a
morphism $g\colon X\rightarrow Y$ such that $\mathsf{Ker}g=B$. We may then
define a relation on $X$ by $R=\\{(x,y)\mid g(x)=g(y)\\}$. We will write
$y_{1}Ry_{2}$ if $(y_{1},y_{2})\in R$. Then this relation is an equivalence
relation since:

  * —

$g(x)=g(x)$ hence $xRx$ for all $x\in X$. Thus the relation is reflexive.

  * —

If $xRy$, then $g(x)=g(y)$ and thus $g(y)=g(x)$. Hence $yRx$ and the relation
is symmetric.

  * —

If $xRy$ and $yRz$, then $g(x)=g(y)$ and $g(y)=g(z)$. Therefore $g(x)=g(z)$
and thus $xRz$. Hence the relation is transitive.

Now if $g(x_{1})=g(y_{1})$ and $g(x_{2})=g(y_{2})$ then

$g(d(x_{1},x_{2}))=d(g(x_{1}),g(x_{2}))=d(g(y_{1}),g(y_{2}))=g(d(y_{1},y_{2})).$

Therefore if $(x_{1},y_{1}),(x_{2},y_{2})\in R$, then
$d((x_{1},y_{1}),(x_{2},y_{2}))=(d(x_{1},x_{2}),d(y_{1},y_{2}))\in R$.
Similarly, if $(x_{1},y_{1}),(x_{2},y_{2})\in R$, then
$p((x_{1},y_{1}),(x_{2},y_{2}))=(p(x_{1},x_{2}),p(y_{1},y_{2}))\in R$.
Therefore $R$ is a congruence. We then consider the set $X/R$ and define the
operations $p^{\prime}([x],[y])=[p(x,y)]$ and $d^{\prime}([x],[y])=[d(x,y)]$
on $X/R$ and define the constant $0^{\prime}=[0]$. The operations $p^{\prime}$
and $d^{\prime}$ are well-defined since $R$ is a congruence. Then $X/R$ is a
Słomiński algebra since:

  * —

$d^{\prime}([x],[x])=[d(x,x)]=[0]$ and

  * —

$p^{\prime}(d^{\prime}([x],[y]),[y])=p^{\prime}([d(x,y)],[y])=[x]$.

We will then denote $X/R$ as $X/B$ from now on (since the subalgebra $B$
defines the relation). Define a map $\pi_{B}\colon X\rightarrow X/B$ by
$\pi_{B}(x)=[x]$. It may be verified that $\pi_{B}$ is a morphism — this
follows from the definitions of $d^{\prime}$ and $p^{\prime}$. Now

$\mathsf{Ker}\pi_{B}=\\{x\in X\mid[x]=[0]\\}=\\{x\in X\mid
g(x)=g(0)\\}=\\{x\in X\mid g(x)=0\\}=\\{x\in X\mid x\in B\\}=B.$

Suppose $h\colon X\rightarrow Y$ is a morphism such that
$B\subseteq\mathsf{Ker}h$. Then define a map $h^{\prime}\colon X/B\rightarrow
Y$ by $h^{\prime}([x])=h(x)$. Note that $h^{\prime}$ is well-defined since if
$[x]=[y]$, then $g(x)=g(y)$ and $0=d(g(x),g(y))=g(d(x,y))$. It follows that
$d(x,y)\in B$. Therefore,
$d(h^{\prime}([x]),h^{\prime}([y]))=d(h(x),h(y))=h(d(x,y))=0$ (since
$B\subseteq\mathsf{Ker}h$) and thus $h([x])=h([y])$. Moreover, $h^{\prime}$ is
a morphism since
$h^{\prime}(d([x],[y]))=h^{\prime}([d(x,y)])=h(d(x,y))=d(h(x),h(y))=d(h^{\prime}([x]),h^{\prime}([y]))$
and $h^{\prime}(p([x],[y]))=p(h^{\prime}([x]),h^{\prime}([y]))$ by a similar
argument. It is then clear that $h=h^{\prime}\pi_{B}$ since
$h(x)=h^{\prime}([x])=h^{\prime}\pi_{B}(x)$. Moreover, if
$h=h^{\prime\prime}\pi_{B}$ then $h^{\prime\prime}[x]=h^{\prime}[x]$ and thus
$h^{\prime\prime}=h^{\prime}$. This shows that $\pi_{B}$ is a projection
associated to $B$. ∎

######  Lemma 2.6.

Axiom 4 holds for the Słomiński algebra interpretation.

###### Proof.

Suppose $f\colon X\rightarrow Y$ is a morphism. We may define a mapping
$h\colon X/\mathsf{Ker}f\rightarrow\mathsf{Im}f$ by $h([x])=f(x)$. Note that
$h$ is well-defined since $[x]=[y]\Rightarrow f(x)=f(y)\Rightarrow
h([x])=h([y])$. Moreover, $h$ is a morphism since

$h(d([x],[y]))=h([d(x,y)])=f(d(x,y))=d(f(x),f(y))=d(h([x]),h([y]))$

and $h(p([x],[y]))=p(h([x]),h([y]))$ (by a similar calculation). Moreover $h$
is surjective since for any $y\in\mathsf{Im}f$, there exists $x\in X$ such
that $f(x)=y$ and therefore we have $h([x])=y$. Furthermore, if
$h([x])=h([y])$, then $f(x)=f(y)$ and thus $[x]=[y]$. This shows that $h$ is
injective and therefore $h$ is an isomorphism. Notice that
$\iota_{\mathsf{Im}f}h\pi_{\mathsf{Ker}f}(x)=\iota_{\mathsf{Im}f}h([x])=\iota_{\mathsf{Im}f}(f(x))=f(x)$
and hence $f=\iota_{\mathsf{Im}f}h\pi_{\mathsf{Ker}f}$. ∎

######  Lemma 2.7.

Axiom 5 holds for the Słomiński algebra interpretation.

###### Proof.

Note that we should show that normal subalgebras are stable under projections
and conormal subalgebras are stable under embeddings, which is equivalent to
Axiom 5. However, since all subalgebras are conormal, it suffices to show that
normal subalgebras are stable under projections.

Suppose $f\colon X\rightarrow Y$ is a projection (note that $f$ is surjective
by Lemma 2.5) and $K$ is a normal subalgebra of $X$. Then there exists a
morphism $g\colon X\rightarrow Z$ such the $\mathsf{Ker}g=K$. Define a
relation $R$ on $Y$ by

$R=\\{(y_{1},y_{2})\mid\text{$\exists x_{1},x_{2}\in X$ such that
$f(x_{1})=y_{1}$, $f(x_{2})=y_{2}$, $g(x_{1})=g(x_{2})$}\\}.$

Note that we write $y_{1}Ry_{2}$ if $(y_{1},y_{2})\in R$. Then $R$ is an
equivalence relation since:

  * —

for all $y\in Y$, since $f$ is surjective, there exists $x\in X$ such that
$f(x)=y$ and clearly $g(x)=g(x)$. This shows that $yRy$ for all $y\in Y$ and
hence the relation is reflexive.

  * —

If $yRz$, then exist $w$, $x\in X$ such that $f(w)=y$, $f(x)=z$ and
$g(w)=g(x)$. Then clearly $zRx$ by definition of $R$ and hence the relation is
symmetric.

  * —

If $yRz$ and $zRt$, then there exist $v$, $w$, $x\in X$ such that $f(v)=y$,
$f(w)=z$, $f(x)=t$ and $g(v)=g(w)$, $g(w)=g(x)$. Then $g(v)=g(x)$ and $yRt$
and hence the relation is transitive.

Moreover, we will show that $R$ is a congruence. If $(y,z)$, $(t,s)\in R$,
then there exist $a$, $b$, $c$, $d\in X$ such that $f(a)=y$, $f(b)=z$,
$f(c)=t$, $f(d)=s$ and $g(a)=g(b)$, $g(c)=g(d)$. Then $f(d(a,c))=d(y,t)$,
$f(d(b,d))=d(z,s)$ and $g(d(a,c))=g(d(b,d))$. Therefore we obtain

$(d(y,t),d(z,s))\in R.$

Moreover, $(p(y,t),p(z,s))\in R$ by a similar argument. Therefore $R$ is a
congruence. We may then consider the quotient algebra $Y/R$ with constant
$0^{\prime}=[0]$ and binary operations $p^{\prime}([x],[y])=[p(x,y)]$ and
$d^{\prime}([x],[y])=[d(x,y)]$. The operations $p^{\prime}$ and $d^{\prime}$
are well defined since $R$ is a congruence (note that this quotient is a
Słomiński algebra by the same argument as in the proof of Lemma 2.5).
Moreover, the map $\pi\colon Y\rightarrow Y/R$, given by $\pi(y)=[y]$ is a
morphism as in the proof of Lemma 2.5. Now suppose $y\in\mathsf{Ker}\pi$, then
$yR0$. Hence there exists $x$, $u\in X$ such that $f(x)=y$, $f(u)=0$ and
$g(x)=g(u)$. Then $y=fx\in f(g^{-1}gu)=f(u\vee\ker g)=f(u)\vee
f\mathsf{Ker}g=fK$ by applying Lemma 2.3 and property $(G2)$ of the direct
image map. Conversely, if $y\in fK$, then there exists $x\in K$ such that
$y=fx$. Furthermore, since $x\in K$, we have $gx=0$. It follows that $yR0$
since $f0=0$ and $g0=0$. Therefore $\mathsf{Ker}\pi=fK$ and hence $fK$ is
normal. ∎

######  Remark 2.8.

Słomiński algebras form only a particular type of a semi-abelian variety of
universal algebras. In general, a variety of universal algebras, seen as a
category, is semi-abelian if and only if it is a pointed protomodular [3]
variety and if and only if its algebraic theory contains a unique constant
$0$, binary terms $d_{1},\dots,d_{n}$, and an $(n+1)$-ary term $p$ satisfying
the following identities (see [5]):

$\left\\{\begin{array}[]{l}d_{1}(x,x)=\dots=d_{n}(x,x)=0,\\\
p(d_{1}(x,y),\dots,d_{n}(x,y),y)=x.\end{array}\right.$

Such varieties were first introduced and studied by A. Ursini [36, 37]. As the
reader will notice, Słomiński algebras are given by the case when $n=1$ and
$p$ and $d=d_{1}$ are the only basic operations in the variety. All results
contained in this section easily generalize to algebras in an arbitrary semi-
abelian variety. In fact, semi-abelian varieties are precisely where they
generalize: as it follows from [21], validity of the last part of Axiom 2
would force a variety to be semi-abelian.

##  3\. Embedding and Projection Basics

Now we look at some significant consequences of the axioms from Section 1.
Recall firstly the definitions of a monomorphism and an epimorphism from
category theory, which are related to the notions of embeddings and
projections in this context. A morphism $f\colon X\rightarrow Y$ in a category
is a _monomorphism_ if $fg_{1}=fg_{2}\Rightarrow g_{1}=g_{2}$ for any two
parallel morphisms $g_{1}$ , $g_{2}\colon V\to X$. Dually, a morphism $g\colon
X\rightarrow Y$ in a category is an _epimorphism_ if $f_{1}g=f_{2}g\Rightarrow
f_{1}=f_{2}$ for any $f_{1}$, $f_{2}\colon Y\to Z$.

######  Lemma 3.1.

An embedding is a monomorphism, and dually, a projection is an epimorphism.

###### Proof.

Suppose $\iota_{S}\colon S\rightarrow G$ is the embedding of a subgroup $S$ of
a group $G$. We will show that $\iota_{S}$ is a monomorphism. Suppose
$\iota_{S}f=\iota_{S}f^{\prime}$, for parallel morphisms $f$, $f^{\prime}$
with codomain $S$. It is then clear that $\mathsf{Im}(\iota_{S}f)\subseteq S$,
since $\mathsf{Im}\iota_{S}\subseteq S$. Therefore by applying Axiom 3, we
obtain that the morphism $f$ must be the unique morphism such that it makes
the diagram

---
$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{S}}$$\textstyle{G}$$\textstyle{F\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\iota_{S}f}$

commute. Now $f^{\prime}$ satisfies the same condition. Therefore, since $f$
is unique, we may conclude that $f=f^{\prime}$. The statement that a
projection is an epimorphism is dual to the statement that an embedding is a
monomorphism and hence must also be true. ∎

######  Lemma 3.2.

Given two embeddings, $m$ and $m^{\prime}$, associated to the same subgroup
$S$ of a group $G$, we have $m^{\prime}=mi$ for a unique isomorphism $i$.
Moreover, if $m\colon M\rightarrow G$ is an embedding associated to $S$, then
so is $mi$, for any isomorphism $i$ with codomain $M$.

###### Proof.

It is clear that $\mathsf{Im}m\subseteq S$ and $\mathsf{Im}m^{\prime}\subseteq
S$ since $m$ and $m^{\prime}$ are embeddings. Hence we then obtain
$m=m^{\prime}u$ and $m^{\prime}=mv$ (by the definition of an embedding) for
unique morphisms $u$ and $v$. Therefore $m=m\mathsf{id}_{M}=m^{\prime}u=mvu$
and hence $vu=\mathsf{id}_{M}$ since $m$ is a monomorphism (by Lemma 3.1). We
obtain $uv=\mathsf{id}_{M^{\prime}}$ by a similar argument. Therefore $u$ and
$v$ are isomorphisms. The uniqueness of the isomorphism $u$ in the expression
$m^{\prime}=mu$ follows from the fact that $m^{\prime}$ is a monomorphism. Now
suppose $m\colon M\rightarrow G$ is an embedding associated to $S$ and
$i\colon N\rightarrow M$ is an isomorphism. We will show that $mi$ is an
embedding associated to $S$. Firstly, since $\mathsf{Im}m\subseteq S$, we have
$\mathsf{Im}(mi)\subseteq S$. Now suppose $f\colon U\rightarrow G$ is a
morphism such that $\mathsf{Im}f\subseteq S$. Since $m$ is an embedding, we
obtain $f=mu$ for a unique morphism $u\colon U\rightarrow M$. Now since $i$ is
an isomorphism, there exists a morphism $i^{\prime}$ such that
$ii^{\prime}=\mathsf{id}_{M}$ and $i^{\prime}i=\mathsf{id}_{N}$. Hence
$f=mu=m\mathsf{id}_{M}u=mii^{\prime}u=mi(i^{\prime}u)$. It remains to show
that $i^{\prime}u$ is the unique morphism such that $f=mi(i^{\prime}u)$. If
$f=mij$ for a morphism $j$ with codomain $N$, then $ij=u$ since the morphism
$u$ is unique. Hence $j=\mathsf{id}_{N}j=i^{\prime}ij=i^{\prime}u$ and
therefore the morphism $i^{\prime}u$ is unique. ∎

######  Lemma 3.3.

Given two projections $e$ and $e^{\prime}$ associated to the same subgroup
$S$, we have $e=ie^{\prime}$ for a unique isomorphism $i$. Moreover, if
$e:G\rightarrow H$ is an embedding associated to a subgroup $S$, then so is
the composite $ie$ for any isomorphism $i$ with domain $H$.

Note Lemma 3.3 is dual to Lemma 3.2 and hence also holds. Some useful
consequences of Lemmas 3.2 and 3.3 are:

  * —

For a conormal subgroup $S$ of a group $G$, the embeddings associated to $S$
are the composites $\iota_{S}i$ where $i$ is any isomorphism with codomain
$S$.

  * —

For a group $G$, $\mathsf{id}_{G}$ is an embedding associated to $G$ and a
projection associated to $1$. Therefore we must have
$\iota_{G}=\mathsf{id}_{G}i$ for a unique isomorphism $i$. Hence
$\iota_{G}=\mathsf{id}_{G}i=i$ and therefore $\iota_{G}$ is an isomorphism.
Dually $\pi_{1}$ is also an isomorphism.

######  Lemma 3.4.

Any embedding is an embedding associated to its image. Moreover, the image of
an embedding associated to a conormal subgroup $S$ is the subgroup $S$.

###### Proof.

Suppose $f$ is an embedding associated to a subgroup $A$. We will show that
$f$ is also an embedding associated to $\mathsf{Im}f$. Suppose $g$ is a
morphism such that $\mathsf{Im}g\subseteq\mathsf{Im}f$. Then we have
$g=g^{\prime}f$ (for a unique morphism $g^{\prime}$) since
$\mathsf{Im}g\subseteq\mathsf{Im}f\subseteq A$ and $f$ is an embedding
associated to $A$.

For the last part suppose $S$ is a conormal subgroup of a group $G$. Then
there must exist a morphism $f\colon H\rightarrow G$ such that
$\mathsf{Im}f=S$. By the universal property of $\iota_{S}$, we have
$f=\iota_{S}g$ for a unique morphism $g\colon H\rightarrow S/1$. Hence
$S=fH=\iota_{S}gH$ and therefore $S\subseteq\mathsf{Im}\iota_{S}$. Moreover,
by the definition of the embedding $\iota_{S}$, we have
$\mathsf{Im}\iota_{S}\subseteq S$. Therefore $\mathsf{Im}\iota_{S}=S$. ∎

######  Lemma 3.5.

Any projection is a projection associated to its kernel. Moreover, the kernel
of a projection associated to a normal subgroup $S$ of $G$ is the subgroup
$S$.

###### Proof.

The result is dual to Lemma 3.4 and hence must also be true. ∎

######  Lemma 3.6.

A morphism is an embedding if and only if it has trivial kernel.

###### Proof.

Suppose a morphism $f\colon A\rightarrow B$ is an embedding associated to a
conormal subgroup $S$. Then we must have $f=\iota_{S}i$ for an isomorphism
$i\colon A\rightarrow S/1$. Accordingly, there must exist a morphism
$i^{\prime}\colon S/1\rightarrow A$ such that $ii^{\prime}=\mathsf{id}_{S/1}$.
Now by Lemma 3.4, $\mathsf{Im}\iota_{S}=S$. Moreover, by Axiom 4, $f$
factorizes as
$f=\iota_{\mathsf{Im}f}h\pi_{\mathsf{Ker}f}=\iota_{S}h\pi_{\mathsf{Ker}f}$
(since $\mathsf{Im}f=S$ by Lemma 3.4) where $h\colon
S/\mathsf{Ker}f\rightarrow S$ is an isomorphism. Now by Lemma 3.1, $f$ is a
monomorphism and noting that
$f\mathsf{id}_{A}=f=\iota_{S}h\pi_{\mathsf{Ker}f}=\iota_{S}\mathsf{id}_{S/1}h\pi_{\mathsf{Ker}f}=\iota_{S}ii^{\prime}h\pi_{\mathsf{Ker}f}=fi^{\prime}h\pi_{\mathsf{Ker}f}$,
we have $i^{\prime}h\pi_{\mathsf{Ker}f}=\mathsf{id}_{A}$. Hence
$\mathsf{Ker}f=\mathsf{id}_{A}\mathsf{Ker}f=i^{\prime}h\pi_{\mathsf{Ker}f}\mathsf{Ker}f=i^{\prime}h1=1$
and therefore $f$ has trivial kernel.

Conversely, suppose $\mathsf{Ker}f$ is trivial, while $f$ factorizes as
$f=\iota_{\mathsf{Im}f}h\pi_{\mathsf{Ker}f}=\iota_{\mathsf{Im}f}h\pi_{1}$,
where $h$ is an isomorphism. Then if
$\mathsf{id}_{G}=\iota_{G}\bar{h}\pi_{1}$, we have that $\bar{h}\pi_{1}$ is an
isomorphism by applying Lemma 3.2 (since both $\mathsf{id}_{G}$ and
$\iota_{G}$ are embeddings associated to $G$). Hence $\pi_{1}$ is an
isomorphism. Therefore by Lemma 3.2, $f$ is an embedding associated to
$\mathsf{Im}f$. ∎

######  Lemma 3.7.

A morphism is a projection if and only if its image is the largest subgroup of
the codomain.

###### Proof.

The result is dual to Lemma 3.6 and hence must also be true. ∎

######  Lemma 3.8.

A morphism is both an embedding and a projection if and only if it is an
isomorphism.

###### Proof.

Given a morphism $f\colon G\rightarrow H$, by Axiom 4, we may factorize $f$ as
$f=\iota_{\mathsf{Im}f}h\pi_{\mathsf{Ker}f}$ where $h$ is an isomorphism. If
$f$ is both an embedding and a projection, then by Lemma 3.6 and Lemma 3.7,
$\mathsf{Ker}f=1$ and $\mathsf{Im}f=H$. Now by a consequence of Lemma 3.2 and
Lemma 3.3, we obtain that $\iota_{H}$ and $\pi_{1}$ are isomorphisms and hence
$f=\iota_{H}h\pi_{1}$ is an isomorphism. Conversely, suppose $f$ is an
isomorphism. Then there must exist a morphism $f^{\prime}\colon H\rightarrow
G$ such that $f^{\prime}f=\mathsf{id}_{G}$. Therefore
$1=f^{\prime}1=f^{\prime}(ff^{-1}1)=\mathsf{id}_{G}f^{-1}1=f^{-1}1$ and hence
$f$ has trivial kernel. Now by applying duality, we obtain that the image of
$f$ is the largest subgroup of $H$. Therefore, by Lemma 3.6 and Lemma 3.7, we
obtain that $f$ is both an embedding and a projection. ∎

######  Lemma 3.9.

A morphism is an embedding if and only if the corresponding direct image map
is injective. Dually, a morphism is a projection if and only if the
corresponding direct image map is surjective.

###### Proof.

Suppose $f_{*}\colon\mathsf{Sub}X\rightarrow\mathsf{Sub}Y$ is injective. Then
$f$ must have trivial kernel. It is then an embedding by Lemma 3.6.
Conversely, suppose $f$ is an embedding. By Lemma 3.6, we know that $f$ has
trivial kernel. Suppose $f(U)=f(V)$ for two subgroups $U$ and $V$ of $X$. Then
by Axiom 2 we have
$U\vee\mathsf{Ker}f=f^{-1}f(U)=f^{-1}f(V)=V\vee\mathsf{Ker}f$. Since $f$ has
trivial kernel, it follows that $U=V$ and the direct image corresponding to
$f$ is injective. For the second part, notice that the image of morphism $f$
is a projection if and only if its image is the largest subgroup of the
codomain (by Lemma 3.7). If the direct image map corresponding to $f$ is
surjective, then its image must be the largest subgroup of the codomain and
hence, must be a projection. If $f$ is a projection, it follows that
$ff^{-1}S=S\wedge\mathsf{Im}f=S$, for any subgroup $S$ of the codomain (since
$\mathsf{Im}f$ must be the largest subgroup of the codomain). Hence the direct
image map $f_{*}$ corresponding to $f$ is surjective. ∎

######  Lemma 3.10.

Consider a morphism $f\colon X\rightarrow Y$ such that $f=me$, where $m\colon
Z\rightarrow Y$ is an embedding and $e\colon X\rightarrow Z$ is a projection.
If $f=m^{\prime}e^{\prime}$, where $m^{\prime}\colon Z^{\prime}\rightarrow Y$
is an embedding and $e^{\prime}\colon X\rightarrow Z^{\prime}$ is a
projection, then there is an isomorphism $i\colon Z\to Z^{\prime}$ making the
following diagram commute:

$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\scriptstyle{m}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e^{\prime}}$$\scriptstyle{e}$$\textstyle{Y}$$\textstyle{Z^{\prime}.\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{m^{\prime}}$

###### Proof.

Firstly,

$\mathsf{Im}m=mZ=meX=\mathsf{Im}f=m^{\prime}e^{\prime}X=m^{\prime}Z^{\prime}$

and hence there exists a morphism $i\colon Z\rightarrow Z^{\prime}$ such that
$m=m^{\prime}i$. Moreover, since $m$ and $m^{\prime}$ are embeddings, we
obtain that $i$ is an embedding. Furthermore, we have
$m^{\prime}ie=me=m^{\prime}e^{\prime}$ and therefore, since $m^{\prime}$ is a
monomorphism, we obtain $e^{\prime}=ie$. Lastly, since $e$ and $e^{\prime}$
are projections, we have that $i$ is a projection. Therefore $i$ is an
isomorphism and moreover, this isomorphism makes the diagram commute. ∎

As noted in [12], the property established in the following lemma is in fact
equivalent to Axiom 5 (under the rest of the axioms). The proof of this lemma
(as well as the subsequent lemma) is identical to the one given in [12].

######  Lemma 3.11.

Normal subgroups are stable under direct images along projections and conormal
subgroups are stable under inverse images along embeddings.

###### Proof.

Let $N$ be a normal subgroup of $G$ and let $p\colon G\rightarrow H$ be a
projection. Then by Axiom 5, the subgroup $N\vee\mathsf{Ker}p$ is normal. Now
consider the projection $p^{\prime}$ associated to $N\vee\mathsf{Ker}p$, whose
kernel must be $N\vee\mathsf{Ker}p$ itself. Now clearly
$\mathsf{Ker}p\subseteq N\vee\mathsf{Ker}p$ and by the universal property of
the projection $p$, we must have $p^{\prime}=vp$ for some morphism $v$. Then,

$\begin{split}pN&=pN\wedge\mathsf{Im}p=pp^{-1}pN=p(N\vee\mathsf{Ker}p)\\\
&=p(\mathsf{Ker}p^{\prime})=p(p^{-1}v^{-1}1)=pp^{-1}(\mathsf{Ker}v)\\\
&=\mathsf{Ker}v\wedge\mathsf{Im}p=\mathsf{Ker}v\wedge
H=\mathsf{Ker}v.\end{split}$

Note that $\mathsf{Im}p=H$ by Lemma 1.3. Hence $pN$ is normal since
$\mathsf{Ker}v$ is normal. The rest of the proof follows by duality.∎

######  Lemma 3.12.

Consider a morphism $f\colon A\rightarrow B$ and subgroups $X\lhd Y$ of $B$
(see Definition 1.6). If $f^{-1}Y$ is conormal, then $f^{-1}X\lhd f^{-1}Y$.

###### Proof.

Notice that
$\mathsf{Im}(f\iota_{f^{-1}Y})=ff^{-1}Y=Y\wedge\mathsf{Im}f\subseteq Y$. This
implies that there exists a morphism $j\colon Y\rightarrow B$ such that
$f\iota_{f^{-1}Y}=\iota_{Y}j$. Then we have
$\mathsf{Ker}(\pi_{\iota_{Y}^{-1}X}j)=j^{-1}\iota_{Y}^{-1}X=\iota_{f^{-1}Y}^{-1}f^{-1}X$.
Therefore $\iota_{f^{-1}Y}^{-1}f^{-1}X$ is a normal subgroup of $f^{-1}Y$ and
hence $f^{-1}X\lhd f^{-1}Y$.∎

######  Lemma 3.13.

Any two normal subgroups $N$ and $R$ of a group $G$ give rise to a commutative
diagram

$\textstyle{Z}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{y}$$\textstyle{G,\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\scriptstyle{n}$$\scriptstyle{r}$

where $n$, $p$ and $r$ are projections associated to $N$, $N\vee R$ and $R$,
respectively. Moreover, in this diagram, $x$ and $y$ are projections and
$y^{-1}xS=rn^{-1}S$ for any subgroup $S$ of $X$.

###### Proof.

Firstly note that $N\subseteq N\vee R=\mathsf{Ker}p$ and hence, by Axiom 3,
there exists a unique morphism $x\colon X\rightarrow Z$ such that $p=xn$. By a
similar argument, we may obtain the morphism $y$. We then obtain the desired
commutative diagram. We also deduce that $y$ is a projection since $p=yr$,
where $p$ and $r$ are both projections. Similarly $x$ is a projection.

Moreover, by commutativity, we obtain $N\vee R=r^{-1}\mathsf{Ker}y$ and hence
$\mathsf{Ker}y=r(N\vee R)=rN\vee rR=rN$ (noting that $\mathsf{Im}r=Y$). Then:

$\displaystyle
y^{-1}xS=y^{-1}xnn^{-1}S=y^{-1}yrn^{-1}S=rn^{-1}S\vee\mathsf{Ker}y=rn^{-1}S\vee
rN=r(n^{-1}S\vee N)=rn^{-1}S.$

Here, the last equality is obtained by noting that $N=\mathsf{Ker}n\subseteq
n^{-1}S$. ∎

######  Lemma 3.14.

Whenever $A\vee B\subseteq S$, where $S$ is a conormal subgroup of a group,
then we have:

$\iota_{S}^{-1}(A\vee B)=\iota_{S}^{-1}A\vee\iota_{S}^{-1}B.$

###### Proof.

Note that $A\subseteq S$ (since $A\subseteq A\vee B\subseteq S$) and
$B\subseteq S$ (since $B\subseteq A\vee B\subseteq S$). Hence $A\wedge S=A$
and $B\wedge S=B$. Therefore:

$\displaystyle\iota_{S}^{-1}A\vee\iota_{S}^{-1}B$
$\displaystyle=(\iota_{S}^{-1}A\vee\iota_{S}^{-1}B)\vee 1$
$\displaystyle=(\iota_{S}^{-1}A\vee\iota_{S}^{-1}B)\vee\mathsf{Ker}\iota_{S}$
[$\iota_{S}$ is injective]
$\displaystyle=\iota_{S}^{-1}\iota_{S}(\iota_{S}^{-1}A\vee\iota_{S}^{-1}B)$
$\displaystyle=\iota_{S}^{-1}(\iota_{s}\iota_{S}^{-1}A\vee\iota_{S}\iota_{S}^{-1}B)$
$\displaystyle=\iota_{S}^{-1}((A\wedge S)\vee(B\wedge S))$
$\displaystyle=\iota_{S}^{-1}(A\vee B).$

##  4\. The Pyramid

In this section, we consider the question when composition of direct and
inverse image maps of morphisms does produce a new morphism. This gives us a
tool for constructing new morphisms, which proves to be useful for diagram
lemmas such as the Snake Lemma.

A diagram

$\textstyle{X_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}}$$\textstyle{X_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\textstyle{X_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{3}}$$\textstyle{X_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{4}}$$\textstyle{X_{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n}}$$\textstyle{X_{n}}$

of groups and morphisms is called a zigzag. Note that the arrowheads have been
deliberately left out as they are allowed to appear either on the left or on
the right. The groups in a zigzag will be called the nodes of the zigzag. The
opposite zigzag of a given zigzag is simply the zigzag reflected horizontally.
A subgroup $T$ of $X_{n}$ is obtained from a subgroup $S$ of $X_{0}$ by
chasing it forward along the zigzag when:

$T=f_{n}^{\circ}\cdots f_{4}^{\circ}f_{3}^{\circ}f_{2}^{\circ}f_{1}^{\circ}S,$

where $f_{i}^{\circ}=f_{i}$ if the arrowhead in the $i$-th position appears on
the right and $f_{i}^{\circ}=f_{i}^{-1}$ when the arrowhead in the $i$-th
position appears on the left. Note that we may obtain a subgroup $V$ of
$X_{0}$ from a subgroup $U$ of $X_{n}$ by chasing it backwards along the
zigzag in a dual manner. In a zigzag, when all the arrows pointing to the left
are isomorphisms, we say the zigzag is collapsible and the composite
$f_{n}^{\circ}\cdots f_{4}^{\circ}f_{3}^{\circ}f_{2}^{\circ}f_{1}^{\circ}$ is
said to be the induced (homo)morphism, where each $f_{i}^{\circ}$ is defined
as before; however, in this case, $f_{i}^{-1}$ represents the inverse of the
isomorphism $f_{i}$.

From a given a zigzag, we are able to construct a pyramid of morphisms — see
Figure 1,

---
$\textstyle{X_{0}^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{0}^{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}^{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{2}^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{0}^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{2}^{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{3}^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{0}^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{3}^{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{4}^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{0}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{2}^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{4}^{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{0}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{2}^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{3}^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n-1}^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{0}^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{2}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{3}^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{4}^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n-1}^{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n}^{n}}$
Figure 1. A Pyramid of Morphisms

which is a commutative diagram where the base of the pyramid is the given
zigzag. In the pyramid, each upward pointing arrow is a projection and each
downward pointing arrow is an embedding. We reproduce the construction of the
pyramid from [12].

Given a zigzag

$\textstyle{X_{0}^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{2}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{3}^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{4}^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n-1}^{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n}^{n}}$

we construct the triangles in the base of the pyramid

$\textstyle{X^{i}_{i-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{i-1}_{i-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{i}}$$\textstyle{X^{i}_{i}}$

by decomposing the morphism $f_{i}$ into an embedding $m$, associated to the
image of $f_{i}$, composed with a projection $e$, associated to the kernel of
$f_{i}$. The fact that any morphism can be decomposed in this manner follows
from Axiom 4. Note that since the direction is not indicated in the zigzag,
the possibilities for the triangles in the base are

$\textstyle{X^{i}_{i-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{m}$$\textstyle{X^{i-1}_{i-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{i}}$$\scriptstyle{e}$$\textstyle{X^{i}_{i}}$
and
$\textstyle{X^{i}_{i-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{m}$$\textstyle{X^{i-1}_{i-1}}$$\textstyle{X^{i}_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces.}$$\scriptstyle{f_{i}}$$\scriptstyle{e}$

We then obtain the base

$\textstyle{X_{0}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{2}^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{3}^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n-1}^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{0}^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{2}^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{3}^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{4}^{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n-1}^{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n}^{n}}$

of the pyramid. So far the upward pointing arrows are projections and the
downward pointing arrows are embeddings. To build the rest of the pyramid from
this base, we consider a wedge

$\textstyle{X^{i}_{j-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{i+1}_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{i}_{j}}$

and describe how to build a diamond

$\textstyle{X^{i+1}_{j-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{i}_{j-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{i+1}_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{i}_{j}}$

(by describing how to construct the top wedge in the diamond). If both arrows
in the wedge are pointing upward, they must be projections and we may complete
it to a projection diamond by Lemma 3.13. Dually, if both arrows point
downward, they must be embeddings and we may complete it to an embedding
diamond by the dual of Lemma 3.13.

There are still two remaining cases for the given wedge, which are

$\textstyle{X^{i}_{j-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{i+1}_{j}}$$\textstyle{X^{i}_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$
and
$\textstyle{X^{i}_{j-1}}$$\textstyle{X^{i+1}_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{i}_{j}.\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

In both cases, we consider the composite of the arrows in the wedge as
indicated by the dotted arrows in the wedges

$\textstyle{X^{i}_{j-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{i+1}_{j}}$$\textstyle{X^{i}_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$
and
$\textstyle{X^{i}_{j-1}}$$\textstyle{X^{i+1}_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{i}_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

respectively. Then, in both cases, we construct the top wedge (and hence the
diamond) by decomposing the dotted arrow as an embedding composed with a
projection (as we did to build the triangles in the base). This decomposition
is again guaranteed by Axiom 4.

It is then possible to construct the pyramid by repeating the construction of
the diamonds to obtain every next layer until we get to the top, which will
complete the construction.

Note that in each case of constructing the diamond, we have that the upward
pointing arrows are projections and the downward pointing arrows are
embeddings. This property is also true for the construction of the triangles
in the base of the pyramid. Therefore, all arrows pointing upward in the
pyramid are projections and all arrows pointing downward in the pyramid are
embeddings.

$\textstyle{X_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g_{0}}$$\scriptstyle{f_{1}}$$\textstyle{X_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g_{1}}$$\scriptstyle{f_{2}}$$\textstyle{X_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g_{2}}$$\scriptstyle{f_{3}}$$\textstyle{X_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g_{3}}$$\scriptstyle{f_{4}}$$\textstyle{X_{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g_{4}}$$\textstyle{X_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g_{n}}$$\textstyle{X^{\prime}_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{\prime}_{1}}$$\textstyle{X^{\prime}_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{\prime}_{2}}$$\textstyle{X^{\prime}_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{3}}$$\textstyle{X^{\prime}_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{4}}$$\textstyle{X^{\prime}_{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X^{\prime}_{n}}$
Figure 2. Diagram connecting two zigzags

Consider a diagram in Figure 2, where both the top and the bottom rows are
zigzags. A square

$\textstyle{X_{i-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{i}}$$\scriptstyle{g_{i-1}}$$\textstyle{X_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g_{i}}$$\textstyle{X^{\prime}_{i-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{\prime}_{i}}$$\textstyle{X^{\prime}_{i}}$

from the diagram in Figure 2 is said to be commutative when

$\begin{cases}g_{i}f_{i}=f^{\prime}_{i}g_{i-1},&\text{if $f_{i}$ and
$f^{\prime}_{i}$ are both right-pointing,}\\\
g_{i-1}f_{i}=f^{\prime}_{i}g_{i},&\text{if $f_{i}$ and $f^{\prime}_{i}$ are
both left-pointing,}\\\ g_{i-1}=f^{\prime}_{i}g_{i}f_{i},&\text{if $f_{i}$ is
right-pointing and $f^{\prime}_{i}$ is left-pointing,}\\\
g_{i}=f^{\prime}_{i}g_{i-1}f_{i},&\text{if $f_{i}$ is left-pointing and
$f^{\prime}_{i}$ is right-pointing.}\end{cases}$

######  Lemma 4.1.

When in a commutative diagram

$\textstyle{X_{i-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{i}}$$\scriptstyle{g_{i-1}}$$\textstyle{X_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g_{i}}$$\textstyle{X^{\prime}_{i-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{\prime}_{i}}$$\textstyle{X^{\prime}_{i}}$

from the diagram in Figure 2, either the top or the bottom arrows are
isomorphisms, replacing any of these arrows with their inverses (pointing in
the opposite direction), still results in a commutative diagram.

###### Proof.

Let us assume $f_{i}$ is an isomorphism. If both $f_{i}$ and $f^{\prime}_{i}$
are right-pointing, then $g_{i}f_{i}=f^{\prime}_{i}g_{i-1}$. Hence, if we
apply the inverse image map of $f_{i}$ on the right of both sides, we obtain
$g_{i}=f^{\prime}_{i}g_{i-1}f_{i}^{-1}$. This shows that if we replace $f_{i}$
with its inverse image map then the resulting diagram is commutative. If
$f_{i}$ is left-pointing and $f^{\prime}_{i}$ is right-pointing, then
$g_{i}=f^{\prime}_{i}g_{i-1}f_{i}$. If we then apply the inverse image map on
the right of both sides, we obtain $g_{i}f_{i}^{-1}=f^{\prime}_{i}g_{i-1}$.
The remaining two cases follow by a similar argument to the ones above.
Moreover, the case that $f^{\prime}_{i}$ is an isomorphism follows by duality.
∎

######  Lemma 4.2.

In a diagram given by Figure 2, where all squares are commutative, if the top
and bottom rows are collapsible, then $g_{n}f=f^{\prime}g_{0}$ where $f$ and
$f^{\prime}$ are the morphisms induced by the top and the bottom rows,
respectively. Moreover, if all morphisms $g_{i}$ are isomorphisms such that
replacing them with their inverses pointing up retain commutativity in each
square in Figure 2, then the top row is collapsible if and only if the bottom
row is collapsible.

###### Proof.

If we apply Lemma 4.1 to the top and bottom rows, we may then apply
commutativity of each square to deduce $g_{n}f=f^{\prime}g_{0}$. We next
assume all the morphisms $g_{i}$ are isomorphisms such that replacing them
with their inverses pointing up retain commutativity in each square in Figure
2. Suppose the bottom row is collapsible and consider a left-pointing morphism
$f_{i}$ in the top row. If $f^{\prime}_{i}$ is left-pointing (so that, in
particular, it is an isomorphism), then it follows that $f_{i}$ is an
isomorphism (since the remaining three morphisms in the commutative square are
isomorphisms). If $f^{\prime}_{i}$ is right-pointing, then
$g_{i}=f^{\prime}_{i}g_{i-1}f_{i}$ and hence $f_{i}$ is injective. The
commutative square formed by replacing $g_{i}$ and $g_{i-1}$ with their
inverses yields $g_{i-1}^{-1}=f_{i}g_{i}^{-1}f^{\prime}_{i}$. It follows that
$f_{i}$ is surjective and therefore is an isomorphism. Hence the top row is
collapsible. A similar argument shows that if top row is collapsible, then the
bottom row is collapsible. ∎

######  Lemma 4.3.

Given a diagram in Figure 2, where all squares commute and the top and the
bottom rows are collapsible zigzags, if $g_{0}$ and $g_{n}$ are identity
morphisms then both collapsible zigzags induce the same morphism.

###### Proof.

The result follows by setting $g_{0}=\mathsf{id}_{X_{0}}$ and
$g_{n}=\mathsf{id}_{X_{n}}$ and applying Lemma 4.2. ∎

######  Lemma 4.4.

Two pyramids built from the same zigzag admit isomorphisms between the
corresponding nodes so that the diagram formed by the two pyramids and the
isomorphisms is commutative.

###### Proof.

The isomorphisms of the triangles in the base are obtained by identity
morphisms (to connect the zigzags which are the same) and Lemma 3.10 (for the
top node of the triangle). Furthermore, this guarantees commutativity for the
triangles in the base of both pyramids together with the isomorphisms.
Moreover, two arrows in corresponding triangles in the base of the pyramid
point in the same direction due to the construction of these triangles. If we
then assume that the isomorphisms between the nodes in the bottom wedge of a
diamond and the diagram formed by the two corresponding wedges together with
its isomorphisms is commutative, we will show that the top nodes of the
diamonds, formed from these wedges, are isomorphic. Suppose one of the
diamonds considered is a projection diamond, then due to the isomorphisms
connecting the bottom wedges, the remaining diamond must also be a projection
diamond. Moreover, using the commutativity of the bottom wedges and their
isomorphisms, we deduce that the direct image of the join of the kernels of
the bottom wedge is preserved by the direct image of the isomorphism
corresponding to the bottom node. We then conclude that the vertical
projection (with the top node as the codomain) composed with the isomorphism
of the bottom node yields a projection associated to the same subgroup as the
one associated to the vertical projection of the other diamond. Therefore, by
applying Lemma 3.3, we obtain the desired isomorphism. Note that this
isomorphism makes the diagram commutative. A dual argument handles the case of
embedding diamonds. For the remaining two cases, we apply Lemma 3.10 to obtain
the isomorphism of the top node. Moreover, this also yields commutativity. ∎

######  Definition 4.5.

A subquotient is a zigzag in which all left pointing arrows are embeddings and
all right pointing arrows are projections.

######  Lemma 4.6.

The opposite zigzag of a subquotient is collapsible if and only if chasing the
trivial subgroup of the final node backward along the zigzag results in the
trivial subgroup of the initial node.

###### Proof.

The left pointing arrows in the opposite zigzag of a subquotient are
projections. Therefore the opposite zigzag of a subquotient is collapsible if
and only each of the projections has trivial kernel (i.e., they are associated
to the trivial subgroup of the corresponding node). If the opposite zigzag of
a subquotient is collapsible, then it is clear that chasing the trivial
subgroup backward along the zigzag results in the trivial subgroup of the
initial node (since each right pointing arrow would have trivial kernel).
Conversely if we assume that chasing the trivial subgroup of the final
subgroup backward along the zigzag results in the trivial subgroup of the
initial node, then we obtain that chasing the trivial subgroup of the final
node backward along the zigzag up to the node to the right of the initial node
results in the trivial subgroup. This is true since if the first morphism was
left pointing, it must be an embedding and hence it would have trivial kernel.
If it were right pointing, then it must be a projection and hence we may take
the direct image map and obtain the desired conclusion (noting that the
inverse image maps of projections are right inverses of their direct image
maps). By repeating this process, we may deduce that chasing the trivial
subgroup of the final node results in the trivial subgroup at each node.
Therefore each right pointing morphism must be injective and hence all
projections are isomorphisms (since right pointing morphisms are exactly the
projections). Hence, the left pointing arrows in the opposite zigzag are
isomorphisms. ∎

######  Definition 4.7.

Consider a zigzag

$\textstyle{X_{p_{1}}^{q_{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{p_{2}}^{q_{2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{p_{3}}^{q_{3}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{p_{n}}^{q_{n}}}$

obtained from the pyramid. The zigzag is said to be horizontal when
$\\{p_{i+1}-p_{i},q_{i+1}-q_{i}\\}=\\{0,1\\}$ (i.e., when $p_{i+1}=p_{i}$ and
$q_{i+1}-q_{i}=1$ or when $p_{i+1}-p_{i}=1$ and $q_{i+1}=q_{i}$) for each
$i\in\\{1,2,\dots,n-1\\}$. The zigzag is said to be vertical when
$\\{p_{i}-p_{i+1},q_{i+1}-q_{i}\\}=\\{0,1\\}$ (i.e., when $p_{i}=p_{i+1}$ and
$q_{i+1}-q_{i}=1$ or when $p_{i}-p_{i+1}=1$ and $q_{i+1}=q_{i}$) for each
$i\in\\{1,2,\dots,n-1\\}$.

The horizontal zigzag that joins $X_{0}^{0}$ with $X_{n}^{n}$, that runs up
the left side and down the right side of the pyramid, is said to be the
principal horizontal zigzag of the pyramid. The vertical zigzag going up along
the left side of the triangular outline of the pyramid, which joins
$X_{0}^{0}$ and $X_{n}^{0}$, is said to be the left principal vertical zigzag
of the pyramid. Similarly, the vertical zigzag going up the right side of the
triangular outline of the pyramid, which joins $X_{n}^{n}$ with $X_{0}^{n}$,
is said to be the right principal vertical zigzag of the pyramid.

######  Lemma 4.8.

Vertical zigzags are subquotients.

###### Proof.

By definition of a vertical zigzag, the right pointing arrows coincide with
the arrows pointing upward and the left pointing arrows coincide with the
arrows pointing downward. The result then follows by noting that, by
construction, the upward and downward pointing arrows of the pyramid are
projections and embeddings respectively. ∎

######  Lemma 4.9.

Chasing a subgroup from one node to another downward via a vertical zigzag and
then upward via the same zigzag back to the original node gives back the
original subgroup.

###### Proof.

We know that vertical zigzags are subquotients by Lemma 4.8. Hence chasing a
subgroup downward via a vertical zigzag applies direct image maps of
embeddings and inverse image maps of projections to the subgroup. If we then
chase the subgroup upward via the same zigzag, we would further apply inverse
image maps of embeddings and direct image maps of projections. However, since
the direct image map of an embedding is a right inverse of its corresponding
inverse image map and since the inverse image map of a projection is a right
inverse of its corresponding direct image map, we are able to cancel out each
direct image map with its inverse image map and we are left with the original
subgroup. ∎

######  Lemma 4.10.

Chasing a trivial/largest subgroup upward along a vertical zigzag results in a
trivial/largest subgroup.

###### Proof.

Vertical zigzags are subquotients by Lemma 4.8. Therefore, since projections
are surjective, we have that chasing the largest subgroup upward along a
vertical zigzag results in a largest subgroup. Moreover, since embeddings are
injective, chasing a trivial subgroup upward results in a trivial subgroup. ∎

######  Lemma 4.11.

Chasing a trivial/largest subgroup downward/upward along a vertical zigzag
results in a trivial/largest subgroup if and only if the zigzag is collapsible
in the downward/upward direction.

###### Proof.

Vertical zigzags are subquotients by Lemma 4.8. The result then follows by
applying Lemma 4.6 and its dual. ∎

######  Lemma 4.12.

Chasing a subgroup from one node to another forward or backward along a
horizontal zigzag does not depend on the choice of path.

###### Proof.

By Lemma 3.13 and its dual, we note that chasing a subgroup along the top
wedge of a projection or embedding diamond, is equivalent to chasing it along
the bottom. Moreover, this property is clear for the remaining two cases of
diamonds in the pyramid. From this, we deduce that the pyramid is commutative
in the sense that chasing a subgroup along the top wedge of a diamond yields
the same result as chasing along the bottom wedge of the same diamond. Now
consider any horizontal zigzag between a node $X_{p_{i}}^{q_{i}}$ and
$X_{p_{j}}^{q_{j}}$. By applying the commutativity of each diamond, for which
the horizontal zigzag includes the bottom wedge of the diamond, we obtain that
chasing a subgroup along this zigzag yields the same result as chasing the
subgroup along the horizontal zigzag that joins $X_{p_{i}}^{q_{i}}$ with
$X_{p_{i}}^{q_{j}}$ followed by chasing along the horizontal zigzag joining
$X_{p_{i}}^{q_{j}}$ with $X_{p_{j}}^{q_{j}}$. Therefore chasing a subgroup
along a horizontal zigzag between two nodes does not depend on the choice of
path. ∎

######  Definition 4.13.

When the principal horizontal zigzag of a pyramid constructed from a given
zigzag is collapsible, the given zigzag is said to induce a (homo)morphism and
the morphism induced by the principal horizontal zigzag is said to be the
induced (homo)morphism of the given zigzag.

######  Lemma 4.14.

For a given zigzag, the induced morphism is unique when it exists (in
particular, it does not depend on the actual pyramid).

###### Proof.

Consider a zigzag which induces a morphism using a specific pyramid
constructed from the zigzag. If we have any other pyramid constructed from
this zigzag, we are able to connect the two pyramids with isomorphisms to form
a commutative diagram (by Lemma 4.4). Notice that in each commutative square
formed by the isomorphisms, replacing both the isomorphisms with their
inverses preserves commutativity (by the construction of the isomorphisms in
each case). Therefore, by Lemma 4.2, we obtain that the principal horizontal
zigzag (of the second pyramid) is collapsible. Moreover, the morphisms
connecting the initial and final nodes of the zigzag are identity morphisms.
Hence, by applying Lemma 4.3, we obtain that the induced morphisms are equal.
∎

The following lemma shows that for a collapsible zigzag, the notion of an
induced morphism introduced at the start of this section agrees with the one
introduced in Definition 4.13.

######  Lemma 4.15.

When a zigzag is collapsible, the principal horizontal zigzag of the
corresponding pyramid is collapsible, and moreover, the morphism induced by
the first collapsible zigzag is the same as the one induced by the principal
horizontal zigzag.

###### Proof.

We will first show that morphisms in the pyramid pointing north-west or south-
west are isomorphisms. This is true for the triangles in the base since the
south-west and north-west pointing arrows arise from the decomposition of
left-pointing arrows into an embedding of its image composed with a projection
of its kernel. For projection diamonds, if the north-west pointing arrow in
the bottom wedge is an isomorphism, then by commutativity of the diamond
(guaranteed by Lemma 3.13), we obtain that the north-west pointing morphism in
the top wedge is an isomorphism. By duality, if the south-west pointing
morphism in the bottom wedge of an embedding diamond is an isomorphism then
the south-west pointing morphism in the top wedge is an isomorphism. In the
case of the bottom wedge being a north-west pointing morphism composed with a
south-west pointing morphism being isomorphisms, we obtain that the
constructed vertical morphism is an isomorphism and hence the north-west and
south-west pointing morphisms in the top wedge are isomorphisms. Note that the
last case for the diamond does not contain north-west or south-west pointing
morphisms and hence we do not need to consider it. This shows that all south-
west and north-west pointing morphisms in the pyramid are isomorphisms.
Moreover, this shows that the principal vertical zigzag is collapsible and the
zigzag induces a morphism (in the sense of the pyramid). Moreover, by
commutativity of the pyramid, the morphism induced by the principal horizontal
zigzag is the same as the morphism induced by the original zigzag. ∎

######  Lemma 4.16.

When a zigzag and its opposite zigzag are both collapsible, the morphisms
induced by both of these zigzags are isomorphisms and are inverses of each
other.

###### Proof.

Suppose we are given a zigzag

$\textstyle{X_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}}$$\textstyle{X_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\textstyle{X_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{3}}$$\textstyle{X_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{4}}$$\textstyle{X_{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n}}$$\textstyle{X_{n}}$

such that both the zigzag and its opposite

$\textstyle{X_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n}}$$\textstyle{X_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n-1}}$$\textstyle{X_{n-2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n-2}}$$\textstyle{X_{n-3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n-3}}$$\textstyle{X_{n-4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}}$$\textstyle{X_{0}}$

are collapsible. Note that, in the opposite zigzag, each $f_{i}$ points in the
opposite direction to the $f_{i}$ in the original zigzag. Since the original
zigzag is collapsible, the arrows in it that point to the left are
isomorphisms. Moreover, the arrows pointing to the right in the original
zigzag are isomorphisms too, since they point to the left in the opposite
zigzag, which is also collapsible. It then follows that morphisms induced by
the given zigzag and its opposite are isomorphisms that are inverses of each
other. ∎

######  Theorem 4.17 (Homomorphism Induction Theorem, [12]).

For a zigzag to induce a homomorphism it is necessary and sufficient that
chasing the trivial subgroup of the initial node forward along the zigzag
results in the trivial subgroup of the final node, and chasing the largest
subgroup of the final node backward along the zigzag results in the largest
subgroup of the initial node. Moreover, when a zigzag induces a homomorphism,
the induced homomorphism is unique and the direct and inverse image maps of
the induced homomorphism are given by chasing a subgroup forward and backward,
respectively, along the zigzag.

###### Proof.

If we apply commutativity, we may chase subgroups along the principal
horizontal zigzag of the pyramid. Chasing the trivial subgroup upward along
the principal left vertical zigzag results in a trivial subgroup by Lemma
4.10. If we then continue by chasing the trivial subgroup downward along the
right principal vertical zigzag and obtain a trivial subgroup, then by Lemma
4.11, we deduce that the right principal vertical zigzag is collapsible in the
downward direction. Similarly, chasing a largest subgroup upward along a
vertical zigzag results in a largest subgroup by Lemma 4.10. If we then
continue chasing a largest subgroup downward along the left principal zigzag,
by the dual of Lemma 4.6, we obtain that the left principal zigzag is
collapsible in the upward direction. Therefore the principal horizontal zigzag
is collapsible. Hence, the zigzag induces a morphism and moreover, the induced
morphism is unique by Lemma 4.14. Again by commutativity, we may reduce
chasing subgroups forward and backwards along the principal horizontal zigzag
to chasing along the original zigzag (noting that, for a given morphism, its
inverse image coincides with the the direct image of the inverse morphism). ∎

######  Lemma 4.18.

A zigzag induces a homomorphism if and only if the opposite zigzag induces a
homomorphism in the dual category. Moreover, the homomorphisms induced in both
categories are the same.

###### Proof.

It follows from the Homomorphism Induction Theorem that a zigzag induces a
morphism if and only if in the opposite category, chasing the largest subgroup
of the initial node forward yields the largest subgroup of the final node and
chasing the smallest subgroup of the final node backward yields the smallest
subgroup of the initial node. This is equivalent to the opposite of the zigzag
inducing a homomorphism in the opposite category. When this is the case, if we
horizontally reflect a pyramid, constructed from the original zigzag, and
consider it in the opposite category, it becomes a pyramid for the opposite
zigzag in the opposite category. It follows that the induced morphisms are the
same. ∎

######  Theorem 4.19 (Universal Isomorphism Theorem, [12]).

If the largest and smallest subgroups are preserved by chasing them from each
end node of the zigzag to the opposite end node, then it induces a
homomorphism and the induced homomorphism is an isomorphism.

###### Proof.

If we apply the Homomorphism Induction Theorem, Theorem 4.17, we obtain that
the zigzag induces a morphism that is both injective and surjective. Therefore
the zigzag induces an isomorphism. ∎

######  Remark 4.20.

If a zigzag satisfies the assumptions of the Universal Isomorphism Theorem,
then so does the opposite zigzag. Furthermore, it follows from Lemma 4.16,
that the isomorphisms induced by the zigzag and its opposite are inverses of
each other.

The following is an application of the Universal Isomorphism Theorem.

######  Theorem 4.21 ([12]).

Consider any morphism $f\colon A\rightarrow B$, and subgroups
$\mathsf{Ker}f\subseteq W\subseteq X$ of $A$, where $X$ is conormal. Then
$W\lhd X$ if and only if $fW\lhd fX$, and when this is the case, there is an
isomorphism $X/W\approx fX/fW$.

###### Proof.

First note that since $X$ is conormal, the embedding $\iota_{X}$ exists and
moreover, $\iota_{X}X/1=X$. Therefore $fX=\mathsf{Im}(f\iota_{X})$. We then
obtain the commutative diagram

$\textstyle{X/1\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{X}}$$\scriptstyle{\pi_{\mathsf{Ker}(f\iota_{X})}}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{B}$$\textstyle{(X/1)/\mathsf{Ker}(f\iota_{X})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h}$$\textstyle{fX/1.\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota_{fX}}$

The factorization of the morphism $f\iota_{X}$ is obtained by Axiom 4; note
that the morphism $h$ is an isomorphism. If $W\lhd X$, then $\iota_{X}^{-1}W$
is a normal subgroup of $X/1$. Furthermore,
$h\pi_{\mathsf{Ker}(f\iota_{X})}\iota_{X}^{-1}W$ is a normal subgroup of
$fX/1$ by Lemma 3.11. However

$\displaystyle h\pi_{\mathsf{Ker}(f\iota_{X})}\iota_{X}^{-1}W$
$\displaystyle=h\pi_{\mathsf{Ker}(f\iota_{X})}\iota_{X}^{-1}(W\vee\mathsf{Ker}f)$
[$\mathsf{Ker}f\subseteq W$]
$\displaystyle=h\pi_{\mathsf{Ker}(f\iota_{X})}\iota_{X}^{-1}f^{-1}fW$
$\displaystyle=h\pi_{\mathsf{Ker}(f\iota_{X})}\pi_{\mathsf{Ker}(f\iota_{X})}^{-1}h^{-1}\iota_{fX}^{-1}fW$
$\displaystyle=h(h^{-1}\iota_{fX}^{-1}fW\wedge\mathsf{Im}\pi_{\mathsf{Ker}(f\iota_{X})})$
$\displaystyle=hh^{-1}\iota_{fX}^{-1}fW$ [$\pi_{\mathsf{Ker}(f\iota_{X})}$ is
surjective] $\displaystyle=\iota_{fX}^{-1}W.$

Therefore $fW\lhd fX$. Conversely if $fW\lhd fX$, then $\iota_{fX}^{-1}fW$ is
a normal subgroup of $fX/1$. Moreover we have that

$W=W\vee\mathsf{Ker}f=f^{-1}fW\lhd f^{-1}fX=X\vee\mathsf{Ker}f=X,$

where the normality is obtained by Lemma 3.12.

Moreover, whenever these equivalent cases occur, the desired isomorphism
between $X/W$ and $fX/fW$ is induced by the zigzag

$X/W\xleftarrow{\pi_{\iota_{X}^{-1}}W}X/1\xrightarrow{\iota_{X}}A\xrightarrow{f}B\xleftarrow{\iota_{fX}}fX/1\xrightarrow{\pi_{\iota_{fX}^{-1}}fW}fX/fW.\qed$

Lastly, we generalize the result in [12], which gives a necessary and
sufficient condition, based on the relations induced by morphisms, for when a
zigzag of homomorphisms of groups (in the ordinary sense) induces a morphism.
In the more general result presented below, ordinary groups are replaced with
Słomiński algebras.

Consider a morphism $f\colon X\rightarrow Y$. Then $f$ defines a relation
$\bar{f}$ over the sets $X$ and $Y$ given by $x\bar{f}y\Leftrightarrow
f(x)=y$. Furthermore, the opposite relation $\bar{f}^{\mathsf{op}}$ is given
by $y\bar{f}^{\mathsf{op}}x\Leftrightarrow f(x)=y$.

Consider a zigzag

$\textstyle{X_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}}$$\textstyle{X_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{2}}$$\textstyle{X_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{3}}$$\textstyle{X_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{4}}$$\textstyle{X_{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n}}$$\textstyle{X_{n}.}$

We write $f_{i}^{\bullet}=\bar{f_{i}}$ if the arrowhead in the $i$-th place in
the zigzag appears on the right and
$f_{i}^{\bullet}=\bar{f_{i}}^{\mathsf{op}}$ if the arrowhead in the $i$-th
place in the zigzag appears on the left. The _induced relation_ of the zigzag
is the relation given by the composite $f_{n}^{\bullet}\cdots
f_{4}^{\bullet}f_{3}^{\bullet}f_{2}^{\bullet}f_{1}^{\bullet}$ of relations
$f_{i}^{\bullet}$.

The proof of the following theorem is adapted from the proof of the
corresponding result for groups in [12]. Although this falls outside the scope
of the present paper, it should be possible to generalize this result to
arbitrary semi-abelian categories.

######  Theorem 4.22 ([10]).

A zigzag of Słomiński algebra morphisms induces a morphism if and only if the
induced relation is a function. Moreover, when this is the case, the function
is precisely the induced morphism.

###### Proof.

We first show that any two horizontal zigzags between two nodes in the pyramid
induce the same relation. For horizontal zigzags consisting of arrows staying
in the base of the pyramid, the result is clear. It then suffices to show that
for each diamond in the pyramid, the relation induced by the top wedge is the
same as the relation induced by the bottom wedge. We prove this property for
projection diamonds only, since the property for embedding diamonds follows by
duality and the remaining two cases are clear. Consider a projection diamond

$\textstyle{Z}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{y}$$\textstyle{G.\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\scriptstyle{n}$$\scriptstyle{r}$

Then it follows, by commutativity, that the relation induced by the bottom
wedge is smaller than the one induced by the top wedge. To show that the
relation induced by the bottom wedge is bigger than the relation induced by
the top wedge, consider elements $a\in X$ and $b\in Y$ such that $x(a)=y(b)$.
Then there must exist $c\in G$ such that $n(c)=a$ since $n$ is surjective. Now
the element $d(b,r(c))$ must belong to $\mathsf{Ker}y$ since

$y(d(b,r(c)))=d(y(b),yr(c))=d(y(b),xn(c))=d(y(b),x(a))=0.$

Moreover,

$\displaystyle\mathsf{Ker}n\vee\mathsf{Ker}r=\mathsf{Ker}p=\mathsf{Ker}(yr)=r^{-1}\mathsf{Ker}y$
$\displaystyle\Rightarrow
r(\mathsf{Ker}n\vee\mathsf{Ker}r)=rr^{-1}\mathsf{Ker}y$
$\displaystyle\Rightarrow r\mathsf{Ker}n\vee
r\mathsf{Ker}r=\mathsf{Ker}y\wedge\mathsf{Im}r$
$\displaystyle\Rightarrow\mathsf{Ker}y=r\mathsf{Ker}n.$

Therefore there exists $d\in\mathsf{Ker}n$ such that $r(d)=d(b,r(c))$. Let
$e=p(d,c)$. Then

$n(e)=n(p(d,c))=p(n(d),n(c))=p(0,a)=a$

and

$r(e)=r(p(d,c))=p(r(d),r(c))=p(d(b,r(c)),r(c))=b.$

Therefore, the relation induced by the bottom wedge is indeed bigger than the
relation induced by the top wedge and hence we may conclude that the induced
relations are the same. Thus we have shown that any two horizontal zigzags
between two nodes in the pyramid induce the same relation. This means that the
relation induced by a zigzag is the same as the relation induced by the
corresponding principal horizontal zigzag. If the principal horizontal zigzag
is collapsible, then it is clear that the induced relation is the induced
morphism. Conversely, suppose that the induced relation is a function. We will
show that the principal horizontal zigzag is collapsible. Consider the first
left-pointing arrow (which is an embedding) in the left principal vertical
zigzag. Then any element in its codomain must be related to an element in its
domain since the previous morphisms in the left principal vertical zigzag are
surjections and since the element must be related to an element of
$X_{n}^{n}$. Therefore the arrow is surjective and is hence an isomorphism. A
similar argument shows that all left pointing arrows in the zigzag connecting
$X_{n}^{n}$ and $X_{0}^{n}$ are isomorphisms. ∎

##  5\. Classical Diagram Lemmas

In this section, we state and prove the classical diagram lemmas of
homological algebra (found, e.g., in [30, 31]) in the self-dual axiomatic
context described in Section 1. The diagram lemmas will be formulated in terms
of exact sequences and injective and surjective morphisms, where in this
context injectivity and surjectivity refer to the injectivity and surjectivity
of the direct image maps. Notice that injective morphisms are precisely
embeddings and dually, surjective morphisms are precisely projections by Lemma
3.9. Most of these lemmas can be broken up into two dual halves, which, thanks
to the axiomatic context being invariant under duality, allows one to
establish the lemma by proving its one half only.

In our proofs we use an equational (algebraic) method of chasing subgroups,
following the method found in [14]. With certain auxiliary lemmas, it is
possible to develop a more geometric diagram chasing method, similar to the
one used in e.g., [30, 31, 19, 22, 33] (see, in particular, Remark 2.8 in
[22]). In the geometric approach, in a single diagram chase one only makes use
of either the direct image maps, or the inverse image maps, and bottom and top
subgroups respectively, but otherwise does not make use of the order of
subgroup lattices (i.e., the subgroup lattices are viewed as merely pointed
sets). However, it is not known whether the geometric approach can give rise
to criteria for homomorphism induction for group-like structures that would be
invariant under the duality of the corresponding noetherian form (note that
the criteria in our Homomorphism Induction Theorem makes use of direct and
inverse image maps simultaneously). Therefore, in this paper, we leave out
entirely the geometric approach to diagram chasing.

We will use an arrow of the form
$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y}$
to represent an injective morphism in a diagram, and we will use
$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y}$
to represent a surjective morphism.

######  Definition 5.1.

A sequence of groups and morphisms, $G\xrightarrow{f}H\xrightarrow{g}I$, is
said to be exact at $H$ if $\mathsf{Im}f=\mathsf{Ker}g$.

######  Definition 5.2.

A sequence of groups and morphisms, $X_{1}\rightarrow
X_{2}\rightarrow\dots\rightarrow X_{n-1}\rightarrow X_{n}$, is exact if it is
exact at each node in the sequence except the end nodes $X_{1}$ and $X_{n}$.

According to Mac Lane [30], the following diagram lemma was originally
formulated by J. Leicht. In [30] it was stated and proved for the context of
modules.

######  Theorem 5.3 (Four Lemma).

Consider a commutative diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{s}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\scriptstyle{t}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h}$$\scriptstyle{u}$$\textstyle{D\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{v}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x}$$\textstyle{B^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{y}$$\textstyle{C^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{z}$$\textstyle{D^{\prime},}$

where the rows are exact sequences. Then:

  1. (i)

$g(\mathsf{Ker}t)=\mathsf{Ker}u$,

  2. (ii)

$y^{-1}(\mathsf{Im}u)=\mathsf{Im}t$.

In particular, if $t$ is injective, then so is $u$, and when $u$ is
surjective, so is $t$ (Weak Four Lemma).

###### Proof.

(i) holds because

$\displaystyle\mathsf{Ker}u$ $\displaystyle=u^{-1}1\wedge u^{-1}z^{-1}1$
[$u^{-1}1\subseteq u^{-1}z^{-1}1$] $\displaystyle=u^{-1}1\wedge h^{-1}v^{-1}1$
[by commutativity] $\displaystyle=u^{-1}1\wedge h^{-1}1$ [$v$ is injective]
$\displaystyle=u^{-1}1\wedge\mathsf{Im}g$ [by exactness of the top row]
$\displaystyle=gg^{-1}u^{-1}1$ $\displaystyle=gt^{-1}y^{-1}1$ [by
commutativity] $\displaystyle=gt^{-1}xA^{\prime}$ [by exactness of the bottom
row] $\displaystyle=gt^{-1}xsA$ [$s$ is surjective] $\displaystyle=gt^{-1}tfA$
[by commutativity] $\displaystyle=g(fA\vee\mathsf{Ker}t)$
$\displaystyle=gfA\vee g\mathsf{Ker}t$ $\displaystyle=g\mathsf{Ker}t.$
$\displaystyle\text{[by exactness of the top row]}.$

Statement (ii) follows from (i) by duality. ∎

The following classical diagram lemma goes back to [8, 6, 30].

######  Theorem 5.4 (Five Lemma).

Consider a commutative diagram,

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{s}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\scriptstyle{t}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h}$$\scriptstyle{u}$$\textstyle{D\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{v}$$\scriptstyle{m}$$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{w}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x}$$\textstyle{B^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{y}$$\textstyle{C^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{z}$$\textstyle{D^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{n}$$\textstyle{E^{\prime},}$

where the rows are exact sequences. If $t$ and $v$ are isomorphisms, $s$ is
surjective and $w$ is injective, then $u$ is an isomorphism. In more detail,

  1. (i)

If $s$ is surjective and $t$ and $v$ are injective, then $u$ is injective.

  2. (ii)

If $w$ is injective and $t$ and $v$ are surjective, then $u$ is surjective.

###### Proof.

By Lemma 3.8, $t$ and $v$ are injective and surjective. Therefore by applying
both parts of The Four Lemma, we obtain that $u$ is both injective and
surjective, and hence an isomorphism. ∎

######  Definition 5.5.

A sequence of groups and morphisms, $A\xrightarrow{f}B\xrightarrow{g}C$, is
said to be short exact if $f$ is injective, $g$ is surjective and if
$\mathsf{Im}f=\mathsf{Ker}g$.

The $3\times 3$ Lemma appears in Buchsbaum’s early approach [6] to what is now
called an abelian category. In [4], Bourn gives a version in a context which
encompasses semi-abelian categories. The proof given in [14] is identical to
the one below. The formulation of the $3\times 3$ Lemma, in the context of
this paper, along with a partial proof is given in [38].

######  Theorem 5.6 ($3\times 3$ Lemma).

Consider a commutative diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{s}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\scriptstyle{t}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{u}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x}$$\scriptstyle{i}$$\textstyle{B^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{y}$$\scriptstyle{j}$$\textstyle{C^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{k}$$\textstyle{A^{\prime\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{m}$$\textstyle{B^{\prime\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{n}$$\textstyle{C^{\prime\prime}}$

where each column is a short exact sequence.

  1. (i)

If the bottom two rows are short exact, then so is the top,

  2. (ii)

If the top two rows are short exact, then so is the bottom.

  3. (iii)

If the top and bottom rows are short exact and $yx$ is a zero morphism, then
the middle row is short exact (the Middle $3\times 3$ Lemma).

###### Proof.

(i) We will show that $\mathsf{Ker}f=1$ (i.e., $f$ is injective).

$\displaystyle\mathsf{Ker}f$ $\displaystyle=f^{-1}1$
$\displaystyle=f^{-1}t^{-1}1$ [$t$ is injective] $\displaystyle=s^{-1}x^{-1}1$
[by commutativity] $\displaystyle=s^{-1}1$ [the middle row is short exact]
$\displaystyle=1$ [the left column is short exact]

We will show $gB=C$ (i.e., $g$ is surjective).

$\displaystyle C$ $\displaystyle=C\vee\mathsf{Ker}u$ [$\mathsf{Ker}u\subseteq
C$] $\displaystyle=u^{-1}uC$ $\displaystyle=u^{-1}k^{-1}1$ [the right column
is short exact] $\displaystyle=u^{-1}(k^{-1}1\wedge C^{\prime})$
$\displaystyle=u^{-1}(k^{-1}1\wedge\mathsf{Im}y)$ [the middle row is short
exact] $\displaystyle=u^{-1}yy^{-1}k^{-1}1$
$\displaystyle=u^{-1}yj^{-1}n^{-1}1$ [by commutativity]
$\displaystyle=u^{-1}yj^{-1}mA^{\prime\prime}$ [the bottom row is exact]
$\displaystyle=u^{-1}yj^{-1}miA^{\prime}$ [the left column is short exact]
$\displaystyle=u^{-1}yj^{-1}jxA^{\prime}$ [by commutativity]
$\displaystyle=u^{-1}y(xA^{\prime}\vee\mathsf{Ker}j)$
$\displaystyle=u^{-1}y(xA^{\prime}\vee tB)$ [the middle column is short exact]
$\displaystyle=u^{-1}(yxA^{\prime}\vee ytB)$ $\displaystyle=u^{-1}(1\vee ytB)$
[the middle row is short exact] $\displaystyle=u^{-1}ytB$
$\displaystyle=u^{-1}ugB$ [by commutativity] $\displaystyle=gB$ [the right
column is short exact]

We will show $\mathsf{Im}f=\mathsf{Ker}g$ (i.e., the top row is exact).

$\displaystyle\mathsf{Ker}g$ $\displaystyle=g^{-1}1$
$\displaystyle=g^{-1}u^{-1}1$ [the right column is short exact]
$\displaystyle=t^{-1}y^{-1}1$ [by commutativity]
$\displaystyle=t^{-1}xA^{\prime}$ [the middle row is short exact]
$\displaystyle=t^{-1}t(t^{-1}xA^{\prime})$ [the middle column is short exact]
$\displaystyle=t^{-1}(tt^{-1}xA^{\prime})$
$\displaystyle=t^{-1}(\mathsf{Im}x\wedge\mathsf{Im}t)$
$\displaystyle=t^{-1}xx^{-1}tB$ $\displaystyle=t^{-1}xx^{-1}j^{-1}1$ [the
middle column is short exact] $\displaystyle=t^{-1}xi^{-1}m^{-1}1$ [by
commutativity] $\displaystyle=t^{-1}xi^{-1}1$ [the bottom row is short exact]
$\displaystyle=t^{-1}xsA$ [the left column is short exact]
$\displaystyle=t^{-1}tfA$ [by commutativity] $\displaystyle=fA$ [the middle
column is short exact] $\displaystyle=\mathsf{Im}f$

(ii) is dual to (i) and hence follows by duality.

(iii) We will show that $\mathsf{Ker}x=1$.

$\displaystyle\mathsf{Ker}x$ $\displaystyle=x^{-1}1\wedge x^{-1}j^{-1}1$
[$x^{-1}1\subseteq x^{-1}j^{-1}1$] $\displaystyle=x^{-1}1\wedge i^{-1}m^{-1}1$
[by commutativity] $\displaystyle=X^{-1}1\wedge i^{-1}1$ [$m$ is injective]
$\displaystyle=x^{-1}1\wedge\mathsf{Im}s$ [by exactness of the left column]
$\displaystyle=ss^{-1}x^{-1}1$ $\displaystyle=sf^{-1}t^{-1}1$ [by
commutativity] $\displaystyle=s1$ [$f$ and $t$ are injective]
$\displaystyle=1$

The fact that $y$ is surjective follows by duality. We will show that
$\mathsf{Ker}y=\mathsf{Im}x$.

$\displaystyle\mathsf{Im}x$ $\displaystyle=\mathsf{Im}x\vee\mathsf{Im}(xs)$
[$\mathsf{Im}(xs)\subseteq\mathsf{Im}x$] $\displaystyle=\mathsf{Im}x\vee tfA$
[by commutativity] $\displaystyle=\mathsf{Im}x\vee tg^{-1}1$ [top row is
exact] $\displaystyle=\mathsf{Im}x\vee tg^{-1}u^{-1}1$ [$u$ is injective]
$\displaystyle=\mathsf{Im}x\vee tt^{-1}y^{-1}1$ [by commutativity]
$\displaystyle=\mathsf{Im}x\vee(\mathsf{Ker}y\wedge\mathsf{Im}t)$
$\displaystyle=\mathsf{Im}x\vee(\mathsf{Ker}j\wedge\mathsf{Ker}y)$ [middle
column is exact]
$\displaystyle=(\mathsf{Im}x\vee\mathsf{Ker}j)\wedge\mathsf{Ker}y$ [by the
Restricted Modular Law] $\displaystyle=j^{-1}jxA^{\prime}\wedge\mathsf{Ker}y$
$\displaystyle=j^{-1}miA^{\prime}\wedge\mathsf{Ker}y$ [by commutativity]
$\displaystyle=j^{-1}mA^{\prime\prime}\wedge\mathsf{Ker}y$ [$i$ is surjective]
$\displaystyle=j^{-1}n^{-1}1\wedge\mathsf{Ker}y$ [by exactness of bottom row]
$\displaystyle=y^{-1}k^{-1}1\wedge\mathsf{Ker}y$ [by commutativity]
$\displaystyle=\mathsf{Ker}y$

∎

One of the earliest appearances of the Snake Lemma is in [6]. The name is
derived from the shape of the construction of the connecting morphism in the
resulting exact sequence. The Snake Lemma is the main tool used to construct
the long exact homology sequence (see [30] for instance) which allows one to
extract information about higher-order homology objects from lower-order
homologies.

######  Theorem 5.7 (Snake Lemma).

Given a commutative diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\alpha}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\scriptstyle{\beta}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{\prime}}$$\textstyle{B^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g^{\prime}}$$\textstyle{C^{\prime}}$

with the rows exact, if $\mathsf{Ker}\alpha$, $\mathsf{Ker}\beta$,
$\mathsf{Ker}\gamma$ are conormal and $\mathsf{Im}\alpha$, $\mathsf{Im}\beta$,
$\mathsf{Im}\gamma$ are normal, then there exists an exact sequence

$\textstyle{\mathsf{Ker}\alpha\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{f}}$$\textstyle{\mathsf{Ker}\beta\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{g}}$$\textstyle{\mathsf{Ker}\gamma\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta}$$\textstyle{\mathsf{Coker}\alpha\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{f}^{\prime}}$$\textstyle{\mathsf{Coker}\beta\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{g}^{\prime}}$$\textstyle{\mathsf{Coker}\gamma.}$

###### Proof.

Let us first construct each morphism in the required sequence. We will do so
in each case by carrying out chasing required by Theorem 4.17 along an
appropriate zigzag.

To get the morphism $\bar{f}$, consider the zigzag

$\mathsf{Ker}\alpha\xrightarrow{\iota_{\mathsf{Ker}\alpha}}A\xrightarrow{f}B\xleftarrow{\iota_{\mathsf{Ker}\beta}}\mathsf{Ker}\beta.$

$\displaystyle\iota_{\mathsf{Ker}\beta}^{-1}f\iota_{\mathsf{Ker}\alpha}1$
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}1$ $\displaystyle=1$
[$\iota_{\mathsf{Ker}\beta}$ is injective]
$\displaystyle\iota_{\mathsf{Ker}\alpha}^{-1}f^{-1}\iota_{\mathsf{Ker}\beta}\mathsf{Ker}\beta$
$\displaystyle=\iota_{\mathsf{Ker}\alpha}^{-1}f^{-1}\mathsf{Ker}\beta$
[$\mathsf{Im}\iota_{\mathsf{Ker}\beta}=\mathsf{Ker}\beta$]
$\displaystyle=\iota_{\mathsf{Ker}\alpha}^{-1}\alpha^{-1}f^{\prime-1}1$ [by
commutativity]
$\displaystyle=\iota_{\mathsf{Ker}\alpha}^{-1}\mathsf{Ker}\alpha$
[$f^{\prime}$ is injective] $\displaystyle=\mathsf{Ker}\alpha$

To obtain the morphism $\bar{g}$, consider the zigzag

$\mathsf{Ker}\beta\xrightarrow{\iota_{\mathsf{Ker}\beta}}B\xrightarrow{g}C\xleftarrow{\iota_{\mathsf{Ker}\gamma}}\mathsf{Ker}\gamma.$
$\displaystyle\iota_{\mathsf{Ker}\gamma}^{-1}g\iota_{\mathsf{Ker}\beta}1$
$\displaystyle=\iota_{\mathsf{Ker}\gamma}^{-1}1$ $\displaystyle=1$
[$\iota_{\mathsf{Ker}\gamma}$ is injective]
$\displaystyle\iota_{\mathsf{Ker}\beta}^{-1}g^{-1}\iota_{\mathsf{Ker}\gamma}\mathsf{Ker}\gamma$
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}g^{-1}\mathsf{Ker}\gamma$
[$\mathsf{Im}\iota_{\mathsf{Ker}\gamma}=\mathsf{Ker}\gamma$]
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}\beta^{-1}g^{\prime-1}1$ [by
commutativity]
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}\beta^{-1}g^{\prime-1}1\vee\iota_{\mathsf{Ker}\beta}^{-1}\beta^{-1}1$
[$\beta^{-1}1\subseteq\beta^{-1}g^{{}^{\prime}-1}1$]
$\displaystyle=\mathsf{Ker}\beta$

To find the morphism $\delta$, consider the zigzag

$\mathsf{Ker}\gamma\xrightarrow{\iota_{\mathsf{Ker}\gamma}}C\xleftarrow{g}B\xrightarrow{\beta}B^{\prime}\xleftarrow{f^{\prime}}A^{\prime}\xrightarrow{\pi_{\mathsf{Im}\alpha}}\mathsf{Coker}\alpha.$
$\displaystyle\pi_{\mathsf{Im}\alpha}f^{\prime-1}\beta
g^{-1}\iota_{\mathsf{Ker}\gamma}1$
$\displaystyle=\pi_{\mathsf{Im}\alpha}f^{\prime-1}\beta g^{-1}1$
$\displaystyle=\pi_{\mathsf{Im}\alpha}f^{\prime-1}\beta fA$ [$g^{-1}1=fA$]
$\displaystyle=\pi_{\mathsf{Im}\alpha}f^{\prime-1}f^{\prime}\alpha A$ [by
commutativity] $\displaystyle=\pi_{\mathsf{Im}\alpha}\alpha A$ [$f^{\prime}$
is injective] $\displaystyle=1$

Dually,

$\iota_{\mathsf{Ker}\gamma}^{-1}g\beta^{-1}f^{\prime}\pi_{\mathsf{Im}\alpha}^{-1}\mathsf{Coker}\alpha=\iota_{\mathsf{Ker}\gamma}^{-1}g\beta^{-1}f^{\prime}A^{\prime}=\mathsf{Ker}\gamma.$

$\bar{f}^{\prime}$ is constructed dually to $\bar{f}$. Further note that
$\bar{f}^{\prime}$ is induced by the zigzag

$\mathsf{Coker}\alpha\xleftarrow{\pi_{\mathsf{Im}\alpha}}A^{\prime}\xrightarrow{f^{\prime}}B^{\prime}\xrightarrow{\pi_{\mathsf{Im}\beta}}\mathsf{Coker}\beta.$

The morphism $\bar{g}^{\prime}$ constructed dually to $\bar{g}$.

We then have the sequence

$\textstyle{\mathsf{Ker}\alpha\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{f}}$$\textstyle{\mathsf{Ker}\beta\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{g}}$$\textstyle{\mathsf{Ker}\gamma\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta}$$\textstyle{\mathsf{Coker}\alpha\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{f}^{\prime}}$$\textstyle{\mathsf{Coker}\beta\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{g}^{\prime}}$$\textstyle{\mathsf{Coker}\gamma.}$

Let us now show that this sequence is exact. Note that by the Homomorphism
Induction Theorem, to compute the kernels and images of these morphisms, it
suffices to chase the smallest subgroup backward and the largest subgroup
forward, respectively, along the zigzags that were used to construct those
morphisms.

$\displaystyle\mathsf{Im}\bar{f}$
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}f\iota_{\mathsf{Ker}\alpha}\mathsf{Ker}\alpha$
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}f\alpha^{-1}1$
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}f\alpha^{-1}f^{{}^{\prime}-1}1$
[$f^{\prime}$ is injective]
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}ff^{-1}\beta^{-1}1$ [by
commutativity]
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}(\beta^{-1}1\wedge\mathsf{Im}f)$
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}(\beta^{-1}1)\wedge(\iota_{\mathsf{Ker}\beta}^{-1}\mathsf{Im}f)$
$\displaystyle=\mathsf{Ker}\beta\wedge(\iota_{\mathsf{Ker}\beta}^{-1}\mathsf{Ker}g)$
[by exactness of the top row]
$\displaystyle=\iota_{\mathsf{Ker}\beta}^{-1}g^{-1}\iota_{\mathsf{Ker}\gamma}1$
$\displaystyle=\mathsf{Ker}\bar{g}$ $\displaystyle\mathsf{Ker}\delta$
$\displaystyle=\iota_{\mathsf{Ker}\gamma}^{-1}g\beta^{-1}f^{\prime}\pi_{\mathsf{Im}\alpha}^{-1}1$
$\displaystyle=\iota_{\mathsf{Ker}\gamma}^{-1}g\beta^{-1}f^{\prime}\mathsf{Im}\alpha$
$\displaystyle=\iota_{\mathsf{Ker}\gamma}^{-1}g\beta^{-1}\beta fA$ [by
commutativity]
$\displaystyle=\iota_{\mathsf{Ker}\gamma}^{-1}g(fA\vee\beta^{-1}1)$
$\displaystyle=\iota_{\mathsf{Ker}\gamma}^{-1}(gfA\vee g\beta^{-1}1)$
$\displaystyle=\iota_{\mathsf{Ker}\gamma}^{-1}(1\vee g\beta^{-1}1)$ [the top
row is exact] $\displaystyle=\iota_{\mathsf{Ker}\gamma}^{-1}g\beta^{-1}1$
$\displaystyle=\mathsf{Im}\bar{g}$

It follows that $\mathsf{Im}\delta=\mathsf{Ker}\bar{f}$ by applying duality to
the induced morphism (this is possible thanks to Lemma 4.18).

Lastly, $\mathsf{Im}\bar{f}^{\prime}=\mathsf{Ker}\bar{g}^{\prime}$ follows by
duality applied to $\mathsf{Im}\bar{f}=\mathsf{Ker}\bar{g}$. ∎

######  Remark 5.8.

Note that the Snake Lemma proved here is more general than the one found in
[38], which relies on the additional assumptions that the kernel of the
diagonal of the right square is conormal and that the image of the diagonal of
the left square is normal. While it is true, for groups, that the kernel of
any morphism is conormal (since every subgroup of a given group is conormal),
it is not always true that the image of a morphism is normal. We then give an
example in the context of groups in which the image of the diagonal of the
left square is not normal. Consider a commutative diagram of groups

$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{f}$$\textstyle{V\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\scriptstyle{g}$$\textstyle{V/B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h}$$\textstyle{V\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\textstyle{D_{8}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j}$$\textstyle{D_{8}/V.}$

The group $D_{8}$ is the dihedral group of order $8$ given by $D_{8}=\langle
a,b\mid
o(b)=2,o(a)=4,ba=a^{-1}b\rangle=\\{e,a,a^{2},a^{3},b,ab,a^{2}b,a^{3}b\\}$
(where $o(x)$ is the order of the element $x$). $B$ is the subgroup
$B=\\{e,b\\}$ of $D_{8}$, while $V$ is the subgroup $V=\\{e,b,a^{2},a^{2}b\\}$
of $D_{8}$. Note that $B\approx\mathds{Z}_{2}$ (since $B$ is of order 2),
$V\approx\mathds{V}_{4}$ (the Klein-four group), and $B\subseteq V$. The
morphism $f$ is the embedding of $B$ into $V$. The morphism $i$ is similarly
defined to be the embedding of the subgroup $V$ into the group $D_{8}$. Notice
that since $B$ and $V$ are subgroups of index $2$ in $V$ and $D_{8}$
respectively, both $B$ and $V$ are normal in $V$ and $D_{8}$ respectively.
Therefore, we may define the morphisms $g$ and $j$ as projections associated
to these normal subgroups. Observe that the rows are then exact. Lastly, the
morphism $h$ is a zero morphism. Observe that $h$ makes the right square
commute since $ji$ is a zero morphism (since $\mathsf{Im}i=\mathsf{Ker}j$).
Furthermore, since the rows are exact, we may apply the Snake Lemma (from this
paper) to obtain an exact sequence

$1\rightarrow 1\rightarrow V/B\rightarrow V/B\rightarrow D_{8}/V\rightarrow
D_{8}/V.$

Notice that $V/B$ and $D_{8}/V$ are both groups of order $2$ and hence must be
isomorphic to $\mathds{Z}_{2}$. We then obtain the exact sequence

$1\rightarrow
1\rightarrow\mathds{Z}_{2}\rightarrow\mathds{Z}_{2}\rightarrow\mathds{Z}_{2}\rightarrow\mathds{Z}_{2}$

by including the relevant isomorphisms. However, since the normalizer of the
subgroup $B$ of $D_{8}$ is precisely $V$ (this may be easily verified by
calculation), we have that $B$ is normal in $V$ but not in $D_{8}$. Therefore,
the image $\mathsf{Im}(if)=B$ of the left diagonal is not normal. This shows
us that the Snake Lemma in [38] may not be applied to this example, while the
Snake Lemma in this paper may.

##  6\. Exercises

In this section, we propose some exercises for the reader to try out the
technique of proving diagram lemmas illustrated in the previous section on
some known diagram lemmas as well as a few new ones. Some of the exercises in
this section were obtained as new diagram lemmas discovered by the first
author in [10]. However many of the exercises are well-known diagram lemmas.

######  Exercise 6.1 (Incomplete Snail Lemma).

Given a commutative diagram,

$\textstyle{W_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x}$$\scriptstyle{g}$$\textstyle{W_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{y}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{b}$$\scriptstyle{d}$$\textstyle{Y_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e}$$\scriptstyle{a}$$\textstyle{Y_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{Z,}$

where all sequences except possibly the sequence $W_{1}\rightarrow
W_{2}\rightarrow Z$ are exact. Prove that also the sequence ${W_{1}\rightarrow
W_{2}\rightarrow Z}$ is exact.

######  Exercise 6.2.

Consider a commutative diagram:

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{x}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\scriptstyle{y}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{z}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{m}$$\textstyle{B^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{n}$$\textstyle{C^{\prime}.}$

Prove one of the following two statements and deduce the other by duality:

  1. (i)

If $y$ is surjective and the top row is exact, then the bottom row is exact.

  2. (ii)

If $y$ is injective and the bottom row is exact, then the top row is exact.

######  Exercise 6.3 (Spider Lemma).

Consider a commutative diagram,

$\textstyle{V\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{g}$$\textstyle{W}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\scriptstyle{h}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{j}$$\scriptstyle{k}$$\textstyle{Z,}$

where the diagonals are short exact sequences. Prove that if $k$ is an
isomorphism, then so is $f$. Hint: use duality to shorten the proof.

######  Exercise 6.4.

Consider a commutative diagram:

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{b}$$\textstyle{G\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{c}$$\scriptstyle{d}$$\textstyle{D\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\scriptstyle{u}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{q}$$\scriptstyle{v}$$\textstyle{J\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$$\scriptstyle{x}$$\textstyle{H\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{y}$$\textstyle{C}$$\textstyle{I,}$

where the horizontal sequence is exact. Prove the following:

  1. (i)

If $u$ is surjective and $v$ and $y$ are injective, then $x$ is injective.

  2. (ii)

If $a$ and $c$ are surjective and $d$ is injective, then $b$ is surjective.

######  Exercise 6.5 (Diamond Lemma).

Consider a commutative diagram,