∎

11institutetext: K. Spendlove 22institutetext: Mathematical Institute
University of Oxford
Oxford UK
22email<EMAIL_ADDRESS>33institutetext: R.C.A.M. Vandervorst
44institutetext: Department of Mathematics
VU University Amsterdam
The Netherlands
44email<EMAIL_ADDRESS>
# Graded differential groups, Cartan-Eilenberg systems
and conjectures in Conley index theory

Kelly Spendlove  Robert Vandervorst

(Version date: )

###### Abstract

Cartan-Eilenberg systems play an prominent role in the homological algebra of
filtered and graded differential groups and (co)chain complexes in particular.
We define the concept of Cartan-Eilenberg systems of abelian groups over a
poset. Our main result states that a filtered chain isomorphism between free,
${\mathsf{P}}$-graded differential groups is equivalent to an isomorphism
between associated Cartan-Eilenberg systems. An application of this result to
the theory of dynamical systems addresses two open conjectures posed by J.
Robbin and D. Salamon regarding uniqueness type questions for connection
matrices, as detailed in robbin:salamon2  and atm . The main result of this
paper also proves that the connection matrix theories in fran ,
robbin:salamon2  and hms  are equivalent in the setting of vector spaces, as
well as uniqueness of connection matrices for Morse-Smale gradient systems,
cf. Reineck .

###### Keywords:

Graded differential group Cartan-Eilenberg system Chain isomorphism Connection
matrix.

###### MSC:

55Uxx 06D05 37B25 37B35

##  1 Introduction

In the realm of algebraic topology and homological algebra, differential
groups graded by a poset ${\mathsf{P}}$ play a crucial role in the theory of
spectral sequences and spectral systems and their applications in various
fields, in particular dynamical systems. This paper studies the intricate
relationship between ${\mathsf{P}}$-graded differential groups and Cartan-
Eilenberg systems, cf. CE . Our investigation revisits the traditional concept
of Cartan-Eilenberg systems by considering abelian groups over an arbitrary
poset. In particular, an application of our main result to dynamical systems
theory confirms two open conjectures posited by J. Robbin and D. Salamon,
referenced in robbin:salamon2  and atm , which concern the uniqueness of
connection matrices—a pivotal concept in the analysis of the global behavior
of dynamical systems.

In order to state the main results in this paper we start with a brief,
intuitive definition of Cartan-Eilenberg systems over a lattice. Let
${\mathsf{P}}$ be a finite poset and let ${\mathsf{O}}({\mathsf{P}})$ be the
lattice of down-sets in ${\mathsf{P}}$. A Cartan-Eilenberg system E over
${\mathsf{O}}({\mathsf{P}})$ consists of a functor ${\mathsf{E}}$ from the
arrow category of ${\mathsf{O}}({\mathsf{P}})$, which consists of ordered
pairs $(\alpha,\beta)$, with $\alpha\subseteq\beta$, to the category of
abelian groups and an additional structure that accounts for exact triangles
in homology, cf. Sect. 4.1 for a detailed definition. If we unpack the above
informal definition we obtain the application
$(\alpha,\beta)\mapsto{\mathsf{E}}(\alpha,\beta)=E^{\beta}_{\alpha}$. The
functor ${\mathsf{E}}$ yields the homomorphisms $\ell\colon
E^{\beta}_{\alpha}\to E^{\beta^{\prime}}_{\alpha^{\prime}}$ for all
$(\alpha,\beta)\leq(\alpha^{\prime},\beta^{\prime})$,111The order on order
pairs is point wise, i.e. $(\alpha,\beta)\leq(\alpha^{\prime},\beta^{\prime})$
if and only $\alpha\subseteq\alpha^{\prime}$ and
$\beta\subseteq\beta^{\prime}$. and the composition
$E^{\beta}_{\alpha}\xrightarrow[]{\ell}E^{\beta^{\prime}}_{\alpha^{\prime}}\xrightarrow[]{\ell}E^{\beta^{\prime\prime}}_{\alpha^{\prime\prime}}$
is given by $\ell\colon E^{\beta}_{\alpha}\to
E^{\beta^{\prime\prime}}_{\alpha^{\prime\prime}}$ due to the transitivity in
${\mathsf{O}}({\mathsf{P}})$. Moreover there exist connecting homomorphisms
$k\colon E^{\gamma}_{\beta}\to E^{\beta}_{\alpha}$ and $k\colon
E^{\gamma^{\prime}}_{\beta^{\prime}}\to E^{\beta^{\prime}}_{\alpha^{\prime}}$
such that the diagrams222In the special cases
$(\alpha,\beta)\leq(\alpha,\gamma)$ and $(\alpha,\gamma)\leq(\beta,\gamma)$
the morphisms $\ell$ are denoted by $i$ and $j$ respectively.

${E^{\beta}_{\alpha}}$${E^{\gamma}_{\alpha}}$${E^{\gamma}_{\beta}}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{k}$
${E^{\gamma}_{\beta}}$${E^{\beta}_{\alpha}}$${E^{\gamma^{\prime}}_{\beta^{\prime}}}$${E^{\beta^{\prime}}_{\alpha^{\prime}}}$$\scriptstyle{k}$$\scriptstyle{\ell}$$\scriptstyle{\ell}$$\scriptstyle{k}$
(1)

are exact and commutative respectively for all pairs of ordered triples
$(\alpha,\beta,\gamma)\leq(\alpha^{\prime},\beta^{\prime},\gamma^{\prime})$.333An
ordered triple $(\alpha,\beta,\gamma)$ is defined by
$\alpha\subseteq\beta\subseteq\gamma$. The order on ordered triples is again
point wise. By construction $\ell\colon
E^{\beta}_{\alpha}\xrightarrow[\mathrm{id}]{\ell}E^{\beta}_{\alpha}$ is the
identity homomorphism and the exactness of (1)[left] shows that
$E^{\alpha}_{\alpha}=0$ for all $\alpha\in{\mathsf{O}}({\mathsf{P}})$, cf.
HelleRognes . A Cartan-Eilenberg system is _excisive_ if
$E_{\alpha}^{\beta}\cong E_{\alpha^{\prime}}^{\beta^{\prime}}$ for all
$\beta\smallsetminus\alpha=\beta^{\prime}\smallsetminus\alpha^{\prime}$, cf.
Lem. 5. A morphism between Cartan-Eilenberg systems E and $\text{\bf
E}^{\prime}$ is given by homomorphisms $h^{\beta}_{\alpha}\colon
E^{\beta}_{\alpha}\to E^{\prime\beta}_{\alpha}$ which satisfy the commutative
diagram:

${E_{\alpha}^{\beta}}$${E^{\gamma}_{\alpha}}$${E^{\gamma}_{\beta}}$${E_{\alpha}^{\beta}}$${E^{\prime\beta}_{\alpha}}$${E^{\prime\gamma}_{\alpha}}$${E^{\prime\gamma}_{\beta}}$${E^{\prime\beta}_{\alpha}}$$\scriptstyle{i}$$\scriptstyle{h^{\beta}_{\alpha}}$$\scriptstyle{j}$$\scriptstyle{h^{\gamma}_{\alpha}}$$\scriptstyle{k}$$\scriptstyle{h^{\gamma}_{\beta}}$$\scriptstyle{h^{\beta}_{\alpha}}$$\scriptstyle{i^{\prime}}$$\scriptstyle{j^{\prime}}$$\scriptstyle{k^{\prime}}$
(2)

any $\alpha\subseteq\beta\subseteq\gamma$. In this case we write
$h\colon\text{\bf E}\to\text{\bf E}^{\prime}$ as a morphism of Cartan-
Eilenberg systems.

A free, ${\mathsf{P}}$-graded differential group is a free abelian group $C$
with an endomorphism ${\mathrm{d}}_{C}$, with ${\mathrm{d}}_{C}^{2}=0$, and a
${\mathsf{P}}$-grading which yields a decomposition

$C=\bigoplus_{p\in{\mathsf{P}}}G_{p}C,$ (3)

where $G_{p}C\subseteq C$ are free subgroups. A free, ${\mathsf{P}}$-graded
differential group is strict if the restrictions ${\mathrm{d}}\colon G_{p}C\to
G_{p}C$ are trivial for all $p\in{\mathsf{P}}$. A homomorphism of differential
groups is called a chain homomorphism. A chain homomorphism is
${\mathsf{O}}({\mathsf{P}})$-filtered if it preserves the filterings. Precise
definitions are provided in Section 2 and 2.1. As detailed in Section 2.2, a
${\mathsf{P}}$-graded differential group $(C,{\mathrm{d}})$ gives rise to a
Cartan-Eilenberg system $\text{\bf E}(C,{\mathrm{d}}_{C})$ with $E$-terms
$E_{\alpha}^{\beta}=H(G_{\beta\smallsetminus\alpha}C)$, where
$(G_{\beta\smallsetminus\alpha}C,{\mathrm{d}}_{C})$ is a well-defined
differential subgroup and $E_{\alpha}^{\beta}$ is its homology. The main
result of this paper can be phrased as follows:

Theorem A (cf. Thm. 5.1) Two free, finitely generated, strict,
${\mathsf{P}}$-graded differential groups are
${\mathsf{O}}({\mathsf{P}})$-filtered chain isomorphic if and only if there
respective Cartan-Eilenberg systems are isomorphic.

An important application of Theorem A is to the theory of connection matrices
developed by R. Franzosa, who introduces connection matrices in the algebraic
topological theory of Morse representations for dynamical systems, cf. fran
and Sect. 6. Connection matrices contain information about the connection
orbit structure between the Morse sets in a Morse representation and generate
the central algebraic structure referred to as a _module braid_ in fran2 ;
fran , which serves as an invariant for the Morse representation in question.
In Section 7 we prove that Franzosa’s module braids in the setting of Morse
representations are examples of Cartan-Eilenberg systems. The main result in
fran  states that these module braids are always generated, under some mild
conditions, by an appropriately constructed free, graded differential group —
a _connection matrix_. Theorem A shows that two connection matrices assigned
to a Morse representation are conjugate via filtered chain isomorphisms and
thus unique up to filtered conjugacy. A first partial result in this direction
was obtained in atm  and hms . Theorem A settles the two central conjectures
formulated in robbin:salamon2  concerning the similarity of connection
matrices, and completes the result in atm . In Section 7 we give an extensive
account of these conjectures and their proofs. Another consequence of Theorem
A concerns connection matrix theories in the setting of vector spaces.

Theorem B (cf. Thm. 7.1) Let $(D,{\mathrm{d}}_{D})$ be an
${\mathsf{O}}({\mathsf{P}})$-filtered differential vector space. Then, every
strict, ${\mathsf{P}}$-graded differential vector space $(A,{\mathrm{d}}_{A})$
whose Cartan-Eilenberg systems is isomorphic to $\text{\bf
E}(D,{\mathrm{d}}_{D})$, is ${\mathsf{O}}({\mathsf{P}})$-filtered chain
equivalent to $(D,{\mathrm{d}}_{D})$.

As a consequence of Theorem B the connection matrix theories of fran ,
robbin:salamon2  and hms  in the setting of finite dimensional differential
vector spaces are equivalent. Without the restriction of finite dimensionality
the notions in robbin:salamon2  and fran  are equivalent.

Another application of Theorem A (Thm. 5.1) involves the uniqueness of
connection matrices for Morse-Smale gradient flows. A ${\mathsf{P}}$-graded
chain complex $(C,{\mathrm{d}}_{C})$ of ${\mathbb{Z}}_{2}$-vector spaces is
Morse-Smale graded if there exists an order-preserving map
$\mu\colon{\mathsf{P}}\to{\mathbb{Z}}$ such that $\mu(p)=\mu(q)$, $p\neq q$,
then $p~{}\|~{}q$, and $G_{p}C_{k}={\mathbb{Z}}_{2}$. The pair
$({\mathsf{P}},\mu)$ is called a Morse-Smale grading. For such chain complexes
we have the following uniqueness result.

Theorem C (cf. Thm. 7.2) Let $(C,{\mathrm{d}}_{C})$ and
$(C,{\mathrm{d}}^{\prime}_{C})$ be algebraic Morse-Smale graded chain
complexes with Morse-Smale grading $({\mathsf{P}},\mu)$ for both chain
complexes. Then, ${\mathrm{d}}_{C}={\mathrm{d}}^{\prime}_{C}$ if and only if
$\text{\bf E}(C,{\mathrm{d}}_{C})\cong\text{\bf
E}(C,{\mathrm{d}}^{\prime}_{C})$.

Theorem C applies to Morse-Smale gradient flows and provides an algebraic
proof of a result by J. Reineck, cf. Reineck , that connection matrices for
such systems are unique, cf. Sect. 7.3.

###### Remark 1

In this paper we choose to present the results in the category $\mathbf{Ab}$
of abelian groups. The results remain valid if we consider $R$-modules where
the ring $R$ is a principal ideal domain. In partucular, the theory also
applies to vector spaces when $R$ is taken to be a field ${\mathbb{K}}$.

##  2 Graded and filtered differential groups

An _ordered decomposition_ of an abelian group $C$ is a finite poset
$\mathrm{\Pi}=(\mathrm{\Pi},\leq)$ consisting of non-trivial projections
$\pi\colon C\to C$ such that

  1. (i)

$\pi\circ\pi^{\prime}=0$ for all distinct pairs
$\pi,\pi^{\prime}\in\mathrm{\Pi}$;

  2. (ii)

$\sum_{\pi\in\mathrm{\Pi}}\pi=\mathrm{id}$.

Let $({\mathsf{P}},\leq)$ be a finite poset and let
${\mathsf{O}}({\mathsf{P}})$ be the bounded, distributive lattice of down-sets
in ${\mathsf{P}}$. A _${\mathsf{P}}$ -grading_ on $C$ is an order-embedding

$\mathrm{grd}\colon\mathrm{\Pi}\hookrightarrow{\mathsf{P}},$

and the pair $(C,\mathrm{grd})$ is called a _${\mathsf{P}}$ -graded abelian
group_. A (finite) filtering on an abelian group $C$ is a homomorphism

$\mathrm{flt}\colon{\mathsf{O}}({\mathsf{P}})\to\mathsf{Sub}~{}C,$

where $\mathsf{Sub}~{}C$ is the lattice of subgroups of $C$. The image of
$\mathrm{flt}$ is a sublattice consisting of subgroups $F_{\alpha}C\subseteq
C$ ordered by inclusion. Note that $\alpha\mapsto F_{\alpha}C$ being a lattice
homomorphism implies that $F_{\alpha\cup\beta}C=F_{\alpha}C+F_{\beta}C$ and
$F_{\alpha\cap\beta}C=F_{\alpha}C\cap F_{\beta}C$. The pair $(C,\mathrm{flt})$
is called an _${\mathsf{O}}({\mathsf{P}})$ -filtered group_. A graded abelian
group yields a filtering
$\mathrm{flt}\colon{\mathsf{O}}({\mathsf{P}})\to\mathsf{Sub}~{}C$ of $C$ given
by $\alpha\mapsto F_{\alpha}C:=\bigoplus_{\pi\in\mathrm{grd}^{-1}(\alpha)}\pi
C$. We use the convention that the empty sum is $0$. The homomorphism
$\mathrm{flt}$ extends to a meet-semilattice homomorphism
$\beta\smallsetminus\alpha\mapsto
G_{\beta\smallsetminus\alpha}C:=\bigoplus_{\pi\in\mathrm{grd}^{-1}(\beta\smallsetminus\alpha)}\pi
C$. In particular, this assigns the ‘atoms’ $\\{p\\}\mapsto G_{p}C$ and
associated decomposition:
$G_{\beta\smallsetminus\alpha}C=\bigoplus_{p\in\beta\smallsetminus\alpha}G_{p}C,$
with $G_{p}C=\pi C$, if $\mathrm{grd}^{-1}(\\{p\\})=\\{\pi\\}$ and $G_{p}C=0$
if $\mathrm{grd}^{-1}(\\{p\\})=\varnothing$. We refer to the subgroups
$G_{p}C$ as the subgroups of homogeneous elements of degree $p$. Since
$\mathrm{grd}$ need not be onto, some of the subgroups $G_{p}C$ may be the
trivial subgroup $0$ for some $p$ and therefore the subgroups $G_{p}C$
coincide with the components $\operatorname{im}\pi$ whenever $G_{p}C$ is non-
trivial. It is customary to denote the ${\mathsf{P}}$-graded abelian group by
its decomposition $C=\bigoplus_{p\in{\mathsf{P}}}G_{p}C$. A
${\mathsf{P}}$-graded abelian group $(C,\mathrm{grd})$, with
$C=\bigoplus_{p\in{\mathsf{P}}}G_{p}C$, is _free abelian_ if and only if the
components $G_{p}C$ are free abelian. Conversely, a filtered abelian group
also yields a ${\mathsf{P}}$-graded abelian group
$A=\bigoplus_{p\in{\mathsf{P}}}G_{p}A$, by defining $G_{p}A\cong
F_{\beta}C/F_{\alpha}C$, $\beta\smallsetminus\alpha=\\{p\\}$. In general $A$
is not isomorphic to $C$. If the groups $G_{p}A$ are free there exist
subspaces $G_{p}C\subseteq C$, with $C_{p}C\cong G_{p}A$, such that
$F_{\alpha}C=\bigoplus_{p\in\alpha}G_{p}C$, cf. robbin:salamon2 .

A homomorphism $f\colon C\to C^{\prime}$ of ${\mathsf{P}}$-graded abelian
groups that satisfies the property $f\bigl{(}G_{p}C\bigr{)}\subseteq
G_{q}C^{\prime}$ is called _homogeneous_ , or _${\mathsf{P}}$ -graded_. A
homomorphism $f\colon C\to C^{\prime}$ of ${\mathsf{P}}$-graded, or
${\mathsf{O}}({\mathsf{P}})$-filtered groups that satisfies
$f\bigl{(}F_{\alpha}C\bigr{)}\subseteq F_{\alpha}C^{\prime}$ is called
_${\mathsf{O}}({\mathsf{P}})$ -filtered_. Note that homomorphisms of graded
groups may be filtered when the graded groups are regarded as filtered groups,
with $F_{\alpha}C=\bigoplus_{p\in\alpha}G_{p}C$ being the filtering induced by
the ${\mathsf{P}}$-grading on $C$. A homomorphism $f\colon C\to C^{\prime}$ is
an ${\mathsf{O}}({\mathsf{P}})$-filtered isomorphism if

$f(F_{\alpha}C)=F_{\alpha}C^{\prime},\quad\forall\alpha\in{\mathsf{O}}({\mathsf{P}}).$
(4)

If the order structure of the grading, or filtering is clear from context we
refer to (homogeneous) graded and filtered homomorphisms.

###### Remark 2

An isomorphism that preserves the filterings is not necessarily an filtered
isomorphism in the sense of Condition (4) as the following example shows. Let
$C={\mathbb{Z}}$ and $C^{\prime}={\mathbb{Z}}$ and $f\colon C\to C^{\prime}$
is the identity homomorphism. Suppose $C$ is filtered as $0\subseteq
4{\mathbb{Z}}\subseteq{\mathbb{Z}}$ and $C^{\prime}$ is filtered as
$0\subseteq 2{\mathbb{Z}}\subseteq{\mathbb{Z}}$. Clearly $f$ is an isomorphism
but $f(4{\mathbb{Z}})\subsetneq 2{\mathbb{Z}}$. If both maps $f$ and $f^{-1}$
are ${\mathsf{O}}({\mathsf{P}})$-filtered, then
$F_{\alpha}C=f^{-1}f(F_{\alpha}C)\subseteq
f^{-1}(F_{\alpha}C^{\prime})\subseteq F_{\alpha}C$ which yields (4). Thus
filtered isomorphism is an isomorphism for which $f$ and $f^{-1}$ are
filtered.

###  2.1 Differential groups

A _differential group_ $(C,{\mathrm{d}}_{C})$ consists of an abelian group $C$
and endomorphism ${\mathrm{d}}_{C}\colon C\to C$ with
${\mathrm{d}}_{C}^{2}=0$, cf. (MacLane, , II.1). The elements $c\in C$ are
referred to as _chains_ and the endomorphism ${\mathrm{d}}_{C}$ is referred to
as the _differential_. We fix the following notation:
$Z(C):=\ker{\mathrm{d}}_{C}$ are the _cycles_ in $C$ and
$B(C):=\mathrm{im~{}}{\mathrm{d}}_{C}$ are the _boundaries_ in $C$. The
_homology_ of $(C,{\mathrm{d}}_{C})$ is defined as $H(C):=Z(C)/B(C)$. A
homomorphism $f\colon(C,{\mathrm{d}}_{C})\to(A,{\mathrm{d}}_{A})$ of
differential groups is a homomorphism which satisfies
${\mathrm{d}}_{A}f=f{\mathrm{d}}_{C}$ and is called a _chain homomorphism_ ,
or _chain isomorphism_ if $f$ is an isomorphism of abelian groups.
Differential groups and chain homomorphisms form the category
$\mathbf{Ab}_{\mathrm{d}}$. Homology for a differential group defines a
covariant functor ${\mathsf{H}}\colon\mathbf{Ab}_{\mathrm{d}}\to\mathbf{Ab}$
given by $(C,{\mathrm{d}}_{C})\mapsto H(C)$, called the _homology functor_.
The homology may be regarded as a differential group with ${\mathrm{d}}=0$.
Next we consider grading and filtering in combination with the differental
${\mathrm{d}}_{C}$.

###### Definition 1

A triple $\bigl{(}C,{\mathrm{d}}_{C},\mathrm{grd}\bigr{)}$ is called a
${\mathsf{P}}$-graded differential group if

  1. (i)

$(C,{\mathrm{d}}_{C})$ is a differential group;

  2. (ii)

$(C,\mathrm{grd})$ is a ${\mathsf{P}}$-graded abelian group with grading
$\mathrm{grd}\colon\mathrm{\Pi}\hookrightarrow{\mathsf{P}}$;

  3. (iii)

${\mathrm{d}}_{C}$ is ${\mathsf{O}}({\mathsf{P}})$-filtered.

A ${\mathsf{P}}$-graded differential group is _strict_ if

  1. (iv)

${\mathrm{d}}_{C}F_{\alpha}C\subseteq F_{\alpha\smallsetminus\\{p\\}}C$, for
all $p$ maximal in $\alpha$.

A chain homomorphism, or chain isomorphism between graded or filtered
differential groups preserving the filtering is called an
_${\mathsf{O}}({\mathsf{P}})$ -filtered chain homomorphism_, or
_${\mathsf{O}}({\mathsf{P}})$ -filtered chain isomorphism_. The condition on
${\mathrm{d}}_{C}$ to be ${\mathsf{O}}({\mathsf{P}})$-filtered implies that
the restriction ${\mathrm{d}}_{C}\colon G_{q}C\to G_{p}C$ is well-defined. The
latter is non-trivial only if $p\leq q$. If the differential is strict then
the latter inequality is strict, i.e. $p<q$. For any convex set
$\beta\smallsetminus\alpha$ the pair
$(G_{\beta\smallsetminus\alpha}C,{\mathrm{d}}_{C})$ is a differential
subgroup. More generally a triple $(C,{\mathrm{d}}_{C},\mathrm{flt})$ is
called an _${\mathsf{O}}({\mathsf{P}})$ -filtered differential group_ if
$(C,\mathrm{flt})$ is ${\mathsf{O}}({\mathsf{P}})$-filtered and the
differential satisfies ${\mathrm{d}}_{C}F_{\alpha}C\subseteq F_{\alpha}C$ for
all $\alpha\in{\mathsf{O}}({\mathsf{P}})$. The
${\mathsf{O}}({\mathsf{P}})$-filtering induced by a ${\mathsf{P}}$-graded
group automatically satisfies the conditions of a
${\mathsf{O}}({\mathsf{P}})$-filtered differential group via
$F_{\alpha}C:=\bigoplus_{p\in\alpha}G_{p}C$. As before the converse is not
true in general.

###### Example 1

A chain complex $(C,{\mathrm{d}}_{C})$ is an example of a graded differential
group, traditionally denoted by $C=\bigoplus_{k}C_{k}$, where the differential
${\mathrm{d}}_{C}$ is homogeneous of degree $-1$, and thus filtered. This is
equivalent to ${\mathrm{d}}_{C}=\bigoplus_{k}{\mathrm{d}}_{k}$, with
${\mathrm{d}}_{k}\colon C_{k}\to C_{k-1}$ and
${\mathrm{d}}_{k}{\mathrm{d}}_{k+1}=0$. The chain maps are taken be to
homogeneous, i.e. $f\colon C\to C^{\prime}$, $f=\bigoplus_{k}f_{k}$, with
$f_{k}\colon C_{k}\to C^{\prime}_{k}$ and
${\mathrm{d}}^{\prime}_{k}f_{k}=f_{k-1}{\mathrm{d}}_{k}$, for all $k$. The
same holds for cochain complexes.

###### Remark 3

In the notation for graded and filtered differential groups we will omit
$\mathrm{grd}$ and $\mathrm{flt}$ from the notation if there is no ambiguity
about the grading or filtering, and simply write $(C,{\mathrm{d}}_{C})$.

###  2.2 Cartan-Eilenberg systems

In Section 1 a basic definition of Cartan-Eilenberg systems in the category of
abelian groups is given. Filtered differential groups are a prime source for
generating excisive Cartan-Eilenberg systems. Let $(C,{\mathrm{d}}_{C})$ be a
filtered differential group. Since the differential ${\mathrm{d}}_{C}$
preserves the filtering we have the short exact sequence:

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(5)

The associated homology is given by
$E^{\beta}_{\alpha}:=H(F_{\beta}C/F_{\alpha}C)$,
$E^{\gamma}_{\alpha}:=H(F_{\gamma}C/F_{\beta}C)$ and
$E^{\gamma}_{\beta}:=H\bigl{(}F_{\gamma}C/F_{\beta}C\bigr{)}$. This yields the
exact triangles in (1) where the connecting homomorphism $k\colon
H\bigl{(}F_{\beta}C/F_{\alpha}C\bigr{)}\to H(F_{\alpha}C)$ is constructed in
the usual way. All other axioms are readily verified which yields a Cartan-
Eilenberg system denoted $\text{\bf E}(C,{\mathrm{d}}_{C})$. The excisive
property follows from the fact that
$F_{\alpha\cup\beta}C/F_{\alpha}C=(F_{\alpha}C+F_{\beta}C)/F_{\alpha}C\cong
F_{\beta}C/(F_{\alpha}C\cap F_{\beta}C)=F_{\beta}C/F_{\alpha\cap\beta}C$, cf.
Lem. 5. The (excisive) Cartan-Eilenberg system $\text{\bf
E}(C,{\mathrm{d}}_{C})$ is the Cartan-Eilenberg system of the
${\mathsf{O}}({\mathsf{P}})$-filtered differential group
$(C,{\mathrm{d}}_{C})$. Since graded differential groups are filtered
differential groups via $F_{\alpha}C:=\bigoplus_{p\in\alpha}G_{p}C$, we define
the Cartan-Eilenberg system of a ${\mathsf{P}}$-graded differential group as
outlined above. The induced Cartan-Eilenberg system is again denoted by
$\text{\bf E}(C,{\mathrm{d}}_{C})$.

###### Remark 4

For finitely generated, strict ${\mathsf{P}}$-graded differential groups the
homology groups satisfy $H(G_{p}C,{\mathrm{d}})=G_{p}C\cong
E^{\beta}_{\alpha}$, with $\beta\smallsetminus\alpha=\\{p\\}$ for all
$p\in{\mathsf{P}}$.

##  3 Induced chain homomorphisms

In this section we prove two results, Theorems 3.1 and 3.2, concerning chain
homomorphisms that lift homomorphisms on homology. We start with a number of
preliminary lemmas.

###### Lemma 1

Let $D$ be a free abelian group, $(A,{\mathrm{d}}_{A})$ a differential group,
and let $h\colon D\to A$ be a homomorphism such that ${\mathrm{d}}_{A}h=0$ and
$H(h)=0$.444If $D$ is regarded as differential group with ${\mathrm{d}}=0$,
then the condition ${\mathrm{d}}_{A}h=0$ is equivalent to $h$ being a chain
homomorphism. The induced homology map is denoted by $H(h)\colon D\to H(A)$.
We say that $D$ is a _boundaryless_ differential group. Then, $h\colon D\to
B(A)$ and there exists a lift $l\colon D\to A$ such that the diagram

${A}$${D}$${B(A)}$$\scriptstyle{{\mathrm{d}}_{A}}$$\scriptstyle{h}$$\scriptstyle{l}$
(6)

commutes, i.e. $h={\mathrm{d}}_{A}l$.

###### Proof

By the assumption that ${\mathrm{d}}_{A}h=0$ we have that $h\colon D\to Z(A)$.
Observe the short exact sequence for homology

${0}$${B(A)}$${Z(A)}$${H(A)}$${0,}$$\scriptstyle{e}$$\scriptstyle{\rho}$ (7)

where $\rho$ is the projection $x\mapsto[x]\in H(A)$ and $e$ is inclusion.
Consider the diagram:

${A}$${D}$${0}$${B(A)}$${Z(A)}$${H(A)}$${0}$$\scriptstyle{{\mathrm{d}}_{A}}$$\scriptstyle{h}$$\scriptstyle{h}$$\scriptstyle{l}$$\scriptstyle{H(h)}$$\scriptstyle{e}$$\scriptstyle{\rho}$
(8)

From the assumption that $H(h)=\rho h=0$ and the _exactness_ of (7) we
conclude that $h$ maps into $B(A)$ and thus $h\colon D\to B(A)\subseteq Z(A)$.
Since $D$ is a free abelian group it is projective and therefore there exists
a lift $l\colon D\to A$ such that the diagram in (6) commutes. ∎

###### Remark 5

For the action of differentials we omit parentheses and for other
homomorphisms we adopt the usual functional notation, i.e. ${\mathrm{d}}c$ and
$h(c)$. Parentheses are also omitted in compositions of maps.

###### Remark 6

In the remainder of the paper $C\oplus C^{\prime}$ (and $A\oplus A^{\prime}$)
denotes the Cartesian product as direct sum with specifically constructed
differentials to emphasize that this is not the product of differential groups
with the product differential but a product group with a chosen differential.

###### Remark 7

If $(A,{\mathrm{d}}_{A})$ is a chain complex and $h_{k}\colon D\to A_{k}$ with
${\mathrm{d}}_{k}h_{k}=0$, then $h_{k}\colon D\to B_{k}(A)$ and there exists a
lift $l_{k}\colon D\to A_{k+1}$ such that $h_{k}={\mathrm{d}}_{k+1}l_{k}$.

###### Lemma 2

Let $C,C^{\prime}$ be abelian groups and consider the short exact sequence

${0}$${C}$${C\oplus
C^{\prime}}$${C^{\prime}}$${0,}$$\scriptstyle{i}$$\scriptstyle{j}$ (9)

where $i(c)=(c,0)$, $j(c,c^{\prime})=c^{\prime}$. Let ${\mathrm{d}}_{C}$ be a
differential for $C$ and let $C^{\prime}$ be boundaryless. Then, for $i$ and
$j$ to be chain homomorphisms the differentials on $C\oplus C^{\prime}$ are of
the form

${\mathrm{d}}_{C\oplus C^{\prime}}=\begin{pmatrix}{\mathrm{d}}_{C}&\lambda\\\
0&0\end{pmatrix},$ (10)

where $\lambda\colon C^{\prime}\to C$ is a chain homomorphisms, i.e.
${\mathrm{d}}_{C}\lambda=0$. Moreover, for the long exact sequence (exact
triangle)

${H(C)}$${H(C\oplus
C^{\prime})}$${C^{\prime}}$${H(C)}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{H(\lambda)}$
(11)

the canonical connecting homomorphism is given by $H(\lambda)$.

###### Proof

Let us start with the differentials in (10). In order for $i$ and $j$ to be
chain maps we need: $i{\mathrm{d}}_{C}={\mathrm{d}}_{C\oplus C^{\prime}}i$ and
$j{\mathrm{d}}_{C\oplus C^{\prime}}={\mathrm{d}}_{C^{\prime}}j$. By assumption
${\mathrm{d}}_{C\oplus C^{\prime}}$ is given by

${\mathrm{d}}_{C\oplus C^{\prime}}=\begin{pmatrix}\kappa~{}&~{}\lambda\\\
\mu~{}&~{}\nu\end{pmatrix},$

for some homomorphisms $\kappa\colon C\to C$, $\lambda\colon C^{\prime}\to C$,
$\mu\colon C\to C^{\prime}$ and $\nu\colon C^{\prime}\to C^{\prime}$. From the
conditions on $i$ and $j$, and the fact that $C^{\prime}$ is boundaryless
(${\mathrm{d}}_{C^{\prime}}=0$), we derive that
$\bigl{(}{\mathrm{d}}_{C}c,0\bigr{)}=i{\mathrm{d}}_{C}c={\mathrm{d}}_{C\oplus
C^{\prime}}ic=\bigl{(}\kappa c,\mu c\bigr{)}$, which yields that
$\kappa={\mathrm{d}}_{C}$ and $\mu=0$. Moreover, $\nu
c^{\prime}=j{\mathrm{d}}_{C\oplus C^{\prime}}(c,c^{\prime})=0\cdot
j(c,c^{\prime})=0$, which implies that $\nu=0$ and therefore
${\mathrm{d}}_{C\oplus C^{\prime}}$ is given by (10) for some homomorphism
$\lambda\colon C^{\prime}\to C$. Since ${\mathrm{d}}_{C\oplus C^{\prime}}$ is
a differential we obtain the condition ${\mathrm{d}}_{C}\lambda=0$, which
makes $\lambda$ a chain homomorphism.

As for the long exact sequence we follow the standard construction for the
connecting homomorphism. Let $c^{\prime}\in C^{\prime}=Z(C^{\prime})$. By
exactness, $c^{\prime}=j(\tilde{c},c^{\prime})$ for some $\tilde{c}\in C$.
Since $j$ is a chain homomorphism we have that $j{\mathrm{d}}_{C\oplus
C^{\prime}}(\tilde{c},c^{\prime})=0\cdot j(\tilde{c},c^{\prime})$ and thus
${\mathrm{d}}_{C\oplus C^{\prime}}(\tilde{c},c^{\prime})\in\ker
j=\mathrm{im~{}}i$, which implies that ${\mathrm{d}}_{C}\tilde{c}+\lambda
c^{\prime}=c$ for some $c\in C$. Observe that
${\mathrm{d}}_{C}c={\mathrm{d}}_{C}\bigl{(}{\mathrm{d}}_{C}\tilde{c}+\lambda
c^{\prime}\bigr{)}={\mathrm{d}}_{C}^{2}\tilde{c}+{\mathrm{d}}_{C}\lambda
c^{\prime}=0$. Define the connecting homomorphism
$k[c^{\prime}]:=[c]=[{\mathrm{d}}_{C}\tilde{c}+\lambda c^{\prime}]=[\lambda
c^{\prime}]=H(\lambda)[c^{\prime}]$, which establishes the connecting
homomorphism. ∎

###### Remark 8

If $(C,{\mathrm{d}}_{C})$ is a chain complex the short exact sequence in (9)
holds degree wise and yields:

${0}$${C_{k}}$${C_{k}\oplus
C^{\prime}_{k}}$${C^{\prime}_{k}}$${0}$${0}$${C_{k-1}}$${C_{k-1}\oplus
C^{\prime}_{k-1}}$${C^{\prime}_{k-1}}$${0}$$\scriptstyle{i_{k}}$$\scriptstyle{{\mathrm{d}}_{k}}$$\scriptstyle{j_{k}}$$\scriptstyle{\bar{\mathrm{d}}_{k}}$$\scriptstyle{0}$$\scriptstyle{i_{k-1}}$$\scriptstyle{j_{k-1}}$
(12)

In this case
$\bar{\mathrm{d}}_{k}(c,c^{\prime})={\mathrm{d}}_{k}c+\lambda_{k}c^{\prime}$,
where $\lambda_{k}\colon C^{\prime}_{k}\to C_{k-1}$ satisfies
${\mathrm{d}}_{k-1}\lambda_{k}=0$, and for the long exact sequence
$H_{k}(\lambda_{k})$ is the connecting homomorphism. The homology in degree
$k$ is given by the functor $H_{k}$.

###### Lemma 3

Consider the following commutative diagram of differential groups with exact
rows:

${0}$${C}$${C\oplus C^{\prime}}$${C^{\prime}}$${0}$${0}$${A}$${A\oplus
A^{\prime}}$${A^{\prime}}$${0}$$\scriptstyle{i}$$\scriptstyle{f^{\prime}}$$\scriptstyle{j}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{i}$$\scriptstyle{j}$
(13)

where $i(c)=(c,0)$, $j(c,c^{\prime})=c^{\prime}$, $f^{\prime}$ and
$f^{\prime\prime}$ are chain homomorphisms and where $C^{\prime}$ and
$A^{\prime}$ are boundaryless, $C^{\prime}$ is a free abelian group and
$(C,{\mathrm{d}}_{C})$ is a differential group. Then, the differentials on
$C\oplus C^{\prime}$ and $A\oplus A^{\prime}$ are given by Lemma 2 and are of
the form

${\mathrm{d}}_{C\oplus C^{\prime}}=\begin{pmatrix}{\mathrm{d}}_{C}&\lambda\\\
0&0\end{pmatrix},\quad{\mathrm{d}}_{A\oplus
A^{\prime}}=\begin{pmatrix}{\mathrm{d}}_{A}&\lambda^{\prime}\\\
0&0\end{pmatrix},$ (14)

where $\lambda\colon C^{\prime}\to C$ and $\lambda^{\prime}\colon
A^{\prime}\to A$ are chain homomorphisms, i.e. ${\mathrm{d}}_{C}\lambda=0$ and
${\mathrm{d}}_{A}\lambda^{\prime}=0$. Let $K:=jZ(C\oplus C^{\prime})\subseteq
C^{\prime}$ and $K^{\prime}:=jZ(A\oplus A^{\prime})\subseteq A^{\prime}$. If,

$H(f^{\prime})H(\lambda)=H(\lambda^{\prime})f^{\prime\prime},$ (15)

then the map $f^{\prime\prime}$ induces a chain homomorphism
$f^{\prime\prime}\colon K\to K^{\prime}$ by restriction, such that the diagram

${0}$${Z(C)}$${Z(C\oplus C^{\prime})}$${K}$${0}$${0}$${Z(A)}$${Z(A\oplus
A^{\prime})}$${K^{\prime}}$${0}$$\scriptstyle{i}$$\scriptstyle{f^{\prime}}$$\scriptstyle{j}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{i}$$\scriptstyle{j}$
(16)

commutes with exact rows. Moreover, if $C^{\prime}$ and $A^{\prime}$ are
finitely generated free abelian groups, then there exist complemented
subgroups $M\subseteq C^{\prime}$, $M^{\prime}\subseteq A^{\prime}$ and
isomorphisms $\omega\colon M\to K$ and $\omega^{\prime}\colon M^{\prime}\to
K^{\prime}$, such that the diagram

${0}$${Z(C)}$${Z(C\oplus C^{\prime})}$${M}$${0}$${0}$${Z(A)}$${Z(A\oplus
A^{\prime})}$${M^{\prime}}$${0}$$\scriptstyle{i}$$\scriptstyle{f^{\prime}}$$\scriptstyle{\bar{j}}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{i}$$\scriptstyle{\bar{j}}$
(17)

commutes with exact rows, and where $f^{\prime\prime}\colon M\to M^{\prime}$
is a chain homomorphism given by restriction.

###### Proof

By definition $Z(C\oplus
C^{\prime})=\bigl{\\{}(z,z^{\prime})\mid{\mathrm{d}}_{C}z+\lambda
z^{\prime}=0\bigr{\\}}$ and the projection $K$ is characterized by

$K=\bigl{\\{}z^{\prime}\in C^{\prime}\mid{\mathrm{d}}_{C}z+\lambda
z^{\prime}=0\text{~{}for some~{}}z\in C\bigr{\\}}.$ (18)

The same holds for $Z(A\oplus A^{\prime})$ and $K^{\prime}$. With the chain
maps $i$ and $j$ the rows in (16) are exact. Since $f^{\prime}$ is a chain
homomorphism the restriction of $f^{\prime}$ to $Z(C)$ defines the
homomorphism $f^{\prime}\colon Z(C)\to Z(A)$. Moreover, the diagram in (16)
commutes provided that $f^{\prime\prime}$ is a well-defined homomorphism from
$K$ to $K^{\prime}$. Consider the compositions $f^{\prime}\lambda$ and
$\lambda^{\prime}f^{\prime\prime}$. Then,

${\mathrm{d}}_{A}f^{\prime}\lambda=f^{\prime}d_{C}\lambda=0,\quad\text{and}\quad{\mathrm{d}}_{A}\lambda^{\prime}f^{\prime\prime}=0,$
(19)

which follows from the fact that $d_{C}\lambda=0$ and
$d_{A}\lambda^{\prime}=0$. We conclude that $f^{\prime}\lambda\colon
C^{\prime}\to Z(A)$ and $\lambda^{\prime}f^{\prime\prime}\colon C^{\prime}\to
Z(A)$. Define the homomorphism
$h=f^{\prime}\lambda-\lambda^{\prime}f^{\prime\prime}$ acting from
$C^{\prime}$ to $A$. Then, ${\mathrm{d}}_{A}h=0$ and thus $h\colon
C^{\prime}\to Z(A)$. Consider the diagram:

${C^{\prime}}$${0}$${B(A)}$${Z(A)}$${H(A)}$${0}$$\scriptstyle{h}$$\scriptstyle{H(h)}$$\scriptstyle{e}$$\scriptstyle{\rho}$
(20)

Observe that $H(h)=0$. Indeed, since both $f^{\prime}(\lambda c^{\prime})\in
Z(A)$ and $\lambda^{\prime}f^{\prime\prime}(c^{\prime})\in Z(A)$ the homology
class satisfies:

$\displaystyle H(h)[c^{\prime}]$ $\displaystyle=[h(c^{\prime})]=\rho
h(c^{\prime})=\rho f^{\prime}(\lambda
c^{\prime})-\rho\lambda^{\prime}f^{\prime\prime}(c^{\prime})=[f^{\prime}(\lambda
c^{\prime})]-[\lambda^{\prime}f^{\prime\prime}(c^{\prime})]$
$\displaystyle=H(f^{\prime})[\lambda
c^{\prime}]-H(\lambda^{\prime})f^{\prime\prime}(c^{\prime})=H(f^{\prime})H(\lambda)[c^{\prime}]-H(\lambda^{\prime})f^{\prime\prime}(c^{\prime})=0,$

where $[c^{\prime}]=c^{\prime}$, and which uses (15).

Let $z^{\prime}\in K$. We show that $f^{\prime\prime}(z^{\prime})\in
K^{\prime}$. Under restriction of $h$ to $K\subseteq C^{\prime}$ we have again
${\mathrm{d}}_{A}h|_{K}=0$ and $H(h|_{K})=0$. Since $C^{\prime}$ is a free
abelian group and $K\subseteq C^{\prime}$ a subgroup it follows that $K$ is
free and thus projective.555Since abelian groups are ${\mathbb{Z}}$-modules
the objects in the category $\mathbf{Ab}$ are free if and only if they are
projective. Submodules of a free ${\mathbb{Z}}$-module are again free and thus
projective. By Lemma 1 $h|_{K}\colon K\to B(A)$ and there exists a lift
$l\colon K\to A$ such that $h|_{K}={\mathrm{d}}_{A}l$. For $z^{\prime}$ define
an element $a\in A$, given by $a=l(z^{\prime})$, so that
${\mathrm{d}}_{A}a={\mathrm{d}}_{A}l(z^{\prime})=h|_{K}(z^{\prime})$. This
implies that ${\mathrm{d}}_{A}a=f^{\prime}(\lambda
z^{\prime})-\lambda^{\prime}f^{\prime\prime}(z^{\prime})$ and thus
$f^{\prime}(\lambda
z^{\prime})=\lambda^{\prime}f^{\prime\prime}(z^{\prime})+{\mathrm{d}}_{A}a$.
For $z^{\prime}\in K$ the definition of the latter yields a $z\in C$ such that
$(z,z^{\prime})\in Z(C\oplus C^{\prime})$, i.e. ${\mathrm{d}}_{C}z+\lambda
z^{\prime}=0$. From $a$ we define
$\zeta=\bigl{(}f^{\prime}(z)+a,f^{\prime\prime}(z^{\prime})\bigr{)}\in A\oplus
A^{\prime}$. Then, by the choice of $a$ and the fact that $f^{\prime}$ is a
chain homomorphism, we have that:

$\displaystyle{\mathrm{d}}_{A\oplus A^{\prime}}\zeta$
$\displaystyle={\mathrm{d}}_{A}(f^{\prime}(z)+a)+\lambda^{\prime}f^{\prime\prime}(z^{\prime})=f^{\prime}({\mathrm{d}}_{C}z)+\big{(}\lambda^{\prime}f^{\prime\prime}(z^{\prime})+{\mathrm{d}}_{A}a\big{)}$
$\displaystyle=f^{\prime}({\mathrm{d}}_{C}z)+f^{\prime}(\lambda
z^{\prime})=f^{\prime}({\mathrm{d}}_{C}z+\lambda z^{\prime})=0,$

which shows that $\zeta\in Z(A\oplus A^{\prime})$. By definition
$f^{\prime\prime}(z^{\prime})=j(\zeta)\in K^{\prime}$ proving our assertion.
The restriction of $f^{\prime\prime}$ to $K$ is obviously a chain homomorphism
since $C^{\prime}$ and $A^{\prime}$ are boundaryless.

Since $K\subseteq C^{\prime}$ is a subgroup of a free abelian group it is also
free, with $\operatorname{rank}(K)\leq\operatorname{rank}(C^{\prime})<\infty$,
and the basis elements for $K$ can be expressed in terms of a basis for
$C^{\prime}$. To be more precise, using Smith normal form, we can choose a
basis $\\{c^{\prime}_{1},\cdots,c^{\prime}_{r}\\}$ for $C^{\prime}$,
$r=\operatorname{rank}(C^{\prime})$, such that
$\\{\omega_{1}c_{1}^{\prime},\cdots,\omega_{k}c_{k}^{\prime}\\}$ is a basis
for $K$, with
$r^{\prime}=\operatorname{rank}(K)\leq\operatorname{rank}(C^{\prime})=r$. By
construction, the subgroup subgroup $M\subseteq C^{\prime}$, generated by the
indicated elements $\\{c_{1}^{\prime},\cdots,c_{r^{\prime}}^{\prime}\\}$, is
isomorphic to $K$ and an isomorphism $\omega\colon M\to K$ is given as
follows: $c^{\prime}_{i}\mapsto\omega_{i}c^{\prime}_{i}$,
$i=1,\cdots,r^{\prime}$. By construction $M$ allows a complementary subgroup
$M^{c}=[c^{\prime}_{r^{\prime}+1},\cdots,c^{\prime}_{r}]$ such that
$C^{\prime}=M\oplus M^{c}$. The same applies to $K^{\prime}\subseteq
A^{\prime}$ by choosing a basis $\\{a^{\prime}_{1},\cdots,a^{\prime}_{s}\\}$,
$s=\operatorname{rank}(A^{\prime})$ such that
$K^{\prime}=[\omega^{\prime}_{1}a^{\prime}_{1},\cdots,\omega^{\prime}_{s^{\prime}}a^{\prime}_{s^{\prime}}]$,
$s^{\prime}=\operatorname{rank}(K^{\prime})\leq\operatorname{rank}(A^{\prime})=s$.
As before this yields a decomposition $A^{\prime}=M^{\prime}\oplus M^{\prime
c}$, where $M^{\prime}\subseteq A^{\prime}$ and $M^{\prime}$ isomorphic to
$K^{\prime}$ with isomorphism $\omega^{\prime}\colon M^{\prime}\to K^{\prime}$
given by $a^{\prime}_{j}\mapsto\omega^{\prime}_{j}a^{\prime}_{j}$,
$j=1,\cdots,s^{\prime}$. Since $f^{\prime\prime}$ maps from $K$ to
$K^{\prime}$, for $i=1,\cdots,r^{\prime}$ we have,
$f^{\prime\prime}(\omega_{i}c^{\prime}_{i})=\lambda_{i1}\omega^{\prime}_{1}a^{\prime}_{1}+\cdots+\lambda_{is^{\prime}}\omega^{\prime}_{s^{\prime}}a^{\prime}_{s^{\prime}}$,
$f^{\prime\prime}(c^{\prime}_{i})=\mu_{i1}a^{\prime}_{1}+\cdots+\mu_{is}a^{\prime}_{s}$,
$i=1,\cdots,r^{\prime}$, and
$f^{\prime\prime}(\omega_{i}c^{\prime}_{i})=\omega_{i}\mu_{i1}a^{\prime}_{1}+\cdots+\omega_{i}\mu_{is}a^{\prime}_{s}$.
This implies
$f^{\prime\prime}(\omega_{i}c^{\prime}_{i})=\omega_{i}\mu_{i1}a^{\prime}_{1}+\cdots+\omega_{i}\mu_{i\ell}a^{\prime}_{s^{\prime}}$
and thus $\mu_{ij}=0$ for $j=s^{\prime}+1,\cdots,s$ and
$i=1,\cdots,r^{\prime}$, and
$\omega_{i}\mu_{ij}=\lambda_{ij}\omega^{\prime}_{j}$ for
$j=1,\cdots,s^{\prime}$ and $i=1,\cdots,r^{\prime}$. Therefore,
$f^{\prime\prime}\colon M\to M^{\prime}$. As before the diagram is commutative
with exact rows. ∎

###### Remark 9

Equation (15) yields a commutative diagram in homology and a cube of
homomorphisms of which the front and back faces are not commutative in
general:

${C^{\prime}}$${H(C)}$${A^{\prime}}$${H(A)}$$\scriptstyle{H(\lambda)}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{H(f^{\prime})}$$\scriptstyle{H(\lambda^{\prime})}$
${C^{\prime}}$${C}$${C^{\prime}}$${C}$${A^{\prime}}$${A}$${A^{\prime}}$${A}$$\scriptstyle{0}$$\scriptstyle{\lambda}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{{\mathrm{d}}_{C}}$$\scriptstyle{f^{\prime}}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{\lambda}$$\scriptstyle{f^{\prime}}$$\scriptstyle{0}$$\scriptstyle{\lambda^{\prime}}$$\scriptstyle{{\mathrm{d}}_{A}}$$\scriptstyle{\lambda^{\prime}}$

where $C^{\prime}=H(C^{\prime})$, $A^{\prime}=H(A^{\prime})$,
$f^{\prime\prime}=H(f^{\prime\prime})$ and $\lambda$ and $\lambda^{\prime}$
are chain homomorphisms.

###### Remark 10

If $(C,{\mathrm{d}}_{C})$ and $(A,{\mathrm{d}}_{A})$ are chain complexes we
consider the chain homomorphisms $f^{\prime}$ and $f^{\prime\prime}$ degree
wise, i.e. $f^{\prime}_{k}\colon C_{k}\to A_{k}$ and
$f^{\prime\prime}_{k}\colon C^{\prime}_{k}\to A^{\prime}_{k}$ for all $k$. In
this case (15) becomes
$H_{k-1}(f^{\prime}_{k-1})H_{k}(\lambda_{k})=H_{k}(\lambda^{\prime}_{k})f^{\prime\prime}_{k}$,
with $\lambda_{k}\colon C^{\prime}_{k}\to C_{k-1}$ and
$\lambda^{\prime}_{k}\colon A^{\prime}_{k}\to A_{k-1}$. Diagram (17) holds
degree wise with complemented subgroups $M_{k}$ and $M^{\prime}_{k}$ for all
$k$.

###### Lemma 4

Consider the commutative diagram:

${P^{\prime}}$${P^{\prime}\oplus
P^{\prime\prime}}$${P^{\prime\prime}}$${Q^{\prime}}$${Q^{\prime}\oplus
Q^{\prime\prime}}$${Q^{\prime\prime}}$${D^{\prime}}$${D}$${D^{\prime\prime}}$${B^{\prime}}$${B}$${B^{\prime\prime}}$$\scriptstyle{f^{\prime}}$$\scriptstyle{\epsilon^{\prime}}$$\scriptstyle{\epsilon}$$\scriptstyle{\epsilon^{\prime\prime}}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{\eta^{\prime}}$$\scriptstyle{\eta}$$\scriptstyle{\eta^{\prime\prime}}$$\scriptstyle{g^{\prime}}$$\scriptstyle{i_{D}}$$\scriptstyle{g}$$\scriptstyle{j_{D}}$$\scriptstyle{g^{\prime\prime}}$$\scriptstyle{i_{B}}$$\scriptstyle{j_{B}}$
(21)

where $\eta^{\prime}$ is surjective, the sequence
$0\xrightarrow[]{~{}~{}~{}}B^{\prime}\xrightarrow[]{i_{B}}B\xrightarrow[]{j_{B}}B^{\prime\prime}$
is exact and $P^{\prime\prime}$ is a free abelian group. The maps on the top
part of the diagram are the usual inclusion and projection homomorphisms.
Then, there exists a homomorphism $\gamma\colon P^{\prime\prime}\to
Q^{\prime}$ such that for the choice

$f=\begin{pmatrix}f^{\prime}&\gamma\\\ 0&f^{\prime\prime}\end{pmatrix}\colon
P^{\prime}\oplus P^{\prime\prime}\to Q^{\prime}\oplus Q^{\prime\prime},$ (22)

the diagram

${P^{\prime}}$${P^{\prime}\oplus
P^{\prime\prime}}$${P^{\prime\prime}}$${Q^{\prime}}$${Q^{\prime}\oplus
Q^{\prime\prime}}$${Q^{\prime\prime}}$${D^{\prime}}$${D}$${D^{\prime\prime}}$${B^{\prime}}$${B}$${B^{\prime\prime}}$$\scriptstyle{f^{\prime}}$$\scriptstyle{\epsilon^{\prime}}$$\scriptstyle{\epsilon}$$\scriptstyle{f}$$\scriptstyle{\epsilon^{\prime\prime}}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{\eta^{\prime}}$$\scriptstyle{\eta}$$\scriptstyle{\eta^{\prime\prime}}$$\scriptstyle{g^{\prime}}$$\scriptstyle{i_{D}}$$\scriptstyle{g}$$\scriptstyle{j_{D}}$$\scriptstyle{g^{\prime\prime}}$$\scriptstyle{i_{B}}$$\scriptstyle{j_{B}}$
(23)

commutes and $f$ is a lift of $g$ in the sense that $g\epsilon=\eta f$.

###### Proof

We define a homomorphism $\gamma\colon P^{\prime\prime}\to Q^{\prime}$ so that
$f$ given in (22) is a lift of $g$ in the sense that $g\epsilon=\eta f$.
Regardless of the definition of $\gamma$, we already have commutativity on
$P^{\prime}\oplus 0$. Let $(x,0)\in P^{\prime}\oplus P^{\prime\prime}$. Then,
$f(x,0)=\bigl{(}f^{\prime}(x),0\bigr{)}$ and a diagram chase yields

$\eta
f(x,0)=\eta(f^{\prime}(x),0)=i_{B}\eta^{\prime}f^{\prime}(x)=i_{B}g^{\prime}\epsilon^{\prime}(x)=gi_{D}\epsilon^{\prime}(x)=g\epsilon(x,0).$

It remains to define $\gamma$ on $P^{\prime\prime}$. Let $(x,y)\in
P^{\prime}\oplus P^{\prime\prime}$. In order for $f$ to be a lift of $g$ we
need $g\bigl{(}\epsilon(x,y)\bigr{)}=\eta f(x,y)$ and therefore

$\displaystyle g\epsilon(x,y)$ $\displaystyle=\eta
f(x,y)=\eta\bigl{(}f^{\prime}(x)+\gamma y,f^{\prime\prime}(y)\bigr{)}$
$\displaystyle=\eta\bigl{(}f^{\prime}(x),f^{\prime\prime}(y)\bigr{)}+\eta(\gamma
y,0)=\eta\bigl{(}f^{\prime}(x),f^{\prime\prime}(y)\bigr{)}+i_{B}\eta^{\prime}\gamma
y,$

which implies

$i_{B}\eta^{\prime}\gamma
y=g\epsilon(x,y)-\eta\bigl{(}f^{\prime}(x),f^{\prime\prime}(y)\bigr{)}.$ (24)

Using the commutativity with respect to elements $(x,0)\in P^{\prime}\oplus
P^{\prime\prime}$ one derives that
$g\epsilon(x,0)-\eta\bigl{(}f^{\prime}(x),0\bigr{)}=0$ and thus

$\displaystyle g\epsilon(x,y)$
$\displaystyle-\eta\bigl{(}f^{\prime}(x),f^{\prime\prime}(y)\bigr{)}$
$\displaystyle=g\epsilon(x,0)-\eta\bigl{(}f^{\prime}(x),0\bigr{)}+g\epsilon(0,y)-\eta\bigl{(}0,f^{\prime\prime}(y)\bigr{)}$
$\displaystyle=g\epsilon(0,y)-\eta\bigl{(}0,f^{\prime\prime}(y)\bigr{)},$

showing that the expression
$g\epsilon(x,y)-\eta\bigl{(}f^{\prime}(x),f^{\prime\prime}(y)\bigr{)}$ is
independent of $x$. Observe that

$\displaystyle
j_{B}\Bigl{[}g\epsilon(x,y)-\eta\bigl{(}f^{\prime}(x),f^{\prime\prime}(y)\bigr{)}\Bigr{]}$
$\displaystyle=j_{B}g\epsilon(x,y)-j_{B}\eta\bigl{(}f^{\prime}(x),f^{\prime\prime}(y)\bigr{)}$
$\displaystyle=g^{\prime\prime}j_{D}\epsilon(x,y)-\eta^{\prime\prime}f^{\prime\prime}(y)$
$\displaystyle=g^{\prime\prime}\epsilon^{\prime\prime}(y)-\eta^{\prime\prime}f^{\prime\prime}(y)=0,$

using the fact that $f^{\prime\prime}$ is a lift of $g^{\prime\prime}$. Since
$0\xrightarrow[]{~{}~{}~{}}B^{\prime}\xrightarrow[]{i_{B}}B\xrightarrow[]{j_{B}}B^{\prime\prime}$
is _exact_ we have that $\ker j_{B}=\mathrm{im~{}}i_{B}$ and $i_{B}$ is
injective. Consequently,

$\displaystyle b$
$\displaystyle=g\epsilon(x,y)-\eta\bigl{(}f^{\prime}(x),f^{\prime\prime}(y)\bigr{)}$
$\displaystyle=g\epsilon(0,y)-\eta\bigl{(}0,f^{\prime\prime}(y)\bigr{)}\in\mathrm{im~{}}i_{B},$

independent of $x$. Due to the latter and the injectivity of $i_{B}$ this
defines a homomorphism $\beta\colon P^{\prime\prime}\to B^{\prime}$ given by
$\beta
y=i_{B}^{-1}\bigl{(}g\epsilon(0,y)-\eta\bigl{(}0,f^{\prime\prime}(y)\bigr{)}\bigr{)}$.
By assumption we have the _surjective_ homomorphism $\eta^{\prime}\colon
Q^{\prime}\twoheadrightarrow B^{\prime}$. Since $P^{\prime\prime}$ is free
abelian, and therefore projective, there exists a lift $\gamma\colon
P^{\prime\prime}\to Q^{\prime}$ such that the diagram

${Q^{\prime}}$${P^{\prime\prime}}$${B^{\prime}}$$\scriptstyle{\eta^{\prime}}$$\scriptstyle{\beta}$$\scriptstyle{\gamma}$

commutes. Therefore, $\gamma\eta^{\prime}y=\beta y$ and thus
$i_{B}\gamma\eta^{\prime}y=i_{B}\beta y$ which establishes (24) and concludes
the construction of $f$. ∎

The following theorem concerns $C\oplus C^{\prime}$ and $A\oplus A^{\prime}$
not as product of differential groups but with differentials
${\mathrm{d}}_{C\oplus C^{\prime}}$ and ${\mathrm{d}}_{A\oplus A^{\prime}}$
respectively, cf. (14), and with embedded differential groups $C$ and $A$ with
differential ${\mathrm{d}}_{C}$ and ${\mathrm{d}}_{A}$ respectively, and
quotient groups $C^{\prime}$ and $A^{\prime}$ (boundaryless).

###### Remark 11

In the case of chain complexes Lemma 4 can be applied degree wise which yields
homomorphism $f$ in every degree, cf. Thm. 3.1.

###### Theorem 3.1

Consider the following commutative diagram of free differential groups with
exact rows:

${0}$${C}$${C\oplus C^{\prime}}$${C^{\prime}}$${0}$${0}$${A}$${A\oplus
A^{\prime}}$${A^{\prime}}$${0}$$\scriptstyle{i}$$\scriptstyle{f^{\prime}}$$\scriptstyle{j}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{i}$$\scriptstyle{j}$
(25)

where $C^{\prime}$ and $A^{\prime}$ are finitely generated and boundaryless
differential groups, $i$ and $j$ are given in Lemma 3 and $f^{\prime}$ and
$f^{\prime\prime}$ are chain homomorphisms. By Lemma 2 the differentials
${\mathrm{d}}_{C\oplus C^{\prime}}$ and ${\mathrm{d}}_{A\oplus A^{\prime}}$
are given by (14). Assume that the long exact sequences in homology make the
commutative diagram

${H(C)}$${H(C\oplus C^{\prime})}$${C^{\prime}}$${H(C)}$${H(A)}$${H(A\oplus
A^{\prime})}$${A^{\prime}}$${H(A)}$$\scriptstyle{H(f^{\prime})}$$\scriptstyle{g}$$\scriptstyle{H(\lambda)}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{H(f^{\prime})}$$\scriptstyle{H(\lambda^{\prime})}$
(26)

where $g\colon H(C\oplus C^{\prime})\to H(A\oplus A^{\prime})$ is a
homomorphism, and $H(f^{\prime})$ and $f^{\prime\prime}=H(f^{\prime\prime})$
are the induced homomorphism on homology and the rows are exact (exact
triangles). Then, there exists a chain homomorphism $f$ that is a lift for $g$
such that the diagram

${0}$${C}$${C\oplus C^{\prime}}$${C^{\prime}}$${0}$${0}$${A}$${A\oplus
A^{\prime}}$${A^{\prime}}$${0}$$\scriptstyle{f^{\prime}}$$\scriptstyle{f}$$\scriptstyle{f^{\prime\prime}}$
(27)

is a morphism of short exact sequences of differential groups, i.e. the
diagram commutes with short exact rows.

###### Proof

We construct a homomorphism $f\colon C\oplus C^{\prime}\to A\oplus A^{\prime}$
which satisfies two requirements: (i) $f$ is a chain homomorphism which fits
into Diagram (27), and (ii) $f$ induces $g$, i.e. $g=H(f)$. As in the proof of
Lemma 3 define the subgroups $K=jZ(C\oplus C^{\prime})$ and
$K^{\prime}=jZ(A\oplus A^{\prime})$, which are isomorphic to the complemented
subgroups $M$ and $M^{\prime}$ respectively. Given the diagram in (25) and by
the commutativity in (26), which implies (15), Lemma 3 yields the commutative
diagram in (17). Since the rows in Diagram (17) are exact and both $M$ and
$M^{\prime}$ are free and thus projective666Both $M\subseteq C^{\prime}$ and
$M^{\prime}\subseteq A^{\prime}$ and $C^{\prime}$ and $A^{\prime}$ are free
abelian groups. This implies that both $M$ and $M^{\prime}$ are free abelian
and thus projective. the short exact sequences split as:

${0}$${Z(C)}$${Z(C)\oplus M}$${M}$${0}$${0}$${Z(A)}$${Z(A)\oplus
M^{\prime}}$${M^{\prime}}$${0}$$\scriptstyle{f^{\prime}}$$\scriptstyle{f^{\prime\prime}}$
(28)

i.e. $Z(C\oplus C^{\prime})\cong Z(C)\oplus M$ and $Z(A\oplus A^{\prime})\cong
Z(A)\oplus M^{\prime}$. To be more precise, since $M$ is projective we have a
section map $s\colon M\to Z(C\oplus C^{\prime})$ and the diagram

${M}$${0}$${Z(C)}$${Z(C\oplus
C^{\prime})}$${M}$${0}$$\scriptstyle{s}$$\scriptstyle{\mathrm{id}}$$\scriptstyle{i}$$\scriptstyle{j}$

such that $js=\mathrm{id}$. Moreover, $s$ is an isomorphism from $M$ to
$s(M)$. By construction $Z(C\oplus C^{\prime})=\ker
j\oplus\mathrm{im~{}}s=Z(C)\oplus s(M)\cong Z(C)\oplus M$. The homomorphism
$s$ is given by the formula $z^{\prime}\mapsto(\sigma z^{\prime},z^{\prime})$,
for some $\sigma\colon M\to C$ such that ${\mathrm{d}}_{C}\sigma
z^{\prime}+\lambda z^{\prime}=0$, i.e.

${\mathrm{d}}_{C}\sigma=-\lambda,\quad\text{on~{}~{}}M.$ (29)

Summarizing we obtain:

${Z(C)\oplus M}$${0}$${Z(C)}$${Z(C)\oplus s(M)}$${M}$${0}$${Z(C\oplus
C^{\prime})}$$\scriptstyle{\cong}$$\scriptstyle{r}$$\scriptstyle{i}$$\scriptstyle{j}$

where both rows are exact and the isomorphism $r$ is given by

$r=\begin{pmatrix}\mathrm{id}~{}&~{}\sigma\\\
0~{}&~{}\mathrm{id}\end{pmatrix}\colon Z(C)\oplus M\to Z(C\oplus C^{\prime}).$
(30)

Indeed, for $c\in Z(C)$ and $z^{\prime}\in M$, $r(c,z^{\prime})=(c+\sigma
z^{\prime},z^{\prime})$ and ${\mathrm{d}}_{C}(c+\sigma z^{\prime})+\lambda
z^{\prime}={\mathrm{d}}_{C}\sigma z^{\prime}+\lambda z^{\prime}=0$. The map
$r$ extends to an automorphism on $C\oplus C^{\prime}$ by defining
$\sigma\colon C^{\prime}\to C$, where we use the fact that $C^{\prime}=M\oplus
M^{c}$ and define $\sigma|_{M^{c}}=0$. The same analysis applies to $Z(A\oplus
A^{\prime})\cong Z(A)\oplus M^{\prime}$, with automorphism $r^{\prime}$ and
$\sigma^{\prime}\colon A^{\prime}\to A$ with $\sigma^{\prime}|_{M^{\prime
c}}=0$. In order to apply Lemma 4 we conjugate the differentials
${\mathrm{d}}_{C\oplus C^{\prime}}$ and ${\mathrm{d}}_{A\oplus A^{\prime}}$:

$\bar{\mathrm{d}}_{C\oplus C^{\prime}}:=r^{-1}{\mathrm{d}}_{C\oplus
C^{\prime}}r,\quad\text{and}\quad\bar{\mathrm{d}}_{A\oplus
A^{\prime}}:=r^{\prime-1}{\mathrm{d}}_{A\oplus A^{\prime}}r^{\prime}.$

A direct calculation shows that

$\bar{\mathrm{d}}_{C\oplus
C^{\prime}}=\begin{pmatrix}{\mathrm{d}}_{C}&~{}~{}{\mathrm{d}}_{C}\sigma+\lambda\\\
0~{}&~{}0\end{pmatrix}=\begin{pmatrix}{\mathrm{d}}_{C}&~{}~{}\bar{\lambda}\\\
0~{}&~{}0\end{pmatrix}=\begin{pmatrix}{\mathrm{d}}_{C}&~{}0\\\
0~{}&~{}0\end{pmatrix}\quad\text{on $C\oplus M$},$

where $\bar{\lambda}={\mathrm{d}}_{C}\sigma+\lambda=0$ on $M$. Moreover,
$\bar{\mathrm{d}}_{C\oplus C^{\prime}}={\mathrm{d}}_{C\oplus C^{\prime}}$ on
$C\oplus M^{c}$, i.e. $\bar{\lambda}|_{M}=0$ and
$\bar{\lambda}|_{M^{c}}=\lambda$. The same characterization applies to
$\bar{\mathrm{d}}_{A\oplus A^{\prime}}$, with
$\bar{\lambda}^{\prime}={\mathrm{d}}_{A}\sigma^{\prime}+\lambda^{\prime}$.
This yields $\bar{g}\colon H(C\oplus C^{\prime})\to H(A\oplus A^{\prime})$,
where the homology are isomorphic to the homologies in Diagram (26). Now apply
Lemma 4 with Diagrams (26) and (28) as input. This yields the commutatative
diagram:

${Z(C)}$${Z(C)\oplus M}$${M}$${Z(A)}$${Z(A)\oplus
M^{\prime}}$${M^{\prime}}$${H(C)}$${H(C\oplus
C^{\prime})}$${C^{\prime}}$${H(A)}$${H(A\oplus
A^{\prime})}$${A^{\prime}}$$\scriptstyle{f^{\prime}}$
$\scriptstyle{\bar{f}}$$\scriptstyle{\epsilon^{\prime\prime}}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{\eta^{\prime}}$
$\scriptstyle{H(f^{\prime})}$ $\scriptstyle{\bar{g}}$
$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{i}$

where the homomorphism $\bar{f}\colon Z(C)\oplus M\to Z(A)\oplus M^{\prime}$
is given by

$\bar{f}|_{Z(C)\oplus M}=\begin{pmatrix}f^{\prime}~{}&\bar{\gamma}|_{M}\\\
0~{}&f^{\prime\prime}\end{pmatrix}\colon Z(C)\oplus M\to Z(A)\oplus
M^{\prime},$

for some homomorphism $\bar{\gamma}|_{M}\colon M\to Z(A)$, with
${\mathrm{d}}_{A}\bar{\gamma}|_{M}=0$, is a chain homomorphism, since the
differentials on $Z(C)\oplus M$ and $Z(A)\oplus M^{\prime}$ vanish. Moreover,
$\bar{f}$ is a lift for the homology map $\bar{g}\colon H(C\oplus
C^{\prime})\to H(A\oplus A^{\prime})$. Because $f^{\prime}$ and
$f^{\prime\prime}$ are defined on $C$ and $C^{\prime}$ respectively it remains
to extend $\bar{\gamma}|_{M}$ to all of $C^{\prime}$ in order to construct a
chain homomorphism $\bar{f}\colon C\oplus C^{\prime}\to A\oplus A^{\prime}$.
As pointed out above $M$ is free and complemented with $C^{\prime}=M\oplus
M^{c}$, and $M^{c}$ free and thus projective. We now construct $\bar{\gamma}$
on $M^{c}$. As before we write
$\bar{h}=f^{\prime}\bar{\lambda}-\bar{\lambda}^{\prime}f^{\prime\prime}=h+f^{\prime}{\mathrm{d}}_{C}\sigma-{\mathrm{d}}_{A}\sigma^{\prime}f^{\prime\prime}$.
Note that $\bar{h}|_{M}=0$. By (19), ${\mathrm{d}}_{A}\bar{h}=0$ and
$H(\bar{h})=0$, cf. Lem. 3. This applies in particular to the restriction
$\bar{h}|_{M^{c}}$. Since $M^{c}$ is projective we can apply Lemma 1 (using
that the differential on $C^{\prime}$ and thus $M^{c}$ is zero) which yields
the homomorphism $\bar{h}|_{M^{c}}\colon M^{c}\to B(A)$ and a lift
$\bar{\gamma}|_{M^{c}}\colon M^{c}\to A$

${A}$${M^{c}}$${B(A)}$$\scriptstyle{{\mathrm{d}}_{A}}$$\scriptstyle{\bar{h}|_{M^{c}}}$$\scriptstyle{\bar{\gamma}|_{M^{c}}}$

such that ${\mathrm{d}}_{A}\bar{\gamma}|_{M^{c}}=\bar{h}|_{M^{c}}$ and thus
${\mathrm{d}}_{A}\bar{\gamma}=\bar{h}$. The homomorphism $\bar{\gamma}$ is now
defined on all of $C^{\prime}$. This completes the construction of
$\bar{f}\colon C\oplus C^{\prime}\to A\oplus A^{\prime}$. The restriction to
$Z(C)\oplus M$ is a lift for $\bar{g}$ and therefore $\bar{f}$ is a lift for
$\bar{g}$. Verifying that $\bar{f}$ is a chain homomorphism is equivalent to
verifying that $f=r^{\prime}\bar{f}r^{-1}$ is a chain homomorphism. Observe
that $f$ is given by

$f=r^{\prime}\bar{f}r^{-1}=\begin{pmatrix}f^{\prime}~{}&-f^{\prime}\sigma+\sigma^{\prime}f^{\prime\prime}+\bar{\gamma}\\\
0~{}&f^{\prime\prime}\end{pmatrix}=\begin{pmatrix}f^{\prime}~{}&\gamma\\\
0~{}&f^{\prime\prime}\end{pmatrix}\colon C\oplus C^{\prime}\to A\oplus
A^{\prime}$ (31)

With this choice of $f$ Diagram (27) commutes regardless of the choice of
$\gamma$. In order for $f$ to be a chain homomorphism we need
${\mathrm{d}}_{A\oplus A^{\prime}}f=f{\mathrm{d}}_{C\oplus C^{\prime}}$. The
compositions are given by

$\displaystyle{\mathrm{d}}_{A\oplus
A^{\prime}}f=\begin{pmatrix}{\mathrm{d}}_{A}&\lambda^{\prime}\\\
0&0\end{pmatrix}\begin{pmatrix}f^{\prime}&\gamma\\\
0&f^{\prime\prime}\end{pmatrix}=\begin{pmatrix}{\mathrm{d}}_{A}f^{\prime}&{\mathrm{d}}_{A}\gamma+\lambda^{\prime}f^{\prime\prime}\\\
0&0\end{pmatrix}$ $\displaystyle f{\mathrm{d}}_{C\oplus
C^{\prime}}=\begin{pmatrix}f^{\prime}&\gamma\\\
0&f^{\prime\prime}\end{pmatrix}\begin{pmatrix}{\mathrm{d}}_{C}&\lambda\\\
0&0\end{pmatrix}=\begin{pmatrix}f^{\prime}{\mathrm{d}}_{C}&f^{\prime}\lambda\\\
0&0\end{pmatrix}$

This yields the condition:

${\mathrm{d}}_{A}\gamma=f^{\prime}\lambda-\lambda^{\prime}f^{\prime\prime}=h.$
(32)

Now,
${\mathrm{d}}_{A}\gamma=-{\mathrm{d}}_{A}f^{\prime}\sigma+{\mathrm{d}}_{A}\sigma^{\prime}f^{\prime\prime}+{\mathrm{d}}_{A}\bar{\gamma}=-f^{\prime}{\mathrm{d}}_{C}\sigma+{\mathrm{d}}_{A}\sigma^{\prime}f^{\prime\prime}+\bar{h}$.
On $M$ we obtain
${\mathrm{d}}_{A}\gamma=-f^{\prime}{\mathrm{d}}_{C}\sigma+{\mathrm{d}}_{A}\sigma^{\prime}f^{\prime\prime}=f^{\prime}\lambda-\lambda^{\prime}f^{\prime\prime}=h$,
and on $M^{c}$,
${\mathrm{d}}_{A}\gamma=-f^{\prime}{\mathrm{d}}_{C}\sigma+{\mathrm{d}}_{A}\sigma^{\prime}f^{\prime\prime}+\bar{h}=h$,
which realizes (32). By construction the chain homomorphism $f$ is a lift for
$g$ which completes the proof. ∎

###### Remark 12

As for Lemma 4, in the case of chain complexes Theorem 3.1 can be applied
degree wise which yields homomorphism $f$ in every degree.

Consider subgroups $C,C^{\prime}\subseteq\overline{C}$ and
$A,A^{\prime}\subseteq\overline{A}$. The subgroups $C,C^{\prime}$ and
$A,A^{\prime}$ are equipped with differentials
${\mathrm{d}}_{C},{\mathrm{d}}_{C^{\prime}}$ and
${\mathrm{d}}_{A},{\mathrm{d}}_{A^{\prime}}$ respectively. The next result
concerns the sums $C+C^{\prime}$ and $A+A^{\prime}$ as differential groups.

###### Theorem 3.2

Let $C,C^{\prime}\subseteq\overline{C}$ and
$A,A^{\prime}\subseteq\overline{A}$ be subgroups. Consider the following
diagram with exact rows:

${0}$${C\cap C^{\prime}}$${C\times
C^{\prime}}$${C+C^{\prime}}$${0}$${0}$${A\cap A^{\prime}}$${A\times
A^{\prime}}$${A+A^{\prime}}$${0}$$\scriptstyle{\varrho}$$\scriptstyle{\bar{f}}$$\scriptstyle{f^{\prime}\times
f^{\prime\prime}}$$\scriptstyle{\varpi}$$\scriptstyle{f}$$\scriptstyle{\varrho}$$\scriptstyle{\varpi}$
(33)

where $\varrho(c)=(c,-c)$ and $\varpi(c,c^{\prime})=c+c^{\prime}$. We equip
$C,C^{\prime}$ and $A,A^{\prime}$ with differentials
${\mathrm{d}}_{C},{\mathrm{d}}_{C^{\prime}}$ and
${\mathrm{d}}_{A},{\mathrm{d}}_{A^{\prime}}$ respectively. The product
differentials are given by

${\mathrm{d}}_{C\times
C^{\prime}}={\mathrm{d}}_{C}\times{\mathrm{d}}_{C^{\prime}}=\begin{pmatrix}{\mathrm{d}}_{C}&0\\\
0&{\mathrm{d}}_{C^{\prime}}\end{pmatrix},\quad\text{and}\quad{\mathrm{d}}_{A\times
A^{\prime}}={\mathrm{d}}_{A}\times{\mathrm{d}}_{A^{\prime}}=\begin{pmatrix}{\mathrm{d}}_{A}&0\\\
0&{\mathrm{d}}_{A^{\prime}}\end{pmatrix}.$

The homomorphisms $f^{\prime}\colon C\to A$ and $f^{\prime\prime}\colon
C^{\prime}\to A^{\prime}$ are chain homomorphisms with respect to
${\mathrm{d}}_{C}$ and ${\mathrm{d}}_{A}$, and ${\mathrm{d}}_{C^{\prime}}$ and
${\mathrm{d}}_{A^{\prime}}$ respectively. Then,

  1. (i)

the maps $\varrho$ and $\varpi$ are chain homomorphisms if and only if
${\mathrm{d}}_{C\cap
C^{\prime}}c={\mathrm{d}}_{C}c={\mathrm{d}}_{C^{\prime}}c$ for all $c\in C\cap
C^{\prime}$ and
${\mathrm{d}}_{C+C^{\prime}}(c+c^{\prime}):={\mathrm{d}}_{C}c+{\mathrm{d}}_{C^{\prime}}c^{\prime}$
for all $c\in C$ and all $c^{\prime}\in C^{\prime}$, and the same for
${\mathrm{d}}_{A\cap A^{\prime}}$ and ${\mathrm{d}}_{A+A^{\prime}}$;

  2. (ii)

Diagram (33) is commutative and the maps $\bar{f}$ and $f$ are chain
homomorphisms if and only if $\bar{f}(c)=f^{\prime}(c)=f^{\prime\prime}(c)$
for all $c\in C\cap C^{\prime}$ and
$f(c+c^{\prime}):=f^{\prime}(c)+f^{\prime\prime}(c^{\prime})$ for all $c\in C$
and all $c^{\prime}\in C^{\prime}$.

Given the commutative diagrams in homology:

${H(C)}$${H(C+C^{\prime})}$${H(A)}$${H(A+A^{\prime})}$$\scriptstyle{i}$$\scriptstyle{H(f^{\prime})}$$\scriptstyle{g}$$\scriptstyle{i}$
${H(C^{\prime})}$${H(C+C^{\prime})}$${H(A^{\prime})}$${H(A+A^{\prime})}$$\scriptstyle{i}$$\scriptstyle{H(f^{\prime\prime})}$$\scriptstyle{g}$$\scriptstyle{i}$
(34)

where $i([c]_{C})=[c]_{C+C^{\prime}}$ and
$i([c^{\prime}]_{C^{\prime}})=[c^{\prime}]_{C+C^{\prime}}$, and analogously
for $A,A^{\prime}$ and $A+A^{\prime}$. Then,

  1. (iii)

the chain homomorphism $f$ lifts $g$, i.e. $g=H(f)$.

###### Proof

Let us start with the observation that the sum of $f^{\prime}$ and
$f^{\prime\prime}$ is well-defined under the assumption that
$f^{\prime}=f^{\prime\prime}$ on $C\cap C^{\prime}$. The same applies to the
sum of differentials ${\mathrm{d}}_{C}$ and ${\mathrm{d}}_{C^{\prime}}$, and
${\mathrm{d}}_{A}$ and ${\mathrm{d}}_{A^{\prime}}$. Consider
$c+c^{\prime}=\bar{c}+\bar{c}^{\prime}$, $c,\bar{c}\in C$ and
$c^{\prime},\bar{c}^{\prime}\in C^{\prime}$. This implies that
$c-\bar{c}=c^{\prime}-\bar{c}^{\prime}\in C\cap C^{\prime}$. We show that
$f^{\prime}(c)+f^{\prime\prime}(c^{\prime})=f^{\prime}(\bar{c})+f^{\prime\prime}(\bar{c}^{\prime})$.
Consider,

$f^{\prime}(c)-f^{\prime}(\bar{c})+f^{\prime\prime}(c^{\prime})-f^{\prime\prime}(\bar{c}^{\prime})=f^{\prime}(c-\bar{c})-f^{\prime\prime}(c^{\prime}-\bar{c}^{\prime})=0,$

since $c-\bar{c}=c^{\prime}-\bar{c}^{\prime}\in C\cap C^{\prime}$ and
$f^{\prime}=f^{\prime\prime}$ on $C\cap C^{\prime}$, which proves $f\colon
C+C^{\prime}\to A+A^{\prime}$ is well-defined. Similarly,
${\mathrm{d}}_{C+C^{\prime}}\colon C+C^{\prime}\to C+C^{\prime}$ and
${\mathrm{d}}_{A+A^{\prime}}\colon A+A^{\prime}\to A+A^{\prime}$ are well-
defined.

For the diagrams:

${C\cap C^{\prime}}$${C\times C^{\prime}}$${C\cap C^{\prime}}$${C\times
C^{\prime}}$$\scriptstyle{\varrho}$$\scriptstyle{{\mathrm{d}}_{C\cap
C^{\prime}}}$$\scriptstyle{{\mathrm{d}}_{C\times
C^{\prime}}}$$\scriptstyle{\varrho}$ ${C\times
C^{\prime}}$${C+C^{\prime}}$${C\times
C^{\prime}}$${C+C^{\prime}}$$\scriptstyle{\varpi}$$\scriptstyle{{\mathrm{d}}_{C\times
C^{\prime}}}$$\scriptstyle{{\mathrm{d}}_{C+C^{\prime}}}$$\scriptstyle{\varpi}$

to commute we obtain ${\mathrm{d}}_{C\times
C^{\prime}}(c,-c)=\bigl{(}{\mathrm{d}}_{C}c,-{\mathrm{d}}_{C^{\prime}}c\bigr{)}=\bigl{(}{\mathrm{d}}_{C\cap
C^{\prime}}c,-{\mathrm{d}}_{C\cap C^{\prime}}c\bigr{)}$ for all $c\in C\cap
C^{\prime}$, which is equivalent to ${\mathrm{d}}_{C\cap
C^{\prime}}={\mathrm{d}}_{C}={\mathrm{d}}_{C^{\prime}}$. By the same token the
second diagram yields
${\mathrm{d}}_{C+C^{\prime}}(c+c^{\prime})={\mathrm{d}}_{C}c+{\mathrm{d}}_{C^{\prime}}c^{\prime}$
which is well-defined by our above observation. The same arguments apply to
${\mathrm{d}}_{A},{\mathrm{d}}_{A^{\prime}}$, ${\mathrm{d}}_{A\cap
A^{\prime}}$ and ${\mathrm{d}}_{A+A^{\prime}}$ concluding the proof of (i).

Consider the diagrams:

${C\cap C^{\prime}}$${C\times C^{\prime}}$${A\cap A^{\prime}}$${A\times
A^{\prime}}$$\scriptstyle{\varrho}$$\scriptstyle{\bar{f}}$$\scriptstyle{f^{\prime}\times
f^{\prime\prime}}$$\scriptstyle{\varrho}$ ${C\times
C^{\prime}}$${C+C^{\prime}}$${A\times
A^{\prime}}$${A+A^{\prime}}$$\scriptstyle{\varpi}$$\scriptstyle{f^{\prime}\times
f^{\prime\prime}}$$\scriptstyle{f}$$\scriptstyle{\varpi}$

In order for the first diagram to commute we chase the diagram which yields
the condition $(f^{\prime}\times
f^{\prime\prime})(c,-c)=\bigl{(}f^{\prime}(c),-f^{\prime\prime}(c^{\prime})\bigr{)}=\bigl{(}\bar{f}(c),-\bar{f}(c)\bigr{)}$
for all $c\in C\cap C^{\prime}$. This condition is equivalent to
$\bar{f}(c)=f^{\prime}(c)=f^{\prime\prime}(c)$ for all $c\in C\cap
C^{\prime}$. As for the second diagram we obtain
$f(c+c^{\prime})=f^{\prime}(c)+f^{\prime\prime}(c^{\prime})$ for all $c\in C$
and $c^{\prime}\in C^{\prime}$, and $f$ is well-defined by the above
observation. It remains to show that $\bar{f}$ and $f$ satisfy the chain
homomorphism property. Since $f^{\prime}$ and $f^{\prime\prime}$ are chain
homomorphisms, and the differential coincide on the intersections, $\bar{f}$
is a chain homomorphism. For $f$ we have

$\displaystyle f\bigl{(}{\mathrm{d}}_{C+C^{\prime}}(c+c^{\prime})\bigr{)}$
$\displaystyle=f\bigl{(}{\mathrm{d}}_{C}c+{\mathrm{d}}_{C^{\prime}}c^{\prime}\bigr{)}=f^{\prime}({\mathrm{d}}_{C}c)+f^{\prime\prime}({\mathrm{d}}_{C^{\prime}}c^{\prime})$
$\displaystyle={\mathrm{d}}_{A}f^{\prime}(c)+{\mathrm{d}}_{A^{\prime}}f^{\prime\prime}(c^{\prime})={\mathrm{d}}_{A+A^{\prime}}\bigl{(}f^{\prime}(c)+f^{\prime\prime}(c^{\prime})\bigr{)}$
$\displaystyle={\mathrm{d}}_{A+A^{\prime}}f(c+c^{\prime}),$

proving (ii).

As for (iii) we argue as follows. From (34) we have that
$g\bigl{(}[c]_{C+C^{\prime}}\bigr{)}=[f^{\prime}(c)]_{A+A^{\prime}}$, $c\in C$
and
$g\bigl{(}[c^{\prime}]_{C+C^{\prime}}\bigr{)}=[f^{\prime\prime}(c^{\prime})]_{A+A^{\prime}}$,
$c^{\prime}\in C^{\prime}$. Consequently,

$\displaystyle g\bigl{(}[c+c^{\prime}]_{C+C^{\prime}}\bigr{)}$
$\displaystyle=g\bigl{(}[c]_{C+C^{\prime}}\bigr{)}+g\bigl{(}[c^{\prime}]_{C+C^{\prime}}\bigr{)}=[f^{\prime}(c)]_{A+A^{\prime}}+[f^{\prime\prime}(c^{\prime})]_{A+A^{\prime}}$
$\displaystyle=[f^{\prime}(c)+f^{\prime\prime}(c^{\prime})]_{A+A^{\prime}}=[f(c+c^{\prime})]_{A+A^{\prime}}$
$\displaystyle=H(f)\bigl{(}[c+c^{\prime}]_{C+C^{\prime}}\bigr{)},$

which completes the proof. ∎

###### Remark 13

As before Theorem 3.2 can be carried out degree wise which is of particular
interest for chain complexes.

###### Remark 14

The Lemma 1, 2 and 4 hold for $R$-modules over any commutative ring $R$ and
replacing free abelian with projective. Lemma 3 requires submodules of
projective modules to be projective which dictates $R$ to be an hereditary
ring. Theorem 3.1 can therefore also be stated for $R$-modules over an
hereditary ring $R$ with free abelian replace by projective. Theorem 3.2 holds
for $R$-modules over any commutative ring $R$. Consequently, all of the above
mentioned lemmas and theorems hold if we consider rings $R$ that are principal
ideal domains. An example of a principal ideal domain is a field
${\mathbb{K}}$ which implies that all of the above results remain valid for
${\mathbb{K}}$-vector spaces. If for example $R={\mathbb{Z}}$ we retrieve the
category of abelian groups. More general principal ideals are for instance
polynomial rings ${\mathbb{K}}[x]$ over a field ${\mathbb{K}}$.

###### Remark 15

Chain complexes are the primary sources for differential groups, especially in
their applications to dynamical systems. As noted in Example 1, a chain
complex is a graded differential group with an off-diagonal differential. This
characteristic ensures that the homology is also graded:
$H(C)=\bigoplus_{k}H_{k}(C)$, where
$H_{k}(C)=\ker{\mathrm{d}}_{k}/\mathrm{im~{}}{\mathrm{d}}_{k+1}$. The chain
maps for chain complexes are homogeneous and induce homogeneous homomorphisms
on homology: $H(f)=\bigoplus H_{k}(f_{k})$ with $H_{k}(f_{k})\colon
H_{k}(C)\to H_{k-1}(C)$. An ${\mathsf{O}}({\mathsf{P}})$-filtering is defined
by $\alpha\mapsto F_{\alpha}C_{k}$ for all $k$ — degree wise filtering — which
is a filtering on $C$, but not vice versa. Chain homomorphisms are assumed to
be filtered degree wise: $f_{k}(F_{\alpha}C_{k})\subseteq
F_{\alpha}C^{\prime}_{k}$ for all $k$. The morphisms for the associated
Cartan-Eilenberg systems are homogeneous with respect to the integer grading
induced by the chain complexes. The assertions of Theorems 3.1 and 3.2 remain
valid if the differential groups are chain complexes. The techniques used in
the proofs of Theorems 3.1 and 3.2 allow us to construct the chain maps
separately in each degree. In the previous paragraphs we indicated the
restriction to chain complexes in Remarks 7, 8, 10, 11, 12 and 13. For
simplicity of exposition, we do not use the specific grading of chain
complexes.

##  4 Properties of excisive Cartan-Eilenberg systems

In Section 1 we introduced the notion of Cartan-Eilenberg system over a
general poset. In this section we will provide a formal definition and derive
some essential properties needed to prove the main theorem.

###  4.1 A categorical definition

Let $({\mathsf{Q}},\leq)$ be a poset, or more generally a pre-order. We may
regard ${\mathsf{Q}}$ as small (thin) category, denoted $\text{\bf
Q}_{\mathbf{1}}$, where the objects in the category are the elements in
${\mathsf{Q}}$ and the order relations $\alpha\leq\beta$ account for the
morphisms, or arrows, i.e. $\alpha\leq\beta$ yields the arrow
$\alpha\to\beta$. The _arrow category_ of ${\mathsf{Q}}$ consists of pairs
$(\alpha,\beta)$, with $\alpha\leq\beta$, and unique morphisms
$(\alpha,\beta)\to(\gamma,\delta)$ for $\alpha\leq\gamma$ and
$\beta\leq\delta$, and is denoted by $\text{\bf Q}_{\mathbf{2}}$. The latter
corresponds to commutative diagrams in ${\mathsf{Q}}$. The (functor)
categories $\text{\bf Q}_{\mathbf{n}}$ consist of concatenations of $n-1$
arrows and are given by $n$-tuples777The entries $\alpha_{i}$ in
$(\alpha_{1},\cdots,\alpha_{n})$ satisfy $\alpha_{1}\leq\cdots\leq\alpha_{n}$
and are not necessarily distinct elements of ${\mathsf{Q}}$.
$(\alpha_{1},\cdots,\alpha_{n})\in{\mathsf{Q}}\times\cdots\times{\mathsf{Q}}$,
$\alpha_{i}\in{\mathsf{Q}}$, with the additional requirement that
$\alpha_{1}\leq\cdots\leq\alpha_{n}$.888The notation $\text{\bf
Q}_{\mathbf{n}}$ incates the functor category of functors
$F\colon{\mathbf{n}}\to{\mathsf{Q}}$, where ${\mathbf{n}}$ is the category
$\bullet\to\bullet\to\cdots\to\bullet\to\bullet$ consisting of $n+1$ objects
and $n$ arrows. The objects are identified with nested $n$-tuples in
${\mathsf{Q}}$ and the morphisms are given by the induced order on $n$-tuples.
An object in $\text{\bf Q}_{\mathbf{n}}$ may also be regarded as a path of
length $n$ in the directed graph induced by the poset ${\mathsf{Q}}$.
Following HelleRognes  we consider the covariant functors $\mathsf{p}_{0}$,
$\mathsf{p}_{1}$ and $\mathsf{p}_{2}$ acting from $\text{\bf Q}_{\mathbf{3}}$
to $\text{\bf Q}_{\mathbf{2}}$, given by

$(\alpha,\beta,\gamma)\xmapsto{\mathsf{p}_{0}}(\alpha,\beta),\quad(\alpha,\beta,\gamma)\xmapsto{\mathsf{p}_{1}}(\alpha,\gamma),\text{~{}~{}and~{}~{}}(\alpha,\beta,\gamma)\xmapsto{\mathsf{p}_{2}}(\beta,\gamma),$

respectively, and the natural transformations
${\mathsf{i}}\colon\mathsf{p}_{0}\Rightarrow\mathsf{p}_{1}$ and
${\mathsf{j}}\colon\mathsf{p}_{1}\Rightarrow\mathsf{p}_{2}$ whose components
are given by

$(\alpha,\beta)\xmapsto{{\mathsf{i}}}(\alpha,\gamma)\text{~{}~{}and~{}~{}}(\alpha,\gamma)\xmapsto{{\mathsf{j}}}(\beta,\gamma),$

respectively.

###### Definition 2

A _Cartan-Eilenberg system_ 999Cartain-Eilenberg systems can be formulated in
any abelian category such as $R$-modules or ${\mathbb{K}}$-vector spaces. over
a poset ${\mathsf{Q}}$ consists of a covariant functor
${\mathsf{E}}\colon\text{\bf Q}_{\mathbf{2}}\to\mathbf{Ab}$ and a natural
transformation
${\mathsf{k}}\colon{\mathsf{E}}\mathsf{p}_{2}\Rightarrow{\mathsf{E}}\mathsf{p}_{0}$
between the composite functors ${\mathsf{E}}\mathsf{p}_{2}$ and
${\mathsf{E}}\mathsf{p}_{0}$, called the _connecting homomorphism_ , such that

${{\mathsf{E}}\mathsf{p}_{0}}$${{\mathsf{E}}\mathsf{p}_{1}}$${{\mathsf{E}}\mathsf{p}_{2}}$$\scriptstyle{{\mathsf{E}}{\mathsf{i}}}$$\scriptstyle{{\mathsf{E}}{\mathsf{j}}}$$\scriptstyle{{\mathsf{k}}}$

is an exact triangle, where the natural transformations
${\mathsf{E}}{\mathsf{i}}$ and ${\mathsf{E}}{\mathsf{j}}$ are the right
whiskerings of ${\mathsf{E}}$ and ${\mathsf{i}}$, and ${\mathsf{E}}$ and
${\mathsf{j}}$ respectively. A Cartan-Eilenberg system over ${\mathsf{Q}}$ is
denoted by $\text{\bf E}=\bigl{(}\text{\bf
Q}_{\mathbf{2}},{\mathsf{E}},{\mathsf{k}}\bigr{)}$.

###### Remark 16

In CE  the poset ${\mathsf{Q}}$ are the extended integers. The above
definition is based on the extension in HelleRognes  for linearly ordered
posets.

Morphism between Cartan-Eilenberg systems E and $\text{\bf E}^{\prime}$ are
given as natural transformations
${\mathsf{h}}\colon{\mathsf{E}}\Rightarrow{\mathsf{E}}^{\prime}$ between the
functors ${\mathsf{E}}$ and ${\mathsf{E}}^{\prime}$ such that

${{\mathsf{E}}\mathsf{p}_{2}}$${{\mathsf{E}}\mathsf{p}_{0}}$${{\mathsf{E}}^{\prime}\mathsf{p}_{2}}$${{\mathsf{E}}^{\prime}\mathsf{p}_{0}}$$\scriptstyle{{\mathsf{k}}}$$\scriptstyle{{\mathsf{h}}\mathsf{p}_{2}}$$\scriptstyle{{\mathsf{h}}\mathsf{p}_{0}}$$\scriptstyle{{\mathsf{k}}^{\prime}}$
(35)

is commutative, where ${\mathsf{h}}\mathsf{p}_{2}$ and
${\mathsf{h}}\mathsf{p}_{1}$ are the left whiskerings of ${\mathsf{h}}$ and
$\mathsf{p}_{2}$ and $\mathsf{p}_{0}$ respectively. The components are given
in (2).

###  4.2 Incomparable convex sets

We restrict here to excisive Cartan-Eilenberg systems over a finite
distributive lattice ${\mathsf{Q}}={\mathsf{O}}({\mathsf{P}})$, where
${\mathsf{P}}$ is a finite poset. Recall that a Cartan-Eilenberg system is
excisive if $\ell\colon
E_{\alpha\cap\beta}^{\alpha}\xrightarrow[]{\cong}E_{\beta}^{\alpha\cup\beta}$
is an isomorphism for all $\alpha,\beta\in{\mathsf{O}}({\mathsf{P}})$.

###### Lemma 5

Let $(\alpha,\beta),(\alpha^{\prime},\beta^{\prime})\in\text{\bf
Q}_{\mathbf{2}}$ with
$\beta\smallsetminus\alpha=\beta^{\prime}\smallsetminus\alpha^{\prime}$. Then,
$E^{\beta}_{\alpha}\cong E^{\beta^{\prime}}_{\alpha^{\prime}}$.

###### Proof

Define $\tilde{\alpha}=\alpha\vee\alpha^{\prime}$ and
$\tilde{\beta}=\beta\vee\beta^{\prime}$. Then,
$\alpha,\alpha^{\prime}\leq\tilde{\alpha}$,
$\beta,\beta^{\prime}\leq\tilde{\beta}$ and
$\beta\smallsetminus\alpha=\beta^{\prime}\smallsetminus\alpha^{\prime}=\tilde{\beta}\smallsetminus\tilde{\alpha}$.
Observe that
$(\beta\cap\tilde{\beta})\smallsetminus(\alpha\cup\tilde{\alpha})=\beta\smallsetminus\tilde{\alpha}=\tilde{\beta}\smallsetminus\tilde{\alpha}$.
Consequently,
$(\beta\cup\tilde{\alpha})\smallsetminus\tilde{\alpha}=\tilde{\beta}\smallsetminus\tilde{\alpha}$
and therefore $\tilde{\beta}=\beta\cup\tilde{\alpha}$. Similarly,
$\beta\smallsetminus(\beta\cap\tilde{\alpha})=\beta\smallsetminus\alpha$ which
implies that $\alpha=\beta\cap\tilde{\alpha}$. By the excisive property we
obtain
$E^{\beta}_{\alpha}=E^{\beta}_{\tilde{\alpha}\cap\beta}\xrightarrow[]{\ell}E^{\tilde{\alpha}\cup\beta}_{\tilde{\alpha}}=E^{\tilde{\beta}}_{\tilde{\alpha}},$
By the same token one proves that
$E^{\beta^{\prime}}_{\alpha^{\prime}}=E^{\beta^{\prime}}_{\tilde{\alpha}\cap\beta^{\prime}}\xrightarrow[]{\ell^{\prime}}E^{\tilde{\alpha}\cup\beta^{\prime}}_{\tilde{\alpha}}=E^{\tilde{\beta}}_{\tilde{\alpha}},$
and thus $(\ell^{\prime})^{-1}\ell\colon
E^{\beta}_{\alpha}\xrightarrow[]{\cong}E^{\beta^{\prime}}_{\alpha^{\prime}}$.
∎

In an excisive Cartan-Eilenberg system the isomorphisms between $E$-terms
yield additional relations on the maps $i$ and $j$.

###### Lemma 6

Consider non-trivial triples
$(\alpha,\beta,\gamma),(\alpha,\beta^{\prime},\gamma)$, i.e.
$\alpha\subsetneq\gamma$, such that
$\beta\smallsetminus\alpha=\gamma\smallsetminus\beta^{\prime}$ and
$\beta^{\prime}\smallsetminus\alpha=\gamma\smallsetminus\beta$. Then, in the
exact triangles

${E^{\beta}_{\alpha}}$${E^{\gamma}_{\alpha}}$${E^{\gamma}_{\beta}}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{k}$
${E^{\beta^{\prime}}_{\alpha}}$${E^{\gamma}_{\alpha}}$${E^{\gamma}_{\beta^{\prime}}}$$\scriptstyle{i^{\prime}}$$\scriptstyle{j^{\prime}}$$\scriptstyle{k^{\prime}}$
(36)

the homomorphisms $i,i^{\prime}$ and $j,j^{\prime}$ satisfy:
$\ell^{-1}j^{\prime}i=\mathrm{id}$ on $E_{\alpha}^{\beta}$ and
$ji^{\prime}\ell^{\prime-1}=\mathrm{id}$ on $E_{\beta}^{\gamma}$, where
$\ell\colon E^{\beta}_{\alpha}\to E^{\gamma}_{\beta^{\prime}}$ and
$\ell^{\prime}\colon E^{\beta^{\prime}}_{\alpha}\to E^{\gamma}_{\beta}$ are
the isomorphisms by the excisive property.

###### Proof

An excisive Cartan-Eilenberg system yields the isomorphisms $\ell\colon
E^{\beta}_{\alpha}\to E^{\gamma}_{\beta^{\prime}}$ and $\ell^{\prime}\colon
E^{\gamma}_{\beta}\to E^{\beta^{\prime}}_{\alpha}$. By the definition of the
functor ${\mathsf{E}}$ and the categorical interpretation ordered pairs,
$j^{\prime}i$ is the homomorphism $\ell\colon E^{\beta}_{\alpha}\to
E^{\gamma}_{\beta^{\prime}}$, which implies that $j^{\prime}i$ is an
isomorphism and $\ell^{-1}j^{\prime}i=\mathrm{id}$ on $E^{\beta}_{\alpha}$.
The same applies to isomorphism $\ell^{\prime}\colon
E^{\beta^{\prime}}_{\alpha}\to E^{\gamma}_{\beta}$ and
$ji^{\prime}{\ell^{\prime}}^{-1}=\mathrm{id}$ on $E^{\gamma}_{\beta}$. ∎

###### Remark 17

For non-trivial triples
$(\alpha,\beta,\gamma),(\alpha,\beta^{\prime},\gamma)$, such that
$\beta\smallsetminus\alpha=\gamma\smallsetminus\beta^{\prime}$ and
$\beta^{\prime}\smallsetminus\alpha=\gamma\smallsetminus\beta$, the convex
sets $\xi=\beta\smallsetminus\alpha$ and $\eta=\gamma\smallsetminus\beta$ are
_incomparable_ , or _parallel_ , i.e. $\xi$ and $\eta$ are incomparable with
respect induced order by ${\mathsf{P}}$. They satisfy the additional property
that $\xi\cap\eta=\varnothing$ and
$\xi\cup\eta=\zeta=\gamma\smallsetminus\alpha$, as well as
$\beta\cap\beta^{\prime}=\alpha$ and $\beta\cup\beta^{\prime}=\gamma$. In
pariticular, $\beta\neq\beta^{\prime}$.

For the $E$-terms of the triples
$(\alpha,\beta,\gamma),(\alpha,\beta^{\prime},\gamma)$ in Lemma 6 the
homomorphisms $i,i^{\prime}$ are injective and the homomorphisms
$j,j^{\prime}$ are surjective. In particular, using Lemma 6,

${0}$${E_{\alpha}^{\beta}}$${E^{\gamma}_{\alpha}}$${E^{\gamma}_{\beta}}$${0,}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{i^{\prime}\ell^{\prime-1}}$

which shows that the exact sequence is right split and thus a split exact
sequence with $E^{\gamma}_{\alpha}\cong E^{\beta}_{\alpha}\oplus
E^{\gamma}_{\beta}$. The same construction follows by using the other triple.
The connecting homomorphisms $k$ and $k^{\prime}$ are trivial proving that the
exact triangle in (36) are in fact split exact sequences. These consideration
hold for excisive Cartan-Eilenberg systems in general. If we consider Cartan-
Eilenberg systems generated by a ${\mathsf{P}}$-graded differential group we
can derive properties of the differential for the above case of triples
describing incomparable convex sets. As before consider triples
$(\alpha,\beta,\gamma),(\alpha,\beta^{\prime},\gamma)$, such that
$\beta\smallsetminus\alpha=\gamma\smallsetminus\beta^{\prime}$ and
$\beta^{\prime}\smallsetminus\alpha=\gamma\smallsetminus\beta$. Let
$C=\bigoplus_{p\in{\mathsf{P}}}G_{p}C$, then (5) yields the exact sequences

${G_{\gamma\smallsetminus\beta^{\prime}}C}$${G_{\beta\smallsetminus\alpha}C\oplus
G_{\gamma\smallsetminus\beta}C}$${0}$${G_{\beta\smallsetminus\alpha}C}$${G_{\gamma\smallsetminus\alpha}C}$${G_{\gamma\smallsetminus\beta}C}$${0}$${0}$${G_{\beta^{\prime}\smallsetminus\alpha}C}$${G_{\gamma\smallsetminus\alpha}C}$${G_{\gamma\smallsetminus\beta^{\prime}}C}$${0}$${G_{\gamma\smallsetminus\beta}C}$${G_{\beta^{\prime}\smallsetminus\alpha}C\oplus
G_{\gamma\smallsetminus\beta^{\prime}}C}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{i^{\prime}}$$\scriptstyle{j^{\prime}}$
(37)

Since $i,i^{\prime}$ are chain homomorphisms by the assumption that the
differential ${\mathrm{d}}_{C}$ is ${\mathsf{O}}({\mathsf{P}})$-filtered we
obtain the following splitting lemma for the differential ${\mathrm{d}}_{C}$.

###### Lemma 7

Under the assumption that
$\beta\smallsetminus\alpha=\gamma\smallsetminus\beta^{\prime}$ and
$\beta^{\prime}\smallsetminus\alpha=\gamma\smallsetminus\beta$ the
differential ${\mathrm{d}}_{C}\colon G_{\gamma\smallsetminus\alpha}C\to
G_{\gamma\smallsetminus\alpha}C$ is given by:

${\mathrm{d}}_{G_{\gamma\smallsetminus\alpha}C}=\begin{pmatrix}{\mathrm{d}}_{G_{\beta\smallsetminus\alpha}C}&0\\\
0~{}&{\mathrm{d}}_{G_{\gamma\smallsetminus\beta}C}\end{pmatrix}\colon
G_{\beta\smallsetminus\alpha}C\oplus G_{\gamma\smallsetminus\beta}\to
G_{\beta\smallsetminus\alpha}C\oplus G_{\gamma\smallsetminus\beta},$ (38)

where ${\mathrm{d}}_{G_{\beta\smallsetminus\alpha}C}$ and
${\mathrm{d}}_{G_{\gamma\smallsetminus\beta}C}$ are the restrictions of
${\mathrm{d}}_{C}$ to $G_{\beta\smallsetminus\alpha}C$ and
$G_{\gamma\smallsetminus\beta}$ respectively.

###### Proof

If we apply the first part of the proof of Lemma 2 to both exact sequences in
(37) we obtain that the off diagonal terms are zero completing the proof. ∎

###### Remark 18

For Cartan-Eilenberg systems generated by ${\mathsf{P}}$-graded differential
groups the conclusion of Lemma 6 follows immediately since
$ji^{\prime}=\mathrm{id}$ and $j^{\prime}i=\mathrm{id}$ as indicated in
Diagram (37). The direct sum decomposition is immediate from the form of the
differential.

##  5 Strict representations

Two ${\mathsf{O}}({\mathsf{P}})$-filtered chain homomorphisms
$f,g\colon(C,{\mathrm{d}}_{C})\to(A,{\mathrm{d}}_{A})$ are said to be
${\mathsf{O}}({\mathsf{P}})$-filtered chain homotopic if there is an
${\mathsf{O}}({\mathsf{P}})$-filtered homomorphism $h\colon C\to A$ such that
$f-g=h{\mathrm{d}}_{C}+{\mathrm{d}}_{A}h$; such a map $h$ is called an
${\mathsf{O}}({\mathsf{P}})$-filtered chain homotopy from $f$ to $g$.
Notation: $f^{\prime}\sim f$. In this case $(C,{\mathrm{d}}_{C})$ and
$(A,{\mathrm{d}}_{A})$ are _${\mathsf{O}}({\mathsf{P}})$ -filtered chain
homotopic_. Notation: $(C,{\mathrm{d}}_{C})\sim(A,{\mathrm{d}}_{A})$. An
${\mathsf{O}}({\mathsf{P}})$-filtered chain homomorphism
$f\colon(C,{\mathrm{d}}_{C})\to(A,{\mathrm{d}}_{A})$ is an
${\mathsf{O}}({\mathsf{P}})$-filtered chain (homotopy) equivalence if there is
an ${\mathsf{O}}({\mathsf{P}})$-filtered chain homomorphism
$f^{\prime}\colon(A,{\mathrm{d}}_{A})\to(C,{\mathrm{d}}_{C})$ such that
$f^{\prime}f\sim\mathrm{id}_{C}$ and $ff^{\prime}\sim\mathrm{id}_{A}$. In this
case $(C,{\mathrm{d}}_{C})$ and $(A,{\mathrm{d}}_{A})$ are
_${\mathsf{O}}({\mathsf{P}})$ -filtered chain equivalent_. Notation:
$(C,{\mathrm{d}}_{C})\simeq(A,{\mathrm{d}}_{A})$. Two differential groups
$(C,{\mathrm{d}}_{C})$ and $(A,{\mathrm{d}}_{A})$ are
_${\mathsf{O}}({\mathsf{P}})$ -filtered chain isomorpic_ if there exist
${\mathsf{O}}({\mathsf{P}})$-filtered chain homomorphisms
$f\colon(C,{\mathrm{d}}_{C})\to(A,{\mathrm{d}}_{A})$ and
$f^{\prime}\colon(A,{\mathrm{d}}_{A})\to(C,{\mathrm{d}}_{C})$ such that
$f^{\prime}f=\mathrm{id}_{C}$ and $ff^{\prime}=\mathrm{id}_{A}$. In this case
$f$ is called an ${\mathsf{O}}({\mathsf{P}})$-filtered chain isomorphism.
Notation: $(C,{\mathrm{d}}_{C})\cong(A,{\mathrm{d}}_{A})$.

In this section we establish the fundamental equivalence result in the setting
of strict, ${\mathsf{P}}$-graded differential groups. A ${\mathsf{P}}$-graded
group $C=\oplus_{p\in{\mathsf{P}}}G_{p}C$ is free if and only if the groups
$G_{p}C$ are free for all $p\in{\mathsf{P}}$.

###### Theorem 5.1

Two free, finitely generated, strict, ${\mathsf{P}}$-graded differential
groups $(C,{\mathrm{d}}_{C})$ and $(A,{\mathrm{d}}_{A})$ are
${\mathsf{O}}({\mathsf{P}})$-filtered chain isomorphic if and only if
$\text{\bf E}(C,{\mathrm{d}}_{C})\cong\text{\bf E}(A,{\mathrm{d}}_{A})$.

###### Proof

If $(C,{\mathrm{d}}_{C})$ and $(A,{\mathrm{d}}_{A})$ are
${\mathsf{O}}({\mathsf{P}})$-filtered chain isomorphic then the associated
Cartan-Eilenberg systems are isomorphic which follows directly from the
definition of Cartan-Eilenberg system for a graded differential group. For the
converse we argue by induction using the lattice structure of
${\mathsf{O}}({\mathsf{P}})$.

Let $q\in{\mathsf{P}}$ be a minimal element and let $\alpha=\\{q\\}$. Since
$\text{\bf E}(C,{\mathrm{d}}_{C})\cong\text{\bf E}(A,{\mathrm{d}}_{A})$ we
have $F_{\alpha}C=G_{q}C$, $F_{\alpha}A=G_{q}A$, and a chain isomorphism
$f\colon G_{q}C\xrightarrow[]{\cong}G_{q}A$, with
$f(F_{\alpha}C)=F_{\alpha}A$, which uses the fact that $(C,{\mathrm{d}}_{C})$
and $(A,{\mathrm{d}}_{A})$ are finitely generated and strict, cf. Rmk. 4.
Since the differentials on $G_{q}C$ and $G_{q}A$ vanish $H(f)=f$. This holds
for all minimal elements $q\in{\mathsf{P}}$, which concludes the first step in
the induction. Let $\alpha\in{\mathsf{O}}({\mathsf{P}})$ and suppose
$f^{\prime}\colon F_{\alpha}C\to F_{\alpha}A$ is an
${\mathsf{O}}({\mathsf{P}})$-filtered chain isomorphism with

$f^{\prime}(F_{\alpha^{\prime}}C)=F_{\alpha^{\prime}}A,\quad\forall\alpha^{\prime}\subseteq\alpha.$
(39)

Let $q\in{\mathsf{P}}\smallsetminus\alpha$ be a minimal element. Define
$\beta=\bigl{\downarrow}q$, which is a join-irreducible down-set in
${\mathsf{O}}({\mathsf{P}})$ with unique immediate predecessor
$\beta^{\dagger}\subseteq\alpha$. This implies that
$\beta=\beta^{\dagger}\cup\\{q\\}$. The next step is to construct a chain
isomorphism $f\colon F_{\beta}C\xrightarrow[]{\cong}F_{\beta}A$, where
$F_{\beta}C=F_{\beta^{\dagger}}C\oplus G_{q}C$ and
$F_{\beta}A=F_{\beta^{\dagger}}A\oplus G_{q}A$, and which satisfies
$f(F_{\beta}C)=F_{\beta}A$. Since $\text{\bf
E}(C,{\mathrm{d}}_{C})\cong\text{\bf E}(A,{\mathrm{d}}_{A})$ we have a chain
isomorphism $f^{\prime\prime}\colon G_{q}C\xrightarrow[]{\cong}G_{q}A$ and an
isomorphism $g\colon H(F_{\beta^{\dagger}}C\oplus G_{q}C)\to
H(F_{\beta^{\dagger}}A\oplus G_{q}A)$ given by the exact triangles:

${H(F_{\beta^{\dagger}}C)}$${H(F_{\beta^{\dagger}}C\oplus
G_{q}C)}$${G_{q}C}$${E^{\beta^{\dagger}}_{\emptyset}}$${E^{\beta}_{\emptyset}}$${E^{\beta}_{\beta^{\dagger}}}$${E^{\beta^{\dagger}}_{\emptyset}}$${E^{\prime\beta^{\dagger}}_{\emptyset}}$${E^{\prime\beta}_{\emptyset}}$${E^{\prime\beta}_{\beta^{\dagger}}}$${E^{\prime\beta^{\dagger}}_{\emptyset}}$${H(F_{\beta^{\dagger}}A)}$${H(F_{\beta^{\dagger}}A\oplus
G_{q}A)}$${G_{q}A}$$\scriptstyle{H(f^{\prime})}$$\scriptstyle{g}$$\scriptstyle{f^{\prime\prime}}$

Since $(C,{\mathrm{d}}_{C})$ and $(A,{\mathrm{d}}_{A})$ are finitely generated
and strict we have $H(G_{q}C)=G_{q}C$ and $H(G_{q}A)=G_{q}A$. On the chain
level this yields the following diagram:

${0}$${F_{\beta^{\dagger}}C}$${F_{\beta^{\dagger}}C\oplus
G_{q}C}$${G_{q}C}$${0}$${0}$${F_{\beta^{\dagger}}A}$${F_{\beta^{\dagger}}A\oplus
G_{q}A}$${G_{q}A}$${0}$$\scriptstyle{f^{\prime}}$$\scriptstyle{f}$$\scriptstyle{f^{\prime\prime}}$
(40)

As before, since both $(C,{\mathrm{d}}_{C})$ and $(A,{\mathrm{d}}_{A})$ are
finitely generated, strict and free it follows that $G_{q}C$ and $G_{q}A$ are
finitely generated, free, boundaryless abelian groups (and thus projective).
The components $F_{\beta^{\dagger}}C$ and $F_{\beta^{\dagger}}A$ are free
differential groups. By Theorem 3.1 there exists a chain homomorphism $f$ that
induces $g$. The chain homomorphism $f$ is a chain isomorphism via the five
lemma since both $f^{\prime}$ (by induction) and $f^{\prime\prime}$ are chain
isomorphisms. The isomorphism $f$ is filtered by construction, i.e.
$f(F_{\beta}C)\subseteq F_{\beta}A$. The above procedure yields a more precise
statement, i.e. $\beta^{\prime}\subsetneq\beta$ if and only if
$\beta^{\prime}\subseteq{\beta^{\dagger}}$, which implies, by the induction
hypotheses, that
$f(F_{\beta^{\prime}}C)=f^{\prime}(F_{\beta^{\prime}}C)=F_{\beta^{\prime}}A$
for all $\beta^{\prime}\subseteq\beta^{\dagger}\subseteq\alpha$. By the form
of $f$ given in (31), and the fact that $f^{\prime\prime}(G_{q}C)=G_{q}A$, we
have

$\displaystyle f(F_{\beta}C)$
$\displaystyle=f\bigl{(}F_{\beta^{\dagger}}C\oplus
G_{q}C\bigr{)}=\bigl{[}f^{\prime}(F_{\beta^{\dagger}}C)+\gamma(G_{q}C)\bigr{]}\oplus
f^{\prime\prime}(G_{q}C)$ $\displaystyle=F_{\beta^{\dagger}}A\oplus
G_{q}A=F_{\beta}A,$

since $\gamma(G_{q}C)\subseteq F_{\beta^{\dagger}}A$.

It remains to construct $f$ on $F_{\beta^{\prime}}C$ for down-sets
$\beta^{\prime}\subseteq\alpha\cup\\{q\\}$, where
$\alpha\cup\\{q\\}\in{\mathsf{O}}({\mathsf{P}})$. Since every down-set can be
uniquely represented as an irredundant union of join-irreducible down-sets,
the down-set $\beta^{\prime}$ is represented as

$\beta^{\prime}=\bigl{\downarrow}q\cup\bigl{\downarrow}p_{i_{1}}\cup\cdots\cup\bigl{\downarrow}p_{i_{k}}=\beta\cup\gamma,\quad
p_{i_{1}},\cdots,p_{i_{k}}\in\alpha,\quad\beta=\bigl{\downarrow}q,$ (41)

and
$\gamma=\bigl{\downarrow}p_{i_{1}}\cup\cdots\cup\bigl{\downarrow}p_{i_{k}}\subseteq\alpha$.
For $\beta\cup\gamma$ and $\beta\cap\gamma$ we have the following diagram:

${(\beta\cap\gamma)\cup(\beta\smallsetminus\gamma)\cup(\gamma\smallsetminus\beta)}$${(\beta\cap\gamma)\cup(\beta\smallsetminus\gamma)}$${(\beta\cap\gamma)\cup(\gamma\smallsetminus\beta)}$${\beta\cap\gamma}$

where
$(\beta\cap\gamma)\cup(\beta\smallsetminus\gamma)\cup(\gamma\smallsetminus\beta)=\beta\cup\gamma=\beta^{\prime}$.
Consider the decompositions:

$F_{\beta}C=F_{\beta\cap\gamma}C\oplus G_{\beta\smallsetminus\gamma}C,\quad
F_{\gamma}C=F_{\beta\cap\gamma}C\oplus G_{\gamma\smallsetminus\beta}C$

and

$F_{\beta^{\prime}}C=F_{\beta\cap\gamma}C\oplus\Bigl{[}G_{\beta\smallsetminus\gamma}C\oplus
G_{\gamma\smallsetminus\beta}C\Bigr{]}=F_{\beta}C+F_{\gamma}C.$

The same decomposition holds for $F_{\beta}A$, $F_{\gamma}A$ and
$F_{\beta^{\prime}}A$. Since $\beta\smallsetminus\gamma$ and
$\gamma\smallsetminus\beta$ are uncomparable convex sets Lemma 7, with the
triples $(\beta\cap\gamma,\beta,\beta\cup\gamma)$ and
$(\beta\cap\gamma,\gamma,\beta\cup\gamma)$, implies that the differential on
$G_{\beta\smallsetminus\gamma}C\oplus G_{\gamma\smallsetminus\beta}C$ is a
direct sum and thus the differentials are of the form:

${\mathrm{d}}_{F_{\beta}C}=\begin{pmatrix}{\mathrm{d}}_{F_{\beta\cap\gamma}C}&\mathrm{\Lambda}\\\
0&{\mathrm{d}}_{G_{\beta\smallsetminus\gamma}C}\end{pmatrix},\quad{\mathrm{d}}_{F_{\gamma}C}=\begin{pmatrix}{\mathrm{d}}_{F_{\beta\cap\gamma}C}&\mathrm{\Lambda^{\prime}}\\\
0&{\mathrm{d}}_{G_{\gamma\smallsetminus\beta}C}\end{pmatrix},$ (42)

and

${\mathrm{d}}_{F_{\beta^{\prime}}C}=\begin{pmatrix}{\mathrm{d}}_{F_{\beta\cap\gamma}C}&\mathrm{\Lambda}~{}~{}&\mathrm{\Lambda^{\prime}}\\\
0&{\mathrm{d}}_{G_{\beta\smallsetminus\gamma}C}~{}~{}&0\\\
0&0~{}~{}&{\mathrm{d}}_{G_{\gamma\smallsetminus\beta}C}\end{pmatrix},$ (43)

It is readily verified that with $C=F_{\beta}C$, $C^{\prime}=F_{\gamma}C$,
$A=F_{\beta}A$ and $A^{\prime}=F_{\gamma}A$, and
$C+C^{\prime}=F_{\beta^{\prime}}C$ and $A+A^{\prime}=F_{\beta^{\prime}}A$,
Theorem 3.2(i) is satisfied for the differentials. By construction the chain
isomorphisms $f^{\prime}=f|_{F_{\beta}C}$ and
$f^{\prime\prime}=f|_{F_{\gamma}C}$ satisfy Theorem 3.2(ii) and therefore
$f:=f|_{F_{\beta}C+F_{\gamma}C}$ is a well-defined chain homomorphism on
$F_{\beta^{\prime}}C$ and $f(F_{\beta^{\prime}}C)=F_{\beta^{\prime}}A$.
Indeed, $f^{\prime}(F_{\beta}C)=F_{\beta}A$ and
$f^{\prime\prime}(F_{\gamma}C)=F_{\gamma}A$, and
$f(c+c^{\prime})=f^{\prime}(c)+f^{\prime\prime}(c^{\prime})$ for $c\in
F_{\beta}C$ and $c^{\prime}\in F_{\gamma}C$. This implies that

$f(F_{\beta^{\prime}}C)=f(F_{\beta}C+F_{\gamma}C)=F_{\beta}A+F_{\gamma}A=F_{\beta^{\prime}}A.$

and therefore the chain homomorphism is well-defined on
$F_{\alpha\cup\downarrow q}C$ and is filtered, i.e.
$f(F_{\alpha^{\prime}}C)=F_{\alpha^{\prime}}A$ for all
$\alpha^{\prime}\subseteq\bigl{\downarrow}q\cup\alpha$. Since
$f(F_{\beta\cap\gamma}C)=F_{\beta\cap\gamma}A$, the homomorphism $f$ is of the
form

$f^{\prime}=f|_{F_{\beta}C}=\begin{pmatrix}f|_{F_{\beta\cap\gamma}C}&\mathrm{\chi}\\\
0&f|_{G_{\beta\smallsetminus\gamma}C}\end{pmatrix},\quad
f^{\prime\prime}=f|_{F_{\gamma}C}=\begin{pmatrix}f|_{F_{\beta\cap\gamma}C}&\mathrm{\chi^{\prime}}\\\
0&f|_{G_{\gamma\smallsetminus\beta}C}\end{pmatrix}$

Since $f|_{F_{\beta}C}$ and $f|_{F_{\gamma}C}$ are isomorphisms so are
$f|_{G_{\beta\smallsetminus\gamma}C}$ and
$f|_{G_{\gamma\smallsetminus\beta}C}$, and by construction
$f(G_{\beta\smallsetminus\gamma}C)\subseteq G_{\beta\smallsetminus\gamma}A$
and $f(G_{\gamma\smallsetminus\beta}C)\subseteq
G_{\gamma\smallsetminus\beta}A$. Since
$\beta^{\prime}\smallsetminus\beta=\gamma\smallsetminus\beta$ and
$\beta^{\prime}\smallsetminus\gamma=\beta\smallsetminus\gamma$ we have the
exact triangles:

${E^{\beta}_{\varnothing}}$${E^{\beta^{\prime}}_{\varnothing}}$${E^{\beta^{\prime}}_{\beta}\cong
E^{\gamma}_{\beta}}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{k}$
${E^{\gamma}_{\varnothing}}$${E^{\beta^{\prime}}_{\varnothing}}$${E^{\beta^{\prime}}_{\gamma}\cong
E^{\beta}_{\gamma}}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{k}$ (44)

From the isomorphism between Cartan-Eilenberg systems we obtain the
commutative squares:

${E^{\beta}_{\varnothing}}$${E^{\beta^{\prime}}_{\varnothing}}$${E^{\prime\beta}_{\varnothing}}$${E^{\prime\beta^{\prime}}_{\varnothing}}$$\scriptstyle{i}$$\scriptstyle{H(f^{\prime})}$$\scriptstyle{g}$$\scriptstyle{i}$
${E^{\gamma}_{\varnothing}}$${E^{\beta^{\prime}}_{\varnothing}}$${E^{\prime\gamma}_{\varnothing}}$${E^{\prime\beta^{\prime}}_{\varnothing}}$$\scriptstyle{i}$$\scriptstyle{H(f^{\prime\prime})}$$\scriptstyle{g}$$\scriptstyle{i}$

By Theorem 3.2(iii) $f$ lifts $g$ and since $f^{\prime}$ and
$f^{\prime\prime}$ are chain isomorphisms by construction the five lemma
applied to the short exact sequences associated to the exact triangles in (44)
imply that also $f$ is a chain isomorphism, i.e.

${0}$${F_{\beta}C}$${F_{\beta}C+F_{\gamma}C}$${G_{\gamma\smallsetminus\beta}C}$${0}$${0}$${F_{\beta}A}$${F_{\beta}A+F_{\gamma}A}$${G_{\gamma\smallsetminus\beta}A}$${0}$$\scriptstyle{\cong}$$\scriptstyle{f^{\prime}}$$\scriptstyle{f}$$\scriptstyle{\cong}$

and similarly for

${0}$${F_{\gamma}C}$${F_{\beta}C+F_{\gamma}C}$${G_{\beta\smallsetminus\gamma}C}$${0}$${0}$${F_{\gamma}A}$${F_{\beta}A+F_{\gamma}A}$${G_{\beta\smallsetminus\gamma}A}$${0}$$\scriptstyle{\cong}$$\scriptstyle{f^{\prime\prime}}$$\scriptstyle{f}$$\scriptstyle{\cong}$

which completes the construction of $f$ on $F_{\beta^{\prime}}C$ for down-sets
$\beta^{\prime}\subseteq\alpha\cup\\{q\\}$. This process terminates after
finitely many steps. ∎

###### Remark 19

If $h$ is a morphism between E and $\text{\bf E}^{\prime}$ such that
$h|_{E^{\beta}_{\alpha}}$ is an isomorphism for all
$\beta\smallsetminus\alpha=\\{p\\}$, $p\in{\mathsf{P}}$, then $h$ is a
isomorphism on all $E$-terms $E^{\beta}_{\alpha}$, for all
$\alpha\subseteq\beta$. This is a direct consequence of the five lemma.

##  6 Equivalence to Franzosa’s module braids

In fran  a data structure is developed to relate Conley indices of Morse sets
in a Morse representation, cf. lsa3 . In this section we show that Franzosa’s
data structure is equivalent to a Cartan-Eilenberg system.

###  6.1 Module braids

The starting point in fran  is a finite poset $({\mathsf{P}},\leq)$. We recall
some notation from fran  to relate elements in ${\mathsf{Co}}({\mathsf{P}})$,
the meet-semilattice of convex sets in ${\mathsf{P}}$, and formulate these
concepts in terms of lattice theory. Two elements
$\xi,\eta\in{\mathsf{Co}}({\mathsf{P}})$ are _adjacent_ if and only if there
exists an ordered triple $(\alpha,\beta,\gamma)$ such that
$\xi=\beta\smallsetminus\alpha$ and $\eta=\gamma\smallsetminus\beta$. Then,
$\xi\cap\eta=0$ and the union satisfies
$\xi\eta:=\xi\cup\eta=\gamma\smallsetminus\alpha$ which is a _decomposition_
of convex sets. For every convex set $\xi\in{\mathsf{Co}}({\mathsf{P}})$
assign a differential group $(D_{\xi},{\mathrm{d}}_{\xi})$ such that

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is exact.101010In fran  the slightly more general assumption of weak exactness
is assumed. The results in this section remain true under the assumption of
weak exactness. Weak exactness in fran  can be avoided by using Borel-Moore
homology for example. If $\xi$ and $\eta$ are incomparable then
$j^{\prime}i=\mathrm{id}$ on ξ where $i\colon D_{\xi}\to D_{\xi\eta}$ and
$j^{\prime}\colon D_{\eta\xi}\to D_{\xi}$. Franzosa refers to the above
structure as a ‘chain complex braid’. The prime example of a chain complex
braid is given by a ${\mathsf{O}}({\mathsf{P}})$-filtered differential group
$(D,{\mathrm{d}}_{D})$ and is given in (5). As in Section 4.2 one proves that
$G_{\beta\smallsetminus\alpha}D\cong
G_{\beta^{\prime}\smallsetminus\alpha^{\prime}}D$ for
$\beta\smallsetminus\alpha=\beta^{\prime}\smallsetminus\alpha^{\prime}=:\xi$
along with the other properties. The homology is denote by
$E_{\xi}:=H(D_{\xi},{\mathrm{d}}_{\xi})$ and satisfies the following
properties:

###### Definition 3

A _module braid_ B over a meet-semilattice ${\mathsf{Co}}({\mathsf{P}})$ is a
collection of abelian groups (modules) $E_{\xi}\in\mathbf{Ab}$, indexed by
$\xi\in{\mathsf{Co}}({\mathsf{P}})$, such that

  1. (i)

for every pair of _adjacent_ $\xi,\eta\in{\mathsf{Co}}({\mathsf{P}})$ the
triangles

${E_{\xi}}$${E_{\xi\eta}}$${E_{\eta}}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{k}$
(45)

are exact;

  2. (ii)

for _incomparable_ elements $\xi,\eta\in{\mathsf{Co}}({\mathsf{P}})$, cf. Rmk.
17, both triangles

${E_{\xi}}$${E_{\xi\eta}}$${E_{\eta}}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{k}$
${E_{\eta}}$${E_{\eta\xi}}$${E_{\xi}}$$\scriptstyle{i^{\prime}}$$\scriptstyle{j^{\prime}}$$\scriptstyle{k^{\prime}}$
(46)

are exact and $j^{\prime}i=\mathrm{id}$ on $E_{\xi}$ and
$ji^{\prime}=\mathrm{id}$ on $E_{\eta}$.

  3. (iii)

for every triple of adjacent elements111111A triple $\xi,\eta,\zeta$ of
adjacent convex sets is equivalent to a ordered quadruple
$(\alpha,\beta,\gamma,\delta)$ such that $\xi=\beta\smallsetminus\alpha$,
$\eta=\gamma\smallsetminus\beta$ and $\zeta=\delta\smallsetminus\gamma$. These
convex sets are mutually disjoint and their union is
$\delta\smallsetminus\alpha$. $\xi,\eta,\zeta\in{\mathsf{Co}}({\mathsf{P}})$
the ‘braid diagram’

${E_{\xi}}$${E_{\zeta}}$${E_{\xi\eta}}$${E_{\xi\eta\zeta}}$${E_{\eta}}$${E_{\eta\zeta}}$${E_{\zeta}}$${E_{\xi}}$${E_{\xi\eta}}$${E_{\eta}}$${E_{\xi\eta\zeta}}$$\scriptstyle{i}$$\scriptstyle{i}$$\scriptstyle{k}$$\scriptstyle{k}$$\scriptstyle{j}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{j}$$\scriptstyle{i}$$\scriptstyle{k}$$\scriptstyle{j}$$\scriptstyle{k}$$\scriptstyle{k}$$\scriptstyle{k}$$\scriptstyle{i}$$\scriptstyle{i}$$\scriptstyle{j}$$\scriptstyle{i}$
(47)

commutes.

###### Remark 20

The terminology module braid is best explained since the exact triangles in
(45) correspond to the strands in the braid diagram in (47). In the spirit of
triangulated categories the braid diagram can also be reshaped as a
commutative diagram of exact triangles:

${E_{\xi}}$${E_{\eta}}$${E_{\eta\zeta}}$${E_{\xi\eta}}$${E_{\zeta}}$${E_{\xi\eta\zeta}}$$\scriptstyle{i}$$\scriptstyle{i}$$\scriptstyle{k}$$\scriptstyle{i}$$\scriptstyle{k}$$\scriptstyle{j}$$\scriptstyle{j}$$\scriptstyle{i}$$\scriptstyle{k}$$\scriptstyle{k}$$\scriptstyle{j}$$\scriptstyle{j}$
(48)

The dashed ‘strand’ corresponds to the middle exact triangle in (48), which
makes the diagram commutative.

###### Lemma 8

Let E be a Cartan-Eilenberg system over ${\mathsf{O}}({\mathsf{P}})$, with
${\mathsf{P}}$ a finite poset. Then, the diagram

${E^{\beta}_{\alpha}}$${E^{\gamma}_{\beta}}$${E^{\delta}_{\beta}}$${E^{\gamma}_{\alpha}}$${E^{\delta}_{\gamma}}$${E^{\delta}_{\alpha}}$$\scriptstyle{i}$$\scriptstyle{i}$$\scriptstyle{k}$$\scriptstyle{i}$$\scriptstyle{k}$$\scriptstyle{j}$$\scriptstyle{j}$$\scriptstyle{i}$$\scriptstyle{k}$$\scriptstyle{k}$$\scriptstyle{j}$$\scriptstyle{j}$
(49)

commutes for every ordered quadruple $(\alpha,\beta,\gamma,\delta)$ in
${\mathsf{O}}({\mathsf{P}})$.

###### Proof

The commutative diagram follows by defining the dashed triangle by considering
the compositions:
$E^{\gamma}_{\beta}\xrightarrow[]{k}E^{\beta}_{\alpha}\xrightarrow[]{i}E^{\delta}_{\alpha}\xrightarrow[]{j}E^{\delta}_{\beta}$,
$E^{\delta}_{\beta}\xrightarrow[]{k}E^{\beta}_{\alpha}\xrightarrow[]{i}E^{\delta}_{\alpha}\xrightarrow[]{j}E^{\delta}_{\gamma}$,
and
$E^{\delta}_{\gamma}\xrightarrow[]{k}E^{\gamma}_{\alpha}\xrightarrow[]{j}E^{\gamma}_{\beta}$.
The first two compositions are equal to the homomorphisms $i\colon
E^{\gamma}_{\beta}\to E^{\delta}_{\beta}$ and $j\colon E^{\delta}_{\beta}\to
E^{\delta}_{\gamma}$, since nested ordered pairs correspond to unique
homomorphisms. The latter composition is the connecting homomorphism $k\colon
E^{\delta}_{\gamma}\to E^{\gamma}_{\beta}$ due to the commutative squares for
$k$ in (1)[right]. This implies that the dashed triangle is the exact triangle
for $(\beta,\gamma,\delta)$. ∎

The diagram in (49) is a reshaped version of the octahedral diagram in
triangulated categories and is a structure present in Cartan-Eilenberg
systems, cf. (Matschke, , Lem. 4.8). The octahedral diagram can also be used
in the definition of a Cartan-Eilenberg system by removing the commutative
squares in (1)[right] and adding some additional conditions. This idea lies at
the heart of proving that Franzosa’s module braids correspond to excisive
Cartan-Eilenberg systems.

###### Theorem 6.1

A module braid B over a meet-semilattice ${\mathsf{Co}}({\mathsf{P}})$ for
some finite poset ${\mathsf{P}}$ is equivalent to an excisive Cartan-Eilenberg
system E over ${\mathsf{O}}({\mathsf{P}})$.

###### Proof

We start with the observation that an excisive Cartan-Eilenberg system we
choose $E$-terms over ${\mathsf{Co}}({\mathsf{P}})$ by setting
$E_{\xi}=E^{\beta}_{\alpha}\cong E^{\beta^{\prime}}_{\alpha^{\prime}}$, with
$\xi=\beta\smallsetminus\alpha=\beta^{\prime}\smallsetminus\alpha^{\prime}$.
The exact triangles for E yield the exact triangle for B in (i). Property (ii)
follows from Lemma 6 and the octahedral diagram (module braid diagram) is
given by Lemma 8, which establishes Property (iii). We conclude that an
excisive Cartan-Eilenberg system ${\mathsf{O}}({\mathsf{P}})$ defines a module
braid over ${\mathsf{Co}}({\mathsf{P}})$.

For the converse we want to show that (i)-(iii) in the definition of module
braid define an excisive Cartan-Eilenberg system in the sense that the
commutative squares in (1)[right] exist. Define $E^{\beta}_{\alpha}:=E_{\xi}$
for all ordered pairs $(\alpha,\beta)$ such that
$\beta\smallsetminus\alpha=\xi$. In (45) choose $\xi=\alpha$ and
$\eta=\beta\smallsetminus\alpha$ for any the ordered pair $(\alpha,\beta)$,
which establishes the associated exact triangles in (1)[left], together with
the maps $i$, $j$ and $k$. The next step is to define the maps $\ell\colon
E^{\beta}_{\alpha}\to E^{\beta^{\prime}}_{\alpha^{\prime}}$ and $\ell\colon
E^{\gamma}_{\beta}\to E^{\gamma^{\prime}}_{\beta^{\prime}}$ in (1)[right] for
ordered triples
$(\alpha,\beta,\gamma)\leq(\alpha^{\prime},\beta^{\prime},\gamma^{\prime})$.
Consider the octahedral diagrams in (48) in the following two
configurations:121212The notation
$\genfrac{}{}{0.0pt}{}{\alpha}{~{}}\genfrac{}{}{0.0pt}{}{\undergroup{\subseteq}}{\eta}\genfrac{}{}{0.0pt}{}{\beta}{~{}}$
indicates the convex set $\eta=\beta\smallsetminus\alpha$.

$\begin{aligned}
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&\genfrac{}{}{0.0pt}{}{\alpha}{~{}}\genfrac{}{}{0.0pt}{}{\undergroup{\subseteq}}{\xi}\genfrac{}{}{0.0pt}{}{\beta}{~{}}\genfrac{}{}{0.0pt}{}{\undergroup{\subseteq}}{\eta^{\prime}}\genfrac{}{}{0.0pt}{}{\beta^{\prime}}{~{}}\genfrac{}{}{0.0pt}{}{\undergroup{\subseteq}}{\zeta^{\prime}}\genfrac{}{}{0.0pt}{}{\gamma^{\prime}}{~{}}\end{aligned}\quad\quad\leavevmode\hbox
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