# Essential cohomology modules

Runak H. Mustafa  Mathematics Department, Faculty of Science
Soran University
44008, Soran, Erbil Kurdistan Region, Iraq<EMAIL_ADDRESS>and
Ismael Akray  Mathematics Department, Faculty of Science
Soran University
44008, Soran, Erbil Kurdistan Region, Iraq<EMAIL_ADDRESS>
###### Abstract.

In this article, we give a generalization to injective modules by using
$e$-exact sequences introduced by Akray in [1] and name it $e$-injective
modules and investigate their properties. We reprove both Baer criterion and
comparison theorem of homology using $e$-injective modules and $e$-injective
resolutions. Furthermore, we apply the notion $e$-injective modules into local
cohomology to construct a new form of the cohomology modules call it essential
cohomology modules (briefly $e$-cohomology modules). We show that the torsion
functor $\Gamma_{a}(-)$ is an $e$-exact functor on torsion-free modules. We
seek about the relationship of $e$-cohomology within the classical cohomology.
Finally, we conclude that they are different on the vanishing of their
$i_{th}$ cohomology modules.

###### Key words and phrases:

essential exact sequence, homology module, essential injective module, local
cohomology module

###### 2020 Mathematics Subject Classification:

13C11, 46M18, 13D45

##  1\. Introduction

Exact sequences play an important roles in module theory and homological
algebra. Some notions such as injectivity, projectivity, flatness and derived
functors have been defined and analyzed by exact sequence approach.
Generalizing exact sequences gives possibilities to generalize some related
notions which are defined by exact sequences. An exact sequence $0\rightarrow
A\stackrel{{\scriptstyle i}}{{\to}}B\stackrel{{\scriptstyle p}}{{\to}}C\to 0$
is split if there exists a morphism $j:C\rightarrow B$ (or $f:B\rightarrow A$)
such that $pj=I_{C}$ (or $\>fi=I_{A})$. In 1972, R. S. Mishra introduced a
generalization for split sequence where a semi-sequence
$M_{i-1}\stackrel{{\scriptstyle f_{i-1}}}{{\to}}M_{i}\stackrel{{\scriptstyle
f_{i}}}{{\to}}M_{i+1}$ is called semi-split if $Ker(f_{i})$ is a direct
summand of $M_{i}$ [7]. So a semi-split is split if and only if it is exact.
In 1999, Davvaz and parnian-Goramaleky introduced a generalization for exact
sequences called it a $U$-exact sequence, where a sequence of $R$-modules and
$R$-homomorphisms $\dots\to M_{i-1}\stackrel{{\scriptstyle
f_{i-1}}}{{\to}}M_{i}\stackrel{{\scriptstyle f_{i}}}{{\to}}M_{i+1}\to\dots$ is
a $U_{i+1}$-exact at $M_{i}$ if $Im(f_{i})=f_{i+1}^{-1}(U_{i+1})$, where
$U_{i+1}$ is a submodule of $M_{i+1}$ [5]. Also, a short sequence $0\to
A\stackrel{{\scriptstyle f}}{{\to}}B\stackrel{{\scriptstyle g}}{{\to}}C\to 0$
of $R$-modules is a $U$-exact if it is ${0}$-exact at $A$, $U$-exact at $B$
and ${0}$-exact at $C$, equivalently, if $f$ is monic, $g$ is epic and
$Imf=g^{-1}(U)$ for a submodule $U$ of $C$. Clearly, a $U$-exact sequence is
exact if and only if $U=0$. Furthermore, they defined a $V$-coexact sequence
as a dual notion of $U$-exact sequence, where a sequence $0\to
A\stackrel{{\scriptstyle f}}{{\to}}B\stackrel{{\scriptstyle g}}{{\to}}C\to 0$
is $V$\- coexact if $f$ is monic, $g$ is epic and $f(V)=Ker(g)$, where $V$ is
a submodule of $A$. A submodule $N$ of an $R$-module $M$ is called essential
or large in $M$ if it has non-zero intersection with every non-zero submodule
of $M$ and denoted by $N\leqslant_{e}M$. Akray in 2020 [1] introduced another
generalization to exact sequences of modules and instead of the equality of
$Im(f)$ with $Ker(g)$ they took $Im(f)$ as a large (essential) submodule of
$Ker(g)$ in a sequence $0\to A\stackrel{{\scriptstyle
f}}{{\to}}B\stackrel{{\scriptstyle g}}{{\to}}C\to 0$ and called it essential
exact sequence or simply $e$-exact sequence. Equivalently, a sequence of
$R$-modules and $R$-morphisms $\dots\to N_{i-1}\stackrel{{\scriptstyle
f_{i-1}}}{{\to}}N_{i}\stackrel{{\scriptstyle f_{i}}}{{\to}}N_{i+1}\to\dots$ is
said to be essential exact (e-exact) at $N_{i}$, if
$Im(f_{i-1})\leqslant_{e}Ker(f_{i})$ and to be e-exact if it is e-exact at
$N_{i}$ for all $i$. In particular, a sequence of $R$-modules and
$R$-morphisms $0\to L\stackrel{{\scriptstyle
f_{1}}}{{\to}}M\stackrel{{\scriptstyle f_{2}}}{{\to}}N\to 0$ is a short
e-exact sequence if and only if
$Ker(f_{1})=0,Im(f_{1})\leqslant_{e}Ker(f_{2})$ and $Im(f_{2})\leqslant_{e}N$.
They studied some basic properties of $e$-exact sequences and established
their connection with notions in module theory and homological algebra [2].
Also, F. Campanini and A. Facchini were worked on $e$-exact sequences and
studied the relation of $e$-exactness with some related functors like the
functor defined on the category of $R$-modules to the spectral category of
$R$-modules and the localization functor with respect to the singular torsion
theory [4].

In this paper, we continue to observe further properties of $e$-exact
sequences and we will restrict our discussion to their applications on both
injective modules and the torsion functor of local cohomology.The local
cohomology was introduced by Grothendieck in a seminar in Harvard 1961 and
written up by Hartshorne in 1967. Next, this subject was studied by Hartshorne
and numerous authors even in the recent years see [3] and [6].

In section two, we introduce the notion essential injective module briefly
$e$-injective module. We prove the Baer Criterion for e-injectives (Theorem
2.2). We prove that a product of $R$-modules is an $e$-injective if and only
if each of its components is an $e$-injective(Proposition 2.6). Also, we show
that a torsion-free $R$-module $E$ is an $e$-injective if and only if
$Hom(\quad,E)$ is a contravariant e-exact functor (Proposition 2.7). Moreover,
we reprove the Comparison Theorem of homology for e-injectives (Theorem 2.13).

Section three devoted for discussing the application of $e$-exact sequence to
local cohomology modules. We construct cohomology modules by using
$e$-injective resolutions and called the $i_{th}$-local cohomology of an
$R$-module $M$ by ${}_{e}H^{i}_{a}(M)$. We prove that ${}_{e}H^{0}_{a}(\>)$ is
naturally equivalent to $r\Gamma_{a}(\>)$ on torsion-free $R$-modules (Theorem
3.7). We study some cases of vanishinig of $e$-cohomology modules (Theorem 3.4
and Theorem 3.5). Furthermore, we show that by an example that not necessary
an ${a}$-torsion $R$-module $M$ has zero $e$-cohomology modules
${}_{e}H^{i}_{a}(M)=0$, for $i>0$ as this is the case in local cohomology
(Example 3.11).

##  2\. Essential injective modules

In this section we introduce essential injective module and investigate some
properties and results on such topic. We begin with their definition.

######  Definition 2.1.

We call an $R$-module $E$ essential injective briefly e-injective if it
satisfies the following condition: for any monic $f_{1}:A_{1}\rightarrow
A_{2}$ and any map $f_{2}:A_{1}\rightarrow E$ of $R$-modules, there exist
$0\neq r\in R$ and $f_{3}:A_{2}\rightarrow E$ such that $f_{3}f_{1}=rf_{2}$.

${E}$${0}$${A_{1}}$${A_{2}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{3}}$

In this case, we say the map $f_{3}$ is essentially extends to the map
$f_{2}$.

Now, we give one of our main results in this section which is the Baer
criterion for $e$-injectives.

######  Theorem 2.2.

[e-Baer Criterion] An $R-$module $E$ is e-injective if and only if for a
nonzero ideal a of $R$, every $R$-map $f:\textbf{a}\to E$ can be extends
essentially to $g:R\to E$ in the sence that there exists $0\neq r\in R$ with
$gi=rf$, where $i:\textbf{a}\to R$ is an inclusion map as in the following
diagram.

${E}$${0}$a${R}$$\scriptstyle{f}$$\scriptstyle{i}$$\scriptstyle{g}$

###### Proof.

Since any ideal of $R$ can be considered as a submodule of the $R$-module $R$,
the existence of an extension $g$ of $f$ is just a special case of the
definition of e-injectivity of $E$. Suppose we have the following diagram,
where $A$ is a submodule of $B$ and $i$ is the inclusion map:

${E}$${0}$${A}$${B}$$\scriptstyle{f}$$\scriptstyle{i}$$\scriptstyle{g}$

Let $X$ be the set of all ordered pairs $(A^{\prime},g^{\prime})$, where
$A\subseteq A^{\prime}\subseteq B$ and $g^{\prime}:A^{\prime}\to E$
essentially extend $f$: that is, $g^{\prime}|_{A}=rf$ for nonzero r in $R$.
Partially ordered $X$ by defining
$(A^{\prime},g^{\prime})\preceq(A^{\prime\prime},g^{\prime\prime})$, which
means $A^{\prime}\subseteq A^{\prime\prime}$ and $g^{\prime\prime}$
essentially extends $g^{\prime}$, so the chains in $X$ have upper bounds in
$X$, hence using Zorn Lemma, there exists a maximal element $(A_{0},g_{0})$ in
$X$. If $A_{0}=B$, we are done, if it is not, there is $b\in B$ with $b\notin
A_{0}$ . Define $\textbf{a}=\\{r\in R:rb\in A_{0}\\}$, it is clear that a is
an ideal in $R$. Define $h:\textbf{a}\to E$, by $h(r)=g_{0}(rb)$. By
hypothesis, there is a map $h^{*}:R\to E$ essentially extends $h$. Finally,
define $A_{1}=A_{0}+<b>$ and $g_{1}:A_{1}\to E$ by
$g_{1}(a_{0}+rb)=g_{0}(a_{0})+rh^{*}(1)$, where $a_{0}\in A_{0}$ and $r\in R$.
Let us show that $g_{1}$ is well-defined. If
$a_{0}+rb=a_{0}^{\prime}+r^{\prime}b$, then $(r-r^{\prime})\in\textbf{a}$.
Therefore, $g_{0}((r-r^{\prime})b)$ and $h(r-r^{\prime})$ are defined, and we
have
$g_{0}(a_{0}^{\prime}-a_{0})=g_{0}((r-r^{\prime})b)=h(r-r^{\prime})=h^{*}(r-r^{\prime})=(r-r^{\prime})h^{*}(1)$.
Thus $g_{0}(a_{0}^{\prime})-g_{0}(a_{0})=rh^{*}(1)-r^{\prime}h^{*}(1)$ and
$g_{(}a_{0}^{\prime})+r^{\prime}h^{*}(1)=g_{0}(a_{0})+rh^{*}(1)$ as desired.
Clearly, $g_{1}(a_{0})=g_{0}(a_{0})$ for all $a_{0}\in A_{0}$, so that the map
$g_{1}$ essentially extends $g_{0}$. We conculde that
$(A_{0},g_{0})\prec(A_{1},g_{1})$ contradicting the maximality of
$(A_{0},g_{0})$. Therefore, $A_{0}=B$, the map $g_{0}$ is an essentially
extends $f$ and $E$ is an e-injective. ∎

The following example shows that an $e$-injective module may not be injective.

######  Example 2.3.

The $\mathbb{Z}$-module $\mathbb{Z}$ is $e$-injective module, but it is not
injective. We can show that $\mathbb{Z}$ is an $e$-injective by using $e$-Baer
Criterion. Let $f_{1}:n\mathbb{Z}\to\mathbb{Z}$ be an inclusion map defined as
$f_{1}(x)=sx;s\in\mathbb{Z}$ and $f_{2}:n\mathbb{Z}\to\mathbb{Z}$ defined as
$f_{2}(x)=mx;m\in\mathbb{Z}$ where $(m,s)=1$. Then by talking $f_{3}=f_{2}$,
we have $f_{3}\circ f_{1}(x)=f_{3}(f_{1}(x))=f_{3}(sx)=msx=sf_{2}(x)$

${\mathbb{Z}}$${0}$${n\mathbb{Z}}$${\mathbb{Z}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{3}}$

It’s esay to see that every submodule of $\mathbb{Z}$-module $\mathbb{Z}$ is
$e$-injective. On an integral domain, every injective is divisble but this is
not the case for e-injectives, since $\mathbb{Z}$ as $\mathbb{Z}$-module is
$e$-injective while not divisble.

######  Definition 2.4.

[1] An $e$-exact sequence $0\rightarrow A\stackrel{{\scriptstyle
i}}{{\to}}B\stackrel{{\scriptstyle p}}{{\to}}C\to 0$ is $e$-split if there
exist $0\neq s\in R$ and a morphism $j:C\rightarrow B$ (or $f:B\rightarrow A$)
such that $pj=sI_{C}$ (or $\>fi=sI_{A})$.

######  Proposition 2.5.

If an e-exact sequence $0\rightarrow A\stackrel{{\scriptstyle
i}}{{\rightarrow}}B\stackrel{{\scriptstyle p}}{{\rightarrow}}C\rightarrow 0$
is $e$-split, then there exists $0\neq r\in R$ such that $rB\cong A\bigoplus
C$.

###### Proof.

We show that $rB=Imi\bigoplus Imj$ there exists $0\neq r\in R$ and $j:C\to B$
with $pj=rI_{C}$. For any $b\in B$, $rb-jp(b)\in Kerp$ and by $e$-exactness,
there exists $a\in A$ and $0\neq s\in R$ such that $i(a)=s(rb-jp(b))$, that is
$srb=i(a)+sjp(b)$. Hence $srB=Imi+Imj$. Now, if $i(x)=rz=j(y)$ for $x\in A$
and $y\in C$, then $0=pi(x)=p(rz)=pj(y)=ry$ and so $rz=j(y)=0$ which implies
that $Imi\cap Imj=0$. Therefore $rB=Imi\bigoplus Imj\cong A\bigoplus C$. ∎

In the following proposition we generalize [8, Proposition 3.38] to
$e$-injective $R$-modules.

######  Proposition 2.6.

A direct product of $R$-modules is an e-injective if and only if each $E_{i}$
is an e-injective $R$-module.

###### Proof.

Suppose that $E=\prod E_{i}$ , $k_{i}:E_{i}\to E$ is the injection and
$p_{i}:E\to E_{i}$ is the projection such that $p_{i}k_{i}=I_{E_{i}}$. Since
$E$ is an e-injective, there exist a homomorphism $h:C\to E$ and $0\neq r\in
R$ such that $hj=r(k_{i}\circ f_{i})$. Now, $g\circ j=(p_{i}\circ h)\circ
j=p_{i}\circ(h\circ j)=p_{i}(r_{i}\circ f_{i}))=r((p_{i}\circ k_{i})\circ
f_{i})=rf_{i}$.

${E}$${E_{i}}$${0}$${B}$${C}$$\scriptstyle{k_{i}}$$\scriptstyle{p_{i}}$$\scriptstyle{f_{i}}$$\scriptstyle{j}$$\scriptstyle{g=p_{i}oh}$$\scriptstyle{h}$

Conversely, consider we have the following diagram in which $E=\prod E_{k}$

${E_{k}}$${E}$${0}$${A}$${B}$$\scriptstyle{P_{k}}$$\scriptstyle{f}$$\scriptstyle{i}$$\scriptstyle{g}$$\scriptstyle{g_{k}}$

Let $p_{k}:E\rightarrow E_{k}$ be the $k_{th}$ projection, so that
$p_{k}f:A\rightarrow E_{k}$, since $E_{k}$ is an e-injective then there exist
$0\neq r\in R$ and $g_{k}:B\rightarrow E_{k}$ such that $g_{k}i=r(p_{k}f)$.
Now, we define $g:B\to E$ by $g:b\mapsto<g_{k}(b)>$ the map $g$ does
essentially extends $f$, since $g(b)=g(i(a))=<g_{k}i(a)>=r<p_{k}f(a)>=rf(a)$.
∎

Recall from [1] the triangle diagram of $R$-modules an $R$-morphism as the
following are $e$-commute if and only if there exists $0\neq r\in R$ such that
$g\circ i=rf$.

${B}$${0}$${A_{1}}$${A_{2}}$$\scriptstyle{f}$$\scriptstyle{i}$$\scriptstyle{g}$

As well too, a squar diagram as the following are $e$-commute if and only if
there exists $0\neq r\in R$ such that $qf=rgt$

${A_{1}}$${A_{2}}$${B_{1}}$${B_{2}}$$\scriptstyle{f}$$\scriptstyle{t}$$\scriptstyle{q}$$\scriptstyle{g}$

######  Proposition 2.7.

Let $E$ be a torsion-free $R$-module. Then $E$ is an e-injective if and only
if $Hom(\quad,E)$ is a contravariant e-exact functor.

###### Proof.

Suppose that $0\rightarrow A\stackrel{{\scriptstyle
i}}{{\rightarrow}}B\stackrel{{\scriptstyle p}}{{\rightarrow}}C\rightarrow 0$
is a short e-exact sequence. Since $E$ is a torsion-free, $Hom(\quad,E)$ is a
left e-exact functor by [1, Theorem 2.7]. It remains to show that
$Hom(\quad,E)$ is a right e-exact, which means
$Hom(B,E)\stackrel{{\scriptstyle i^{*}}}{{\rightarrow}}Hom(A,E)\rightarrow 0$
is an e-exact. For this purpose, let $0\neq f\in Hom(A,E)$. $e$-injectivity of
$E$ implies that there exist $g:B\rightarrow E$ and $0\neq r\in R$ such that
$i^{(}*g)=gi=rf$. Thus we have $Imi^{*}\leqslant_{e}Hom(A,E)$. Therefore
$Hom(\quad,E)$ is an e-exact functor. Conversely, if $Hom(\quad,E)$ is an
e-exact contravariant functor so the sequence $0\rightarrow
Hom(C,E)\stackrel{{\scriptstyle
p^{*}}}{{\rightarrow}}Hom(B,E)\stackrel{{\scriptstyle
i^{*}}}{{\rightarrow}}Hom(A,E)\rightarrow 0$ is an e-exact sequence. Then for
all $0\neq r\in R$ and $0\neq f\in Hom(A,E)$ there exists $g\in Hom(B,E)$ such
that $i^{*}g=rf$ which implies that $gi=rf$, that is, the diagram below

${E}$${0}$${A}$${B}$$\scriptstyle{f}$$\scriptstyle{i}$$\scriptstyle{g}$

is an e-commute and $E$ is an e-injective. ∎

######  Proposition 2.8.

Let $E$ be a torsion-free $e$-injective $R$-module. Then any e-exact sequence
$0\to E\to B\to C\to 0$ $e$-splits.

###### Proof.

Let $E$ be an $e$-injective $R$-module and the sequence $0\rightarrow
E\stackrel{{\scriptstyle i}}{{\to}}B\stackrel{{\scriptstyle p}}{{\to}}C\to 0$
be an e-exact. Then by Proposition 2.7 $0\rightarrow
Hom(C,E)\stackrel{{\scriptstyle
p^{*}}}{{\rightarrow}}Hom(B,E)\stackrel{{\scriptstyle
i^{*}}}{{\rightarrow}}Hom(E,E)\rightarrow 0$ is an e-exact sequence. Since
$I_{E}\in Hom(E,E)$, there exist $g\in Hom(B,E)$ and $0\neq r\in R$ such that
$i^{*}(g)=gi=rI_{E}$ and so the given sequence $e$-splits. ∎

######  Theorem 2.9.

Given an e-commutative diagram of $R$-modules having $e$-exact rows and
torsion-free module $C^{\prime\prime}$:

${A^{\prime}}$${A}$${A^{\prime\prime}}$${0}$${C^{\prime}}$${C}$${C^{\prime\prime}}$${0}$$\scriptstyle{i}$$\scriptstyle{f}$$\scriptstyle{p}$$\scriptstyle{g}$$\scriptstyle{h}$$\scriptstyle{j}$$\scriptstyle{q}$

there exist a unique map $h:A^{\prime\prime}\to C^{\prime\prime}$ making the
augmented diagram $e$-commute.

###### Proof.

If $a^{\prime\prime}\in A^{\prime\prime}$, then there exist $a\in A$ and
$0\neq r\in R$ such that $p(a)=ra^{\prime\prime}$. Define
$h(a^{\prime\prime})=rqg(a)$. We must shows that $h$ is well defined, that is,
if $u\in A$ satiesfies $p(u)=ra^{\prime\prime}$. Now, $p(a)=p(u)$ implies
$p(a-u)=0$, so $a-u\in Kerp$ by $e$-exactness, there exist $0\neq S\in R$ and
$a^{\prime}\in A$ such that $i(a^{\prime})s(a-u)$. Thus,
$rsqg(a-u)=rqg(i(a^{\prime}))=qjf=0$, because $qj=0$. Therefore, $h$ is well-
defined. Suppose that $h^{\prime}:A^{\prime\prime}\to C^{\prime\prime}$ be
another map satisfies $rh^{\prime}p=qg$ and $a\in A$. Then
$rh^{\prime}p(a)=qg=rhp(a)$ and so $h$ is unique.

∎

######  Theorem 2.10.

Given an e-commutative diagram of $R$-modules having $e$-exact rows

${0}$${A^{\prime}}$${A}$${A^{\prime\prime}}$${0}$${C^{\prime}}$${C}$${C^{\prime\prime}}$$\scriptstyle{i}$$\scriptstyle{f}$$\scriptstyle{p}$$\scriptstyle{g}$$\scriptstyle{h}$$\scriptstyle{j}$$\scriptstyle{q}$

there exist a unique map $f:A^{\prime}\to C^{\prime}$ making the augmented
diagram $e$-commute.Moreover, if $g$ and $h$ are isomorphism and $A$ and
$A^{\prime\prime}$ are torsion-free, then $A^{\prime}\cong rC^{\prime}$ for
some non-zero elements $0\neq r\in R$.

###### Proof.

Let $a^{\prime},u^{\prime}\in A^{\prime}$, define
$f(a^{\prime})=rqg(i(a^{\prime}))$. We must shows that $f$ is well defined,
let $i(a^{\prime})=i(u^{\prime})$ implies $qg(i(a^{\prime}-u^{\prime}))=0$, so
$g(i(a-u))\in Ker\>q$ by $e$-exactness, there exist $0\neq r\in R$ and
$c^{\prime}\in C^{\prime}$ such that
$j(c^{\prime})=rg(i(a^{\prime}-u^{\prime}))$. Thus,
$rqg(i(a^{\prime}-u^{\prime}))=qj(c^{\prime})=0$, because $qj=0$. Therefore,
$f$ is well-defined. Suppose that $f^{\prime}:A^{\prime}\to C^{\prime}$ be
another map satisfies $jf^{\prime}=rgi$. Then
$jf^{\prime}(a^{\prime})=rgi(a^{\prime})=jf(a^{\prime})$ this implies that
$(f^{\prime}(a^{\prime})-f(a^{\prime}))\in Kerj=0$ and so $f$ is unique.
Let $a^{\prime}\in Ker\>f$. Then $f(a^{\prime})=0$ and by $e$-commutativity
there exists $0\neq r\in R$ such that $jf(a^{\prime})=0=rgi(a^{\prime})$. Thus
$r.i(a^{\prime})\in Ker\>g=0$ and $A$ is torsion-free so $i(a^{\prime})=0$
implies $a^{\prime}\in Ker\>i=0$. Suppose that $c^{\prime}$ is a non-zero
element of $C^{\prime}$. Since $j(c^{\prime})\in C$ and $g$ is onto, then
there exist $a\in A$ such that $g(a)=j(c^{\prime})$, $e$-commutativity gives
us $qg(a)=rhp(a)$, for some $0\neq r\in R$. Now,
$qg(a)=qj(c^{\prime})=0=rhp(a)$, so $rp(a)\in Ker\>h=0$ torsion-freeness of
$A^{\prime\prime}$ gives us $a\in Ker\>p$, then there exist $0\neq s\in R$ and
$a^{\prime}\in A^{\prime}$ such that $i(a^{\prime})=sa$. By $e$-commutativity
there exist $0\neq t\in R$ such that $jf(a^{\prime})=tgi(a^{\prime})$ so
$jf(a^{\prime})=tg(i(a^{\prime}))=tsg(a)=tsj(c^{\prime})$. We obtain
$(f(a^{\prime})-tsc^{\prime})\in Ker\>j=0$ and so
$f(a^{\prime})=tsc^{\prime}$. Therefore, $A^{\prime}\cong rC^{\prime}$ for
some non-zero element $0\neq r\in R$. ∎

######  Lemma 2.11.

Consider the commutative diagram of $R$-modules and $R$-morphisms, where $R$
is a domain, $B$ and $B^{\prime\prime}$ are torsion-free $R$-modules

${0}$${0}$${0}$${0}$${A}$${A^{\prime}}$${A^{\prime\prime}}$${0}$${0}$${B}$${B^{\prime}}$${B^{\prime\prime}}$${0}$${0}$${C}$${C^{\prime}}$${C^{\prime\prime}}$${0}$${0}$${0}$${0}$$\scriptstyle{f}$$\scriptstyle{i}$$\scriptstyle{f^{\prime}}$$\scriptstyle{j}$$\scriptstyle{p}$$\scriptstyle{g}$$\scriptstyle{i^{\prime}}$$\scriptstyle{g^{\prime}}$$\scriptstyle{j^{\prime}}$$\scriptstyle{p^{\prime}}$$\scriptstyle{h}$$\scriptstyle{h^{\prime}}$

If the columns and the first and the third rows are $e$-exact, then the middle
row is also $e$-exact.

###### Proof.

To proof the middle row is $e$-exact, we have to check the following three
conditions:
1) $Ker(g)=0$. Take $b\in Ker(g)$. Then $j^{\prime}g(b)=0=hi^{\prime}(b)$ and
$i^{\prime}(b)\in Ker(h)$. Since $Ker(h)=0$, so $i^{\prime}(b)=0$ and $b\in
Ker(i^{\prime})$ and as $Im(i)\leqslant_{e}Ker(i^{\prime})$, then there exist
a non-zero element $r$ belongs to $R$ and $a\in A$ such that $i(a)=rb$ and so
$gi(a)=g(rb)=rg(b)=0=jf(a)$, so $f(a)\in Ker(j)$. Since $Ker(j)=0$, $f(a)=0$
and $a\in Ker(f)=0$, which means $a=0$ this implies $rb=0$ and $b=0$, since
$B$ is torsion-free. Therefore, $g$ is monic.

2) $Im(g)\leqslant_{e}Ker(g^{\prime})$. First to prove $Im(g)\subseteq
Ker(g^{\prime})$. Let $b^{\prime}\in Im(g)$. Then there exists $b\in B$ such
that $g(b)=b^{\prime}$ and
$j^{\prime}g(b)=hi^{\prime}(b)=j^{\prime}(b^{\prime})$, which means that
$j^{\prime}(b^{\prime})\in Im(h)\subseteq Ker(h^{\prime})$ and so
$p^{\prime}g^{\prime}(b^{\prime})=h^{\prime}j^{\prime}(b^{\prime})=0$. Hence
$g^{\prime}(b^{\prime})\in Ker(p^{\prime})$ and as
$Im(p)\leqslant_{e}Ker(p^{\prime})$, then there exist $a^{\prime\prime}\in
A^{\prime\prime}$ and a non-zero element $s$ belongs to $R$ such that
$p(a^{\prime\prime})=sg^{\prime}(b^{\prime})=s(g^{\prime}g(b))=0$ and
$sg^{\prime}(b^{\prime})=0$, so $g^{\prime}(b^{\prime})=0$, since
$B^{\prime\prime}$ is torsion-free. Thus $b^{\prime}\in Ker(g^{\prime})$ and
so $Im(g)\subseteq Ker(g^{\prime})$. Now, for essentiality, take $b^{\prime}$
to be a non-zero element of $Ker(g^{\prime})$. Then
$0=g^{\prime}(b^{\prime})=g^{\prime}j(a^{\prime})=pf^{\prime}(a^{\prime})$ and
$f^{\prime}(a^{\prime})\in Ker(p)$. Since $Ker(p)=0,f^{\prime}(a^{\prime})=0$
so $a^{\prime}\in Ker(f^{\prime})$ and as $Im(f)\leqslant_{e}Ker(f^{\prime})$
then there exist a non-zero element $r$ belongs to $R$ and $a\in A$ such that
$f(a)=ra^{\prime}$. Now,
$jf(a)=j(ra^{\prime})=rj(a^{\prime})=rb^{\prime}=gi(a)=g(b)$. Therefore,
$Im(g)\leqslant_{e}Ker(g^{\prime})$.

3) $Im(g^{\prime})\leqslant_{e}B^{\prime\prime}$. Let $b^{\prime\prime}$ be a
non-zero element of $B^{\prime\prime}$ and $p^{\prime}(b^{\prime\prime})\in
C^{\prime\prime}$. Then there exist $c^{\prime}\in C^{\prime}$ and a non-zero
element $r$ belongs to $R$ such that
$h^{\prime}(c^{\prime})=rp^{\prime}(b^{\prime\prime})$ and as
$Im(j^{\prime})\leqslant_{e}C^{\prime}$, then there exist a non-zero element
$s$ belongs to $R$ and $b^{\prime}\in B^{\prime}$ such that
$j^{\prime}(b^{\prime})=sc^{\prime}$. Now, we have
$p^{\prime}g^{\prime}(b^{\prime})=h^{\prime}j^{\prime}(b^{\prime})=h^{\prime}(sc^{\prime})=sh^{\prime}(c^{\prime})=srp^{\prime}(b^{\prime\prime})$
and $p^{\prime}(g^{\prime}(b^{\prime})-srb^{\prime\prime})=0$ which means
$(g^{\prime}(b^{\prime})-srb^{\prime\prime})\in Ker(p^{\prime})$ and as
$Im(p)\leqslant_{e}Ker(p^{\prime})$ then there exist a non-zero element $k\in
R$ and $a^{\prime\prime}\in A^{\prime\prime}$ such that
$p(a^{\prime\prime})=k(g^{\prime}(b^{\prime})-srb^{\prime\prime})$. Also, we
have $f^{\prime}(a^{\prime})=ta^{\prime\prime}$, for a non-zero element $t$
belongs to $R$ and $a^{\prime\prime}\in A^{\prime\prime}$, because
$Im(f^{\prime})\leqslant_{e}A^{\prime\prime}$. Thus
$pf^{\prime}(a^{\prime})=p(ta^{\prime\prime})=tp(a^{\prime\prime})=k(g^{\prime}(b^{\prime})-srb^{\prime\prime})=g^{\prime}j(a^{\prime})=g^{\prime}(b^{\prime})$
which means $g^{\prime}(kb^{\prime}-b^{\prime})=ksrb^{\prime\prime}$.
Therefore, $Im(g^{\prime})\leqslant_{e}B^{\prime\prime}$. ∎

######  Lemma 2.12.

Consider the commutative diagram of $R$-modules and $R$-morphisms, where $R$
is a domain, $C$, $C^{\prime}$ and $C^{\prime\prime}$ are torsion-free
$R$-modules

${0}$${0}$${0}$${0}$${A}$${A^{\prime}}$${A^{\prime\prime}}$${0}$${0}$${B}$${B^{\prime}}$${B^{\prime\prime}}$${0}$${0}$${C}$${C^{\prime}}$${C^{\prime\prime}}$${0}$${0}$${0}$${0}$$\scriptstyle{f}$$\scriptstyle{i}$$\scriptstyle{f^{\prime}}$$\scriptstyle{j}$$\scriptstyle{p}$$\scriptstyle{g}$$\scriptstyle{i^{\prime}}$$\scriptstyle{g^{\prime}}$$\scriptstyle{j^{\prime}}$$\scriptstyle{p^{\prime}}$$\scriptstyle{h}$$\scriptstyle{h^{\prime}}$

If the columns and the above rows are $e$-exact, then the last one is also
$e$-exact.

###### Proof.

To proof the third row is $e$-exact, we have to check the following three
conditions:
1) $Ker(h)=0$. Take $c\in Ker(h)$, then $h(c)=0$ and
$hi^{\prime}(b)=j^{\prime}g(b)$ and and as $Im(i^{\prime})\leqslant_{e}C$ then
there exist $b\in B$ and a non-zero element $r$ belongs to $R$ such that
$i^{\prime}(b)=rc$. Then $hi^{\prime}(b)=h(rc)=rh(c)=0=j^{\prime}g(b)$ so
$g(b)\in Ker(j^{\prime})$ and as $Im(j)\leqslant_{e}Ker(j^{\prime})$, then
there exist a non-zero element $s$ belongs to $R$ and $a^{\prime}\in
A^{\prime}$ such that $j(a^{\prime})=sg(b)$ and so
$g^{\prime}j(a^{\prime})=sg^{\prime}g(b)=0=pf^{\prime}(a^{\prime})$, so
$f^{\prime}(a^{\prime})\in Ker(p)$. Since $Ker(p)=0$,
$f^{\prime}(a^{\prime})=0$ and $a^{\prime}\in Ker(f^{\prime})$ and also by
essentiality there exist $a\in A$ and a non-zero elemeny $t$ belongs to $R$
such that $f(a)=ta^{\prime}$. Then
$jf(a)=j(ta^{\prime})=tj(a^{\prime})=tsg(b)=gi(a)$ and so $g(tsb-i(a))=0$
which means $tsb-i(a)\in Ker(g)=0$ this implies $tsb=i(a)$ and so
$tsi^{\prime}(b)=i^{\prime}i(a)=0$, then $i^{\prime}(b)=0$ since $C$ is
torsion-free. Thus $rc=0$ and so $c=0$. Therefore, $h$ is monic.

2) $Im(h)\leqslant_{e}Ker(h^{\prime})$. First to prove $Im(h)\subseteq
Ker(h^{\prime})$. Let $c^{\prime}\in Im(h)$. Then there exists $c\in C$ such
that $h(c)=c^{\prime}$ and from essentiality there exist $b^{\prime}\in
B^{\prime}$ and a non-zero element $r$ belongs to $R$ such that
$j^{\prime}(b^{\prime})=rc^{\prime}$ and so
$h^{\prime}j^{\prime}(b^{\prime})=h^{\prime}(rc)=rh^{\prime}(c^{\prime})=rh^{\prime}(h(c))=p^{\prime}g^{\prime}(b^{\prime})$
and also by commutativity and essentiality
$g^{\prime}j(a^{\prime})=g^{\prime}(tb^{\prime})=pf^{\prime}(a^{\prime})$ for
a non-zero $t$ belongs to $R$ and $b^{\prime}\in Ker(j^{\prime})$, which means
$Im(g^{\prime})\subseteq Im(p)\subseteq Ker(p^{\prime})$ and so
$p^{\prime}g^{\prime}(tb^{\prime})=0$. Therefore
$h^{\prime}(rc^{\prime})=rh^{\prime}(h(c))=tp^{\prime}g^{\prime}(b^{\prime})=0$
and so $rh^{\prime}(h(c))=0$ so $h^{\prime}(h(c))=0$, since $C^{\prime\prime}$
is torsion-free. Hence $c^{\prime}=h(c)\in Ker(h^{\prime})$. Now, for
essentiality, take $c^{\prime}$ to be a non-zero element of $Ker(h^{\prime})$.
Then $h^{\prime}j^{\prime}(b)=p^{\prime}g^{\prime}(b)$ and as
$Im(j^{\prime})\leqslant_{e}C^{\prime}$ there exist a non-zero element $r$
belongs to $R$ and $b^{\prime}\in B^{\prime}$ such that
$j^{\prime}(b^{\prime})=rc^{\prime}$. Hence $hi^{\prime}(b)=j^{\prime}g(b)$,
since $Im(i^{\prime})\leqslant_{e}C$ and $Im(g)\leqslant_{e}Ker(g^{\prime})\in
B^{\prime}$ so there are non-zero elements $c\in C,b\in B$ and a non-zero
elements $s$ and $k$ belongs to $R$ such that
$hi^{\prime}(b)=h(sc)=sh(c)=j^{\prime}(kb^{\prime})=krc^{\prime}$. Therefore,
$sh(c)=kr(c^{\prime})$ put $k=s$ so $s(h(c)-rc^{\prime})=0$ and so
$h(c)=rc^{\prime}$, since $C^{\prime}$ is torsion-free. Then
$Im(h)\leqslant_{e}Ker(h^{\prime})$.

3) $Im(h^{\prime})\leqslant_{e}C^{\prime\prime}$. Let $c^{\prime\prime}$ be a
non-zero element of $C^{\prime\prime}$ and as
$Im(p^{\prime})\leqslant_{e}C^{\prime\prime}$, there exist
$b^{\prime\prime}\in B^{\prime\prime}$ and a non-zero element $r$ belong to
$R$ such that $p^{\prime}(b^{\prime\prime})=rc^{\prime\prime}$ and as
$Im(g^{\prime})\leqslant_{e}B^{\prime\prime}$, then there exist a non-zero
element $s$ belong to $R$ and $b^{\prime}\in B^{\prime}$ such that
$g^{\prime}(b^{\prime})=sb^{\prime\prime}$ and also we have
$Im(j^{\prime})\leqslant_{e}C^{\prime}$, there exist a non-zero element $t$
belong to $R$ and $b^{\prime}\in B^{\prime}$ such that
$j^{\prime}(b^{\prime})=tc^{\prime}$. Hence
$h^{\prime}j^{\prime}(b^{\prime})=h^{\prime}(tc^{\prime})=th^{\prime}(c^{\prime})=p^{\prime}g^{\prime}(b^{\prime})=p^{\prime}(sb^{\prime\prime})=sp^{\prime}(b^{\prime\prime})=src^{\prime\prime}$,
put $s=t$, so $t(h^{\prime}(c^{\prime}-rc^{\prime\prime}))=0$ this implies
$h^{\prime}(c^{\prime}-rc^{\prime\prime})=0$, since $C^{\prime\prime}$ is
torsion-free. Thus, $h^{\prime}(c^{\prime})=rc^{\prime\prime}$ and
$Im(h^{\prime})\leqslant_{e}C^{\prime\prime}$. ∎

Recall from [1] that an e-injective resolution of an $R-$module $A$ is an
e-exact sequence $0\to
A\stackrel{{\scriptstyle\eta}}{{\to}}E^{0}\stackrel{{\scriptstyle
d^{0}}}{{\to}}E^{1}\stackrel{{\scriptstyle d^{1}}}{{\to}}...\to
E^{n}\stackrel{{\scriptstyle d^{n}}}{{\to}}E^{n+1}\to...$ where each $E^{i}$
is an e-injective $R$-module. Let $f,g:X\to E$ be two chain maps. Then $f$ is
$e$-homotopic to $g$ if there are maps $s^{n+1}:X^{n+1}\to E^{n}$ and non-zero
elements $s$ and $r$ in $R$ such that
$r(h^{n}-f^{n})=s^{n+1}d^{n}+sd^{n-1}s^{n}$ for all $n$. Now, we are in a
position to prove the new form of comparsion theorem by using $e$-injectivity
and $e$-exact sequences.

######  Theorem 2.13.

[Comparison Theorem for e-injectives]
Suppose that we have the following diagram:

${0}$${A}$${E^{0}}$${E^{1}}$${E^{2}}$${...}$${0}$${A^{\prime}}$${X^{0}}$${X^{1}}$${X^{2}}$${...}$$\scriptstyle{\varepsilon}$$\scriptstyle{d^{0}}$$\scriptstyle{d^{1}}$$\scriptstyle{\varepsilon^{\prime}}$$\scriptstyle{f}$$\scriptstyle{d^{\prime
0}}$$\scriptstyle{f^{0}}$$\scriptstyle{f^{1}}$$\scriptstyle{d^{\prime
1}}$$\scriptstyle{f^{2}}$

where the rows are complexes. If each $E^{n}$, in the top row is e-injective
and the bottom row is e-exact, then there exists a chain map
$f:X^{A^{\prime}}\to E^{A}$, (the dashed arrows) making the completed diagram
e-commute. Furthermore, any two such chain maps are e-homotopic.

###### Proof.

We prove the existence of $f^{n}$ by induction $n\geqslant 0$. For the base
step $n=0$, consider the following diagram:

${E^{0}}$${0}$${A^{\prime}}$${X_{0}}$$\scriptstyle{\varepsilon\circ
f}$$\scriptstyle{\varepsilon^{\prime}}$$\scriptstyle{f_{0}}$

Since $\varepsilon^{\prime}$ is monic and $E^{0}$ is e-injective, there exist
$f^{0}:X^{0}\to E^{0}$ and $0\neq r\in R$ such that
$f^{0}\varepsilon^{\prime}=r(\varepsilon\circ f)$. For the inductive step,
consider we have $f^{n-1}$ and $f^{n}$ and the following diagram:

${E^{n-1}}$${E^{n}}$${E^{n+1}}$${...}$${X^{n-2}}$${X^{n-1}}$${X^{n}}$${X^{n+1}}$$\scriptstyle{d^{n-1}}$$\scriptstyle{d^{n}}$$\scriptstyle{d^{n+1}}$$\scriptstyle{d^{\prime
n-1}}$$\scriptstyle{f^{n-1}}$$\scriptstyle{d^{\prime
n}}$$\scriptstyle{f^{n}}$$\scriptstyle{f^{n+1}}$

Since $d^{n}f^{n}(d{{}^{\prime
n-1}}X^{n-1})=r(d^{n}d^{n-1}f^{n-1}(X^{n-1}))=0$,
$Imd^{{}^{\prime}n-1}\subseteq Kerd^{n}f^{n}$ and as
$Imd^{{}^{\prime}n-1}\leqslant_{e}Kerd^{n}$,we have the following diagram

${E^{n+1}}$${0}$${Kerd^{\prime
n}}$${X^{n+1}}$$\scriptstyle{d^{n}f^{n}}$$\scriptstyle{d^{\prime
n}}$$\scriptstyle{f^{n+1}}$

in which $d^{\prime n}$ is monic and $E^{n+1}$ is e-injective, so there exist
$r_{n}\in R\quad$ and $f^{n+1}:X^{n+1}\to E^{n+1}$ such that
$f^{n+1}d^{{}^{\prime}n}=r_{n}d^{n}f^{n}$. Now, to show the uniqueness of $f$
up to e-homotopy. If $h:X^{A^{\prime}}\to E^{A^{\prime}}$, is another chain
map with $h^{0}\varepsilon^{\prime}=r\varepsilon f$, we construct the terms
$s^{n}:X^{n+1}\to E^{n}$ of an e-homotopy $s=s^{n}$ by induction on $n\geq 0$
that we will show that
$s(h^{n}-f^{n})=s^{n+1}d^{n^{\prime}}+r^{\prime}d^{n+1}s^{n}$ for
$s,r^{\prime},p$ and $q$ in $R$. We define $s^{0}:X^{0}\to A$ to be the zero
map. Now to show that $Im(r(h^{n}-f^{n})-r^{\prime}d^{n-1}s^{n})\subseteq
Kerd^{n}$, we have $d^{n}(r(h^{n}-f^{n})-r^{\prime}d^{n-1}s^{n}\\\
=d^{n}(r(h^{n}-f^{n})-d^{n}(r^{\prime}d^{n-1}s^{n})$
$=rd^{n}(h^{n}-f^{n})-r^{\prime}d^{n}(d^{n-1}s^{n})$
$=rd^{n}(h^{n}-f^{n})-r^{\prime}d^{n}(p(h^{n}-f^{n})-qs^{n+1}d^{\prime n})$
$=rd^{n}(h^{n}-f^{n})-r^{\prime}pd^{n}(h^{n}-f^{n})-r^{\prime}qd^{n}(s^{n+1}d^{\prime
n})$
$=rd^{n}(h^{n}-f^{n})-r^{\prime}pd^{n}(h^{n}-f^{n})-r^{\prime}qd^{n}(md^{n-1}s^{n})=0$,
put $r^{\prime}p=r$
Therefore, we have the following diagram:

${E^{n}}$${0}$${Kerd^{\prime
n}}$${X^{n+1}}$$\scriptstyle{r(h^{n}-f^{n})-r^{\prime}d^{n-1}s^{n}}$$\scriptstyle{d^{\prime
n}}$$\scriptstyle{s^{n+1}}$

since $E^{n}$ is e-injective and $d^{\prime n}$ monic, there exist $s^{n+1}$,
$s\in R$ such that
$s^{n+1}d^{n^{\prime}}=s(r(h^{n}-f^{n}-r^{\prime}d^{n-1}s^{n}))$. Therefore,
$sr(h^{n}-f^{n})=s^{n+1}d^{n^{\prime}}+sr^{\prime}d^{n-1}s^{n}$. Hence $f$ and
$h$ are e-homotopic. ∎

######  Proposition 2.14.

Let $I^{\prime\bullet}$ and $I^{\prime\prime\bullet}$ be two e-injective
resolutions of $M^{\prime}$ and $M^{\prime\prime}$ respectively. Suppose that
$0\to M^{\prime}\to M\to M^{\prime\prime}\to 0$ is an e-exact sequence. Then
there exists an e-injective resolution $I^{\bullet}$ such that the following
diagram

${0}$${M^{\prime}}$${M}$${M^{\prime\prime}}$${0}$${0}$${I^{\prime\bullet}}$${I^{\bullet}}$${I^{\prime\prime\bullet}}$${0}$$\scriptstyle{i}$$\scriptstyle{\delta^{\prime}}$$\scriptstyle{p}$$\scriptstyle{\delta^{\prime\prime}}$

is e-commutative in which the bottom row is an e-exact sequence of complexes.

###### Proof.

Consider the following diagram in which the rows are e-exact

${0}$${M^{\prime}}$${M}$${I^{\prime\prime 0}}$${I^{\prime\prime
1}}$${I^{\prime\prime 2}}$${...}$${0}$${M^{\prime}}$${I^{\prime
0}}$${I^{\prime 1}}$${I^{\prime 2}}$${I^{\prime
3}}$${...}$$\scriptstyle{i}$$\scriptstyle{id_{m}}$$\scriptstyle{S^{\prime\prime}op}$$\scriptstyle{f^{0}}$$\scriptstyle{f^{1}}$$\scriptstyle{S^{\prime\prime
0}}$$\scriptstyle{s^{\prime\prime
1}}$$\scriptstyle{f^{2}}$$\scriptstyle{f^{3}}$$\scriptstyle{S^{\prime-1}}$$\scriptstyle{S^{\prime
0}}$$\scriptstyle{S^{\prime 1}}$$\scriptstyle{S^{\prime 2}}$

By the comparison theorem for e-injectives (Theorem 2.13), there exist maps
(dashed arrows) that make all squares e-commute. Now, we define
$I^{n}=I^{\prime}{{}^{n}}\bigoplus I^{\prime\prime}{{}^{n}},\delta^{-1}:M\to
I^{0}$ by $\delta^{-1}(m)=(-f^{0}(m),\delta^{\prime\prime-1}\circ p(m))$ and
$\delta^{n}:I^{n}\to I^{n+1}$ by $\delta^{n}(a,b)=(\delta^{\prime
n}(a)+(-1)^{n}f^{n+1}(b),\delta^{\prime\prime n}(b))$,

${0}$${M^{\prime}}$${M}$${M^{\prime\prime}}$${0}$${0}$${I^{\prime
0}}$${I^{0}}$${I^{\prime\prime}}$${0}$${0}$${I^{\prime
1}}$${I^{1}}$${I^{\prime\prime
1}}$${0}$$\scriptstyle{i}$$\scriptstyle{\delta^{\prime-1}}$$\scriptstyle{p}$$\scriptstyle{\delta^{-1}}$$\scriptstyle{\delta^{\prime\prime-1}}$$\scriptstyle{i^{0}}$$\scriptstyle{\delta^{\prime
0}}$$\scriptstyle{p^{0}}$$\scriptstyle{\delta^{0}}$$\scriptstyle{\delta^{\prime\prime
0}}$$\scriptstyle{i^{1}}$$\scriptstyle{f^{1}}$

since $Ker\delta^{\prime\prime-1}=0$ and $Ker\delta^{\prime-1}=0$,
$Ker\delta^{-1}=0$. Thus $\delta^{-1}$ is monic and from Proposition 2.13, we
get $I^{n}$ is e-injective for each $n\geqslant 0$. To prove that
$Im\delta^{-1}\leqslant_{e}Ker\delta^{0}$ we must show first that
$Im\delta^{-1}\subseteq Ker\delta^{0}$. Thus
$\delta^{0}(\delta^{-1}(m))=\delta^{0}(-f^{0}(m),\delta^{\prime\prime-1}\circ
p(m))=\delta^{\prime}(-f^{0}(m)+(-1)^{0}f^{1}(\delta^{\prime\prime-1}\circ
p(m)),\delta^{\prime\prime 0}(\delta^{\prime\prime-1}\circ p(m)))=0$, which
implies that $Im\delta^{-1}\subseteq Ker\delta^{0}$. Now, for essentiality,
let $0\neq(a,b)\in Ker\delta^{0}$. Then $0=\delta^{0}(a,b)=(\delta^{\prime
0}(a)+(-1)^{0}f^{1}(b),\delta^{\prime\prime 0}(b))$, so $\delta^{\prime
0}(a)+f^{1}(b)=0$ and $\delta^{\prime\prime 0}(b)=0$. Since
$Im\delta^{\prime\prime-1}\leqslant_{e}Ker\delta^{\prime\prime 0}$, there
exists $0\neq r\in R$ such that $\delta^{\prime\prime-1}(m^{\prime\prime})=rb$
for some $m^{\prime\prime}$ in $M^{\prime\prime}$ and
$\delta^{\prime\prime-1}(p(m))=rb$ for some $m\in M$. So
$f^{1}(\delta^{\prime\prime-1}(p(m)))=f^{1}(rb)=rf^{1}(b)=-r\delta^{\prime
0}(a)$ and by e-commutity there exists $0\neq q\in R$ such that
$-r\delta^{\prime 0}(a)=q\delta^{\prime 0}f^{0}(m)$ and this implies that
$r\delta^{\prime 0}(a)+q\delta^{\prime 0}f^{0}(m)=0=\delta^{\prime
0}(ra+qf^{0}(m))$. Thus we get $ra+qf^{0}(m)\in Ker\delta^{\prime 0}$, and
since $Im\delta^{\prime-1}\leqslant_{e}Ker\delta^{\prime 0}$, there exists
$0\neq t\in R$ such that $\delta^{\prime-1}(m^{\prime})=tra+tqf^{0}(m)$ for
some $m^{\prime}\in M^{\prime}$. Now,
$\delta^{-1}(tqm-i(m^{\prime}))=(-f^{0}(tqm-i(m^{\prime})),\delta^{\prime\prime-1}\circ
p(tqm-i(m^{\prime}))=(-f^{0}(tqm)+f^{0}(i(m^{\prime}))),\delta^{\prime\prime-1}p(tqm)-\delta^{\prime\prime-1}p(i(m^{\prime})))=tra-\delta^{\prime-1}(m^{\prime})+p\delta^{\prime-1}(m^{\prime}),tqrb)=s(a,b)$,
put $q=p=1$ which proves that $Im\delta^{-1}\leqslant_{e}Ker\delta^{0}$ for
$n\geqslant 0$ and $\delta^{n+1}(\delta^{n}(a,b))=\delta^{n+1}(\delta^{\prime
n}(a)+(-1)^{n}f^{n+1}(b)),\delta^{\prime\prime n}(b))\\\ =\delta^{\prime
n+1}(\delta^{\prime
n}(a)+(-1)^{n}f^{n+1}(b)+(-1)^{n+1}f^{n+2}(\delta^{\prime\prime
n}(b)),\delta^{\prime\prime n+1}(\delta^{\prime\prime n}(b)))=0$. This proves
that $Im\delta^{n}\subseteq Ker\delta^{n+1}$. Now let $(a,b)\in
Ker\delta^{n+1}$. Then $\delta^{\prime\prime n+1}(b)=0$ and $\delta^{\prime
n+1}(a)+(-1)^{n+1}f^{n+2}(b)=0$. It follows that $\delta^{\prime\prime
n}(c)=rb$, $0\neq r\in R$ for some $c\in I^{\prime\prime n}$. Then
$(-1)^{n}y\delta^{\prime n+1}f^{n+1}(c)=(-1)^{n}f^{n+2}(\delta^{\prime\prime
n}(c))=r\delta^{\prime n+1}(a)$ so that $\delta^{\prime
n+1}(ra-(-1)^{n+1}yf^{n+1}(c))=0$ and $ra-(-1)^{n}yf^{n+1}(c)\in
Ker\delta^{\prime n+1}$. Since $Im\delta^{\prime
n}\leqslant_{e}Ker\delta^{\prime n+1}$, there exists $0\neq q\in R$ such that
$\delta^{\prime n}(d)=q(ra-(-1)^{n}yf^{n+1}(c))$ for some $d\in I^{\prime n}$.
Thus $\delta^{n}(d,c)=(\delta^{\prime
n}(d)+(-1)^{n}f^{n+1}(c),\delta^{\prime\prime n}(c))=r(a,b)$, put $q=y=1$,
which proves that $Im\delta^{n}\leqslant_{e}Ker\delta^{n+1}$. ∎

##  3\. The cohomology regarding to $e$-exact sequences

In this section all rings are Noetherian domain and all modules are unitary
$R$-modules. We want to describe the right derived functors to the additive
covariant functor (torsion functor) $\Gamma_{a}$ of local cohomology with
e-injective resolutions and then we present new forms of some theorems of
cohomology with $e$-exact sequences. Let $a$ be an ideal of $R$. The following
is a useful result that shows the functor $\Gamma_{a}(\>)$ is an $e$-exact
under suitable condition.

######  Lemma 3.1.

Let $0\to L\stackrel{{\scriptstyle f}}{{\to}}M\stackrel{{\scriptstyle
g}}{{\to}}N\to 0$ be an e-exact sequence of $R$-modules and $R$-homomorphisms.
Then
$0\to\Gamma_{a}(L)\stackrel{{\scriptstyle\Gamma_{a}(f)}}{{\to}}\Gamma_{a}(M)\stackrel{{\scriptstyle\Gamma_{a}(g)}}{{\to}}\Gamma_{a}(N)$
is an e-exact sequence. Furthermore, if $N$ is a torsion-free module, then the
functor $\Gamma_{a}(\>)$ is an $e$-exact.

###### Proof.

It is clear that $ker(\Gamma_{a}(f))=0$. To show that
$Im(\Gamma_{a}(f))\leqslant_{e}Ker(\Gamma_{a}(g))$. Let $0\neq y\in
Ker(\Gamma_{a}(g))$. Then $y\in\Gamma_{a}(M)$ and $g(y)=0$, that is there
exist $n\in N$ such that $a^{n}y=0$. Now, since $Imf\leqslant_{e}Kerg$, there
exists $0\neq r\in R$ such that $f(l)=ry$, for some $l\in L$. Since
$f(a^{n}l)=a^{n}f(l)=a^{n}(ry)=0,a^{n}l=0$ and $l\in\Gamma_{a}(L)$. Now, to
show that $Im(\Gamma_{a}(g))\leqslant_{e}\Gamma_{a}(N)$. Let
$y\in\Gamma_{a}(g)\cap Rx$, for $0\neq x\in\Gamma_{a}(N)$. Then $y=g(m)=rx$
for $m\in\Gamma_{a}(M)$ and $r\in R$. By hypothesis, $rx\neq 0$ and so $y\neq
0$ which guarantees the $e$-exactness of the sequence
$0\to\Gamma_{a}(L)\stackrel{{\scriptstyle\Gamma_{a}(f)}}{{\to}}\Gamma_{a}(M)\stackrel{{\scriptstyle\Gamma_{a}(g)}}{{\to}}\Gamma_{a}(N)\to
0$.

∎

######  Definition 3.2.

The right e-derived functors $R^{n}T$ of a covariant functor $T$ are defined
on an $R$-module $A$ by
$(R^{n}T)(A)=H^{n}(TE^{A})=\frac{ker(Td^{n})}{Im(Td^{n-1})}$, where $E:0\to
A\to E^{0}\stackrel{{\scriptstyle d^{0}}}{{\to}}E^{1}\stackrel{{\scriptstyle
d^{1}}}{{\to}}\dots$ is an e-injective resolution of $A$ and $E^{A}$ is its
deleted $e$-injective resolution.

Define the $n_{th}$ $e$-cohomolgy module ${}_{e}H^{n}_{a}(M)$ of $M$ with
respect to an ideal a as the $n_{th}$ right e-derived functor of the torsion
functors $\Gamma_{a}(\>)$, that is
${}_{e}H^{n}_{a}(M)=R^{n}\Gamma_{a}(M)=\frac{ker(\Gamma_{a}(d^{n}))}{Im(\Gamma_{a}(d^{n-1}))}$.
To calculate ${}_{e}H^{n}_{a}(M)$ with e-injective resolution $0\to
M\stackrel{{\scriptstyle\alpha}}{{\to}}I^{0}\stackrel{{\scriptstyle
d^{0}}}{{\to}}I^{1}\to\dots\to I^{n}\stackrel{{\scriptstyle
d^{n}}}{{\to}}I^{n+1}\to\dots$ for any $R$-module $M$ we do the following:
Apply the functor $\Gamma_{a}$ to the deleted complex of $M$
$I^{E}:0\stackrel{{\scriptstyle d^{-1}}}{{\to}}I^{0}\stackrel{{\scriptstyle
d^{0}}}{{\to}}I^{1}\to\dots\to I^{n}\stackrel{{\scriptstyle
d^{n}}}{{\to}}\dots$ and obtain
$0\to\Gamma_{a}(I^{0})\stackrel{{\scriptstyle\Gamma_{a}(d^{0})}}{{\to}}\Gamma_{a}(I^{1})\to\dots\to\Gamma_{a}(I^{n})\to\dots$
then take the $n_{th}$ $e$-cohomology module of this complex, the result is
${}_{e}H^{n}_{a}(M)=\frac{Ker(\Gamma_{a}(d^{n}))}{Im(\Gamma_{a}(d^{n-1}))}=(R^{n}T)(M)=R^{n}(\Gamma_{a}(M))=H^{n}(TE^{M})$.

######  Theorem 3.3.

The right e-derived functors for $\Gamma_{a}$ are additive functors for every
integer $n$.

###### Proof.

Let $f:M\to M^{{}^{\prime}}$ be a morphism. Then by Theorem 2.13, there is a
chain map $\breve{f}:E^{M}\to E^{M^{{}^{\prime}}}$ over $f$, where $E^{M}$ and
$E^{M^{{}^{\prime}}}$ are deleted e-injective resolutions for $M$ and
$M{{}^{\prime}}$ respectively. Then
$\Gamma_{a}\breve{f}:\Gamma_{a}E^{M}\to\Gamma_{a}E^{M^{{}^{\prime}}}$ is also
a chain map, and so there is a well-defined map
${}_{e}H^{n}_{a}(f)=(R^{n}\Gamma_{a})f:H^{n}(\Gamma_{a}E^{M})\to
H^{n}(\Gamma_{a}E^{M^{{}^{\prime}}})$, defined by
${}_{e}H^{n}_{a}(f)=(R^{n}\Gamma_{a})f=H^{n}(\Gamma_{a}\breve{f})=(\Gamma_{a}\breve{f})$
and ${}_{e}H^{n}_{a}=R^{n}\Gamma_{a}$ is also an additive covariant functor,
because
${}_{e}H^{n}_{a}(f+g)=(R^{n}\Gamma_{a})(f+g)=H^{n}(\Gamma_{a}(f+g))=H^{n}(\Gamma_{a}(f)+\Gamma_{a}(g))=H^{n}(\Gamma_{a}(f))+H^{n}(\Gamma_{a}(g))=_{e}H^{n}_{a}(f)+_{e}H^{n}_{a}(g)$.
Therefore the right e-derived functors are additive functors for every integer
n. ∎

######  Proposition 3.4.

Let $A$ be any $R$\- module. Then ${}_{e}H^{n}_{a}(A)=(R^{n}\Gamma_{a})A=0$,
for all negative integers $n$.

###### Proof.

Suppose that $E:0\to A\to E^{0}\stackrel{{\scriptstyle
d^{0}}}{{\to}}E^{1}\to\dots$ be an e-injective resolution for $A$. Then the
deleted complex of $A$ is $E^{A}:0\to E^{0}\stackrel{{\scriptstyle
d^{0}}}{{\to}}E^{1}\to\dots$. After applying $\Gamma_{a}$ on the deleted
complex, we get $\Gamma_{a}E^{n}=0$ for all negative integers n, because the
$n_{th}$ term of $E^{A}$ is zero when $n$ is negative. Hence
${}_{e}H^{n}_{a}(A)=R^{n}\Gamma_{a}(A)=0$ for all negative integers n. ∎

######  Corollary 3.5.

Let $E$ be an e-injective $R$-module. Then
${}_{e}H^{n}_{a}(E)=(R^{n}\Gamma_{a})(E)={0}$, for all $n\geq 1$.

###### Proof.

Since $E$ is an e-injective module, the e-injective resolution of $E$ is $0\to
E\stackrel{{\scriptstyle 1_{E}}}{{\to}}E\to 0$. The corresponding deleted
e-injective resolution $E^{E}$ is the complex $0\to E\to 0$. Hence, the
$n_{th}$ term of $\Gamma_{a}(E^{E})$ is ${0}$ for all $n\geq 1$ and so
${}_{e}H^{n}_{a}(E)=(R^{n}\Gamma_{a})(E)=H^{n}(\Gamma_{a}E^{E})={0}$ for all
$n\geq 1$. ∎

######  Theorem 3.6.

Let $0\to L\stackrel{{\scriptstyle f}}{{\to}}M\stackrel{{\scriptstyle
g}}{{\to}}N\to 0$ be an e-exact sequence of R-modules and R-homomorphisms and
$N$ torsion-free. Then, for each $i\in N_{0}$, there are a connectig
homomorphisms
${{}_{e}H^{i}_{a}(N)}\stackrel{{\scriptstyle\sigma}}{{\to}}{{}_{e}H^{i+1}_{a}(L)}$
and these connecting homomorphisms make the resulting long sequence
$0\to{{}_{e}H^{0}_{a}(L)}\stackrel{{{}_{e}H^{0}_{a}(f)}}{{\to}}{{}_{e}H^{0}_{a}(M)}\stackrel{{{}_{e}H^{0}_{a}(g)}}{{\to}}{{}_{e}H^{0}_{a}(N)}\to{{}_{e}H^{1}_{a}(L)}\to\dots\to{{}_{e}H^{i}_{a}(L)}\stackrel{{{}_{e}H^{i}_{a}(f)}}{{\to}}{{}_{e}H^{i}_{a}(M)}\to{{}_{e}H^{i}_{a}(N)}\stackrel{{\scriptstyle\sigma^{*}}}{{\to}}{{}_{e}H^{i+1}_{a}(L)}\to\dots$
$e$-exact.

###### Proof.

By applying $\Gamma_{a}$ to an e-exact sequence $0\to L\stackrel{{\scriptstyle
f}}{{\to}}M\stackrel{{\scriptstyle g}}{{\to}}N\to 0$ we obtain an $e$-exact
sequence
$0\to\Gamma_{a}(L)\stackrel{{\scriptstyle\Gamma_{a}(f)}}{{\to}}\Gamma_{a}(M)\stackrel{{\scriptstyle\Gamma_{a}(g)}}{{\to}}\Gamma_{a}(N)\to
0$ by Lemma 3.1 and by [2, Theorem 3.1] there is a connecting homomorphism
$\sigma_{n}:H^{n}(\Gamma_{a}(N))\to H^{n+1}(\Gamma_{a}(L))$ and by [2, Theorem
3.2] there is a long $e$-exact sequence of $R$-modules and $R$-morphisms
$0\to{{}_{e}H^{0}_{a}(L)}\stackrel{{{}_{e}H^{0}_{a}(f)}}{{\to}}{{}_{e}H^{0}_{a}(M)}\stackrel{{{}_{e}H^{0}_{a}(g)}}{{\to}}{{}_{e}H^{0}_{a}(N)}\to{{}_{e}H^{1}_{a}(L)}\to\dots\to{{}_{e}H^{i}_{a}(L)}\stackrel{{{}_{e}H^{i}_{a}(f)}}{{\to}}{{}_{e}H^{i}_{a}(M)}\to{{}_{e}H^{i}_{a}(N)}\stackrel{{\scriptstyle\sigma^{*}}}{{\to}}{{}_{e}H^{i+1}_{a}(L)}\to\dots$
∎

######  Theorem 3.7.

For any torsion-free $R$-module $M$, ${}_{e}H^{0}_{a}(M)$ is naturally
equivalent to $r\Gamma_{a}(M)$ for some $r\neq 0\in R$.

###### Proof.

Let $E:0\to
A\stackrel{{\scriptstyle\sigma}}{{\to}}E^{0}\stackrel{{\scriptstyle
d^{0}}}{{\to}}E^{1}\stackrel{{\scriptstyle d^{1}}}{{\to}}E^{2}\to\dots$ be an
e-injective resolution for an $R$-module $A$ and $E^{A}:0\to
E^{0}\stackrel{{\scriptstyle d^{0}}}{{\to}}E^{1}\stackrel{{\scriptstyle
d^{1}}}{{\to}}E^{2}\to\dots$ be the deleted $e$-injective resolution for $A$.
When we apply $\Gamma_{a}(\>)$ on the deleted $e$-injective resolution we get
$0\to\Gamma_{a}(E^{0})\stackrel{{\scriptstyle
d^{0^{*}}}}{{\to}}\Gamma_{a}(E^{1})\stackrel{{\scriptstyle
d^{1^{*}}}}{{\to}}\Gamma_{a}(E^{2})\to\dots$. Then by the definition of
e-cohomology, we have ${}_{e}H^{0}_{a}(A)=H^{0}(\Gamma_{a}(E^{A}))=Kerd^{0*}$.
But the left e-exactness of $\Gamma_{a}(\>)$ gives an e-exact sequence
$0\to\Gamma_{a}(E^{0})\stackrel{{\scriptstyle
d^{0^{*}}}}{{\to}}\Gamma_{a}(E^{1})\stackrel{{\scriptstyle
d^{1^{*}}}}{{\to}}\Gamma_{a}(E^{2})\to\dots$. We define
$\sigma^{*}:\Gamma_{a}(A)\to Kerd^{0*}$. Since
$Im\sigma^{*}\leqslant_{e}Kerd^{0*}$, $\sigma^{*}$ is well-defined and
$\Gamma_{a}(\>)$ a left e-exact functor, $\sigma^{*}$ is monic. Now, we want
to prove that $\sigma^{*}$ is epic. Let $x\in Kerd^{0*}$. Then
$d^{0*}(x)=d^{0}(x)=0$. Therefore $x\in Kerd^{0}$. By e-exactness of the
e-injective resolution we have $Im\sigma\leqslant_{e}Kerd^{0}$, so there exist
$a^{\prime}\in A$ and $0\neq r\in R$ such that $\sigma(a^{\prime})=rx\neq 0$.
Now, we define $f:r\Gamma_{a}(A)\to A$ by $f(ry)=a^{\prime}$. Let
$y_{1},y_{2}\in\Gamma_{a}(A)$ with $ry_{1}=ry_{2}$. Then
$\sigma(a_{1}^{\prime})=\sigma(a_{2}^{\prime})$. By monicness of $\sigma$ we
have $a_{1}^{\prime}=a_{2}^{\prime}$ and so $f$ is well-defined. Now we have
$rx=\sigma(a^{\prime})=\sigma(rf(y))=r\sigma(f(y))=r\sigma^{*}(f(y))$ which is
equivalent to $\sigma^{*}(f(y))=x$. Hence $\sigma^{*}$ is an isomorphism and
since ${}_{e}H^{0}_{a}(A)=Kerd^{0*}$. Therefore ${}_{e}H^{0}_{a}(\>)$ is
isomorphic to $r\Gamma_{a}(\>)$ for some nonzero $r$ in $R$. ∎

######  Corollary 3.8.

If $0\to L\stackrel{{\scriptstyle f}}{{\to}}M\stackrel{{\scriptstyle
g}}{{\to}}N\to 0$ is an e-exact sequence of R-modules where $N$ is torsion-
free, then there is a long e-exact sequence
$0\to{{}_{e}H^{0}_{a}(L)}\stackrel{{\scriptstyle\Gamma_{a}(f)}}{{\to}}{{}_{e}H^{0}_{a}(M)}\stackrel{{\scriptstyle\Gamma_{a}(g)}}{{\to}}{{}_{e}H^{0}_{a}(N)}\stackrel{{\scriptstyle\sigma}}{{\to}}{{}_{e}H^{1}_{a}(L)}\stackrel{{{}_{e}H^{1}_{a}(f)}}{{\to}}{{}_{e}H^{1}_{a}(M)}\stackrel{{{}_{e}H^{1}_{a}(g)}}{{\to}}\dots$.
In addition, if $L$, $M$ and $N$ are torsion-free modules, then there is a
long e-exact sequence $0\to
r_{1}\Gamma_{a}(L)\stackrel{{\scriptstyle\Gamma_{a}(f)}}{{\to}}r_{2}\Gamma_{a}(M)\stackrel{{\scriptstyle\Gamma_{a}(g)}}{{\to}}r_{3}\Gamma_{a}(N)\stackrel{{\scriptstyle\sigma}}{{\to}}{{}_{e}H^{1}_{a}(L)}\stackrel{{{}_{e}H^{1}_{a}(f)}}{{\to}}{{}_{e}H^{1}_{a}(M)}\stackrel{{{}_{e}H^{1}_{a}(g)}}{{\to}}\dots$
for some nonzero $r_{1},r_{2},r_{3}$ in $R$.

###### Proof.

Directly follows from Theorem 3.6 and Theorem 3.7. ∎

######  Theorem 3.9.

Given an e-commutative diagram of R-modules having e-exact rows where
$A^{\prime\prime}$ and $C^{\prime\prime}$ are torsion-free:

${0}$${A^{\prime}}$${A}$${A^{\prime\prime}}$${0}$${0}$${C^{\prime}}$${C}$${C^{\prime\prime}}$${0}$$\scriptstyle{i}$$\scriptstyle{f}$$\scriptstyle{p}$$\scriptstyle{g}$$\scriptstyle{h}$

Then there is an e-commutative diagram with e-exact rows,

${\dots\to{{}_{e}H^{n}_{a}(A^{\prime})}}$${{{}_{e}H^{n}_{a}(A)}}$${{{}_{e}H^{n}_{a}(A^{\prime\prime})}}$${{{}_{e}H^{n+1}_{a}(A^{\prime})}}$${\dots}$${\dots\to{{}_{e}H^{n}_{a}(C^{\prime})}}$${{{}_{e}H^{n}_{a}(C)}}$${{{}_{e}H^{n}_{a}(C^{\prime\prime})}}$${{{}_{e}H^{n+1}_{a}(C^{\prime})}}$${\dots}$$\scriptstyle{i_{*}}$$\scriptstyle{f_{*}}$$\scriptstyle{p_{*}}$$\scriptstyle{g_{*}}$$\scriptstyle{\sigma}$$\scriptstyle{h_{*}}$$\scriptstyle{f_{*}}$$\scriptstyle{j_{*}}$$\scriptstyle{q_{*}}$$\scriptstyle{\sigma^{*}}$

###### Proof.

By Proposition 2.14, we have an e-exact sequence of deleted complexes $0\to
E^{A^{\prime}}\to E^{A}\to E^{A^{\prime\prime}}\to 0$. If $T=\Gamma_{a}$, then
$0\to TE^{A^{\prime}}\to TE^{A}\to TE^{A^{\prime\prime}}\to 0$ is still an
e-exact by Lemma 3.1. By [2, Remark 3.3] there is an $e$-commutative diagram
of $R$-modules and $R$-morphisms

${\dots\to
H^{n-1}(\Gamma_{a}(A^{\prime}))}$${H^{n-1}(\Gamma_{a}(A))}$${H^{n-1}(\Gamma_{a}(A^{\prime\prime}))}$${H^{n}(\Gamma_{a}(A^{\prime}))}$${\dots}$${\dots\to
H^{n-1}(\Gamma_{a}(C^{\prime}))}$${H^{n-1}(\Gamma_{a}(C))}$${H^{n-1}(\Gamma_{a}(C^{\prime\prime}))}$${H^{n}(\Gamma_{a}(C^{\prime}))}$${\dots}$$\scriptstyle{i_{*}}$$\scriptstyle{f_{*}}$$\scriptstyle{p_{*}}$$\scriptstyle{g_{*}}$$\scriptstyle{\sigma}$$\scriptstyle{h_{*}}$$\scriptstyle{f_{*}}$$\scriptstyle{j_{*}}$$\scriptstyle{q_{*}}$$\scriptstyle{\sigma^{*}}$

and the rest will achieved by applying the definition of $e$-cohomology
${}_{e}H^{n}_{a}(\>)=H^{n}(\Gamma_{a}(E^{A}))$. ∎

######  Corollary 3.10.

Let $M$ be an $a$-torsion $R$-module. Then there exists an $e$-injective
resolution of $M$ in which each term is an $a$-torsion $R$-module.

In local cohomology, every $a$-torsion $R$-module $M$ has zero $i_{th}$ local
cohomology modules, that is $H^{i}_{a}(M)=0$, for all $i>0$. while in
$e$-cohomology this is not true in general. To be more precise we present the
following example.

######  Example 3.11.

Consider the $e$-injective resolution
$0\to\frac{\mathbb{Z}}{2\mathbb{Z}}\stackrel{{\scriptstyle
f}}{{\to}}\frac{\mathbb{Z}}{8\mathbb{Z}}\stackrel{{\scriptstyle
g}}{{\to}}\frac{\mathbb{Z}}{16\mathbb{Z}}\to{0}$ for the $\mathbb{Z}$-module
$\frac{\mathbb{Z}}{2\mathbb{Z}}$ where $f(1+2\mathbb{Z})=4+8\mathbb{Z}$ and
$g(n+8\mathbb{Z})=8n+16\mathbb{Z}$. Since each term of the resolution is
$2\mathbb{Z}$-torsion while the $e$-cohomology module
$H^{1}_{2\mathbb{Z}}(\frac{\mathbb{Z}}{2\mathbb{Z}})=\frac{\mathbb{Z}}{8\mathbb{Z}}$
is non-zero.

At the end of the paper and as a future work one can do research on the
$e$-cohomology dimension and its connection with the cohomolgy dimension.

Acknowlegments: We would like to thank the referees for their thoughtful
comments and efforts towards improving the manuscript.

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