# A Categorical Development of Right Derived Functors

Skyler Marks  Skyler Marks Boston University

###### Abstract

Category theory is the language of homological algebra, allowing us to state
broadly applicable theorems and results without needing to specify the details
for every instance of analogous objects. However, authors often stray from the
realm of pure abstract category theory in their development of the field,
leveraging the Freyd-Mitchell embedding theorem or similar results, or
otherwise using set-theoretic language to augment a general categorical
discussion. This paper seeks to demonstrate that - while it is not necessary
for most mathematicians’ purposes - a development of homological concepts can
be contrived from purely categorical notions. We begin by outlining the
categories we will work within, namely Abelian categories (building off
additive categories). We continue to develop cohomology groups of sequences,
eventually culminating in a development of right derived functors. This paper
is designed to be a minimalist construction, supplying no examples or
motivation beyond what is necessary to develop the ideas presented.

##  1 Categorical Background

###  1.1 Set - Theoretic Preface

In order to leverage the full power of category theory, we endeavor to work in
a general category with no set-theoretic language. As such, we will not
restrict ourselves to working within a small category; the classes of objects
and morphisms within the category may be a proper class. However, we will need
some machinery to work with these classes. As such, the axioms of Zermelo -
Fraenkel + choice set theory (ZFC) will make for a clumsy formalism for our
purposes (as a general class is not treated). To rectify this issue, we turn
to the conservative extension of ZFC, von Neumann - Bernays - Gödel set theory
(NBG), for our treatment of classes. More detail can be found in [3]. A
category, for the purpose of this paper, will be thought of as a (possibly
proper) class of objects together with (possibly proper) classes of morphisms
for each pair of objects (pairing allows for this, as each object can be
represented as a set). Most categories of algebraic objects are not small (or
even essentially small) categories; thus, although considering general classes
is often irrelevant to a mathematician’s daily work, it is not entirely
without justification.

###  1.2 Additive Categories

The additive category forms the foundation for abelian categories, which in
turn comprise the realm in which general homological algebra is performed.
There are many formulations of an additive category, most of which specify
outright that the classes $\mathrm{Hom}(A,B)$ (morphisms between objects $A$
and $B$) should be abelian groups. This amounts to adding a structure to the
category, which (although easy to work with) is somewhat undesirable. Our
definition shies away from this measure, instead characterizing additive
categories as ones which include certain objects. These certain objects will
allow us to induce an abelian group on the hom-classes of an additive
category, without specifying any additional structure.

Perhaps the most central of these objects is a biproduct for any two elements
$A$ and $B$, generalizing the notion of a direct sum of abelian groups or the
carteisan product of two sets to an arbitrary category. Such a set theoretic
product is generally a set of pairs $(a,b)$, with $a$ belonging to the first
set of the product and $b$ belonging to the second. We generalize this notion
to a categorical framework, using morphisms which project off the product
(effectively mapping $(a,b)$ to either $a$ or $b$). Then, as is standard with
category theory, we can dualize this definition.

######  Definition 1.1.

The binary product $A\times B$ in a category $\mathscr{A}$ is an object in
$\mathscr{A}$ together with morphisms $\pi_{1}:A\times B\to A$ and
$\pi_{2}:A\times B\to B$ such that for any object $C$ in $\mathscr{A}$ with
maps $f:C\to A$ and $f^{\prime}:C\to B$ there is a unique map $g:C\to A\times
B$ such that $\pi_{1}\circ g=f$ and $\pi_{2}\circ g=f^{\prime}$. The binary
coproduct $A\amalg B$ is the dual of this; an object in $\mathscr{A}$ with
maps $\iota_{1}:A\to A\amalg B$ and $\iota_{2}:B\to A\amalg B$ such that for
any object $C$ in $\mathscr{A}$ with maps $f:A\to C$ and $f^{\prime}:B\to C$
there is a unique map $g:A\amalg B\to C$ such that $g\circ\iota_{1}=f$ and
$g\circ\iota_{2}=f^{\prime}$.

Some examination shows that the coproduct in the category of sets is the
disjoint union, and that the coproduct in the category of groups is the
internal direct product. Of note, however, is that in the category of groups
the internal direct sum is isomorphic to the external direct sum; speaking
categorically, the product is also a coproduct. We generalize this notion as a
biproduct, which will form the foundation for our construction of abelian
categories.

######  Definition 1.2.

For objects $A$ and $B$ in a category $\mathscr{A}$ a biproduct is an object
of $\mathscr{A}$ which is both a product and a coproduct with

$\pi_{1}\circ\iota_{1}=1_{A}\mathrm{\ and\ }\pi_{2}\circ\iota_{2}=1_{B}$ (1)

We will use this object to construct a ”sum” of two morphisms. The
associativity of the product, crucial to the proof that the constructed sum
forms a group, is given by a repeated application of the definition; for more
detail, see [1]. We record this as a lemma for reference later.

######  Lemma 1.1.

The product $(A\oplus B)\oplus C$ is naturally isomorphic to $A\oplus(B\oplus
C)$.

In order to leverage this this construction towards a sum of two morphisms, we
demonstrate the existence of a ”biproduct of morphisms” with a construction
generalizing the notion of the homomorphism $(x,y)\mapsto(f(x),g(y))$ mapping
between direct sums of groups, for homomorphisms $f$ and $g$. Speaking in
categorical terms, the only way to refer to the ”components” of the product is
by use of the canonical projections; thus, we accomplish this generalization
by constructing a morphism which commutes with the canonical projections off
the biproducts.

######  Lemma 1.2.

For each pair for each pair of morphism $f:A\to B$ and $g:A^{\prime}\to
B^{\prime}$, there is a unique morphism $(f\oplus g):A\oplus A^{\prime}\to
B\oplus B^{\prime}$ such that $\pi_{1}\circ(f\oplus g)=f\circ p_{1}$ and
$\pi_{2}\circ(f\oplus g)=g\circ p_{2}$, where $\pi_{1}$ and $\pi_{2}$ are the
canonical projections from $B\oplus B^{\prime}$ to $B$ and $B^{\prime}$
respectively, and $p_{1},p_{2}$ are the projections from $A\oplus A^{\prime}$
to $A$ and $A^{\prime}$ respectively, all as given by the definition of a
categorical biproduct.

###### Proof.

The product $B\oplus B^{\prime}$ projects onto $B$ and $B^{\prime}$, and is
universal with respect to that property. That is, for any other object $C$ and
any other morphism $\pi^{\prime}_{1}:C\to B,\pi^{\prime}_{2}:C\to B^{\prime}$
there must be a unique morphism $\phi:C\to B\oplus B$ such that
$\pi^{\prime}_{1}=\pi_{1}\circ\phi$ and $\pi^{\prime}_{2}=\pi_{2}\circ\phi$.
Letting $C=A\oplus A^{\prime}$, $\pi^{\prime}_{1}=f\circ p_{1}$ and
$\pi^{\prime}_{2}=g\circ p_{2}$ gives us a unique morphism $\phi=(f\oplus
g):A\oplus A^{\prime}\to B\oplus B^{\prime}$ such that $\pi_{1}\circ(f\oplus
g)=f\circ p_{1}$, and $\pi_{2}\circ(f\oplus g)=g\circ p_{2}$, as intended. ∎

To leverage the duality inherent within categorical proofs, we demonstrate
that this morphism also commutes with inclusions.

######  Lemma 1.3.

The morphism $f\oplus g$ also satisfies $f\oplus g\circ i_{1}=\iota_{1}\circ
f$ and $f\oplus g\circ i_{2}=\iota_{2}\circ g$, where $i_{1}$ and $i_{2}$ are
the inclusions into $A\oplus A$ and $\iota_{1}$ and $\iota_{2}$ the inclusions
into $B\oplus B$.

###### Proof.

The morphism $f\oplus g$ is the unique morphism satisfying
$\pi_{1}\circ(f\oplus g)=f\circ p_{1}$ and $\pi_{2}\circ(f\oplus g)=f\circ
p_{2}$, as above. However, if we apply an inclusion on the left and right of
each of these equations we obtain:

$\iota_{1}\circ\pi_{1}\circ(f\oplus g)\circ i_{1}=\iota_{1}\circ f\circ
p_{1}\circ i_{1}\text{ and }\iota_{2}\circ\pi_{2}\circ(f\oplus g)\circ
i_{2}=\iota_{2}\circ f\circ p_{2}\circ i_{2}$

Which simplify by (1) to

$(f\oplus g)\circ i_{1}=\iota_{1}\circ f\text{ and }(f\oplus g)\circ
i_{2}=\iota_{2}\circ f$

As intended. Thus $f\oplus g$ must be the unique morphism (with uniqueness
given by an argument dual to lemma 1.2) satisfying this property. ∎

The final constructions the biproduct contributes are the diagonal map,
generalizing a group homomorphism $x\mapsto(x,x)$, and its dual the
codiagonal. The codiagonal is especially powerful, effectively generalizing
the notion of addition. Note, especially, that existence of these maps follows
from the existence of the biproduct.

######  Definition 1.3.

The diagonal map $d:A\to A\oplus A$ is the unique morphism given by
substituting the $A$ and the identity maps $1_{A}:A\to A$ into the categorical
definition of a product, such that $\pi_{1}\circ d=\pi_{2}\circ d=1_{A}$. The
codiagonal map $d^{\prime}:A\amalg A\to A$ is the dual of this notion for a
coproduct, with $d^{\prime}\circ\iota_{1}=d^{\prime}\circ\iota_{2}=1_{A}$.

These constructions, together with a few standard categorical definitions,
allow us to state the definition of an additive category.

######  Definition 1.4.

An additive category $\mathscr{A}$ satisfies three axioms:

  1. 1.

The category $\mathscr{A}$ has an object which is both initial and final,
called a zero object or 0. A morphism which factors through the zero object is
called a zero morphism, often denoted 0 as well by abuse of notation.

  2. 2.

For each pair of objects $X,Y$ in $\mathscr{A}$, there is a biproduct $X\oplus
Y$ which satisfies the definitions for both products and coproducts in such a
way that cannonical projections commute with cannonical inclusions. This
condition allows us to define an operation $+$ on the class
$\mathrm{Hom}_{\mathscr{A}}(X,Y)$ as the composition

$A\overset{d}{\longrightarrow}A\oplus A\overset{f\oplus
g}{\longrightarrow}B\oplus B\overset{d^{\prime}}{\longrightarrow}B$ (2)

Where $d$ is the diagonal map, $f,g$ is the map given in lemma 1.2, and
$d^{\prime}$ is the codiagonal map.

  3. 3.

For every object $A$ in the category $\mathscr{A}$, there is a morphism
$-1_{A}$ such that $1_{A}+(-1_{A})=0_{A}$, where $1_{A}$ is the identity map
on $A$, and $0_{A}$ is the morphism that factors through 0 (unique, as 0 is
initial and final). This condition ensures that $\mathrm{Hom}(A,B)$ is an
abelian group.

We will show that composition of morphisms distributes over the operation
defined by equation (2), and that the first two conditions give
$\mathrm{Hom}(A,B)$ under (2) the structure of a commutative monoid. From
this, we will demonstrate that condition 3. suffices to turn the hom-classes
into abelian groups. Before we can embark on the proof of this fact, however,
we must prove some results about the natures of the objects which we seek to
work with.

In any ring, the zero object (the identity of the additive group) satisfies
$0a=0b$ for any two objects $a$ and $b$. We will see that the zero object of a
category reflects this property, and indeed, does so uniquely. This fact will
be useful in our endeavor to demonstrate that zero is an additive identity.

######  Lemma 1.4.

The zero morphism $0:B\to B$ ”equalizes any two morphisms $f$ and $g$” - that
is, $0\circ f=0\circ g$ for any $f,g:A\to B$, for any $B$, and is the unique
morphism from $B$ to $B$ which equalizes any two morphisms from any $A$ to
$B$.

###### Proof.

For any two objects $A,B$ in $\mathscr{A}$, consider the unique morphism
$0:B\to B$ which factors through the zero object. Let $f,g:A\to B$. Then
$0\circ f$ is a morphism mapping $A$ to $B$ which factors through 0, and thus
is unique in this respect. However, $0\circ g$ is another such morphism, so
$0\circ f=0\circ g$. Conversely, suppose $z:B\to B$ has the property $z\circ
f=z\circ g$ for every $f,g:A\to B$, for any $A$. Then, in particular when
$A=B$ and $f=1_{B},g=0:B\to B$ we have $z\circ 1_{B}=0$ and thus $z=0$. Thus
$0$ is the unique morphism which equalizes every pair of morphisms with $B$ as
their target. ∎

The lemma above finds use in the following proof, which in turn will be used
to prove that the zero morphism is the additive identity for the monoid
$(\mathrm{Hom}_{\mathscr{A}}(A,B),+)$. Intuitively, we can think of this lemma
as stating that a projection followed by an inclusion is a morphism which
retains no information; the only information which retains no information
about morphisms composed with it is the zero morphism.

######  Lemma 1.5.

For a biproduct $B\oplus B$ in an additive category $\mathscr{A}$ with
canonical projections $\pi_{1},\pi_{2}$ and inclusions $\iota_{1},\iota_{2}$,
we have $\pi_{1}\circ\iota_{2}=\pi_{2}\circ\iota_{1}=0$.

###### Proof.

Let any pair of morphisms $f,g:A\to B$ for arbitrary $A$. Consider the
composition $\pi_{1}\circ\iota_{2}\circ f$. By lemma 1.3 we can rewrite this
as $\pi_{1}\circ g\oplus f\circ\iota_{2}$. But by lemma 1.2 we can rewrite
this again as $g\circ\pi_{1}\circ\iota_{2}$, which we can rewrite again by
lemma 1.2 applied once more as $\pi_{1}\circ g\oplus g\circ\iota_{2}$, which
by lemma 1.3 equals $\pi_{1}\circ\iota_{2}\circ g$. Thus
$\pi_{1}\circ\iota_{2}\circ f=\pi_{1}\circ\iota_{2}\circ g$, and so
$\pi_{1}\circ\iota_{2}:B\to B$ equalizes any pair of morphisms from any object
$A$ into $B$. Thus by lemma 1.4, $\pi_{1}\circ\iota_{2}=0$. A symmetric
argument holds for $\pi_{2}\iota_{1}$. ∎

The following lemma, and its dual (recorded below for completeness) form the
foundation for proving that composition distributes over $+$.

######  Lemma 1.6.

For any map $f:A\to B$ we have $(f\oplus f)\circ d=d\circ f$ where $(f\oplus
f)$ is as in lemma 1.2 and $d$ is the diagonal map.

###### Proof.

From the definition of $(f\oplus f)$ we have that it is the unique morphism
which satisfies $\pi_{1}\circ(f\oplus f)=\pi_{2}\circ(f\oplus f)=f\circ
p_{1}$, and from the definition of the diagonal we have that it is a morphism
which satisfies $\pi_{1}\circ d=\pi_{2}\circ d=1_{A}$. Thus we see that the
composition $(f\oplus f)\circ d$ is a morphism with $\pi_{1}\circ(f\oplus
f)\circ d=\pi_{2}\circ(f\oplus f)\circ d=f$. Substituting $A$ projecting via
$f$ twice onto $B$ into the definition for the biproduct $B\oplus B$
demonstrates that this morphism is unique with respect to this property.
However, $\pi_{1}\circ d\circ f=\pi_{2}\circ d\circ f=f$ by definition of $d$.
Thus $(f\oplus f)\circ d=d\circ f$. ∎

######  Lemma 1.7.

For any map $f:A\to B$ we have $d_{B}^{\prime}\circ(f\oplus f)=f\circ
d_{A}^{\prime}$, where $d^{\prime}_{A}$ is the codiagonal $d^{\prime}:A\oplus
A\to A$ and $d^{\prime}_{B}$ the codiagonal $d^{\prime}:B\oplus B\to B$.

###### Proof.

Dual to lemma 1.6 by lemma 1.3. ∎

We now have all the materials required to embark on the main proof of this
section. Although this result is mentioned offhand in [2] and in the exercises
of [1], the proof thereof is the author’s original work. We begin by proving
composition destributes over addition, which will assist us greatly in our
further proof.

######  Theorem 1.8.

For objects $A,B$ in an additive category $\mathscr{A}$:

  1. 1.

Composition of functions distributes over to $+$ defined by (2).

  2. 2.

The class $\mathrm{Hom}(A,B)$ has the structure of an abelian group with
operation $+$.

###### Proof.

The class $\mathrm{Hom}(A,B)$ is closed under the operation defined by (2), as
categories are closed under composition and each of the maps in the
composition exists for any two objects $A,B$ and any two morphisms $f,g:A\to
B$.

To show that composition distributes over addition, consider $(f+g)\circ h$
for $f,g:A\to B$, and $h:C\to A$.

$C\xrightarrow{h}A\xrightarrow{d}A\oplus A\xrightarrow{(f\oplus g)}B\oplus
B\xrightarrow{d^{\prime}}B$ (3)

By lemma 1.6 this is the same as

$C\xrightarrow{d}C\oplus C\xrightarrow{(h\oplus h)}A\oplus
A\xrightarrow{(f\oplus g)}B\oplus B\xrightarrow{d^{\prime}}B$

Which is exactly

$C\xrightarrow{d}C\oplus C\xrightarrow{(f\circ h\oplus g\circ h)}B\oplus
B\xrightarrow{d^{\prime}}B$

Which is the definition of $f\circ h+g\circ h$. Thus $(f+g)\circ h=f\circ
h+g\circ h$. Bilinearity in the other argument of the composition follows from
a dual argument invoking lemma 1.7. Next, suppose $f,g,h$ morphisms in
$\mathrm{Hom}(A,B)$. Consider the expression $(f+g)+h$, defined to be the
composition:

$A\xrightarrow{d}A\oplus A\xrightarrow{(f+g)\oplus h}B\oplus
B\xrightarrow{d^{\prime}}B$

But $f+g$ is defined to be $d^{\prime}\circ f\oplus g\circ d$, so we can
rewrite as:

$A\xrightarrow{d}A\oplus A\xrightarrow{d\oplus 1_{A}}(A\oplus A)\oplus
A\xrightarrow{(f\oplus g)\oplus h}(B\oplus B)\oplus
B\xrightarrow{d^{\prime},1_{B}}B\oplus B\xrightarrow{d^{\prime}}B$

But by lemma 1.1, specifically the naturality of the isomorphism, this is
equivalent to.

$A\xrightarrow{d}A\oplus A\xrightarrow{1_{A}\oplus d}A\oplus(A\oplus
A)\xrightarrow{f\oplus(g\oplus h)}B\oplus(B\oplus
B)\xrightarrow{1_{B},d^{\prime}}B\oplus B\xrightarrow{d^{\prime}}B$

Which is (by a symmetric argument to the above) equivalent to $f+(g+h)$, so +
is associative.

Now let $A$ be an object in $\mathscr{A}$. Consider the morphism $1_{A}\oplus
0\circ d$. Note that, by lemma 1.2, and the definition of $d$, $\pi_{1}\circ
1_{A}\oplus 0\circ d=1_{A}$ and $\pi_{2}\circ 1_{A}\oplus 0\circ d=0$.
Moreover, by substituting $A$ into the definition for $A\oplus A$ with the
maps $1_{A}$ and $0$, we obtain that this morphism must be unique. But
$\iota_{1}$ is another morphism with this property by lemma 1.5, so
$\iota_{1}=1_{A}\oplus 0\circ d$. Symmetrically, $\iota_{2}=0\oplus 1_{A}\circ
d$. Then $1_{A}+0=d^{\prime}\circ 1_{A}\oplus 0\circ
d=d^{\prime}\circ\iota_{1}=1_{A}$, by definition. Symmetrically,
$0+1_{A}=1_{A}$. This together with distributivity yield that $f+0=f$ for any
$A,B$ and any $f:A\to B$, with 0 the unique zero morphism from $A$ to $B$.
Associativity and identity give $\mathrm{Hom}(A,B)$ the structure of a monoid;
it will suffice to show that this monoid contains inverses for each morphism,
and that $+$ is commutative.

Let $f:A\to B$ for arbitrary objects $A,B$, and consider
$-f\coloneq(-1_{A})\circ f$. Then note that
$f+(-f)=f\circ(1_{A}+(-1_{A}))=f\circ 0=0$ by distributivity. This gives each
$\mathrm{Hom}$-class the structure of a group with operation $+$, which
composition distributes over. Finally, consider that if
$f,g\in\mathrm{Hom}(A,B)$ we have $-1_{B}\circ(f+g)=-1_{B}\circ f+-1_{B}\circ
g=-f+-g$, but also that $-1_{B}\circ(f+g)$ is the unique inverse of $f+g$,
which can also be written as $-g+-f$. Thus $-g+-f=-f+-g$ for every $f,g$, and
$\mathrm{Hom}(A,B)$ is an abelian group. ∎

Perhaps not obviously, this the operation defined by (2) is the only operation
on morphisms satisfying these properties. We direct the reader to Chapter 8.II
of [1] for a detailed proof of this fact, recording the theorem for
completeness.

######  Theorem 1.9.

Any addition of morphisms over which composition distributes is given by the
addition defined in (2).

####  1.2.1 Additive Functors

As we have now fully defined a particular type of categories, the necessary
rhythm of a mathematics text dictates we now define a corresponding type of
functors. For us, these functors take the form of additive functors, and (as
we will see) induce group homomorphisms on the morphisms in an additive
category.

######  Definition 1.5.

An additive functor is one which preserves biproducts.

This somewhat minimal definition contains all the information we require to
construct a group homomorphism; intuitively, since we constructed our group
operation using only the biproduct, this is reasonable.

######  Lemma 1.10.

Consider additive categories $\mathscr{A}$ and $\mathscr{B}$ and functor
$F:\mathscr{A}\to\mathscr{B}$. For morphisms $f$ and $g$, $F(f+g)=F(f)+F(g)$
if and only if $F$ is an additive functor.

###### Proof.

Consider $F(f+g)=F(d^{\prime}\circ(f\oplus g)\circ d)=F(d^{\prime})\circ
F(f\oplus g)\circ F(d)$. But since $F(A\oplus A)=F(A)\oplus F(A)$, we must
have that $F(d^{\prime})=\hat{d^{\prime}}$ where $\hat{d^{\prime}}:F(B)\to
F(B)\oplus F(B)$ is the appropriate codiagonal, as if $\hat{\iota}_{i}$ is the
$i$th projection off the biproduct $F(B)\oplus F(B)$, we have that
$\hat{d^{\prime}}\circ\hat{\iota_{1}}=F(d^{\prime})\circ
F(\iota_{1})=F(d^{\prime}\circ\iota_{1})=1_{B}$, and similarly
$\hat{d^{\prime}}\circ\hat{\iota_{2}}=F(d^{\prime})\circ
F(\iota_{2})=F(d^{\prime}\circ\iota_{2})=1_{B}$. Dually, $F(d)$ is the
diagonal map. Finally, a similar argument says $F(f\oplus g):A\oplus A\to
B\oplus B$ is a morphism with $F(f\oplus g)\circ\hat{\iota_{i}}=F(f\oplus
g)\circ F(\iota_{i})=F(f\oplus g\circ\iota_{i})=F(f)$ or $F(g)$ for $i=1,2$
respectively. This means $F(f\oplus g)=F(f)\oplus F(g)$, so
$F(f+g)=F(d^{\prime})\circ F(f\oplus g)\circ F(d)=\hat{d^{\prime}}\circ
F(f)\oplus F(g)\circ\hat{d}=F(f)+F(g)$, as intended.

Conversely, suppose $F(f+g)=F(f)+F(g)$ for any suitable morphisms $f,g$.
Consider $F(A\oplus B)$ with any $C$ projecting into $F(A)$ and $F(B)$ via
$f_{1}$ and $f_{2}$. Clearly $F(\pi_{1})$ and $F(\pi_{2})$ are morphisms from
$F(A\oplus B)$ to $F(A)$ and $F(B)$ respectively. Then $F(\pi_{1})\circ
F(\iota_{1})\circ f_{1}+F(\pi_{1})\circ F(\iota_{2})\circ
f_{2}=F(\pi_{1}\circ\iota_{1})\circ f_{1}+F(\pi_{1}\circ\iota_{2})\circ
f_{2}=F(1)\circ f_{1}+F(0)\circ f_{2}=F(0)\circ f_{1}=F(f_{1})$. A symmetric
argument holds for $f_{2}$, and a dual argument holds for inclusions. Finally,
$F(\pi_{i})\circ F(\iota_{i})=F(\pi_{i}\circ\iota_{i})=F(1)=1$, so $F(A\oplus
B)$ is a biproduct of $F(A)$ and $F(B)$ with projections $F(\pi_{1})$ and
$F(\pi_{2})$ and inclusions $F(\iota_{1})$ and $F(\iota_{2})$. Thus $F(A\oplus
B)$ is a biproduct of $F(A)$ and $F(B)$, and so $F$ preserves biproducts. ∎

###  1.3 Abelian Categories

We extend the notion of an additive category to capture more of the
information we require from an abelian group. In particular, we begin by
extending the notion of a kernel of a morphism. In a group, we are blessed
with the ability to consider the kernel as a subset of the domain of a
morphism; however, in a general category we have no such luxury. Indeed, the
only way to define a kernel elegantly is by specifying it through a universal
property pertaining to a certain morphism. The abstract nature of this
construction, however, is rewarded by the ability to dualize it to construct a
cokernel. We will come to understand that this cokernel encodes some intrinsic
information about quotient objects, allowing for a very neat extension of many
set-theoretic statements of algebra.

######  Definition 1.6.

Consider a morphism $f:A\to B$ in a category $\mathscr{A}$ with a zero object.
A kernel $\ker f:\operatorname{Ker}f\to A$ of $f$, if it exists, is a morphism
such that $f\circ\ker f=0$ and every morphism $g$ with $f\circ g=0$ factors
uniquely as $\ker f\circ a$. A cokernel
$\operatorname{cok}f:B\to\operatorname{Cok}f$, if it exists, is a morphism
with $\operatorname{cok}f\circ f=0$, and for any other $g$ with $g\circ f=0$,
$g$ factors uniquely as $a\circ\operatorname{cok}f$

For now, we refer to ”a kernel”; more development is necessary to demonstrate
that it is unique up to isomorphism. It is important to note that we associate
the object $\operatorname{Ker}f$ with any kernel $\ker f$, so that the
”kernel” we generally consider is effectively a pair $(\ker
f,\operatorname{Ker}f)$. Of course, the expression $\ker f$ alone captures
this information; the source and target of each morphism is data given in the
morphism itself.

To fully extend the notion of a group, we will need analogous qualifiers to
”injective” and ”surjective”. These will allow us to translate theorems proven
in the language of sets into the language of categories. These come in the
form of the terms monic and epi:

######  Definition 1.7.

A morphism $m:A\to B$ in a category $\mathscr{A}$ is monic if for any two
morphisms $f,g:A\to B$, whenever $m\circ f=m\circ g$ we can deduce that $f=g$
\- that is, when $m$ can be ”canceled on the left”. Similarly, a morphism
$e:A\to B$ in $\mathscr{A}$ is epi if for $f,g:A\to B$ we have $f\circ
e=g\circ e\implies f=g$, that is, $e$ can be canceled on the right.

When a category has a zero object, there is an alternate, equivalent
definition of monics and epis in terms of which morphisms they send to zero.
Aside from providing useful insight into how monics and epis function, this
equivalent definition is often useful when working with kernels and cokernels,
and will ultimately allow use the notion of an ”exact sequence” to fully
characterize monics and epis. Furthermore, this lemma continues to demonstrate
that monics generalize injective homomorphisms - a homomorphism is injective
if and only if it sends only the zero object to zero.

######  Lemma 1.11.

In an additive category, a morphism $m$ is monic if and only if $m\circ
f=0\implies f=0$, and a morphism $e$ is epi if and only if $f\circ e=0\implies
f=0$.

###### Proof.

First suppose $m$ is monic. Then $m\circ f=0=m\circ 0\implies f=0$ by
definition of monic. Conversely, suppose $m$ has the property $m\circ
f=0\implies f=0$ for any suitably composable $f$. Then consider $g,h$ with
$m\circ h=m\circ g$. Let $f=h-g$. Then $m\circ h-m\circ g=0$ by the definition
of $-m\circ g$. However, $0=m\circ h-m\circ g=m\circ(h-g)$ (by bilenarity).
Our assumption then implies that $h-g=0$. Then $h=g$ by adding $g$ to both
sides, and $m$ is monic. A symmetric argument holds for epis. ∎

######  Corollary 1.11.1.

A morphism $m$ is monic if and only if every kernel of it is zero. A morphism
$e$ is epi if and only if every cokernel of it is zero.

###### Proof.

Let $m$ be a monic. Clearly, $m\circ 0=0$; it will suffice to show that no
nonzero morphism $g$ satisfies $m\circ g=0$, but this follows from lemma 1.11.
Conversely, suppose $m$ has kernel 0. Then any morphism $f$ with $m\circ f=0$
factors through 0, and is therefore 0 by lemma 1.11. A dual argument proves
the lemma for epis. ∎

######  Corollary 1.11.2.

The kernel and cokernel of every isomorphism is 0.

As we are working with kernels and cokernels often, it is useful to manipulate
them, almost algebraically. We will use the following lemma frequently to
assist in translation of theorems from set-theoretic kernels to categorical
kernels. Intuitively, it establishes that every kernel of $f$ ”includes” into
the source of $f$ by way of our definition of monics; remarkably, this is
given exclusively by the universal property of kernels. A powerful element of
our definition for abelian categories will be inverting this conditional, so
that every monic is indeed the kernel of some morphism, and every epi a
cokernel.

######  Lemma 1.12.

In an additive category, every kernel is monic, and every cokernel is epi.

###### Proof.

Let $f:A\to B$ be a morphism in an additive category $\mathscr{A}$, and let
$f$ have a kernel $k:K\to A$. Suppose $k$ is not monic. Then there exist $g$
and $g^{\prime}$ such that $k\circ g=k\circ g^{\prime}$ with $g\neq
g^{\prime}$. However, $f\circ(k\circ g)=(f\circ k)\circ g=0\circ g=0$ and
$f\circ(k\circ g^{\prime})=(f\circ k)\circ g^{\prime}=0\circ g^{\prime}=0$, so
$k\circ g=k\circ g^{\prime}$ is a morphism which sends $g$ to zero and which
factors through $k$ in two ways, contradicting the definition of a kernel. A
symmetric argument holds for cokernels. ∎

The following lemma ensures that it is meaningful to refer to ”the kernel” of
a morphism, at least up to isomorphism. Although we have been referring to the
unique kernel till now, we have not used it’s uniqueness in any proof (so this
argument is not circular).

######  Lemma 1.13.

The kernel and cokernel are unique up to isomorphism.

###### Proof.

Suppose there are two kernels of a morphism $f$, $k$ and $k^{\prime}$. Then
$k$ must factor uniquely through $k^{\prime}$ as $k=k^{\prime}\circ i$, and
$k^{\prime}$ must factor uniquely through $k$ as $k^{\prime}=k\circ i$.
Substitution gives $k^{\prime}=k^{\prime}\circ i\circ i^{\prime}$. But the
kernel $k^{\prime}$ is monic by 1.12, so we can cancel to obtain $i\circ
i^{\prime}=1$. A symmetric argument shows the symmetric composition is the
identity, so $i$ is an isomorphism. Thus the two kernels differ only by a
factor of an isomorphism. A dual argument proves that the cokernel is unique
up to isomorphism. ∎

######  Corollary 1.13.1.

The object $\operatorname{Ker}f$ and $\operatorname{Cok}f$ are unique up to
isomorphism.

We can now define the main construction of this chapter, the abelian category.
This category will allow us to pursue variants of many proofs designed for
abelian groups, rings, fields, modules, vector spaces, and so on.

######  Definition 1.8.

An abelian category is an additive category $\mathscr{A}$ which satisfies the
following criterion:

  1. 1.

Every morphism in $\mathscr{A}$ has a kernel and a cokernel.

  2. 2.

Every monic is a kernel, and every epi a cokernel.

We will see that condition 2 effectively encodes the first isomorphism theorem
into our definition of abelian categories, by replacing subgroups with monics
and quotient groups with cokernels and epis.

######  Definition 1.9.

A subobject is a monic; a quotient object, an epi.

We often identify the subobject with it’s source, and the quotient object with
it’s target; this leads to the more intuitive notion of a subobject
”including” into an object. Furthermore, we often equate two subobjects which
differ by a factor of an isomorphism. There are times, however, when we wish
to take the quotient of two subobjects; for this we will need the following
definition. Note that the relation ”factors through” takes the place of ”is a
subobject of” in this definition; this is an important theme that will recur.

######  Definition 1.10.

If a monic $a$ factors through a monic $b$ as $a=b\circ f$, define the
quotient of subobjects $b/a$ to be the quotient object $\operatorname{cok}f$.

e have established that factoring is an important relation; we now proceed to
prove a plethora of results allowing us to easily and conveniently work with
this relation. Firstly, we present a factorization lemma motivating our
definition of ”image”, which will succeed the lemma.

######  Lemma 1.14.

Given a morphism $f:A\to B$ in an abelian category which factors as $m\circ e$
for a monic $m$ and epi $e$, then $m$ must factor through
$\ker(\operatorname{cok}f)$ and $e$ must factor through
$\operatorname{cok}(\ker f)$. Furthermore, any morphism which factors as
$g=f\circ h$ factors through $\ker(\operatorname{cok}f)$ and
$\operatorname{cok}(\ker h)$.

###### Proof.

Consider $\operatorname{cok}f\circ m\circ e=0$. By lemma 1.11, this means that
$\operatorname{cok}f\circ m=0$, which means that $m$ must factor through
$\ker(\operatorname{cok}f)$. A symmetric argument shows $e$ factors through
$\operatorname{cok}(\ker f)$. Suppose another morphism $g=f\circ h$; then
$\operatorname{cok}f\circ g=\operatorname{cok}f\circ f\circ h=0\circ h=0$ so
$g$ must factor through $\ker(\operatorname{cok}f)$. A symmetric argument
holds for the symmetric case. ∎

######  Corollary 1.14.1.

Any morphism $f$ in an abelian category factors through both
$\ker(\operatorname{cok}f)$ and $\operatorname{cok}(\ker f)$

###### Proof.

Consider the factorization $f=f\circ 1$ and $f=1\circ f$, where $1$ is an
identity. ∎

It is perhaps intuitive that the image should be represented by our
generalized categorical subobject, and perhaps intuitive again that $f$ should
factor through a representative of this subobject. This intuition is captured
in the following definition. The final result of this section, a stronger
version of lemma 1.14, will build on results developed from lemma 1.14.

######  Definition 1.11.

Consider a morphism $f:A\to B$ in an abelian category $\mathscr{A}$. Then the
subobject $\ker\operatorname{cok}f$ is the image of $f$, denoted
$\operatorname{im}f$.

Note further that
$\operatorname{im}f:\operatorname{Ker}(\operatorname{cok}f)\to B$; we denote
$\operatorname{Ker}(\operatorname{cok}f)$ as $\operatorname{Im}f$. Uniqueness
(up to isomorphism) is given by the uniqueness of the kernel. We will refer to
$\operatorname{im}(f)$ and $\ker(\operatorname{cok}f)$ interchangeably, often
working with $\ker(\operatorname{cok}f)$ for ease of manipulation and
understanding.

The dual definition to this is also useful;

######  Definition 1.12.

Consider a morphism $f:A\to B$ in an abelian category $\mathscr{A}$. Then the
quotient object $\operatorname{cok}\ker f$ is the coimage of $f$, denoted
$\operatorname{coim}g$.

The following lemma is a useful technical result to relate images and kernels
with coimages and cokernels. It will become especially useful during our
discussion of exact sequences, where relating the kernel and image is of vital
importance. The lemma also plays a crucial role in our final theorem of this
section, the factorization of each morphism in an abelian category.

######  Lemma 1.15.

Let $f:A\to B$ be a morphism in an abelian category. Then
$\ker(\operatorname{cok}(\ker f))$ and $\ker f$ differ by an isomorphism, as
do $\operatorname{cok}(\ker(\operatorname{cok}f))$ and $\operatorname{cok}f$.

###### Proof.

Consider that $\ker f$ is a monic, and by the corollary of lemma 1.14 factors
through $\ker(\operatorname{cok}(\ker f))$ uniquely. Conversely, $f\circ\ker
f=0$, so $f$ factors uniquely as $f=g\circ\operatorname{cok}(\ker f)$ by the
definition of cokernel. But then $f\circ\ker(\operatorname{cok}(\ker
f))=g\circ\operatorname{cok}(\ker f)\circ\ker(\operatorname{cok}(\ker
f))=g\circ 0=0$, so by the definition of kernel $\ker(\operatorname{cok}(\ker
f))$ factors uniquely through $\ker f$. Thus $\ker f$ and
$\ker(\operatorname{cok}(\ker f))$ differ by an isomorphism. A symmetric
argument shows $\operatorname{cok}(\ker(\operatorname{cok}f))$ differs by an
isomorphism from $\operatorname{cok}f$. ∎

######  Corollary 1.15.1.

In an abelian category, any monic $m$ differs by an isomorphism from its
image, and any epi $e$ differs by an isomorphism from its coimage.

###### Proof.

Every monic is the kernel of some $f$, so $m=\ker f$ differs by an isomorphism
from $\ker(\operatorname{cok}(\ker f))=\ker(\operatorname{cok}m)$. A symmetric
argument holds for epis. ∎

The final lemma in this section gives perhaps the strongest characterization
of a general morphism possible in an abelian category.

######  Lemma 1.16.

Given a morphism $f:A\to B$ in an abelian category, $f=m\circ e$, with
$m=\operatorname{im}f$ monic and $e=\operatorname{coim}f$ epi.

###### Proof.

Note that $\operatorname{cok}f\circ f=0$, but by the definition of
$\ker(\operatorname{cok}f)$, any morphism $g$ with $\operatorname{cok}f\circ
g=0$ must factor uniquely through $\ker(\operatorname{cok}f)$, so $f$ must
factor through $\ker(\operatorname{cok}f)$. Thus $f=m\circ e$ for
$m=\ker(\operatorname{cok}f)$, and $e$ some unique morphism. A tedious proof
from [1] (Chapter 8, Section 1, Lemma 1) shows that $e$ must be epi.111When
referencing this lemma, note that for a pair of morphisms $f,g:A\to B$, the
equalizer $e$ such that $ea=eb$ is exactly the kernel $k$ of the morphism
$f-g$, as the universal morphism which gives $k\circ(f-g)=0\implies k\circ
f-k\circ g=0\implies k\circ f=k\circ g$, and so any abelian category has
equalizers for every pair of suitable morphisms. Since $m$ is monic, $f\circ
t=0$ if and only if $e\circ t=0$ for any appropriate $t$. Thus $\ker f=\ker
e$, as any morphism $t$ which sends $f$ to zero also sends $e$ to zero. Then
since an epi $e$ is the cokernel of its kernel, $f=m\circ
e=m\circ\operatorname{cok}(\ker e)=\operatorname{cok}(\ker f)$ and we are
done. ∎

######  Corollary 1.16.1.

The object $\operatorname{Ker}(\operatorname{cok}f)$ is the object
$\operatorname{Cok}(\ker f)$ up to isomorphism.

##  2 Homology, Exact Sequences, and Right Derived Functors

###  2.1 Sequences and Homology

We begin by constructing the main object we will work with, a sequence of
objects.

######  Definition 2.1.

A sequence is an ordered collection of objects in an abelian category
$\mathscr{A}$, together with a morphism from each object to it’s successor
called the differentials. The composition of successive differentials must
also be 0.

The crucial element of this construction is that the composition of successive
maps is zero. This, ultimately, allows us to construct quotients; intuitively,
speaking in group- or set-theoretic terms, the image of one map is then a
subset of the kernel of the next. Since both the image and the kernel are
subgroups, we can then form the quotient of the two - which, ultimately, is
the homology group. From a categorical perspective, we must express this by
way of factoring, which the following lemma does neatly.

######  Lemma 2.1.

Let $f,g$ be morphisms in an abelian category. If $f\circ g=0$, then
$\operatorname{im}g$ factors through the kernel of $f$ as
$\operatorname{im}a^{\prime}$, and $\operatorname{coim}f$ factors as
$a^{\prime}\circ\operatorname{cok}g$ for some $a^{\prime}$.

###### Proof.

Note that if $f\circ g=0$, then $f$ factors through $\operatorname{cok}g$ as
$f=a\circ\operatorname{cok}g$ for some morphism $a$. Then
$(a\circ\operatorname{cok}g)\circ\ker(\operatorname{cok}g)=0$, so
$\ker(\operatorname{cok}g)$ factors through
$\ker(a\circ\operatorname{cok}g)=\ker f$ as $\ker(\operatorname{cok}g)=\ker
f\circ a$. Similarly, note that $g$ factors as $g=\ker f\circ a^{\prime}$.
Then $\operatorname{cok}(\ker f)\circ(\ker f\circ a)=0$, so
$\operatorname{cok}(\ker f)$ factors through $\operatorname{cok}(\ker f\circ
a)=\operatorname{cok}g$ as $\operatorname{cok}(\ker
f)=a^{\prime}\circ\operatorname{cok}g$. ∎

We now distinguish between chain complexes and their dual, cochain complexes.
As we build towards a construction of right derived functors, we will focus
primarily on cochain complexes. However, every construction we perform can be
dualized by passing into the opposite category to apply to chain complexes.

######  Definition 2.2.

If objects in the sequence are numbered such that the maps take object $n$ to
object $n-1$, the sequence is called a chain complex. If maps take object $n$
to object $n+1$, the sequence is called a cochain complex.

We will work primarily with cochain complexes, which we will denote by a
capital letter with a superscript bullet, $A^{\bullet}$. For a cochain complex
$A^{\bullet}$, the individual objects will be denoted $A^{1},A^{2},$ etc.
Having defined an object, it is now preferable to define a map between two
such objects.

######  Definition 2.3.

Let $A^{\bullet}$ and $B^{\bullet}$ be cochain complexes with differentials
$d^{n}$ and $\partial^{n}$. A cochain map $f$ is a $(f^{n})_{n\in\mathbb{Z}}$
such that $f^{n}:A^{n}\to B^{n}$ and $f^{n}\circ d^{n}=\partial^{n}\circ
f^{n-1}$.

As we now have objects and morphisms, we can construct a category. What is
perhaps more interesting is that this category inherent the abelian properties
of the category from which it is constructed. This instantly admits the
interesting idea of higher order constructions, chain complexes of chain
complexes which (inductively) will all be abelian categories. Already, we see
the benefits of our general development; although it is beyond the scope of
this paper, we could instantly apply every result proven heretofore to the
category of cochain complexes of an abelian category, without any loss of
rigor.

######  Theorem 2.2.

Given an abelian category $\mathscr{A}$, the category
$\textbf{CCH}(\mathscr{A})$ with objects cochain complexes of objects in
$\mathscr{A}$ and morphisms cochain maps is an abelian category, with
composition defined componentwise.

###### Proof.

The categorical axioms (morphisms, objects, associativity, identity) follow
from the categorical definitions in $\mathscr{A}$, using the fact that
composition is defined componentwise. Similarly, a short check reveals that
$A^{\bullet}\oplus B^{\bullet}\cong(A\oplus B)^{\bullet}$; so
$\textbf{CCH}(\mathscr{A})$ has biproducts, and the zero complex is one in
which every object is the zero object of $\mathscr{A}$. Inverses then follow
from the inverses in the additive category $\mathscr{A}$; define
$(-f)^{n}=-(f^{n})$, so that $(f-f)=f^{n}-f^{n}=0$. We have thus fulfilled the
definition of an additive category $\textbf{CCH}(\mathscr{A})$

It will suffice to show that $\textbf{CCH}(\mathscr{A})$ has kernels and
cokernels, and that all kernels are monic and all cokernels are epi. Consider
a cochain map $f:A^{\bullet}\to B^{\bullet}$. By definition, $f^{n}\circ\ker
f^{n}=0$. Similarly, if the $n$th component of any other cochain map sends $f$
to zero when composed on the right, that component then factors uniquely
through $\ker f^{n}$. As such, we construct the kernel of $f$ componentwise
with $(\ker f)^{n}=\ker f^{n}$, and this satisfies the universal property of
kernels. A dual construction yields a dual definition for cokernels. It will
suffice to show that monics are cochain maps $f$ where each $f^{n}$ is monic.
But if each $f^{n}$ is monic, then for any suitably composable chain maps
$g,h$ we have $f\circ g=f\circ h\implies(f\circ g)^{n}=(f\circ h)^{n}\implies
f^{n}\circ g^{n}=f^{n}\circ h^{n}\implies g^{n}=h^{n}$, as $f^{n}$ monic.
However, two cochain maps which agree in every component are the same, so
$g=h$ and $f$ is monic. A symmetric argument holds for epis, and since kernels
are monic and cokernels are epi in $\mathscr{A}$, the proposition holds. ∎

We now develop what is perhaps the main construction in this paper, the
cohomology object. If we were working in the category of abelian groups, or
the category of $R$-modules, the kernel of the differential would be a
subgroup or sub-module of the image, and we could take the quotient of the
two. As we are working in a general category, we replace the sets ”image” and
”kernel” with the morphisms $\ker(\operatorname{cok}d^{n})$ and $\ker
d^{n+1}$; we replace ”is a subset of” with ”factors through”, and we replace
quotients with cokernels of an appropriate morphism, or a quotient object of
two subobjects.

######  Definition 2.4.

Let $A^{\bullet}$ be a cochain complex with differentials $d^{n}$. Then for
each $d^{n}$, lemma 2.1 states that $\ker(\operatorname{cok}f)$ factors as
$\ker(\operatorname{cok}f)=\ker f\circ a$. The cohomology object
$H^{n}(A^{\bullet})$ is the quotient object $\operatorname{cok}a=\ker
d^{n+1}/\operatorname{im}d^{n}$

Note the cokernel $\operatorname{cok}a$ effectively measures how much
$\ker(\operatorname{cok}d^{n})$ differs from $\ker d^{n+1}$. Moreover, the
cokernel in general is a sort of ”universal epi”; in set theoretic terms, a
”universal surjection”. As such, it makes intuitive sense that it represents
the quotient; it specifies the largest object which sends a morphism to zero,
capturing all information left after a morphism has acted. We will see that
the homology group is truly a representation of the failure of a sequence to
be exact.

######  Definition 2.5.

A sequence is exact at an object $B$ with morphisms $f:A\to B$ and $g:B\to C$
if the subobject associated with $\ker\operatorname{cok}f$ is equal to the
subobject associated with $\ker g$, and a sequence is said to be exact if it
is exact at every object.

As hinted at above, we see that exactness represents when
$\ker(\operatorname{cok}d^{n})$ ”is effectively the same as” (differs by a
factor of an isomorphism from) $\ker d^{n+1}$; a homology group ”measures”
this failure. A simple way to verify this assertion is to examine the effect
of homology on an exact sequence:

######  Lemma 2.3.

If a cochain complex is exact at $A^{n}$, the cohomology object
$H^{n}(A^{\bullet})$ is the zero morphism.

###### Proof.

It will suffice to show that, for any two monics $f$ and $g$, if $f$ differs
by an isomorphism from $g$, $f/g=0$. Note that if $f=g\circ i$ for an
isomorphism $i$, by the definition of a quotient, $f/g=\operatorname{cok}i$.
But $\operatorname{cok}i=0$ by corollary 1.11.2, and we are done. ∎

We wish, ultimately, to turn cohomology into a functor. Ultimately, given two
cochain complexes $A^{\bullet}$ and $B^{\bullet}$, with a cochain map
$f=(f^{n})$ between them, we seek to find morphisms from the $n$th cohomology
of the first cochain map to the $n$th cohomology of the second in a functorial
manner. We go about this by a construction drawing very heavily on one from
[5].

######  Lemma 2.4.

Given cochain complexes $A^{\bullet}$ and $B^{\bullet}$ and a cochain map
$f:A^{\bullet}\to B^{\bullet}$, we can construct the cohomology map
$H^{n}(f):H^{n}(A^{\bullet})\to H^{n}(B^{\bullet})$.

When examining the following proofs, it may be useful to refer to a diagram…

Figure 1: Constructing the Homology of a Morphism

---
$\textstyle{\operatorname{Im}d^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{\operatorname{im}d^{n}}$$\textstyle{\operatorname{Ker}d^{n+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{cok}a}$$\scriptstyle{\ker
d^{n+1}}$$\textstyle{\operatorname{Cok}a}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
H^{n}(A^{\bullet})}$$\textstyle{A^{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{coim}d^{n}}$$\scriptstyle{f^{n-1}}$$\scriptstyle{d^{n}}$$\textstyle{A^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{n-1}}$$\scriptstyle{d^{n+1}}$$\textstyle{A^{n+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f^{n-1}}$$\scriptstyle{d^{n+2}}$$\textstyle{...}$$\textstyle{B^{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{coim}\partial^{n}}$$\scriptstyle{\partial^{n}}$$\textstyle{B^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial^{n+1}}$$\textstyle{B^{n+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial^{n+2}}$$\textstyle{...}$$\textstyle{\operatorname{Im}\partial^{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a^{\prime}}$$\scriptstyle{\operatorname{im}\partial^{n}}$$\textstyle{\operatorname{Ker}\partial^{n+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{cok}a^{\prime}}$$\scriptstyle{\ker\partial^{n+1}}$$\textstyle{\operatorname{Cok}a^{\prime}}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
H^{n}(B^{\bullet})}$

###### Proof.

Let $d^{n}$ denote the differential of $A^{\bullet}$, and $\partial^{n}$ the
differential of $B^{\bullet}$. By lemma 2.1, we know that
$\operatorname{im}d^{n}$ factors as $\ker d^{n+1}\circ a$ for some
$a:\operatorname{Im}d^{n}=\operatorname{Cok}(\ker
d^{n})\to\operatorname{Ker}d^{n+1}$. Similarly,
$\operatorname{im}\partial^{n}$ factors as $\ker\partial^{n+1}\circ
a^{\prime}$ for some
$a^{\prime}:\operatorname{Cok}(\ker\partial^{n})\to\operatorname{Ker}\partial^{n+1}$
by lemma 2.1. In particular, note that
$H^{n}(A^{\bullet})=\operatorname{Cok}a$ and
$H^{n}(B^{\bullet})=\operatorname{Cok}a^{\prime}$.

We will begin by constructing maps $\alpha:\ker d^{n+1}\to\ker\partial^{n+1}$
and $\beta:\operatorname{Im}d^{n}\to\operatorname{Im}\partial^{n}$ which
commute with $a$ and $a^{\prime}$. This will eventually allow us to construct
our maps between cohomology objects. Note that the map $\partial^{n+1}\circ
f^{n}\circ\ker d^{n+1}=f^{n+1}\circ d^{n+1}\circ\ker d^{n+1}=0$. Thus,
$f^{n}\circ\ker d^{n+1}$ factors as $\ker\partial^{n+1}\circ\alpha$ where
$\alpha:\ker d^{n+1}\to\ker\partial^{n+1}$. Similarly, note that the map
$\operatorname{cok}\partial^{n}\circ\partial^{n}\circ f^{n-1}$ is zero
identically. However, by the definition of a chain map this is equivalent to
$\operatorname{cok}\partial^{n}\circ f^{n}\circ d^{n}$. Since $d^{n}$ factors
as $\ker d^{n+1}\circ a\circ\operatorname{coim}d^{n}$ (by lemma 1.16 and the
definition of $a$; see figure 1 for a visual representation) we have that

$\operatorname{cok}\partial^{n}\circ f^{n}\circ
d^{n}=\operatorname{cok}\partial^{n}\circ f^{n}\circ\ker d^{n+1}\circ
a\circ\operatorname{coim}d^{n}$

But since this whole expression was zero from it’s construction, and
$\operatorname{coim}d^{n}$ is epi, we have that
$\operatorname{cok}\partial^{n}\circ f^{n}\circ\ker d^{n+1}\circ a$ is zero.
Finally, this means that $f^{n}\circ\ker d^{n+1}\circ a$ factors as
$\operatorname{im}f\circ\beta$. Finally, we observe that
$\ker\partial^{n+1}\circ\alpha\circ a=f^{n+1}\circ\alpha\circ
a=\ker(\operatorname{cok}\partial^{n})\circ\beta=\ker\partial^{n+1}\circ
a^{\prime}\circ\beta$. However, since $\ker\partial^{n+1}$ is a monic, we can
cancel to obtain $a^{\prime}\circ\beta=\alpha\circ a$.

This allows us to note that $\operatorname{cok}a^{\prime}\circ\alpha\circ
a=\operatorname{cok}a^{\prime}\circ a^{\prime}\circ\beta=0$, so by the
definition of a cokernel we establish that
$\operatorname{cok}a^{\prime}\circ\alpha$ factors uniquely as
$h\circ\operatorname{cok}a$, where
$h:\operatorname{Cok}a\to\operatorname{Cok}a^{\prime}$. Define $H^{n}(f)=h$ to
obtain the desired result. ∎

We will refer back to this lemma frequently in our next one, demonstrating
that such a construction is sufficiently ”nice” to demonstrate that cohomology
is an additive functor. The following theorem is also loosely based on
proposition 3.1 in [5], dualized for cohomology.

######  Theorem 2.5.

$H^{n}:\textbf{CCH}(\mathscr{A})\to A$ is an additive functor.

###### Proof.

First we must show that $H^{n}$ takes identity morphisms to identity
morphisms. Consider $H^{n}(1)$ for an identity morphism $1$. Then, referring
back to figure 1 and the definitions in lemma 2.4, we see that $\alpha$ in the
definition of $H^{1}(1)$ must be a morphism with $\ker
d^{n+1}=\ker\partial^{n+1}\circ\alpha$. But in this case,
$\ker\partial^{n+1}=\ker d^{n+1}$, so $\ker d^{n+1}=\ker d^{n+1}\circ\alpha$.
But $\ker d^{n+1}$ is an epi; canceling gives us $1=\alpha$. Then
$\operatorname{cok}a^{\prime}\circ\alpha=H^{n}(1)\circ\operatorname{cok}a$, so
$\operatorname{cok}a^{\prime}=H^{n}(1)\circ\operatorname{cok}a$. But since
$d^{n}=\partial^{n}$, $a=a^{\prime}$; since
$\operatorname{cok}a=\operatorname{cok}a^{\prime}$ is an epi, canceling gives
$H^{n}(1)=1_{H^{n}(A^{\bullet})}$.

Now we must show that $H^{n}$ preserves composition. Consider $H^{n}(f\circ
g)$, for cochain maps $f:A\to B$ and $g:B\to C$, denoting the differentials of
$A,B$, and $C$ as $d,\delta,$ and $\partial$ respectively. Consider that
$H^{n}(g\circ f)$ is the unique homomorphism with
$\operatorname{cok}a^{\prime\prime}\circ\alpha_{g}\circ\alpha_{f}=H^{n}(f\circ
g)\circ\operatorname{cok}a$, where $\alpha_{f}$ and $\alpha_{g}$ are the
$\alpha$ given in the proof of lemma 2.4, applied to $f$ and $g$ respectively.
Similarly, $a:\ker(\operatorname{cok}d^{n})\to\ker d^{n+1}$,
$a^{\prime}:\ker(\operatorname{cok}\delta^{n})\to\ker\delta^{n+1}$, and
$a^{\prime\prime}:\ker(\operatorname{cok}\partial^{n})\to\ker\partial^{n+1}$,
as in proof of lemma 2.4. But then $H^{n}(f)\circ
H^{n}(g)\circ\operatorname{cok}a=H^{n}(f)\circ
a^{\prime}\circ\alpha_{g}=a^{\prime\prime}\circ\alpha_{f}\circ\alpha_{g}$, so
$H^{n}(f\circ g)=H^{n}(f)\circ H^{n}(g)$.

Finally, it will suffice to show $H^{n}(f+g)=H^{n}(f)+H^{n}(g)$. But if all
morphisms are as in the proof of lemma 2.4, with $\alpha$ for $f$ and
$\alpha^{\prime}$ for $g$ (similarly for $\beta$), we have that
$\operatorname{cok}a^{\prime}\circ(\alpha+\alpha^{\prime})\circ\operatorname{cok}(\ker
d^{n})=\operatorname{cok}a^{\prime}\circ\alpha\circ\operatorname{cok}(\ker
d^{n})+\operatorname{cok}a^{\prime}\circ\alpha^{\prime}\circ\operatorname{cok}(\ker
d^{n})$ by bilenearity. Both of these are equal to
$\operatorname{cok}a^{\prime}\circ a^{\prime}\circ(\beta+\beta^{\prime})=0$;
as such, another application of distributivity yields that
$\operatorname{cok}a^{\prime}\circ\alpha+\operatorname{cok}a^{\prime}\circ\alpha^{\prime}=\operatorname{cok}a^{\prime}\circ(\alpha+\alpha^{\prime})=H^{n}(f+g)\circ\operatorname{cok}a$.
A final application of distributivity shows that
$(H^{n}(f)+H^{n}(g))\circ\operatorname{cok}a=H^{n}(f)\circ\operatorname{cok}a+H^{n}(g)\circ\operatorname{cok}a$,
the definition of $H^{n}(f)$ simplifies this to
$\operatorname{cok}a^{\prime}\circ\alpha+\operatorname{cok}a^{\prime}\circ\alpha^{\prime}$,
and the rest follows by uniqueness. ∎

We have now fully defined cohomology as an additive functor, a very powerful
characterization and one that will lead directly to the construction of
derived functors, and the proof of several instances of their well-
definedness. The next tool that we will need to complete our definition is an
injective resolution. This will allow us to replace a single, complicated
object with a simple resolution which we understand well, a technique which
forms the basis for derived functors.

###  2.2 Injective Resolutions

####  2.2.1 More on Exact Sequences

Before we construct our injective resolutions, we will need a few more lemmas
to prove further statements. The first of these dualizes the definition of an
exact sequence:

######  Lemma 2.6.

If a sequence $A\xrightarrow{f}B\xrightarrow{g}C$ is exact at $B$, then
$\operatorname{cok}f=a\circ\operatorname{cok}(\ker g)$

###### Proof.

By lemma 2.1, $\operatorname{cok}(\ker g)$ factors through
$\operatorname{cok}f$. By exactness, $\ker g=\ker(\operatorname{cok}f)\circ i$
for an isomorphism $i$; since
$\operatorname{cok}f\circ\ker(\operatorname{cok}f)\circ i=0$, we have that
$\operatorname{cok}f$ must factor through
$\operatorname{cok}(\ker(\operatorname{cok}f)\circ i)=\operatorname{cok}(\ker
g)$ (and must do so as $a\circ\operatorname{cok}(\ker g)$). ∎

Finally, we give one more lemma allowing us to characterize monics and epis
using only exact sequences.

######  Lemma 2.7.

The sequence

$0\xrightarrow{0}A\xrightarrow{f}B$

is exact if and only if $f$ is monic. The sequence

$A\xrightarrow{f}B\xrightarrow{0}0$

is exact if and only if $f$ is epi.

###### Proof.

Suppose $0\xrightarrow{0}A\xrightarrow{f}B$ is exact. Then
$\ker\operatorname{cok}0$ differs by an isomorphism from $\ker f$. Clearly the
cokernel of 0 is the identity on $A$, as every morphism composed with zero is
zero, and each morphism which factors through $A$ factors through $1_{A}$. But
the kernel of an identity is 0 by corollary 1.11.2 thus the kernel of $f$
differs by an isomorphism from zero, and as such is zero. By corollary 1.11.1,
this means that $f$ is a monic. Conversely, suppose $f$ is monic. Then the
kernel of $f$ is zero by corollary 1.11.1. But the cokernel of $0$ is $1_{A}$,
as shown above, and the kernel of $1_{A}$ is zero by corollary 1.11.2. Thus,
$\ker\operatorname{cok}0=0=\ker f$, and the sequence is exact. A dual argument
holds for epis. ∎

We now define projective and injective objects and resolutions. Although this
definition may seem obscure, it is necessary to build resolutions of objects
in a way that captures some representation their internal structure.

######  Definition 2.6.

A projective object $P$ is one which, for any epi $f:B\to C$ together with a
map $\gamma:P\to Y$, there is a $\beta:P\to B$ such that $f\circ\beta=\gamma$.
Similarly, an injective object $I$ is one which, for any monic $g:A\to B$ and
any $\alpha:A\to I$ there is a $\beta:B\to I$ such that $\beta\circ g=\alpha$.
An abelian category $\mathscr{A}$ has ”enough projectives” if and only if for
any object $A$ there is a epi $f:P\to A$ with $P$ projective. Similarly,
$\mathscr{A}$ has enough injectives if and only if for any object $A$ there is
a monic $f:A\to I$ with $I$ injective.

######  Definition 2.7.

An injective (resp. projective) resolution of an object $A$ in an abelian
category $\mathscr{A}$ is an exact cochain complex (resp. chain complex)

$0\to A\to I^{0}\to I^{1}\to I^{2}\to I^{3}\to...$ $\text{(respectively
}...\to P_{3}\to P_{2}\to P_{1}\to A\to 0\text{ )}$

with each $I^{i}$ injective ($P_{i}$ projective).

Given minimal conditions outlined in the above definitions, we can always form
injective resolutions. Eventually, this will guarantee the existence of the
derived functor; more generally, it permits a plethora of techniques (most of
which are beyond this paper) which can be used to simplify complicated
objects.

######  Theorem 2.8.

If an abelian category $\mathscr{A}$ has enough injectives, we can form an
injective resolution of any object $A$.

###### Proof.

Since $\mathscr{A}$ has enough injectives, there is a monic $d^{0}:A\to I^{0}$
for some $I^{0}$. Consider the cokernel
$\operatorname{cok}d^{0}:I^{0}\to\operatorname{Cok}d^{0}$. This cokernel has a
target $\operatorname{Cok}d^{0}$, which in turn admits a monic
$a^{0}:\operatorname{Cok}d^{0}\to I^{1}$, as $\mathscr{A}$ has enough
injectives. Define $d^{1}$ to be the composition
$a^{0}\circ\operatorname{cok}d^{0}$, and proceed inductively. Supposing
$d^{n}$ exists, let $a^{n}$ be the monic mapping $\operatorname{Cok}d^{n}$ to
some injective which we denote $I^{n+1}$, as given by the ”enough injectives”
condition. Define $d^{n+1}:I^{n}\to I^{n+1}$ to be $d^{n+1}\coloneq
a^{n}\circ\operatorname{cok}d^{n}$. This inductively defines an injective
resolution:

$0\xrightarrow{0}A\xrightarrow{d^{0}}I^{0}\xrightarrow{d^{1}}I^{1}\xrightarrow{d^{2}}I^{2}\xrightarrow{d^{3}}...$

With each $I$ injective, and $d^{n+1}\circ
d^{n}=a^{n}\circ\operatorname{cok}d^{n}\circ d^{n}=a^{n}\circ 0=0$. These
conditions ensure that $A\to I^{\bullet}$ is a cochain complex. Since $d^{0}$
is monic, lemma 2.7 guarantees that the sequence is exact at $A$. To show it
is exact at each $I^{n}$, note that each $\operatorname{im}d^{n}$ factors
through $\ker d^{n+1}$ by lemma 2.1. Conversely,
$d^{n+1}=a^{n}\circ\operatorname{cok}d^{n}$ with $a^{n}$ monic and
$\operatorname{cok}d^{n}$ epi, so $\operatorname{cok}d^{n}$ factors as
$f\circ\operatorname{coim}d^{n+1}$ by lemma 1.14. But then
$\operatorname{cok}d^{n}\circ\ker
d^{n+1}=f\circ\operatorname{coim}d^{n+1}\circ\ker d^{n+1}=0$, so $\ker
d^{n+1}$ factors through $\operatorname{im}d^{n}$. Thus $\ker d^{n+1}$ and
$\operatorname{im}d^{n}$ differ by an isomorphism, and so $\ker d^{n+1}$
differs by an isomorphism from $\operatorname{im}d^{n}$. Therefore
$I^{\bullet}$ is an exact cochain complex of injective objects, or an
injective resolution. ∎

A dual construction can be made for projective resolutions; however, this
paper does not concern them, so it is omitted for brevity. There is no
guarantee that this injective resolution is unique, which means it is
impossible to use it to define a functor. However, our next lemma provides a
way of relating two such injective resolutions. First, however, we must define
a certain equivalence of cochain maps, which will allow us to compare such
maps. Eventually, this equivalence will prove to induce equality under
homology.

######  Definition 2.8.

Let $f,g:A^{\bullet}\to B^{\bullet}$, where $A^{\bullet}$ has differentials
$d^{n}$ and $B^{\bullet}$ has differentials $\partial$. Two cochain maps $f$
and $g$ are homotopic if there is an $s^{i}:I^{i}\to J^{i-1}$ such that
$f^{n}-g^{n}=\partial^{n}\circ s^{n-1}+s^{n}\circ d^{n+1}$.

######  Theorem 2.9 (Comparison Lemma, adopted from [4]).

For objects $A$ and $B$ in an abelian category with enough injectives, and
complexes of injective resolutions $A\to I^{\bullet}$ and $B\to J^{\bullet}$,
a map $f:A\to B$ induces a cochain map $f^{i}:I^{i}\to J^{i}$ with $f^{n}\circ
d^{n}=\partial^{n}\circ f^{n-1}$ (where $d^{n}$ is the differential of
$I^{\bullet}$ and $\partial^{n}$ the differential of $J^{\bullet}$). Any two
cochain maps $f^{i}$ and $g^{i}$ induced by $f$ are homotopic.

###### Proof.

For the base case $n=0$, the differentials $d^{0}:A\to I^{0}$ and
$\partial^{0}:B\to J^{0}$ are monic by lemma 2.7, so applying the definition
of an injective object to $d^{0}$ and the composition $\partial^{0}\circ f$
gives a morphism $f^{0}:I^{0}\to J^{0}$ with $f^{0}\circ
d^{0}=\partial^{0}\circ f$. We then construct each subsequent $f^{n}$
inductively, assuming $f^{i}$ for $-1\leq i\leq n$ are constructed such that
$f^{n}\circ d^{n}=\partial^{n}\circ f^{n-1}$.

Note that $\partial^{n+1}\circ f^{n}\circ
d^{n}=\partial^{n+1}\circ\partial^{n}\circ f^{n-1}=0$, so
$(\partial^{n+1}\circ f^{n})$ factors through $\operatorname{cok}(d^{n})$,
which by the exactness of the injective resolution and lemma 2.1 factors
through $\operatorname{cok}(\ker(d^{n+1}))$. But then we can write
$(\partial^{n+1}\circ f^{n})$ as $\eta\circ\operatorname{cok}(\ker d^{n+1})$
for some morphism $\eta:\operatorname{Ker}(\operatorname{cok}d^{n+1})\to
J^{n+1}$ by lemma 2.6. Applying the definition of the injective object
$J^{n+1}$ with morphisms
$\eta:\operatorname{Ker}(\operatorname{cok}d^{n+1})\to J^{n+1}$ and monic
$\ker(\operatorname{cok}d^{n+1}):\operatorname{Ker}(\operatorname{cok}d^{n+1})\to
I^{n+1}$ gives a morphism $\beta:I^{n+1}\to J^{n+1}$ with
$\beta\circ\ker(\operatorname{cok}d^{n+1})=\eta$, which (composing on the
right) gives
$\beta\circ\ker(\operatorname{cok}d^{n+1})\circ\operatorname{cok}(\ker
d^{n+1})=\beta\circ d^{n+1}=\eta\circ\operatorname{cok}(\ker
d^{n+1})=\partial^{n+1}\circ f^{n}$. Defining $f^{n+1}\coloneq\beta$ gives us
our intended construction.

It will thus suffice to show uniqueness up to homotopy equivalence. Let
$f^{i}$ and $g^{i}$ be maps from $I^{i}$ to $J^{i}$, satisfying the necessary
conditions. Construct terms of a map $s^{i}$ inductively. The base case can be
shown easily by letting $s^{-1}$ and $s^{0}$ both be zero. Inductively, assume
that $f^{n}-g^{n}=\partial^{n}\circ s^{n-1}+s^{n}\circ d^{n+1}$.

Note that $(f^{n+1}-g^{n+1}-\partial^{n+1}\circ s^{n})\circ
d^{n+1}=(f^{n+1}-g^{n+1})\circ d^{n+1}-\partial^{n+1}\circ s^{n}\circ
d^{n+1}$. By the definition of $s^{n}$ we can rewrite as
$(f^{n+1}-g^{n+1})\circ d^{n+1}-\partial^{n+1}(f^{n}-g^{n}-\partial^{n}\circ
s^{n-1})$. Expanding, we obtain $f^{n+1}\circ d^{n+1}-g^{n+1}\circ
d^{n+1}-\partial^{n+1}\circ f^{n}+\partial^{n+1}\circ
g^{n}+\partial^{n+1}\circ\partial^{n}\circ s^{n-1}$. The last term cancels,
and we re-arange (by the fact $\mathrm{Hom}(I^{n},J^{n+1})$ is an abelian
group) to obtain:

$f^{n+1}\circ d^{n+1}-\partial^{n+1}\circ f^{n}-g^{n+1}\circ
d^{n+1}+\partial^{n+1}\circ g^{n}$

But by the definition of a cochain map, $f^{n+1}\circ
d^{n+1}=\partial^{n+1}\circ f^{n}$, and a similar equivalence holds for $g$,
so the whole sum is zero. Thus, $\alpha\coloneq
f^{n+1}-g^{n+1}-\partial^{n+1}\circ s^{n}$ factors by the definition of a
cokernel as $\alpha=a\circ\operatorname{cok}d^{n+1}$. But by exactness and
lemma 2.6, $\alpha=\eta\circ\operatorname{cok}\ker d^{n}$. We then apply the
definition of an injective object, noting that
$\eta:\operatorname{Ker}\operatorname{cok}d^{n+1}\to J^{n}$, and that
$\ker(\operatorname{cok}d^{n+1}):\operatorname{Ker}\operatorname{cok}d^{n+1}\to
I^{n+1}$ is an injection. Thus the definition of injectivity gives a morphism
$\beta:I^{n+1}\to J^{n}$ with
$\beta\circ\ker(\operatorname{cok}d^{n+1})=\eta$. Composing
$\operatorname{cok}(\ker d^{n+1})$ on the right gives us that
$\beta\circ\ker(\operatorname{cok}d^{n+1})\circ\operatorname{cok}(\ker
d^{n+1})=\eta\circ\operatorname{cok}(\ker d^{n+1})$. But then lemma 1.16 and
the definition of $\eta$ gives us $\beta\circ
d^{n+1}=f^{n+1}-g^{n+1}-\partial^{n+1}\circ s^{n}$, or that $\beta\circ
d^{n+1}+\partial^{n}\circ s^{n-1}=f^{n}-g^{n}$. We thus define
$s^{n}\coloneq\beta$, and we have constructed a cochain homotopy inductively.
∎

###  2.3 Derived Functors

This brings us to our final construction, the derived functor. Briefly, the
right derived functor is a method of ”fixing” or ”extending” certain functors
which take monics to monics, but do not always take epis to epis. However,
beyond this, the right derived functor is a method of characterizing many
types of cohomology; group cohomology or $\operatorname{Ext}(A,-)$, for
example, is the right derived functor of $\mathrm{Hom}(A,-)$. A dual
construction, the left derived functor, is possible by passing into the
opposite category and performing the same construction; we will omit a
rigorous proof.

######  Definition 2.9.

Choose and fix an injective resolution $I^{\bullet}$. Given an (additive)
functor $F:\mathscr{A}\to\mathscr{B}$ where $\mathscr{A}$ is an abelian
category with enough injectives and $\mathscr{B}$ is an abelian category, the
$i$th right derived functor $R^{i}F(A)$ for an object $A$ is given by
$R^{i}F(A)=H^{i}(F(0\to A\to I^{\bullet}))$. The $i$th right derived functor
likewise acts on a morphism by the rule $R^{i}F(f)=H^{n}(f^{\bullet})$, where
$f^{\bullet}$ is the cochain map induced by on injective resolutions
$I^{\bullet}$ of $A$, and $J^{\bullet}$ of $B$.

######  Theorem 2.10.

Given a left exact functor $F$ and a choice of injective resolution
$I^{\bullet}$ for every object $A$, the right derived functor $R^{i}F$
obtained using $I^{\bullet}$ is an additive functor.

Referring to figure 1 frequently during the following proof is advised.

###### Proof.

First, we must show that $R^{i}F(f)$ is well defined. The comparison lemma
states that any two cochain maps $f$ and $f^{\prime}$ derived from a morphism
$F(g)$ for some $g:A\to B$ will be cochain homotopic; given a cochain map $f$
induced by a map $g$, the cochain map $F(f)$ will also be a cochain map over
$F(g)$ as $F(f)^{0}\circ F(d^{0})=F(f^{0}\circ d^{0})=F(\partial^{n}\circ
g)=F(\partial^{n})\circ F(g)$. Thus, it suffices to show that
$H^{n}(d^{n}\circ s^{n-1}+s^{n}\circ\partial^{n+1})=H^{n}(d^{n}\circ
s^{n-1})+H^{n}(s^{n}\circ\partial^{n+1})=0$. But
$\alpha=a^{\prime}\circ\operatorname{cok}(\ker\partial^{n})\circ
s^{n-1}\circ\ker d^{n+1}$ is a morphism satisfying
$\ker\partial^{n+1}\circ\alpha=f^{n}\circ\ker d^{n+1}$, where
$f^{n}=\partial^{n}\circ s^{n-1}$ (as seen in figure 1). Then
$H^{n}(\partial^{n}\circ
s^{n-1})\circ\operatorname{cok}a=\operatorname{cok}a^{\prime}\circ
a^{\prime}\circ\operatorname{cok}(\ker\partial^{n})\circ s^{n-1}\circ\ker
d^{n+1}=0$. Since $\operatorname{cok}a$ is an epi, lemma 1.11 states that
$H^{n}(\partial^{n}\circ s^{n-1})=0$.

Now consider an $\alpha$ as constructed in figure 1 corresponding to
$s^{n}\circ d^{n+1}$. Noting that $\ker\partial^{n+1}\circ\alpha=s^{n}\circ
d^{n+1}\circ\ker d^{n+1}=0$; by lemma 1.11, this means $\alpha=0$. But then
$\operatorname{cok}a^{\prime}\circ\alpha=H^{n}(s^{n}\circ
d^{n+1})\circ\operatorname{cok}a=0$. Since $\operatorname{cok}a$ is epi, a
final reference to lemma 1.11 gives us that $H^{n}(s^{n}\circ d^{n+1})=0$.
Thus we have show than any map which is homotopic to zero has homology zero,
and therefore shown that $R^{i}F(f)$ is well defined.

Now we must show that $R^{i}F$ is a functor. First, $R^{i}F(1)=H^{i}(i)$,
where $i$ is homotopic to the identity. But by the above, this means that
$R^{i}F(1)$ is the identity, as in tended. Similarly, given two composable
morphisms $f$ and $g$, $R^{i}F(f\circ g)=H^{i}(F(f)^{\prime}\circ
F(g)^{\prime})$, where $F(f)^{\prime}$ and $F(g)^{\prime}$ denote the chain
maps over $F(f)$ and $F(g)$, respectively. However, theorem 2.5 shows that
$H^{i}(F(f)^{\prime}\circ F(g)^{\prime})=H^{i}(F(f)^{\prime})\circ
H^{i}(F(g)^{\prime})$.

Finally, we must show that $R^{i}F$ is additive. By theorem 1.10, it will
suffice to show that for $f,g:A\to B$ morphisms,
$R^{i}F(f+g)=R^{i}F(f)+R^{i}F(g)$. But since $F$ is additive,
$R^{i}F(f+g)=H^{i}(F(f+g)^{\prime})=H^{i}(F(f)^{\prime}+F(g)^{\prime})$ (where
primes again denote passage to the the chain map over a morphism). Since
homology is an additive functor, we then have
$R^{i}F(f+g)=H^{i}(F(f)^{\prime})+H^{i}(F(g)^{\prime})$, as intended. ∎

It is tedious and inelegant to choose an injective resolution every time a
derived functor is to be computed; moreover, it makes computation difficult.
Our following theorem shows that this choice is immaterial, as computing the
derived functor twice with different injective resolutions yields two
isomorphic results. This completes our construction.

######  Theorem 2.11.

Let $R^{n}F(A)$ denote the right derived functor of an object $A$ computed
using an injective resolution $I^{\bullet}$, and $\hat{R}^{n}F(A)$ denote the
right derived functor of the same object computed with injective resolution
$J^{\bullet}$. Then $R^{n}F(A)\cong\hat{R}^{n}F(A)$.

###### Proof.

The identity map on $A$ induces a cochain map $i$ from $J^{\bullet}$ to
$I^{\bullet}$ and a cochain map $i^{-1}$ from $I^{\bullet}$ to $J^{\bullet}$.
Then $i\circ i^{-1}$ is a cochain map from $I^{\bullet}$ to $I^{\bullet}$.
Applying the functor $F$ gives a cochain map $F(i):F(I^{\bullet})\to
F(I^{\bullet})$ induced by the identity $F(1_{A})=1_{F(A)}$ on $F(A)$, which
is homotopic to the identity on $F(I^{\bullet})$ (as the identity on
$F(I^{\bullet})$ is also a map induced by the identity of $F(A)$), so
$R^{n}F(i)\circ R^{n}F(i^{-1})=H^{n}(F(i))=1_{F(I^{\bullet})}$. A symmetric
argument shows that $R^{n}F(i^{-1})\circ R^{n}F(i)=1_{F(J^{\bullet})}$, and so
$R^{n}F(i)$ is an isomorphism. ∎

This construction concludes our paper. As promised, we have omitted most
motivation and examples for the sake of brevity, directing the reader to [4]
for a less categorical development with more concrete examples, [6] for a more
comprehensive treatment of the homological implications of derived functors,
and [1] for more examples and theorems regarding abelian categories.

## References

  * [1] Saunders Mac Lane  “Categories for the Working Mathematician”  Springer Science & Business Media, 2013
  * [2] Jacob Lurie  “Higher Algebra”, 2017  URL: https://www.math.ias.edu/~lurie/papers/HA.pdf
  * [3] Elliott Mendelson  “Introduction to Mathematical Logic”  Van Nostrand Reinhold Company, 1979
  * [4] Joseph J Rotman  “An Introduction to Homological Algebra”  Springer-Verlag, 2009
  * [5] Joshua Ruiter  “Homology in an Abelian Category”, 2019
  * [6] Charles A Weibel  “An Introduction to Homological Algebra”  Cambridge University Press, 1995