Categories and Functors

Category theory has been around for quite some time now and pervades modern mathematics; it

is not really an area of mathematics so much as an area of meta-mathematics. It describes frameworks in which mathematics can be, and usually is, done.

More than that, in many cases category theory has encouraged, or at least facilitated, a change

of perspective. When one studies mathematics, one often wants to study mathematical objects: sets, groups, topological spaces, Lie groups, etc. In category theory, however, at least as much emphasis is placed on maps between objects as on the objects themselves, and in many cases it turns out to be much more fruitful to make the maps the center of focus. Another great success of category theory has been to make ideas like “canonical” precise.

Categories and functors first appeared in the work of Eilenberg-MacLane in algebraic topology in the 1940s.

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(Remember that Algebraic topology : is a branch of  mathematics that uses tools from abstract algebra to study topological spaces. The basic goal is to find algebraic invariants that classify  topological spaces up to homeomorphism, though usually most classify up to homotopy equivalence.Although algebraic topology primarily uses algebra to study topological problems, using topology to solve algebraic problems is sometimes also possible. Algebraic topology, for example, allows for a convenient proof that any subgroup of a free group is again a free group. )Roughly speaking, homological algebra consists of (A) that part of algebra that is fundamental in building the foundations of algebraic topology, and (B) areas that arise naturally in studying (A).This makes studying Categories and functors a branch of homological algebra.Also Projective modules, injective modules and flat modules are branches in studying homological algebra.The concepts of Categories and functors had far wider applications. Many different mathematical topics may be interpreted in terms of categories so that the techniques and theorems of the theory of categories may be applied to these topics. For example. two proofs in disparate areas frequently

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use "similar .. methods. Categorical algebra provides a means of precisely expressingthese similarities. Consequently it is frequently possible to provide a proof in a categoricalsetting, which has as special cases the previously known results from two different areas. This unification process provides a means of comprehending wider areas of mathematics as well as new topics whose fundamentals are expressible in categorical terms.The notions of category and functor, which include the concept of a class. This concept is intended to generalize the concept of a set. The class concept is like the set concept, only some what broader.Besides in set theory it is not possible to carry out the operations over classes which can lead to problems such as Russell’s paradox. All sets are classes and all the elementary set operations, like union, intersection, formation of function, etc.,can be carried out for classes as well.

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1- Categories

There are ideas that seem to occur regardless of the particular structure under consideration.Category theory focuses on principles that are common to all algebraic systems.

Imagine a set theory whose primitive terms, instead of set and element, are set and function.How could we define bijection, cartesian product, union, and intersection? Category theory will force us to think in this way. Now categories are the context for discussing general properties of systems such as groups, rings, vector spaces, modules, sets, and topological spaces, in tandem with their respective transformations: homomorphisms, functions, and continuous maps.

Why we study Categories:

There are two basic reasons for studying categories:

- The first is that they are needed to define functors and natural transformations (which we will do in the later);- Categories will force us to regard a module ( a group, a ring, a vector space, a set or a topological space ) not in isolation, but in a context serving to relate it to all other modules ( groups, rings, vector spaces, sets or topological spaces).

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Definition. A category C consists of three ingredients: a class obj(C) of objects, a set of morphisms Hom(A, B) for every ordered pair (A, B) of objects, and compositionHom(A, B) × Hom(B,C) → Hom(A,C), denoted by( f, g) → g f, for every ordered triple A, B,C of objects. [We often write f : A → B to denote that f ∈ Hom(A, B).] These ingredients are subject to the following axioms:(i) the Hom sets are pairwise disjoint; that is, each morphism has a unique domain and a unique target;So, Hom (A,B)∩ Hom(C,D) = ∅; for (A,B) ≠(C,D).One can force pairwise disjointness by labeling morphisms f ∈ Hom(A, B) by A f B.(ii) for each object A, there is an identity morphism 1A ∈ Hom(A, A) such that f 1A = f and 1B f = f for all f : A → B;(iii) composition is associative: Given morphisms

then h(g f ) = (hg) f.

Remark: The important notion, in this circle of ideas, is not category but functor, which will be introduced in the next section. Categories are necessary because they are an essentialingredient in the definition of functor. A similar situation occurs in linear algebra: Linear

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transformation is the important notion, but we must first consider vector spaces in order todefine it.

The following examples will explain certain fine points of the definition of category.

Example . The empty category ∅ has no objects and no arrows.

Example . The category denoted 1 consist of an object * and identity arrow 1*.

Example . The category denoted 2 has two objects, say x, y and an arrow between them f : x → y.

Example . The discrete category consist only of objects and corresponding identity arrows.

Example . Consider a monoid (M, ·, e) (i.e. a set with associative binary operation '·' and the identity element e). For every m ∈ M we can define a map φm : M → M by setting φm (a) = m · a. Now

(M, ·, e) can be associated with one object category. This object is M itself and arrows are elements of M. The identity arrow is e. We can compose arrows using operation · and this composition is associative as stated in monoid axioms.

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Examples

(1) C = Sets.- The objects in this category are sets. - The morphisms are functions.- The composition is the usual composition of functions.A standard result of set theory is that if A and B are sets, then Hom(A, B), the class of all functions from A to B, is a set. That Hom sets are pairwise disjoint is just the reflection of the definition of equality of functions : In order that two functions be equal, they must, first, have the same domains and the same targets (and, of course, they must have the same graphs).

(2) C = Groups.- The objects are groups.- The morphisms are homomorphisms. - The composition is the usual composition.

(3) C = CommRings.- The objects are commutative rings. - The morphisms are ring homomorphisms. - The composition is the usual composition.

(4) C = RMod- The objects in this category are R-modules, where R is a commutative ring.- The morphisms are R-homomorphisms.

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- The composition is the usual composition. We denote the sets Hom(A, B) in RMod by HomR(A, B).If R = Z, then we often write ZMod = Ab, to remind ourselves that Z-modules are just abelian groups. When we deal with non commutative rings, then we will denote the category of left R-modules by RMod and the category of right R-modules by ModR.

(5) C = PO(X).If X is a partially ordered set, regard it as a category.- The objects are the elements of X.- The Hom sets are either empty or have only one element:

- The composition is given by

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Note that Hom( x , x ) ∅, for all x in the set X, since x ≤ x. So

We insisted, in the definition of category, that Hom(A, B) be a set, but we left open the possibility that it be empty. The category PO(X) is an example in which this possibility occurs. [Not every Hom set in a category C can be empty, for Hom(A, A) ∅ for every object A ∈ C because it contains the identity morphism 1A.]

(6) C = C(G).If G is a group, then the following description defines a category C(G): There is only one object, denoted by ∗, Hom(∗, ∗) = G, and Composition: Hom(∗, ∗) × Hom(∗, ∗) → Hom(∗, ∗); that is, G × G → G, is the given multiplication in G. The category C(G) has an unusual property. Since ∗ is merely an object, not a set, there are no functions ∗ → ∗ defined on it; thus, morphisms here are not functions. Another curious property of this category is another consequence of there being only one object: there are no proper subobjects here.

(7) There are many interesting non algebraic examples of categories. For example, C =Top, the category with objects all topological spaces,

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morphisms all continuous functions, and usual composition.

Here is how to translate isomorphism into categorical language.

Definition. A morphism f : A → B in a category C is an equivalence (or an isomorphism) if there exists a morphism g : B → A in C with g f = 1A and f g = 1B.The morphism g is called the inverse of f .

Remarks:- An inverse of an equivalence (or an isomorphism) in a category C is unique.

- Identity morphisms in a category are always equivalences.

Examples

- If C = PO(X), where X is a partially ordered set, then the only equivalences are identities.

- If C = C(G), where G is a group (see Example (6)), then every morphism is an equivalence.

- If C = Sets, then equivalences are bijections.

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- If C = Groups, C = RMod, or C = CommRings, then equivalences are isomorphisms.

- If C = Top, then equivalences are homeomorphisms.

Definition. A category C is pre-additive if every Hom(A, B) is equipped with a binary operation making it an (additive) abelian group for which the distributive laws hold: for all f, g ∈ Hom(A, B),(i) If p : B → C, then p( f + g) = p f + p g ∈ Hom(A,C);(ii) If q : D → A, then ( f + g)q = f q + g q ∈ Hom( D,B).

Example Groups does not have the structure of apre-additive category.Proof: Let G be not abelian group and let f, g : G → G be homomorphisms, then the function h(x)= f (x)g(x) may not be a homomorphism.

Example. Every ring defines a pre-additive category with one object. The morphisms are given by elements of the ring, composition is given by multiplication and addition of morphisms is the same as addition in the ring. Conversely, the hom-set of a pre-additive category with one object is a

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ring. In fact, for any pre-additive category A, the Hom(X, X) is a ring for any object X in A.

Remark: A category is defined in terms of objects and morphisms; its objects need not be sets, and its morphisms need not be functions [C(G) in Example (6) is such a category].We now give ourselves the exercise of trying to describe various constructions in Sets orin RMod so that they make sense in arbitrary categories.

Subcategories

DefinitionA category D is said to be a subcategory of a category C provided that:

(1) obj(D) is a subclass of obj(C).

(2) Whenever A;B ∈ obj(D), HomD(A;B) is a subset of HomC(A;B).

(3) Whenever f : A → B and g : B → G in D, thecomposite g f : A → G given by C is in HomD(A;G) and is the composite g f given by D.

(4) For each A ∈ D, C's identity morphism 1A is in HomD(A;A) as D's identity morphism.

If also HomD(A;B) = HomC(A;B) for all A;B ∈ obj(D), D is said to be a full subcategory of C.

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Notice that subcategories of C can be identified purely in terms of their objects and morphisms, because C already gives the rest of the structure. And full subcategories can be identified from just the objects! They hypothetically leave all morphisms that they can.

Examples

The category of finite sets forms a full subcategory of the category of sets.

The category whose objects are sets and whose morphisms are bijections forms a non-full subcategory of the category of sets.

The category of abelian groups forms a full subcategory of the category of groups.

The category of rings (whose morphisms are unit-preserving ring homomorphisms) forms a non-full subcategory of the category of rings.

A monoid is a set X together with a binary operation, written like multiplication:

xy for x; y∈ X, which is associative and has a unit element e ∈X, satisfying ex = xe = x for all x ∈ X.

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Such a monoid is a category with one object, and an arrow x for every x ∈ X.Since every group is a monoid, no two groups can be the same monoid and every group homomorphism is a homomorphism of the monoids, Grp is a subcategory of Mon. However, Mon is not a subcategory of Set, because a set can be many different monoids. Mon is a subcategory of Semgrp [the semigroups] because a semigroup can't have more than one identity element.

Every monoid homomorphism of groups is automatically a group homomorphism, so Grp is a full subcategory of Mon. However, there exist maps of monoids preserving multiplication which don't map 1 to 1, hence Mon is not a full subcategory of Semgrp.

Since the isomorphisms in a category C are closed under defined composition and involve all identity morphisms, one can form a subcategory of C by keeping precisely the isomorphisms.

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Properties of Morphisms

In the category of sets, a function f is injective iff f (x) = f (y) implies x = y. But in an abstract category, we don’t have any elements to work with; a morphism f : A → B can be regarded as simply an arrow from A to B. How do we generalize injectivity to an arbitrary category? We must give a definition that does not depend on elements of a set. Since a function f : B → C is also injectiveif and only if f is left cancellable, that is, if g , h : A → B are functions such that f g = f h, then g = h. Indeed, if f is injective, let g , h : A → B be such that f g = f h. Then for any a ∈ A we see that f g(a) = f h(a), so f( g(a)) = f( h(a)) gives g(a) = h(a). Hence, g = h. Conversely, suppose that f is left cancellable and let x, y ∈ B be such that f (x) = f (y) . Next, let A={ a } and define g , h : A → B by g(a) = x and h(a) = y. Since f g = f h implies that g = h, we see thatf g(a) = f(x) = f(y) = f h(a) gives x= g(a) =h(a)= y and so f is injective. One can also show that a morphism f : A → B in Set is a surjective function if and only if whenever g,h : B → C are functions such that g f = h f , then g = h, that is, if f is right cancellable. These observations lead to the following definition.

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Definition. A morphism f in a category C is said to be monic if f is left cancellable. Likewise, f is said to be epic if f is right cancellable. A bimorphism in C is a morphism in C that is monic and epic. We will refer to a category whose objects are sets as a concrete category. Often the sets in a concrete category will have additional structure and its morphisms will, in some sense, be structure preserving. For example, Set, Ab and ModR are examples of concrete categories. A monic (an epic, a bimorphism) morphism may not be as“well behaved” in all concrete categories as they are in Set.

Example 4. The Category DivAb of Divisible Abelian Groups. Remember that an abelian group G is said to be divisible, if for every y∈ G and for every positive integer n, there is an x ∈ G such that nx ¿ y. We can form a category DivAb whose objects are divisible abelian groups and whose morphisms are group homomorphismsf : G → H . The rule of composition for DivAb is composition of group homomorphisms.

Example 6. The Category Rng of Rings. The class O of objects consists of rings notnecessarily having an identity and if R, S ∈ O, then Hom(R, S) is the set of all ring homomorphisms from R to S. The rule of composition is the

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composition of ring homomorphisms. If this category is denoted by Rng and if Ringdenotes the category of all rings with an identity and identity preserving ring homomorphisms, then Ring is a subcategory of Rng.

THEOREM : Let C be a category and f : A → B, g : B → C morphisms in C.(1) If f and g are monic, g f is monic.(2) If g f is monic, then f is monic.(3) If f and g are epic, g f is epic.(4) If g f is epic, then g is epic.Proof (1) If g f h = g f k with h; k : D → A, then f h = f k since g is monic, hence h = k since f is monic.(2) If f h = f k with h; k : D → A, then g f h = g f k, so that h = k since g f is monic.(3) and (4) have essentially the same proof with the arrows reversed.

The following two examples illustrate that fact that a monic (an epic, a bimorphism) morphism in a concrete category may fail to be an injection (a surjection, a bijection).

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Examplesi)

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ii)

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Definition. A morphism f : A → B in a category C is said to be a section (retraction) if there is a morphism g : B → A in C such that gf = idA ( fg = idB) .An isomorphism is a morphism that is a section and a retraction.If f : A → B is a section (retraction), then it is easy to show that f is monic (epic). However, the converse does not hold, as is shown by the following two examples.

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If an arrow indicates implication, then each implication given in the table below holds in a concrete category.

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Proposition: In R-mod, given f ∈ Hom(M , N)1- f is mono if and only if f is injective.2- f is epic if and only if f is surjective.Proof:1: ←¿ f is an injective R-linear map → f is injective as a set function → f is an injection in Sets → f is a monomorphism in Sets → f is a monomorphism in R-mod → f is monic.

→) If f is monic, then let i : ker( f ) → M be the inclusion map and 0 : ker( f ) → M be the constant zero map. It is clear that f i = 0 = f 0. Hence, i : ker( f ) → M = 0 : ker( f ) → M. Thus, Im(i) = Im(0). But, Im(0) = {0M} andIm(i)= ker( f ). This means that ker( f ) = {0M}. That is, f is injective.

2: ←¿f is a surjective R-linear map → f is a surjection in Sets → f is an epimorphism in Sets → f is an epimorphism in R-mod → f is epic.

→) If f is epic, then let π : N → N / ℑ(f ) be the natural projection and 0 : N → N /ℑ( f ) the constant zero map. As before, πf =0 f =0. Thus π : N → N / ℑ(f )= 0 : N →N /ℑ( f ), which means that Im(π) = Im(0). But, Im(π) = N / ℑ(f ) andIm(0) = {0N / ℑ(f )}. Hence, N / ℑ(f )=¿ {0N /ℑ(f )}, and so N = ℑ(f ). This shows that f is surjective.In the category R-mod, the zero module {0} has the property that for any R module M, there is a unique module homomorphism from M to {0} and a

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unique module homomorphism from {0} to M. Here is a generalization of this idea.

Definitions and Comments

Definition: Let A be an object in a category. If for every object B, there is a unique morphism from A to B, then A is said to be an initial object. If for every object B there is a unique morphism from B to A, then A is said to be a terminal ( final )object. A zero object is both initial and terminal. This can be confusing because the categorical use of the word “zero” might not coincide with the conventional terminology in that category. In Abelian groups, the identity element is typically called 0, and so there’s no confusion. In general groups, the identity element is usually called 1.

Examples1- In the category of sets, there is only one initial object, the empty set. The terminal objects are singletons {x}, there’s one and only one function, namely the one that takes every element of  the set A to x, and consequently there are no zero objects.( We can prove that the empty set is the unique initial object in the category of Sets as follows:Suppose f:∅→A is a function. Then, f ⊂ ∅×A=∅. Also, ∅ ⊂ f so  f=∅ ⟹ f=∅.

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This shows that: for C= category of Sets,

Hom (∅,A)={∅}, ∀ A ∈ Obj(C). Hence there is a unique morphism( function) from ∅ to B. Thus, ∅ is an initial object in C. Suppose that B is initial object in C. Then, B≅∅ ⟹ |B|=|∅|=0. Since ∅⊂B and |B|=|∅|<∞, therefore B=∅. Thus ∅ is unique initial object in the category of Sets.)

2- In the category of groups and group homomorphisms, the trivial group is both  initial and final. A group has to have an identity element, so the smallest possible group, the trivial group, is the group that only has an identity element. A group homomorphism must take the identity of one group to the identity of the other, so there’s a unique homomorphism from the trivial group to any other group: the only element of the trivial group is mapped to the identity element of the other group. Also, there’s a unique homomorphism from any group to the trivial group: map everything to the one element of the trivial group. So the category of groups has a zero object , which is the trivial group consisting of the identity alone .

3- PreordersYou can think of a preordered set as a category by saying there exists a morphism between objects A and B if and only if A ≤ B. A minimum

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element is an initial object and a maximum element is a final object. This example shows that a category may or may not have an initial object, and it may or may not have a final object. To see this, pick two real numbers a and b with a< b and consider the intervals (a, b), (a, b], [a, b), and [a, b].

4-RingsSo far initial objects have been the smallest thing in their category: the empty set, the one-element group, the minimum element in a pre-order. Things are different in the category of rings and ring homomorphisms. (Here we assume rings must have a multiplicative identity element. Unfortunately there’s not universal agreement on this definition.)

The zero ring consists of one element, the additive identity 0. Because there’s only one element, the additive identity is also the multiplicative identity.

The zero ring is not initial in the category of rings! There is no homomorphism between the zero ring and any non-zero ring because a ring homomorphism must preserve both the additive and multiplicative identities. That is, the 0 (additive identity) of one ring must go to the 0 of the other, and the 1 (multiplicative identity) of one ring to the 1 of the other. Since 0 = 1 in the zero ring, but not in any non-zero ring, there cannot be a

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homomorphism from the zero ring to a non zero ring: the only element of the zero ring would have to go to two different elements, the 0 and the 1 of the other ring.The integers are initial in the category of rings. To see that there exists a unique ring homomorphism from the integers Z to any other ring R, note that 0 must go to the 0 of R, and 1 must go to the 1 of R. The image of every other integer follows.( Note that f(0)=0 and f(n)=n f(1) for every n in Z )The zero ring is terminal even though it is not initial. Since the initial and terminal objects are different, there is no zero object.

5- FieldsThere is no field homomorphism between two fields with different characteristics. Pick any field F and let p be its characteristic. F cannot be initial because there are only homomorphisms from F to other fields of characteristic p. Similarly, F cannot be final because there are only homomorphisms from fields of characteristic p into F. So fields are an example of a category with no initial or final objects.However, if we restrict ourselves to the category of fields with a given characteristic, then there are initial objects. The rational numbers are initial in fields of characteristic 0, and the integers mod p are initial in any field of characteristic p.

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6- Any topological space X can be viewed as a category by taking the open sets as objects, and a single morphism between two open sets U and V if and only if U ⊂ V. The empty set is the initial object of this category, and X is the terminal object. This is a special case of the case "partially ordered set", mentioned above. Take P :=the set of open subsets.

We are going to prove that any two initial objects are isomorphic, and similarly for terminal objects.This will be a good illustration of the duality principle, to be discussed next.

Definition: DualityIf C is a category, the opposite or dual category Cop

has the same objects as C. The morphisms are those of C with arrows reversed; thus f : A → B is a morphism of Cop if and only if f : B → A is a morphism of C.

If the composition gf is permissible in C, then fg is permissible in Cop.

Remark: To see how the duality principle works, let us first prove that if A and B are initial objects of C, then A and B are isomorphic. There is a unique morphism f : A → B and a unique morphism g : B → A. But 1A : A → A and 1B : B → B, and it follows that gf = 1A and fg = 1B. The point is that

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we need not give a separate proof that any two terminal objects are isomorphic. We have just proved the following:If A and B are objects in a category C, and for every object D of C, there is a unique morphism from A to D and there is a unique morphism from B to D, then A and B are isomorphic.Our statement is completely general; it does not involve the properties of any specific category. If we go through the entire statement and reverse all the arrows, equivalently, if we replace C by Cop, we get: If A and B are objects in a category C, and for every object D of C, there is a unique morphism from D to A and there is a unique morphism from D to B, then A and B are isomorphic.In other words, any two terminal objects are isomorphic. If this is unconvincing, just go through the previous proof, reverse all the arrows, and interchange fg and gf. We say that initial and terminal objects are dual. Similarly, monic and epic morphisms are dual.

If zero objects exist in a category, then we have zero morphisms as well. If Z is a zero object and A and B arbitrary objects, there is a unique f : A → Z and a unique g : Z → B. The zero morphism from A to B, denoted by 0AB, is defined as gf, and it is independent of the particular zero object chosen. Note that since a zero morphism goes through a

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zero object, it follows that for an arbitrary morphism h, we have h0 = 0h = 0.

Kernels and Cokernels

If f : A → B is an R-module homomorphism, then its kernel is, as we know, {x ∈ A: f(x) = 0 }. The cokernel of f is defined as the quotient group B / Im(f). Thus f is injective iff its kernel is 0, and f is surjective if and only if its cokernel is 0. We will generalize these notions to an arbitrary category that contains zero objects. The following diagram indicates the setup for kernels.

We take C to be the kernel of the module homomorphism f, with i the inclusion map. Iffg = 0, then the image of g is contained in the kernel of f, so that g actually maps into C.Thus there is a unique module homomorphism h: D → C such that g = ih; simply take

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h(x) = g(x) for all x. The key to the generalization is to think of the kernel as the morphism i. This is reasonable because C and i essentially encode the same information. Thus a kernel of the morphism f : A → B is a morphism i : C → A such that:(1) f i = 0.(2) If g : D → A and fg = 0, then there is a unique morphism h: D → C such that g = ih.Thus any map killed by f can be factored through i.If we reverse all the arrows in the above diagram and change labels for convenience, we get an appropriate diagram for cokernels.

We take p to be the canonical map of B onto the cokernel of f, so that C = B / Im(f). If gf = 0, then the image of f is contained in the kernel of g, so by the factor theorem, there is a unique homomorphism h such that g = hp. In general, a cokernel of a morphism f : A → B is a morphism p: B → C such that:(1) pf = 0.(2) If g : B → D and gf = 0, then there is a unique morphism h: C → D such that g = hp.

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Thus any map that kills f can be factored through p.Since going from kernels to cokernels simply involves reversing arrows, kernels and cokernels are dual. Note, however, that in an arbitrary category with 0, kernels and cokernels need not exist for arbitrary morphisms. But every monic has a kernel and (by duality) every epic has a cokernel as shown below.

Proposition: Show that in a category with 0, every monic has a kernel and every epic has a cokernel.Proof: A kernel of a monic f : A → B is 0, realized as the zero map from a zero object Z to A. For f0 = 0, and if fg = 0, then fg = f0; since f is monic, g = 0. But then g can be factored through 0ZA . Similarly, a cokernel of the epic f : A → B is the zero map from B to a zero object.

Proposition: The category R-mod has kernels and cokernels.Proof: Let M, N be R-modules, and f : M →N be an R-linear map. It can be shown (easily?) that ker(f) with the inclusion map i : ker(f) → M is the kernel of f in the categorical sense. Similarly, N / ℑ(f ) with the natural projection map π : N → N / ℑ(f ) mapping n → n+ℑ(f ) is the cokernel of f.

Proposition: In R-mod category:1) Every monic is the kernel of its cokernel.2) Every epic is the cokernel of its kernel.

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Proof: Let φ : M →N be monic in R-mod. Then φ is injective by a previous proposition. We know that the cokernel of φ is N / ℑ(φ) with natural projection map π : N → N / ℑ(φ). Now, the kernel of π is ker(π ¿= Im(φ)≅ M (since φ is injective), together with the inclusion mapi : ker(π) → N. Hence, there exists an isomorphism∅ : ker(π) → M , note that ∅=φi.

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is the kernel of its cokernel. The proof for the epic case is similar and is omitted.

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Definition: A category C is called an abelian category if it satisfies the following axioms:(AB I) C is an additive category.(AB II) Every morphism in C has a kernel and a cokernel.(AB III) Every monic in C is the kernel of its cokernel.(AB IV) Every epic in C is the cokernel of its kernel.

There are some alternate definitions of an abelian category (for example, look at, www.math.columbia.edu/ lauda/teaching/rankeya.pdf)

The primary example of an abelian category in is R-mod. However, another important example is the category of sheaves of abelian groups on a topological space X. This example is important for algebraic geometry.

The following is an excerpt from Prof. Ravi Vakil's notes on The Foundations of Algebraic Geometry Ch.1 :"The key thing to remember is that if you understand kernel, cokernels, images and soon in R-mod, you can manipulate objects in any abelian category. This is made precise bythe Freyd{Mitchell Embedding Theorem ... The upshot is that to prove something about

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a diagram in some abelian category, we may assume that it is a diagram of modules oversome ring, and we may then "diagram chase" elements. Moreover, any fact about kernels,cokernels, and so on that hols in R-mod holds in any abelian category."

( Source: THE CATEGORY OF MODULES OVER A COMMUTATIVE RING AND ABELIAN CATEGORIES. By RANKEYA DATTA ).

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Products and CoproductsWe all know how to form the cartesian productA × B of two sets A and B, the set of all ordered pairs (a, b) for a ∈ A and b ∈ B. We also know that often when A and B carry structures of a similar kind, the product A × B can be used with the same kind of structure. Groups, rings, topological spaces, and so on, provide examples of this. In these cases we find that the two projections are arrows in the appropriate category.

We have studied the direct product of groups, rings, and modules. It is natural to try to generalize the idea to an arbitrary category, and a profitable approach is to forget (temporarily) the algebraic structure and just look at the cartesian product A = ∏i

A i of a family of sets Ai, i ∈ I. The key property of a product is that if we are given mapsfi from a set S into the factors Ai, we can lift the fi into a single map f : S → ∏i

A i. The commutative diagram below will explain the terminology.

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In the picture, pi is the projection of A onto the ith factor Ai. If fi(x) = ai, i ∈ I, we take f(x) = (ai, i ∈ I). It follows that pi ◦ f = fi for all i; this is what we mean by lifting the fi to f. (Notice that there is only one possible lifting, i.e., f is unique) . If A is thedirect product of groups Ai and the fi are group homomorphisms, then f will also be a group homomorphism. Similar statements can be made for rings and modules. We can now give a generalization to an arbitrary category.

Definition Let C be a category and {Ai | i ∈ I} a family of objects of C. A product for the family {Ai | i ∈ I}, or simply a product of objects Ai in the category C is an object A, along with a family of morphisms { pi : A → Ai | i ∈ I} , with the following universal mapping property. Given any object S of C and family of morphisms { fi : S → Ai | i ∈ I}, there is a unique morphism f : S → A such that pi f = fi for all i ∈ I.

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In a definition via a universal mapping property, we use a condition involving morphisms,along with a uniqueness statement, to specify an object and morphisms associated with that object. We have already seen this idea in connection with kernels and cokernels.Not every category has products , but if they do exist, they are essentially unique. (The technique for proving uniqueness is also essentially unique) .

PropositionIf (A, pi, i ∈ I) and (B, qi, i ∈ I) are products of the objects Ai, then A and B are isomorphic.Proof. We use the above diagram (1) with S = B and fi = qi to get a morphism f : B → A such that pi f = qi for all i. We use the diagram with S = A, A replaced by B, pi replaced by qi, and fi = pi, to get a morphism h: A → B such that qi h = pi. Thuspi f h = qi h = pi and qi h f = pi f = qi. Butpi1A = pi and qi1B = qi and it follows from the uniqueness condition that f h = 1A and h f = 1B.Formally, we are using the diagram two more times, once with S = A and fi = pi, and once with S = B, A replaced by B, pi replaced by qi, and fi = qi. Thus A and B are isomorphic. ♣

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As a special case the following diagram defines the product of two objects A and B, which is the object P.

That is if X is any object in the category, and f 1: X → A and f 2: X → B  are any morphisms, then there exists a unique morphism g : X →P such that π1 g=f 1  and π2 g=f 2. In other words there is one and only one morphism g such that both triangles in the diagram commute.

Proposition. The cartesian product of non-empty sets is a product in the category of sets.Proof. We will prove it for a product of two sets. Let A and B be any non-empty sets, A × B their cartesian product set, and  π1 , π2 are the canonical coordinate projection maps. We must prove that A × B satisfies the universal property of product. ( Remember that The universal property has two parts: there exists a morphism g and the morphism g is unique.)

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Let  X   be any set and let  f 1: X → A  and f 2: X → B be any functions. We construct the function g : X → A × B by setting g ( x )=( f 1 ( x ) , f 2 ( x )) for every element x in X. This satisfies the first part of universal property: π1 g( x)=π1( f 1 ( x ) , f 2 ( x ))=f 1 (x ) and

π2 g (x)=π2( f 1 ( x ) , f 2 ( x ))=f 2 ( x )

We must prove that if h : X → A × B  is another function with π1 h=f 1 and π2 h= f 2, then g=h.Let x be any element in X, then h(x) is an element (a,b) in A × B, for some a in A and b in B. An easy computation shows that f 1 ( x )=π1 h (x )=π 1 (a ,b )=a and f 2 ( x )=π2 h ( x )=π 2 (a ,b )=b. Therefore h ( x )=(a ,b )=( f 1 (x ) , f 2 ( x ) )=g(x ) for every x in X, and h=g is the unique function required by the second part of universal property. 

Proposition. Intersection of subsets is a product in the power set poset.Proof. We will prove it for a product of two power set poset. Let X be any set, and let A and B be any subsets of X. We must prove that the intersection  A ∩ B has the universal property.Let C be any subset of X such that there are arrows C → A and C → B. By the definition of powerset poset, there's an arrow from C to A if and only if C is a subset of A, in other words if every element of C is

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an element of A. So every element of C is an element of both A and B. The definition of intersection says that an element x belongs to the intersection    A ∩ B  if and only if x belongs to both A and B. Thus C is a subset of    A ∩ B and there is an arrow C → {¿ A ¿ B}A ∩B. Uniqueness and composition property follow from the fact that in any poset there is at most one arrow between any two objects. Proposition. Least common multiple lcm() is a product in the divisibility poset of positive integers.Proof. Let a and b be positive integers. We must prove that the least common multiple of a and b, denoted lcm(a,b), has the universal property.Let c be any positive integer divisible by both a and b. There are integers q and r (quotient and remainder), such that  c = q(lcm(a,b)) + r with  q>0 and  0 ≤ r<lcm(a ,b). Since c and lcm(a,b) are both divisible by a and b, also r must be divisible by them. But r is smaller than lcm(a,b), so r must be zero or otherwise lcm isn't the  least common multiple. Thus c = q(lcm(a,b)) and c is divisible by lcm(a,b). By the definition of poset there is at most one arrow between any two objects, so the uniqueness part is trivial. 

There is also a dual process which is not so clear.

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Given two sets A and B we can form the disjoint union (sum) A + B of the sets, a larger set that includes copies of A and B with minimalinterference. Technically, we tag the elements of A and B to remember their origin, and take the union of the tagged versions of the sets.A + B = (A × {0})∪B ×{1}. We then find that the two embeddings locate disjoint copies of the parent sets within the sum. locate disjoint copies of the parent sets within the sum.

As a special case the following definition defines the coproduct of two objects

Definition: Let C be a category and let X1 and X2 be objects in that category. An object is called the coproduct of these two objects, written X1 ∐ X2 or X1 ⊕ X2 or sometimes simply X1 + X2, if there exist morphisms i1 : X1 → X1 ∐ X2 and i2 : X2 → X1 ∐ X2 satisfying a universal property: for any object Y and morphisms  f1 : X1 → Y and 

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f2 : X2 → Y, there exists a unique morphism  f : X1 ∐ X2 → Y such that  f1 = f ∘ i1 and  f2 = f ∘ i2. That is, the following diagram commutes:

The unique arrow f making this diagram commute may be denoted f1 ∐ f2 or f1 ⊕ f2 or f1 + f2 or [f1, f2]. The morphisms i1 and i2 are called canonical injections, although they need not be injections nor even monic.

For general definition, note that the discussion of diagram (1) indicates that in the categories of groups, abelian groups, rings, and R-modules, products coincide with direct products. But a category can have products that have no connection with a cartesian product of sets.The dual of a product is a coproduct, and to apply duality, all we need to do is reverse all the arrows in (1). The following diagram results.

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We have changed the notation because it is now profitable to think about modules. Supposethat M is the direct sum of the submodules Mj , and ij is the inclusion map of Mj into M. If the fj are module homomorphisms out of the factors Mj and into a module N, the fj can be lifted to a single map f. If xj1 ∈ Mj1, . . . , xjr ∈ Mjr, we takef(xj1 + · · · + xjr) = fj1(xj1) + · · · + fjr(xjr).Lifting means that f ◦ij = fj for all j. We can now give the general definition of coproduct.DefinitionA coproduct for the family { Mj | i ∈ I } of objects in a category C, or simply a coproduct of objects Mj in a category C is an object M, along with a family of morphisms { ij : Mj → M | i ∈ I}, with the following universal mapping property. Given any object N of C and any family of morphisms { fj : Mj → N | i ∈ I}, there is a unique morphism f : M → N such that f ij = fj for all i ∈ I, any two coproducts of a given collection objects are isomorphic. - A coproduct of objects Aj in a category C is

denoted by A = .- The discussion of diagram (2) shows that in the category of R-modules, the coproduct is the direct sum, which is isomorphic to the direct product if there are only finitely many factors.

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- In the category of sets, the coproduct is the disjoint union . To explain what this means, suppose we have setsAj, j ∈ J. We can disjointize the Aj by replacing Aj

by Aj* = {(x, j): x ∈ Aj}. The coproduct is A = ¿ j∈ J A j∗¿¿ , with morphisms ij : Aj → A given by ij(aj) = (aj, j). If for each j we have fj : Aj → B, we define f : A → B by f(aj, j) = fj(aj ).

- Also the coproduct in the category of topological spaces is the disjoint union, where you definethe open sets in the disjoint union to be disjoint unions of opens sets.

- In The product in   of two rings   is just the direct product  . T he coproduct, on the other hand, is the tensor product   (assumed to be over   without qualification). And it turns out that

.

- In the category of Abelian groups, coproducts are

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FUNCTORSAn important use of category theory is to make connections among areas of mathematics. To make these connections, we need to have a method of passing information from one category to another. The concept of a functor meets this requirement, So Functors are homomorphisms of categories.

Definition. Recall that obj(C) denotes the class of all the objects in a category C. If C and D are categories, then a functor T : C → D is a function such that

(i) If A ∈ obj(C), then T (A) ∈ obj(D);

(ii) If f : A → B in C, then T ( f ) : T (A) → T (B) in D;

(iv) For every A ∈ obj(C), T (1A) = 1T (A).

That is the functor T is a rule that assigns each object A in C exactly one object T(A) in D and to each morphism f : A → B of C exactly one

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morphism T ( f ) : T (A) → T (B) in D; such that (iii) and (iv) are satisfied.

Examples(i) If C is a category, then the identity functor 1C : C → C is defined by 1C(A) = A for all objects A, and 1C( f ) = f for all morphisms f.

(ii) If C is a category and A ∈ obj(C), then the Hom functor TA : C → Sets is defined by

TA(B) = Hom(A, B) for all B ∈ obj(C),

and if f : B → D in C, then TA( f ): Hom(A,B)→ Hom(A,D) is given by TA( f ) : h → f h.

We call TA( f ) the induced map, and we denote it by TA( f ) = f ∗ : h → f h.

Because of the importance of this example, we will verify the parts of the definition in detail. First, the very definition of category says that Hom(A, B) is a set. Note that the composite f h makes sense:

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(iii) If R is a commutative ring and A is an R-module, then the Hom functor TA : RMod →Sets has more structure:- If M and N are R-modules, where R is a commutative ring, then HomR(M, N) is an R-module, where addition is defined as follows: If f, g ∈ HomR(M, N), then define f + g : M → N byf + g : m → f (m) + g(m). and scalar multiplication is given byr f : m → f (rm). Moreover, there are distributive laws: If p : M /→ M and q : N → N/, then ( f + g)p = f p + gp and q( f + g) = q f + qgfor all f, g ∈ HomR(M, N).Proof. Verification of the axioms in the definition of R-module is straightforward, but we present the proof of () f = r (r / f ) because it uses commutativity of R.If m ∈ M, then (rr/) f : m → f (rr/m). On the other hand, r (r/f ) : m → (r/ f )(rm) = f (r/rm). Since R is commutative, (rr/ ) f = r (r/f ).- we now show that if f : B → B/ , then the induced map f∗: HomR (A, B) →HomR(A, B/), given by h → f h, is an R-map. First, f∗ is additive: If h, h/ ∈ Hom(A, B), then for all a ∈ A,f∗(h + h/) = f (h + h/) : a → f (ha + h/a)= f ha + f h/a = ( f∗(h) + f∗(h/))(a), so that f∗(h + h/) = f∗(h) + f∗(h/).

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Second, f∗ preserves scalars. Recall that if r ∈ R and h ∈ Hom(A, B), then rh : a → h(ra). Thus, f∗(rh) : a → f (rh)(a) = f h(ra), while r f∗(h) = r f h: a → f h(ra). Therefore, f∗(rh) = (r f )∗(h).In particular, if R is a field, then the HomR’s are vector spaces and the induced maps are linear transformations.

(iv) Let C be a category, and let A ∈ obj(C). Define T : C → C by T (B) = A for every B ∈ obj(C), and T ( f ) = 1A for every morphism f in C. Then T is a functor, called the constant functor at A.

Proposition. If T : C → D is a functor, and if f : A → B is an equivalence in C, then T ( f ) is an equivalence in D.Proof. Since f is an equivalence in C, then there exists a morphism g ( the inverse of f ): B → A in C, such that g f = 1A and f g = 1B . Apply T to the equations, then we have a morphism T(g): T(B) → T(A) in D such that T(g f) = T(1A) and T(f g) =T( 1B). Hence T(g)T( f) = 1T(A) and T(f)T(g) = 1T(B) . Thus T(g) is the inverse of T( f ) and T ( f ) is an equivalence in D.This proposition illustrates, admittedly at a low level, the reason why it is useful to give categorical definitions: Functors can recognize definitions phrased solely in terms of objects, morphisms, and

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diagrams. How could we prove this result in Ab if we regard an isomorphism as a homomorphism that is an injection and a surjection?There is a second type of functor that reverses the direction of arrows.

Definition. If C and D are categories, then a contravariant functor T : C → D is afunction such that(i) if C ∈ obj(C), then T (C) ∈ obj(D);(ii) if f : C →C / in C, then T ( f ) : T (C /) → T (C) in D;

To distinguish them from contravariant functors, the functors defined earlier are called covariant functors.

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