IJETST- Vol.||02||Issue||04||Pages 2178-2183||April||ISSN 2348-9480 2015 International Journal of Emerging Trends in Science and Technology Some Analytical Aspects of Categories Authors Dhanjit Barman Research Scholar Deptt. of Mathematics, Gauhati University, Guwahati, Assam Email: [email protected] Postal Address: C/O Principal, Jalukbari H. S. School, Satmile, Jalukbari, Ghy-14. ABSTRACT In this paper we discuss some analytical aspects of categories. Here we see that if F : C→ D is a covariant functor then image of F will not form a subcategory of D. We provide an example to show it. Also we try to find some results of the category of rings ( Ring), the category of sets (Set), the category of groups (Gp) and category of topological spaces ( Top). We define some functors between categories and discuss their properties. Key words: Category, functor, morphism, monomorphism, epimorphism,bimorphism, isomorphism , balanced, normalcategory. INTRODUCTION and epimorphism and isomorphism from Pareigis Here we discuss some analytical aspects of [11]. categories. We try to find some properties of the category of rings (Ring), category of sets (Set) , MAIN RESULTS category of topological spaces (Top) and category Let us consider, of groups (Gp). Also we find some functors Rng = the category of rings , whose objects are between categories and their properties. If F : rings and morphisms are ring homomorphismms. C→ D is a covariant functor then, in general , its Rngu: the category of rings with unity together image will not form a subcategory of D. with unital (identity preserving) ring homomor- phisms. PRELIMINARIES DivRing: the category of division rings together For notions of category theory we shall in general with unital ring homomorphisms. follow the notation and terminology of Popescu [6]. However, we do deviate somewhat. PROPERTIES For C a category and A, B objects of C, Mor(A, 1. DivRng is full subcategory of Ringu B) denotes the set of morphisms from A to B. 2. Rngu is not full subcategory of Rng as We will also follow Popescu [6] for the definition every ring homomorphism between rings of Preadditive, Additive and Preabelian and with identity is not unital. abelian category. 3. Both DivRng and Rngu are subcategory of For kernel and cokernel we follow MacLane[2]. Rng. We follow the definition of retraction and Proposition 1.1: In DivRng , every epimorphism coretraction from Popescu and Pareigis[11]. is bijective morphism. We shall use the definition of Balanced category Proof: Let f be an epimorphism in DivRng. At from Mitchel [3] Monomorphism from Schubert [5] first we will prove that in DivRng every morphism is monomorphism. Dhanjit Barman www.ijetst.in Page 2178 IJETST- Vol.||02||Issue||04||Pages 2178-2183||April||ISSN 2348-9480 2015 Since the ideals of a division ring are {0} and R. Proof: Let f: A → B is monomorphism. We So every (unital) ring homomorphism in DivRng will show that f is injective. is injective.(by considering the kernel of a unital Suppose to the contrary , f is not injective . So we ring homomorphism of division rings)And so a have a≠a’ such that f(a) = f(a’). monomorphism We will show that functions g,h can be Therefore f is both epimorphism and constructed such that fog = foh monomorphism. Implies g ≠ h (i.e f is not monomrphism). Hence f is bijective morphism. Let us consider the functions g,h : C → A such that fog = foh , where C = {a , a’}. Proposition 1.2: In Rng not all monomorphisms Let us define g(a) = a, g(a’) = a’ and are kernels. h(a) = h(a’) = a. Proof: Let R ∈ obRng be an object in Rng. Let H thus we have f og = foh but g ≠ h be a subring, but not an ideal of R. Hence f is not monomorphism Then’ i’ is not kernel. But ‘i’is monomorphism . Therefore monomorphism implies injective in Set. Proposition 1.3: DivRng has no initial object, no Next we will prove that injective in Set is section. final object, no zero object and no zero morphism. Let f : A →B be injective in Set. Let us define g : Proof: Let R, R’ be two division rings with B →A such that for a fixed a∈ A different characteristics. Then Mor(R , R’) = φ. So g(b) = a’ if f(a’) = b DivRng has no initial object, no final object, no = a if b ∈ B – f(A) . zero object and no zero morphism. Then gof (a’) = g (f (a’)) = g(b) = a’ 1A (a’) Proposition 1.4: Rngu and DivRng are not = > gof = 1A . abelian category. Hence f is section. Proof : since Rngu and DivRng donot have zero B y proposition 1.6 we have F(f) is section in C. morphisms so they canot be additive categories. But every section is monomorphism. And hence both are not abelian categories. Thus F(f ) is monomorphism in C. Proposition 1.5: Rng is also not abelian. Similarly it can be proved that in Set ‘’ f is Proof: Rng has zero morphisms . But then also it Epimorphism iff f is retraction.” is not abelian as sum of two rings is not a ring. By proposition 1.6 F(f) is retraction. Proposition 1.6: Every covariant functor But every retraction in a category is an preserves retraction and section. epimorphism. Proof : Let F : C → D be a covariant functor. Thus F(f) is an epimorphism in C. Let f: A → B be a retraction, then there is a Problem: Provide an example of functor which morphism g :B → A such that fog = 1B. does not preserve monomorphisms. Now F(fog) = f(1B) Soluton: Let us consider the forgetful functor F : = > F(f) o F(g) = 1F(B) [ since F is covariant] DivRng → Rng. = >F(f) is retraction. Here in DivRng, every morphism is Let f: A → B be a section, then there is a monomorphism but the image under F in Ring morphism g : B → A such that gof = 1A. may not be monomorphism. Now F(gof) = f(1A) Let us consider ------ = > F(g)o F(f) = 1F(A) [ since F is covariant] Set = the category of sets together with mapping = >F(f) is section. between them ] Fin Set = the category of finite sets and together Proposition 1.7: Every covariant functor F : Set with maps between them. → C preserves monomorphism and Inj = the category of sets together with the epimorphism. injective maps between them. Dhanjit Barman www.ijetst.in Page 2179 IJETST- Vol.||02||Issue||04||Pages 2178-2183||April||ISSN 2348-9480 2015 Surj = The category of sets together with the (b) = > (d) surjective maps between them. Let F preserves coconstant morphisms. Bij = The category of sets together with the Let Z be a zero object in C. Let g : Z → B be a bijective maps between them. zero morphism. Thus g is both constant and coconstant PROPERTIES: morphisms, by definition. 1. Fin Set is full subcategory of Set. Hence F(g); F(Z) → F(B) is both constant and 2. Inj, Surj and Bij are sub categories of Set. coconstant morphisms in D. 3. Set is not an abelian category as it does not Thus F(Z) is zero object in D. have zero object. 4. Similarly Fin Set, Inj , Surj, Bij are also In the following every category is taken to be a not abelian categories. full subcategory of, Ab, the category of abelian Proposition 1.8: Prove that if F: C → D is a groups. functor between categories with zero object, then a) The category of torsion abelian groups the following conditions are equivalent: (Tor) is an abelian category. a) F preserves constant morphisms; a b) The category of torsion-free abelian morphism f : A → B is constant provided groups (Torfree) is not an abelian that for any pair A′ r⇉s A of morphisms category. we have f o r = f o s. c) The category of fnitely generated abelian b) F preserves coconstant morphisms; (FG-Ab) groups is an abelian category. c) F preserves zero morphisms; d) The category of divisible groups(Div) is d) F preserves zero objects. not an abelian category. Proof: (d) = > (c) Proposition 1.9: T: Ab → Tor is a covariant Let F preserves zero objects i.e. if Z is a zero functor, where T sends every to its torsion object in C then F(Z) is zero object in D. Let g : subgroup and every group homomorphism to its A → B be a zero morphism in C. restrictuion to the torsion subgroup. Then either A or B or both A and B are zero objects. Thus by our assumption either F(A) or Proof: Here T (A) = A’, A’ being torsion F(B) or both are zero objects in D. subgroup of A. Hence F(g) is zero morphism in D. For f ∈ Mor (A, B), g ∈ Mor (C, A), T (f) ∈ Mor (c) = > (b), (a) (A’ , B’) and T (g) ∈ Mor (C’ , A’) T(f) , T(g) Let F preserves zero morphisms. are the restriction to the torsion subgroup such By definition of zero morphism , a zero morphism that T(f)(a’)= a’ and T(g)(c’) = c’ is both constant and coconstant morphism. Then I ) T(fog) (c’)= c’ and {T(f)o T(g)}(c’)= Hence F preserves both constant and coconstant c’ morphisms. Thus T(fog) = T(f) o T(g). (a) = > (d) Similarly it can be shown that Let F preserves constant morphisms. ii) T(1A) = 1T(A). Let Z be a zero object in C. Let g : Z → B be a Hence T is a covariant functor. zero morphism.