sign in sign up
<<
Home , Epimorphism, Monomorphism

PROBLEMS 5 SOLUTIONS OF SOME EXERCISES

1.– Show that in the catgeory AbGroups of abelian groups, a is a if and only if it is injective, and a morphism is an epi- morphism if and only if it is surjective (the latter needs a little more care). Show that the same holds in the FGAbGroups of finitely generated abelian groups (the same proof should work).

Solution: Since two of groups f1, f2 : G1 → G2 are equal if and only if they are equal as maps of sets, we see immediately that a morphism of groups which is a monomorphism (resp. an ) as map of sets is a monomorphism (resp. an epimorphism) as map of groups. Since in the , injective=mono, surjective=epi, an injective morphism of is a monomorphism, a surjective morphism of groups is an epimorphism. Let’s prove the two converses. If f : G1 → G2 is a morphism of abeian groups, which is not injective. Then consider two maps i, z : ker f → G1, defined as follows: i is the canonical inclusion, z is the zero map. Since ker f 6= 0, one has i 6= z. Yet f ◦ i = f ◦ z since both are the zero map: ker f → G2. Hence f is not a monomorphism If f : G1 → G2 is a morphism of abelian groups, which is not surjective. Since G2 is abelian, imf is a normal of G2 and one can consider the factor group G2/imf, which is not the zero group since imf 6= G2. Consider two morphisms of groups G2 → G2/imf, the canonical surjection π and the zero map z. Then π ◦ f = z ◦ f since both are the zero map: G1 → G2/imf. Hence f is not an epimorphism

2.– Let k be a field and let us consider the category FinDimVectk of finite dimensional vector spaces over k (morphisms are linear maps, of course). Show that this category is equivalent to its own opposed category.

Solution: Consider the functor F that to a V attaches its V ∗ (the space of linear forms l on V ), to a f : V → W its transpose tf : W ∗ → v∗. This is a contravariant functor from the category of vector spaces over k FinDimVect. Such a functor can be seen alternatively as a opp covariant functor, say F1, from FinDimVectk to FinDimVectk or equiva- opp lently again a covariant functor F2 from FinDimVectk to FinDimVectk . Now we claim that F1 ◦F2 and F2 ◦F1 which are functors from FinDimVectk opp opp to FinDimVectk, and from FinDimVectk to FinDimVectk respec- opp tively, are isomorphic to the identity functors of FinDimVectk and FinDimVectk respectively. It suffices to do it for the first one, the second been the same functor with all arrows reversed. 1 2 PROBLEMS SET 5 SOLUTIONS OF SOME EXERCISES

The functor F1 ◦ F2 sends a finite dimentional vector space V to its bidual (V ∗)∗, and a morphism f : V → W to ttf :(V ∗)∗ → (W ∗)∗. For every ∗ ∗ V , let θV be the canonical morphism V → (V ) which attaches to v ∈ V ∗ ∗ the linear form on V which sends an element l in V to l(v). The θV ’s define a morphism of functor θ : Id → F1 ◦ F2 because for every linear map f : V → W , the following diagram is commutative, as can be check directly form the definition: f V / W

θV θW  ttf  (V ∗)∗ / (W ∗)∗.

Since the θV are (because V is a finite dimensional vector spaces), one sees easily that θ is an of functors.

3.– Is the category FGAbGroups of finitely generated abelian groups equivalent to its own opposed category? Solution: No. One argument for this is that in the category of FGAbGroups, every epimorphism from an object A to itself is an isomorphism. Indeed, to see that, remember that epimorphism just means surjective by exercise 1, and let f : A → A be a surjective homomorphsim. If f is surjective but not injec- tive, then one has an ascending infinite chain of ker f ⊂ ker f 2 ⊂ ker f 3 ⊂ ... where each inclusion is strict, and this impossible in a finitely generated (exercise). Thus, if FGAbGroups was equivalent to its own oppsed category, the opposed statement, namely that every monomorphism from A to A is an isomorphism would be true. But it is not: the morphsim Z → Z which sends x to 2x, for instance, is injective, hence a monomorphism, but not an isomorphism.

5.– Let C be a category where arbitrary products of two objects exist. Let f : A → A0 and g : B → B0 be two morphisms. Explain how to define naturally a morphism f × g : A × B → A0 × B0.

Solution: By definition, the product A × B comes with two natural map p1 : A × B → A and p2 : A × B → B. Consider the pair of maps f ◦ p1 : 0 0 A × B → A and g ◦ p2 : A × B → B . By the definition of the product A0 × B0 as a limit, this pair of map defines a unique map A × B → A0 × B0 0 0 0 0 such that post-composed with the two p1 : A × B → A and 0 0 0 0 p2 : A × B → B gives respectively f ◦ p1 and f ◦ p2. This is the map we were looking for, and we call it f ◦ g.

6.– Let C and C0 be two catgeories where products of two objects exist, and let F be a functor from C to C0. Lat A and B be two objects in C. Explain why there is a natural map from F (A × B) to F (A) × F (B) in C0. We say that F commute with products if this map is an isomorphism for every objects A and B in C. PROBLEMS SET 5 SOLUTIONS OF SOME EXERCISES 3

Solution: Let p1 and p2 be the two canonical projection A×B → A and A× 0 B → B. Then F (p1) and F (p2) are morphism in C going F (A × B) → F (A) and F (A × B) → F (B). Thus they defines a map F (A × B) → F (A) × F (B) in C0.

8. Let C be a category. For A an object of C, we consider the functor FA from C to the category Sets defined by FA(X) = HomC(A, X). a.– Complete the definition of FA by defining its action on morphisms, and check that it is a functor. Solution: If f : X → Y is a morphism in C, we define a morphism in the category of sets Fa(f): FA(X) = HomC(A, X) → HomC(A, Y ) by sending g ∈ HomC(A, X) to f ◦ g. This cleary makes FA a functor.

b.– Show that A 7→ FA define a contravariant functor from C to the cate- gory Funct(C,Sets) of functors from C to Sets. Solution: Again we have to define the of morphisms by our would- be functor A 7→ FA. If f : A → B is a functor, we want to define a morphism FB → FA (remember we want a contravariant functor) in the category of functors from C to Sets. Consider, for every object X in C the map θf,X : FB(X) = HomC(B,X) → FA(X) = HomC(A, X) which sends g to g ◦f. One sees easily that the θf,X ’s define a morphism of functors θf : FB → FA. One also sees easily that θf 0◦f = θf ◦ θf 0 and that θId = Id, and this completes the definition of the functor A → FA c.– (Yoneda’s lemma) Show that the functor defined in b.– is fully faithful, and therefore essentially injective.

0 Solution: We need to show that if f, f are in HomC(A, B), and θf = θf 0 , 0 then f = f . The meaning of the equality θf = θf 0 between those two morphisms of functors is that for every object X in C, θf,X = θf 0,X as maps from HomC(B,X) to HomC(A, X). In particular θf,B = θf 0,B as maps from HomC(B,B) to HomC(A, B). Applying this to the element IdB we get 0 0 θf,B(IdB) = θf 0,B(IdB), that is using the defintion, IdB ◦f = IdB ◦f or f = f , QED.

In particular, FA determine A up to isomorphism. A functor from C to Sets that is isomorphic to some functor FA is called representable. d.– Show that the forgetful functor from Groups to Sets is representable.

Solution: It is indeed representable by the group Z. To check this, we need to show that the functor G 7→ FZ(G) = Hom(Z,G) (morphism of groups), and G 7→ underlying set of G are isomorphic. For this define θG : Hom(Z,G) → G by sending a morphism of groups f : Z → G to f(1), As seen in class, θG is a bijection. It is clear that the θG’s define a morphism of functor FZ to the functor ”underlying set” which is an isomorphism. e.– Show that the forgetful functor from Finite Groups to Sets is not representable. 4 PROBLEMS SET 5 SOLUTIONS OF SOME EXERCISES

Solution: Suppose this functor is representable by a finite group H. Then in particular, for every finite group G, one has a bijection between the sets Hom(H,G) and G. Let ` be a prime number greater that the order of H. Then certainly Hom(H, Z/`Z) has just one element (the trivial morphism) but Z/`Z has ` ≥ 2 element, so they can nor be in bijection. Contradiction.