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Introduction to Category Theory, I INTRODUCTION TO CATEGORY THEORY, I Homework 2, solutions 1. Find an example to show that the objects and morphisms in the image of a functor F : C!D do not necessarily define a subcategory. Solution: Let C be the following category: there are four objects, say a; b; c; d. In addition to the identity morphisms from an object to itself there are two morphisms; f : a ! b and g : c ! d. Let D be the following category: there are three objects, say 1; 2; 3. In addition to the identity morphisms from an object to itself there are morphisms h: 1 ! 2, k : 2 ! 3 and kh: 1 ! 3. Define F : C!D as follows: for objects, F (a) = 1, F (b) = 2 = F (c) and F (d) = 3, and for morphisms F (f) = h, F (g) = k. Then F is a well-defined functor. However, kh is not an image of a morphism in C. Therefore, the objects and morphisms in the image of F do not form a subcategory. 2. Is there a functor F : Top ! Sets with the property that F (X) equals the set of connected components for every topological space X. Solution: For a topological space X, let F (X) denote the set of connected com- ponents of X. For a continuous function f : X ! Y let F (f): F (X) ! F (Y ) be defined as follows: If Xi is a connected component of X, then F (f)(Xi) equals the connected component of Y that contains Xi. Since continuous functions take connected sets to connected sets, it follows that F (f) is well-defined. Let idX : X ! X be the identity function of X. Then F (idX ): F (X) ! F (X) is the identity function of F (X). Let f : X ! Y and g : Y ! Z be continuous functions. If Xi is a connected component of X, then g(f(Xi)) is connected, and F (g ◦ f)(Xi) = F (g)(F (f)(Xi)). Thus F (g ◦ f) = F (g) ◦ F (f), and F is a functor with the desired properties. 3. Is there a functor F : Group ! Group with the property that F (G) equals G=C(G), for all groups G? Here C(G) is the commutator subgroup of G, i.e., the subgroup of G generated by elements ghg−1h−1 for any g; h 2 G. Solution: Let's first check that C(G) is normal in G, so that G=C(G) is a group: Assume c 2 C(G) and g 2 G. Then gcg−1c−1 2 C(G), and hence also gcg−1 = (gcg−1c−1)c 2 C(G). Thus C(G) is normal. Let H and G be groups and let f : H ! G be a homomorphism. Then f(hgh−1g−1) = f(h)f(g)f(h)−1f(g)−1, for every h; g 2 H. Thus f(C(H)) ⊂ C(G), and it follows that f induces a homomorphism f¯: H=C(H) ! G=C(G): Date: June 20, 2018. 1 2 INTRODUCTION TO CATEGORY THEORY, I If f is the identity homomorphism of G, then clearly f¯ is the identity homomor- phism of G=C(G). If f : H ! G and d: G ! K are homomorphisms, then d ◦ f = d¯◦ f¯: H=C(H) ! K=C(K): It follows that we can define a functor F : Group ! Group as follows: F takes every group G (object) to F (G) = G=C(G) and every group homomorphism f : H ! G (morphism) to F (f) = f¯: H=C(H) ! G=C(G). 4. Is there a functor Z : Group ! Group with the property that Z(G) is the centre of G for all groups G? Solution: Let Σ3 be the symmetric group that permutes three elements, for ex- ample elements a, b c. Then Σ3 has altogether 6 elements. Let G be the cyclic subgroup of Σ3 generated by the element that takes a to b, b to c and c to a. Then G is a normal subgroup of Σ3 (this is easy to check since Σ3 only has 6 elements.) Thus there is the quotient group ∼ Σ3=G = Z2: Assume Z is a functor Z : Group ! Group with the property that Z(G) is the centre of G for all groups G. There are homomorphisms i p Z2 −! Σ3 −! Σ3=G; where i identifies Z2 with the two element subgroup of Σ3, whose non-trivial element switches b and c. Then i(Z2) \ G = feg, where e is the identity element of Σ3. The homomorphism p is the quotient homomorphism. Then p◦i is a group isomorphism. Since Z is a functor, there are homomorphisms Z(i) Z(p) Z(Z2) −! Z(Σ3) −! Z(Σ3=G): Since Z2 is abelian, Z(Z2) = Z2. For Σ3, the centre Z(Σ3) is trivial (easy to check, since there are only 6 elements). Thus we obtain homomorphisms Z(i) Z(p) Z2 −! e −! Z2: Since p ◦ i is an isomorphism, it follows that also Z(p ◦ i) = Z(i) ◦ Z(p) is an isomorphism. Clearly, this is impossible. Thus there does not exist a functor Z taking any group to its centre. 5. Given functors F : D!C and G: E!C, show that there is a category, called the comma category F # G, which has (1) as objects, triples (d 2 obj(D); e 2 obj(E); f : F d ! Ge), and (2) as morphisms (d; e; f) ! (d0; e0; f 0), a pair of morphisms (h: d ! d0; k : e ! e0) so that the square f F d > Ge F h Gk _ _ F d0 > Ge0 f 0 INTRODUCTION TO CATEGORY THEORY, I 3 commutes in C, i.e., so that f 0 ◦ F h = Gk ◦ f. Solution: For a triple (d; e; f) there is the commutative diagram f F d > Ge idF d idGe _ _ F d > Ge: f Consider then triples (d; e; f), (d0; e0; f 0) and (d00; e00; f 00). Assume there are pairs of morphisms (h: d ! d0; k : e ! e0) and (h0 : d0 ! d00; k0 : e0 ! e00) and commutative diagrams f F d > Ge F h Gk _ _ F d0 > Ge0 f 0 and f 0 F d0 > Ge0 F h0 Gk0 _ _ F d00 > Ge00: f 00 Since F and G are functors, F (h0 ◦h) = F (h0)◦F (h) and G(k0 ◦k) = G(k0)◦G(k). Thus combining the diagrams above yields the commutative diagram f F d > Ge F (h0 ◦ h) G(k0 ◦ k) _ _ F d00 > Ge00: f 00 6. Given a partially ordered set (X; ≤) we make a topological space by defining a subset U of X to be open iff for all x; y 2 X, x ≤ y and x 2 U imply y 2 U. This topology on X is called the Alexandroff topology w.r.t. the order ≤. Given two partially ordered sets (X; ≤) and (Y; ≤) , a function f : X ! Y is called monotone increasing for the orderings, if x ≤ y implies f(x) ≤ f(y), for all x; y 2 X. Show that f is monotone increasing if and only if it is continuous for the Alexandroff topologies. Show that this gives a functor Pos ! Top. Solution: Assume f : X ! Y is monotone increasing. Let U be an open subset of Y . Let x 2 f −1(U). Assume y 2 X and x ≤ y. Since f is monotone increasing, it follows that f(x) ≤ f(y). Since U is open, it follows that f(y) 2 U. Thus y 2 f −1(U), and it follows that f −1(U) is open in X. Therefore, f is continuous. 4 INTRODUCTION TO CATEGORY THEORY, I Assume then that f : X ! Y is continuous. Let x; y 2 X and assume x ≤ y. Now U = fz 2 Y j f(x) ≤ zg is an open subset of Y containing f(x). (If z 2 U and z ≤ w, then by reflexivity of the partial order relation, f(x) ≤ w. Thus w 2 U, and it follows that U is open.) Since f is continuous, it follows that f −1(U) is open in X. Since x 2 f −1(U) and x ≤ y, it follows that y 2 f −1(U). Thus f(y) 2 U. But this implies that f(x) ≤ f(y). Therefore, f is monotone increasing. Functoriality: Let F : Pos ! Top be defined as follows: To a partially ordered set (X; ≤) it assigns the topological space X equipped with the Alexandroff topol- ogy. Denote that space by F (X). To a monotone increasing function f it assigns the function F (f) = f itself. Then F (f) is continuous. Now, F (idX ) = idX and for composition F (g ◦ f) = g ◦ f = F (g) ◦ F (f). Thus F is well-defined functor. 7. Let C and D be locally small categories. A functor F : C!D is called faithful, if for each A; B 2 obj(C), the function Hom(A; B) ! Hom(F (A);F (B)); f 7! F (f); is injective. Find an example of a functor F : C!D such that F is faithful but there exist distinct morphisms f1 and f2 in C with F (f1) = F (f2). Solution: Let C consist of two objects a and b, and two morphisms 1a and 1b. Let D consist of the object c and the morphism 1c. Let F : C!D be the functor with F (a) = c = F (b) and F (1a) = 1c = F (1b). Then F is faithful.
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