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INTRODUCTION TO THEORY, I

Homework 2, solutions

1. Find an example to show that the objects and in the of a F : C → D do not necessarily define a . 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 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 X. Solution: For a X, let F (X) denote the set of connected com- ponents of X. For a continuous 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 with the property that F (G) equals G/C(G), for all groups G? Here C(G) is the of G, i.e., the subgroup of G generated by elements ghg−1h−1 for any g, h ∈ G. Solution: Let’s first check that C(G) is normal in G, so that G/C(G) is a group: Assume c ∈ C(G) and g ∈ G. Then gcg−1c−1 ∈ C(G), and hence also gcg−1 = (gcg−1c−1)c ∈ C(G). Thus C(G) is normal. Let H and G be groups and let f : H → G be a . Then f(hgh−1g−1) = f(h)f(g)f(h)−1f(g)−1, for every h, g ∈ 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 , 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 , 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 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 = {e}, where e is the identity element of Σ3. The homomorphism p is the quotient homomorphism. Then p◦i is a group . 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 F : D → C and G: E → C, show that there is a category, called the F ↓ G, which has (1) as objects, triples (d ∈ obj(D), e ∈ 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 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 (X, ≤) we make a topological space by defining a subset U of X to be open iff for all x, y ∈ X, x ≤ y and x ∈ U imply y ∈ U. This 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 ∈ X. Show that f is monotone increasing if and only if it is continuous for the Alexandroff . 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 ∈ f −1(U). Assume y ∈ 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) ∈ U. Thus y ∈ 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 ∈ X and assume x ≤ y. Now U = {z ∈ Y | f(x) ≤ z} is an open subset of Y containing f(x). (If z ∈ U and z ≤ w, then by reflexivity of the partial order relation, f(x) ≤ w. Thus w ∈ U, and it follows that U is open.) Since f is continuous, it follows that f −1(U) is open in X. Since x ∈ f −1(U) and x ≤ y, it follows that y ∈ f −1(U). Thus f(y) ∈ 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 ∈ 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. 8. Let G be a group. Let G-Set be the category of G-sets, i.e., the objects are sets X equipped with an action of G. The morphisms are G-equivariant functions X → Y , where X and Y are G-sets. The fixed point set of a G-set X is the set XG = {x ∈ X | gx = x ∀g ∈ G}. Check that G-Set indeed is a category. Show that there is a functor F : G-Set → Set with F (X) = XG, for every G-set X.

Solution: Let X be a G-set and let 1X be the identity function X → X. Then, for every g ∈ G, and for every x ∈ X, 1X (gx) = gx = g1X (x). Thus 1X is G-equivariant. Let X, Y and Z be G-sets, and let f : X → Y and h: Y → Z be G-equivariant functions. Then, for every g ∈ G and for every x ∈ X, (h ◦ f)(gx) = h(f(gx)) = h(gf(x)) = gh(f(x)) = g(h ◦ f)(x). Thus h◦f is G-equivariant. Since the composition law for functions is associative, composing G-equivariant functions is also an associative operation. It follows that G-Set is a category. Let X and Y be G-sets and let f : X → Y be a G-equivariant function. Assume x ∈ XG. Then, for every g ∈ G, gf(x) = f(gx) = f(x). Thus f(x) ∈ Y G. Define a functor F : G-Set → Set as follows: For objects X in G-Set, F (X) = XG. For morphisms f : X → Y in G-Set, i.e., for G-equivariant functions, let F (f): XG → Y G, F (f)(x) = f(x). Since f(XG) ⊂ Y G, it follows that F (f) is well-defined. INTRODUCTION TO CATEGORY THEORY, I 5

Clearly, F (1X ) = 1XG and for compositions, F (h ◦ f) = F (h) ◦ F (f). Thus F is a well-defined functor.