Category Theory - Homework Randall R. Holmes Auburn University c 2014 by Randall R. Holmes Copyright Last revision: November 11, 2014 HW 1 (Monday, 8/25) 1 1–1 Let X and Y be sets. A relation from X to Y is a subset R of X ×Y . If R is a relation from X to Y , then for x ∈ X and y ∈ Y we write xRy to mean (x, y) ∈ R. Let C be the class of all sets. For sets X and Y , let C(X, Y ) be the set of all relations from X to Y . For sets X, Y , Z, and relations R ∈ C(X, Y ), S ∈ C(Y, Z), define a relation SR ∈ C(X, Z) by putting x(SR)z if and only if xRy and ySz for some y ∈ Y . Prove that C with morphisms and composition so defined is a category. Solution: We need only check associativity of the composition and existence of identity morphisms. (Associativity) Let X, Y , Z, and W be sets and let R ∈ C(X, Y ), S ∈ C(Y, Z), and T ∈ C(Z, W ). For x ∈ X and w ∈ W , we have x[(T S)R]w ⇐⇒ xRy and y(T S)w for some y ∈ Y ⇐⇒ xRy and ySz and zT w for some y ∈ Y , z ∈ Z ⇐⇒ x(SR)z and zT w for some z ∈ Z ⇐⇒ x[T (SR)]w. Therefore, (T S)R = T (SR) and the composition is associative. (Identity) Let X be a set and let 1X ∈ C(X, X) be the identity relation on X: 1X = {(x, x) | x ∈ X}. We claim that 1X is an identity morphism on X. Let Y be a set and let R ∈ C(X, Y ). Let x ∈ X and y ∈ Y . Assume that x(R1X )y. Then x1X z and zRy for some z ∈ X. By the definition of 1X we have z = x, so xRy. Now assume that xRy. Then x1X x and xRy, so x(R1X )y. This shows that R1X = R. A similar argument shows that 1X S = S for every S ∈ C(Y, X). This completes the proof. HW 2 (Wednesday, 9/3) 2 2–1 Let Haus be the full subcategory of Top consisting of all Hausdorff spaces. Prove the following: If α : X → Y is a morphism in Haus and im α is dense in Y (i.e., im α = Y ), then α is epic. (The converse also happens to be true.) Use this result to give an example of a morphism in Haus that is epic but not surjective. Solution: We first show that a continuous map into a Hausdorff space is completely determined by its effect on a dense subset, that is, we establish the following claim: Claim: Let β1 , β2 : Y → Z be continuous maps with Z Hausdorff, let D be a dense subset of Y , and assume that β1 (d) = β2 (d) for all d ∈ D. Then β1 = β2 . Assume to the contrary that β1 (y) 6= β2 (y) for some y ∈ Y . Since Z is Hausdorff, there exist disjoint open sets U1 and U2 in Z with βi (y) ∈ Ui (i = 1, 2). Then y ∈ β1−1 (U1 ) ∩ β2−1 (U2 ) =: U . Since U is a nonempty open subset of Y and D = Y , there exists d ∈ U ∩ D. Then the element β1 (d) = β2 (d) is in U1 ∩ U2 , contradicting that U1 and U2 are disjoint. We conclude that β1 = β2 as claimed. Let α : X → Y be a morphism in Haus and assume that im α is dense in Y . Let β1 , β2 : Y → Z be morphisms in Haus and assume that β1 α = β2 α. In particular, β1 (y) = β2 (y) for all y ∈ im α, so β1 = β2 by the general result established above. Therefore α is epic. The inclusion map α : Q → R is epic (since Q is dense in R), but it is not surjective. HW 3 (Monday, 9/8) 3 2–2 Let C be a category. (a) Prove that a morphism in C is monic and split epic if and only if it is epic and split monic. (b) Prove that if every monic morphism in C is split monic, then every bimorphism in C is an isomorphism. (Note: The dual statement with “epic” replacing “monic” also holds.) Solution: (a) Let α : x → y be a morphism in C. (⇒) Assume that α is monic and split epic. By Theorem 2.4.1, α is epic. Since α is split epic, there exists a morphism β : y → x such that αβ = 1y . We have α(βα) = (αβ)α = 1y α = α = α1x , so βα = 1x since α is monic. Therefore, α is split monic. (⇐) Assume that α is epic and split monic. By Theorem 2.2.1, α is monic. Since α is split monic, there exists a morphism β : y → x such that βα = 1x . We have (αβ)α = α(βα) = α1x = α = 1y α, so αβ = 1y since α is epic. Therefore, α is split epic. Note: The two statements, (⇒) and (⇐), are dual to each other, so it would have sufficed to prove only one of the statements. We gave both proofs in order to give another illustration of the fact that choice of notation can at times somewhat obscure the dual nature of proofs (see comments after Theorem 2.4.1). (b) Assume that every monic morphism in C is split monic. Let α : x → y be a bimorphism. Then α is monic and epic by definition. By our assumption, α is split monic. Then α is split epic as well by part (a), so α is an isomorphism by Theorem 2.5.1. HW 4 (Monday, 9/15) 4 3–1 For groups G1 and G2 let ιi : Gi → G1 × G2 be the injections defined by ι1 (g1 ) = (g1 , e2 ) and ι2 (g2 ) = (e1 , g2 ), where ei is the identity of Gi . Prove that (G1 × G2 , {ιi }) is not, in general, a coproduct of the family {Gi } in the category Grp. Hint: Let G be a nonabelian group and take Gi = G (i = 1, 2). In the definition of coproduct, let αi = 1G : Gi → G (i = 1, 2). Solution: Let G be a nonabelian group (G = S3 , the symmetric group on three elements, for instance). Let Gi = G and put Q = G1 ×G2 . Suppose that (Q, {ιi }) is a coproduct of the family {Gi } in the category Grp. Put αi = 1G : Gi → G (i = 1, 2). There exists a homomorphism γ : Q → G such that γιi = αi (i = 1, 2). For a, b ∈ G we have ab = 1G (a)1G (b) = α1 (a)α2 (b) = γ(ι1 (a))γ(ι2 (b)) = γ((a, e2 ))γ((e1 , b)) = γ((a, b)) = γ((e1 , b)(a, e2 )) = γ((e1 , b))γ((a, e2 )) = γ(ι2 (b))γ(ι1 (a)) = α2 (b)α1 (a) = 1G (b)1G (a) = ba, so that G is abelian, contrary to assumption. We conclude that (Q, {ιi }) is not a coproduct of the family {Gi } in the category Grp. HW 5 (Monday, 9/22) 5 3–2 Let C be a category. Prove that the following are equivalent: (i) Finite products exist in C. (See definition before Theorem 3.7.4.) (ii) C has a terminal object, and for each pair c1 , c2 ∈ C there exists a product of the family {ci }i∈I in C, where I = {1, 2}. Solution: We begin by establishing a claim: Claim: An object c of C is terminal if and only if the pair (c, ∅) is a product of the empty family in C. Let D = Dpr be the auxiliary category in the definition of product of the empty family. Then x 7→ (x, ∅) defines a bijection C → D. Moreover, for each pair x, y ∈ C we have C(x, y) = D((x, ∅), (y, ∅)) (the containment (⊇) always holds and here we get (⊆) as well since there are no commutative diagrams to be satisfied in the definition of morphism in D). It follows that c ∈ C is a terminal object of C if and only if (c, ∅) is a terminal object of D, this latter being the definition of product of the empty family. (i ⇒ ii) Assume that (i) holds. Then C has a product of the empty family of objects and hence a terminal object by the claim. Part (ii) follows. (ii ⇒ i) Assume that (ii) holds. Then C has a terminal object and hence a product of the empty family by the claim. If c is an object of C, then (c, {1c }) is a product of the singleton family {c}. Let I = {1, 2, . . . , n} (n ∈ N ∪ {0}) and let {ci }i∈I be an indexed family of objects of C. We argue by induction on n that there exists a product of this family in C. The cases n = 0 and n = 1 have been handled. Assume that n > 1 and put I 0 = {1, 2, . . . , n − 1}. By the induction hypothesis there exists a product (p0 , {πi0 }) of the family {ci }i∈I 0 . In turn, by the assumption in (ii) there exists a product (p, {β, πn }) of the family {p0 , cn }. For each i ∈ I 0 , put πi = πi0 β : p → ci . (See diagram below.) Claim: The pair (p, {πi }) is a product of the family {ci }i∈I in C. Let D and D0 be the auxiliary categories in the definition of product of the families {ci }i∈I and {ci }i∈I 0 , respectively. Let (x, {αi }i∈I ) be an object of D. Then (x, {αi }i∈I 0 ) is an object of D0 , so there exists a unique morphism γ 0 : x → p0 HW 5 (Monday, 9/22) 6 such that πi0 γ 0 = αi for all i ∈ I 0 : αn > cn d πn γ x γ0 αi / p. β % 0 p πi0 πi ci t Next, since (p, {β, πn }) is a product of the family {p0 , cn }, there exists a unique morphism γ : x → p such that βγ = γ 0 and πn γ = αn . For each i ∈ I 0 we have πi γ = πi0 βγ = πi0 γ 0 = αi . Therefore, πi γ = αi for each i ∈ I, implying that γ : (x, {αi }) → (p, {πi }) is a morphism in D. Finally, let δ : (x, {αi }) → (p, {πi }) be a morphism in D and put δ 0 = βδ. For each i ∈ I 0 we have πi0 δ 0 = πi0 βδ = πi δ = αi , so βδ = δ 0 = γ 0 by the uniqueness of γ 0 . But also, πn δ = αn , so uniqueness of γ (see how this morphism arose in the preceding paragraph) gives δ = γ. Therefore, γ is the unique morphism in D from (x, {αi }) to (p, {πi }) and the proof is complete. HW 6 (Monday, 9/29) 7 3–3 Let K be a field. Prove that coequalizers exist in the category MatK (see Example 1.3.5). Hint: First assume that one of the two given morphisms is the zero matrix. Solution: Let L : n → m be a morphism in MatK (so L is an m × nmatrix). Let P be the matrix with ith row pi , where (p1 , . . . , pq ) is an ordered basis of (col L)⊥ (= orthogonal complement of the column space of L). Then P L = 0 = P 0, so (q, P ) is an object of the auxiliary category D = Dcoeq in the definition of coequalizer of the morphisms L, 0 : n → m. Claim: The pair (q, P ) is a coequalizer of the morphisms L and 0. Let (x, A) ∈ D. We have AL = A0 = 0, so ai ∈ (col L)⊥ for each i, where ai denotes the ithP row of A. Therefore, for each i there exist unique gik ∈ K such that ai = k gik pk . Hence A = GP , where G = [gij ] ∈ MatK (q, x): n L 0 // 6 xO A m P ( G q. Moreover, due to the uniqueness of the gik , the morphism G is the unique one making the triangle in the diagram commutative. Thus, G is the unique morphism in D from (q, P ) to (x, A) and the claim is established. Finally, note that the auxiliary category D = Dcoeq associated with two arbitrary parallel morphisms L1 , L2 : n → m in MatK is identical to the auxiliary category associated with the morphisms L, 0 : n → m, where L = L1 − L2 , so the proof is complete. HW 7 (Monday, 10/6) 8 4–1 Let C be a small category (see Section 4.3). Prove that there exists a faithful functor F : C → Set (so that the pair (C, F ) is concrete). Solution: Define F : C → Set as follows: For c ∈ C let F (c) be the set of all morphisms in C with target c: [ F (c) = C(x, c) x∈C (this is a union of a family of sets indexed by C and since C is a set by assumption it follows that F (c) is indeed a set). For a morphism α : c → d in C define F (α) : F (c) → F (d) by F (α)(γ) = αγ. If α : c → d and β : d → e are morphisms in C, then for every γ ∈ F (c) we have F (βα)(γ) = (βα)γ = β(αγ) = F (β) F (α)(γ) = F (β)F (α)(γ), so F (βα) = F (β)F (α). And for each c ∈ C and γ ∈ F (c), we have F (1c )(γ) = 1c γ = γ = 1F (c) (γ), so F (1c ) = 1F (c) . We conclude that F is a functor. Finally, let α, β : c → d be morphisms in C and assume that F (α) = F (β). Then α = α1c = F (α)(1c ) = F (β)(1c ) = β1c = β. Therefore, F is faithful and the proof is complete. HW 8 (Monday, 10/13) 9 4–2 Let C be a category and let c be a fixed object of C. Prove that the (covariant) Hom functor C(c, − ) : C → Set preserves pullbacks. (Cf. Example 4.1.2.) Solution: Let λi : ai → b (i = 1, 2) be two morphisms in C and let (p, (σ1 , σ2 )) be a pullback of the pair (λ1 , λ2 ). It suffices to establish the following claim. Claim: C(c, p), (σ1∗ , σ2∗ ) is a pullback of the pair λ1∗ , λ2∗ . c X α2 (x) ∃! γ(x) p α1 (x) σ2 α2 γ / # a2 α1 σ1 a1 λ1 λ2 /b. # C(c, p) σ2∗ ' / C(c, a2 ) σ1∗ C(c, a1 ) λ1∗ λ2∗ / C(c, b) . Commutativity of the square on the left implies commutativity of that on the right since functors preserve commutative diagrams (Theorem 4.2.1). This implies that C(c, p), (σ1∗ , σ2∗ ) is an object of the auxiliary category D∗ = Dpb in the definition of pullback of the pair λ1∗ , λ2∗ . Let (X, (α1 , α2 )) be an object of D∗ . Let x ∈ X. We have λ1 ◦ (α1 (x)) = λ1∗ (α1 (x)) = λ1∗ α1 (x) = λ2∗ α2 (x) = λ2∗ (α2 (x)) = λ2 ◦ (α2 (x)), so the pair (c, (α1 (x), α2 (x))) is an object of the auxiliary category D = Dpb in the definition of pullback of the pair (λ1 , λ2 ). Therefore, there exists a unique morphism γ(x) : (c, (α1 (x), α2 (x))) → (p, (σ1 , σ2 )) in D. In particular, γ(x) ∈ C(c, p), so we have a function γ : X → C(c, p). For x ∈ X we have σi∗ γ(x) = σi∗ (γ(x)) = σi ◦ (γ(x)) = αi (x) (i = 1, 2), so σi∗ γ = αi (i = 1, 2), implying that γ is a morphism in D∗ . Let γ 0 : (X, (α1 , α2 )) → C(c, p), (σ1∗ , σ2∗ ) be a morphism in D∗ . For each x∈X σi ◦ (γ 0 (x)) = σi∗ (γ 0 (x)) = σi∗ γ 0 (x) = αi (x) HW 8 (Monday, 10/13) 10 (i = 1, 2), so that γ 0 (x) = γ(x) by the uniqueness of γ(x). Hence γ 0 = γ and the claim is established. HW 9 (Monday, 10/27) 11 5–1 The center Z(C) of a category C is the class of all natural transformations η : 1C → 1C , where 1C is the identity functor on C. Let R be a ring with identity and put C = R Mod. Prove that there is a bijection Z(R) → Z(C), where Z(R) is the center of R, that is, Z(R) = {z ∈ R | zr = rz for all r ∈ R}. z : M → M given by Solution: Let z ∈ Z(R). For M ∈ C the map ηM z (m) = zm is a morphism in C (this uses the fact that z is central). Let ηM α : M → N be a morphism in C: M α N M α z ηM N /M z ηN α /N . For m ∈ M we have z z z z αηM (m) = α(ηM (m)) = α(zm) = zα(m) = ηN (α(m)) = ηN α(m), z = η z α, showing that η z : 1 → 1 is a natural transformation, so that αηM C C N that is, η z ∈ Z(C). Therefore, f : Z(R) → Z(C) given by f (z) = η z is a well-defined map. Let η ∈ Z(C). For M ∈ C and m ∈ M the map ρm : R → M given by ρm (r) = rm is a morphism in C: R ρm M R ρm ηR M /R ηM ρm /M , so naturality of η gives ηR (1)m = ρm (ηR (1)) = ρm ηR (1) = ηM ρm (1) = ηM (ρm (1)) = ηM (m). (1) In particular, letting M = R this equation gives ηR (1)r = ηR (r) = rηR (1) for every r ∈ R, implying ηR (1) ∈ Z(R). Therefore, g : Z(C) → Z(R) given by g(η) = ηR (1) is well-defined. HW 9 (Monday, 10/27) 12 For z ∈ Z(R) we have z gf (z) = g(f (z)) = g(η z ) = ηR (1) = z1 = z, implying gf = 1Z(R) . For η ∈ Z(C), M ∈ C, and m ∈ M we have (using Equation (1) for the last equality) g(η) [f g(η)]M (m) = [f (g(η))]M (m) = ηM (m) = g(η)m = ηR (1)m = ηM (m), implying f g = 1Z(C) . Therefore, f : Z(R) → Z(C) is a bijection as desired. HW 10 (Monday, 11/10) 13 6–1 Prove that a composition of equivalences is an equivalence. Solution: First proof: Let F1 : C → D and F2 : D → E be equivalences. Then there exist functors G1 : D → C and G2 : E → D and natural isomorphisms η 1 : G1 F1 → 1C , θ1 : F1 G1 → 1D , η 2 : G2 F2 → 1D , and θ2 : F2 G2 → 1E . Put F = F2 F1 : C → E and G = G1 G2 : E → C. Define η : GF → 1C by ηx = ηx1 G1 (ηF2 1 (x) ) and note that this is an isomorphism since functors preserve isomorphisms (Theorem 4.2.2). For a morphism α : x → y in C we get F1 (α) : F1 (x) → F1 (y) and hence commutative squares F1 (x) F1 (α) F1 (y) G1 G2 F2 F1 (x) G1 G2 F2 F1 (x) 2 G1 (ηF 1 (x) ) / G1 F1 (x) ηx1 G1 F1 (α) G1 G2 F2 F1 (y) 2 G1 (ηF ) 1 (y) / G1 F1 (y) ηy1 /x α / y x α y which yields the commutative square x α y GF (x) GF (α) ηx α GF (y) /x ηy / y , so that GF ∼ = 1C . Interchanging the roles of Fi and Gi we see that G1 and G2 are equivalences and F G ∼ = 1E as well. Therefore F = F2 F1 is an equivalence. Second proof: Here we use the characterization of equivalence given in Theorem 6.2.1. Let F1 : C → D and F2 : D → E be equivalences and put F = F2 F1 : C → E. Let α, β : x → y be morphisms in C and assume that F (α) = F (β). We HW 10 (Monday, 11/10) 14 have F2 (F1 (α)) = F (α) = F (β) = F2 (F1 (β)) so F1 (α) = F1 (β) since F2 is faithful, and then α = β since F1 is faithful. Therefore, F is faithful. Let x and y be objects of C and let γ : F (x) → F (y) be a morphism. We have γ : F2 (F1 (x)) → F2 (F1 (y)), so using that F2 is full there is a morphism β : F1 (x) → F1 (y) such that F2 (β) = γ. In turn, since F1 is full there is a morphism α : x → y such that F1 (α) = β. Then F (α) = F2 (F1 (α)) = F2 (β) = γ, so F is full. Let e be an object of E. Since F2 is essentially surjective there exists d ∈ D and an isomorphism β : F2 (d) → e. Since F1 is essentially surjective there exists c ∈ C and an isomorphism α : F1 (c) → d. Then βF2 (α) : F (c) → e is an isomorphism (using Theorem 4.2.2), so F is essentially surjective. By Theorem 6.2.1 F = F2 F1 is an equivalence. 6–2 For a functor F : C → D and a small category A define a functor F A : C A → DA by F A (f ) = F f for f ∈ C A and F A (ν)a = F (νa ) for ν a morphism in C A . Prove that a functor F is an equivalence if and only if F A is an equivalence for each small category A. Solution: Let F : C → D be a functor. (⇒) Assume that F is an equivalence so that there exists a functor G : D → C and natural isomorphisms η : GF → 1C and θ : F G → 1D . Let A be a small category. For f ∈ C A define (η A )f : GA F A (f ) → f by ((η A )f )a = ηf (a) . Fix f ∈ C A . For a morphism α : a → b in A, we have commutativity of the diagram on the right a α b GA F A (f )(a) GA F A (f )(α) ((η A )f )a GA F A (f )(b) / f (a) ((η A )f )b f (α) / f (b) since f (α)((η A )f )a = f (α)ηf (a) = ηf (b) GF (f (α)) = ((η A )f )b GA F A (f )(α), HW 10 (Monday, 11/10) 15 where the second equality uses naturality of η. Therefore, (η A )f is a natural transformation (i.e., a morphism in C A ). Next, for a morphism ν : f → g in C A we get commutativity of the diagram on the right f ν g GA F A (f ) GA F A (ν) (η A )f ν GA F A (g) /f (η A )g / g since, for a ∈ A, (ν(η A )f )a = νa ηf (a) = ηg(a) GF (νa ) = ((η A )g GA F A (ν))a , where the second equality uses naturality of η. So η A : GA F A → 1C A is a natural transformation. For each f ∈ C A the natural transformation (η A )f is a natural isomorphism since each component morphism ((η A )f )a = ηf (a) (a ∈ A) is an isomorphism. Therefore, η A is a natural isomorphism (since each of its component morphisms is an isomorphism). Therefore, GA F A ∼ = 1C A . Interchanging the roles of F and G we see that G is an equivalence and F A GA ∼ = 1DA as well. Therefore, F A is an equivalence. (⇐) Assume that F A is an equivalence for each small category A. Let A be the category with one object • and one morphism 1• . We claim that C ∼ = C A . Define a functor J : C → C A by J(c)(•) = c, J(c)(1• ) = 1c for c ∈ C and J(α)• = α for a morphism α in C. Define a functor K : C A → C by K(f ) = f (•) for f ∈ C A and K(ν) = ν• for a morphism ν in C A . For each object c of C and morphism α in C we have KJ(c) = K(J(c)) = J(c)(•) = c, KJ(α) = K(J(α)) = J(α)• = α, and for each object f of C A and each morphism ν in C A we have JK(f )(•) = J(K(f ))(•) = K(f ) = f (•), JK(ν)• = J(K(ν))• = K(ν) = ν• , implying JK(f ) = f and JK(ν) = ν. Therefore KJ = 1C and JK = 1C A , so C ∼ = C A as claimed. HW 10 (Monday, 11/10) 16 Let K : DA → D be the functor obtained by replacing C by D in the definition of K above. For an object c of C and a morphism α in C we have KF A J(c) = K(F J(c)) = F J(c)(•) = F (c), KF A J(α) = F A (J(α))• = F (J(α)• ) = F (α), so F = KF A J. Since a composition of equivalences is an equivalence (Exercise 6–1) the proof is complete. HW 11 (Monday, 11/17) 17 7–1 Prove that for every group G there exists a universal from G to the inclusion functor F : Ab → Grp.

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