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(I) Show That a Morphism Can Have at Most One Inverse Isomorphism. Solution: Let F : X → Y Be a Morphism and Assume G, H: Y → X Are Inverses of F

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(I) Show That a Morphism Can Have at Most One Inverse Isomorphism. Solution: Let F : X → Y Be a Morphism and Assume G, H: Y → X Are Inverses of F INTRODUCTION TO CATEGORY THEORY, I Homework 1, solutions 1. (i) Show that a morphism can have at most one inverse isomorphism. Solution: Let f : X ! Y be a morphism and assume g; h: Y ! X are inverses of f. Thus gf = hf = 1X and fg = fh = 1Y . Then g = g1Y = g(fh) = (gf)h = 1X h = h: (ii) Consider a morphism f : x ! y. Assume there exists a pair of morphisms g; h: y ! x such that gf = 1x and fh = 1y. Show that g = h and f is an isomorphism. Solution: Here g = g1y = g(fh) = (gf)h = 1xh = h: Then hf = gf = 1x and fh = 1y, so that h is the inverse of f. Thus f is an isomorphism. 2. Let C be a category. Let the maximal groupoid of C be the subcategory of C containing all objects of C and only those morphisms that are isomorphisms. Show that the maximal groupoid indeed is a subcategory of C. Solution: Let x be an object of C. Then 1x = 1x1x = 1x so that 1x is its own inverse and an isomorphism. Thus 1x is a morphism in the maximal groupoid. As- sume then that f : x ! y and g : y ! z are morphisms in the maximal groupoid, i.e., they are isomorphisms in C. There are morphisms h: y ! x and d: z ! y such that hf = 1x, fh = 1y, dg = 1y and gd = 1z. Then (gf)(hd) = g(fh)d = g1yd = g(1yd) = gd = 1z and (hd)(gf) = h(dg)f = h1yf = h(1yf) = hf = 1x: Thus dg is an inverse of gf, and it follows that gf is an isomorphism in C, i.e., a morphism in the maximal groupoid. Therefore, the maximal groupoid is a subcategory of C. (The composition of morphisms in the maximal groupoid is the same as in C.) 3. Prove the following Lemma (Riehl, Lemma 1.2.11): (1) If f : x ! y and g : y ! z are monomorphisms, then so is gf : x ! z. (2) If f : x ! y and g : y ! z are morphisms so that gf is monic, then f is monic. Dually: (1) If f : x ! y and g : y ! z are epimorphisms, then so is gf : x ! z. Date: June 12, 2018. 1 2 INTRODUCTION TO CATEGORY THEORY, I (2) If f : x ! y and g : y ! z are morphisms so that gf is epic, then g is epic. Solution: (1) Assume f : x ! y and g : y ! z are monomorphisms. Let h1; h2 : v ! x be morphisms. Assume (gf)h1 = (gf)h2. Then g(fh1) = g(fh2). Since g is monic, it follows that fh1 = fh2. Since f is monic, it follows that h1 = h2. Therefore, gf is a monomorphism.. (2) Assume f : x ! y and g : y ! z are morphisms so that gf is monic. Let h1; h2 : v ! x be morphisms such that fh1 = fh2. Then (gf)h1 = g(fh1) = g(fh2) = (gf)h2. Since gf is monic, it follows that h1 = h2. Thus f is a monomorphism. Dually: (1)Let f : x ! y and g : y ! z be epimorphisms. Let k1; k2 : z ! w be morphisms and assume k1(gf) = k2(gf). Then (k1g)f = k1(gf) = k2(gf) = (k2g)f. Since f is epic, it follows that k1g = k2g. Since g is epic, it follows that k1 = k2. (2) Let f : x ! y and g : y ! z be morphisms so that gf is epic. Let k1; k2 : z ! w be morphisms and assume k1g = k2g. Then k1(gf) = (k1g)f = (k2g)f = k2(gf). Since gf is epic, it follows that k1 = k2. Therefore, g is an epimorphism. 4. Show that the inclusion i: Z ,! Q is both a monomorphism and an epimor- phism in the category Ring of rings. Conclude that map that is both monic and epic does not need to be an isomorphism. Solution: Let R be a ring and let f; g : R ! Z be ring homomorphisms. Assume i ◦ f = i ◦ g. Then f(x) = i(f(x)) = i(g(x)) = g(x), for every x 2 Z Thus f = g, and it follows that i is monic. Assume next that R0 is a ring and p; q : Q ! R0 are ring homomorphisms. Assume p ◦ i = q ◦ i. Then p(n) = q(n) for every n 2 Z. Since p is a ring homomorphism, it follows that 1 1 1 0 = p(1) = p(n · ) = p(n)p( ) R n n 1 1 1 0 0 for every n 2 Z, n 6= 0. Similarly p( n )p(n) = 1R , q( n )q(n) = 1R and q(n)q( n ) = 1R0 , for every n 2 Z, n 6= 0. Then 1 1 1 1 p( ) =p( )1 0 = p( ) q(n)q( ) n n R n n 1 1 = p( )q(n) q( ) n n 1 1 = p( )p(n) q( ) n n 1 =1 0 q( ) R n 1 =q( ); n INTRODUCTION TO CATEGORY THEORY, I 3 for every n 2 Z, n 6= 0. Thus m 1 1 m p( ) = p(m)p( ) = q(m)q( ) = q( ); n n n n for every m; n 2 Z, n 6= 0. Hence p = q, and it follows that i is epic. Clearly, i is not surjective, hence it can not be a ring isomorphism. 5. A morphism f : a ! b in a category C is called regular if there exists a morphism g : b ! a with fgf = f. Show that f is regular, if it has either a left or a right inverse. Prove that every morphism A ! B, where A 6= ;, is regular in Set. Solution: 1) Assume f : a ! b has a right inverse g : b ! a. Then fg = 1b. Thus fgf = (fg)f = 1bf = f; and it follows that f is regular. Assume then g : b ! a is a left inverse of f. Then gf = 1a. Thus fgf = f(gf) = f1x = f; and again it follows that f is regular. 2) Let A and B be sets, A 6= ;, and let f : A ! B be a function. Let f(A) = ff(x) j x 2 Ag: Let f~: A ! f(A); x 7! f(x); and let i: f(A) ! B; y 7! y; be the inclusion. Then f = i ◦ f~. Now, f~ is a surjection, which implies that it has a right inverse: For every y 2 f(A) choose xy 2 A such that f(xy) = y. Defineg ~: f(A) ! A, y 7! xy. ~ Then f ◦ g~ = 1f(A). Since i is an injection, it has a left inverse: Let y0 2 f(A). Define h: B ! f(A) as follows: If y 2 f(A), then h(y) = y. If y 2 B n f(A), then h(y) = y0. Then h ◦ i = 1f(A). Thus h is a left inverse of i, and it follows that i is regular. Let g : B ! A be the composed functiong ~ ◦ h. Then fgf = (i ◦ f~) ◦ (~g ◦ h) ◦ (i ◦ f~) = i ◦ (f~◦ g~) ◦ (h ◦ i) ◦ f~ ~ = i ◦ 1f(A) ◦ 1f(A) ◦ f = i ◦ f~ = f: Therefore, f is regular. 6. Let C be a category defined as follows: The objects of C are triples (A; α; a), where A is a set, α: A ! A is a function and a 2 A. A morphism f :(A; α; a) ! (B; β; b) 4 INTRODUCTION TO CATEGORY THEORY, I is a function f : A ! B satisfying f ◦ α = β ◦ f and f(a) = b: Verify that C is a category. Solution: Let (A; α; a) be an object in C. Then A is a set, a 2 A and α: A ! A is a function. Let 1A : A ! A be the identity function of A. Then 1A(a) = a and α ◦ 1A = 1A ◦ α. Hence 1A :(A; α; a) ! (A; α; a) is a morphism in C. Assume then that f :(A; α; a) ! (B; β; b) and g :(B; β; b) ! (C; γ; c) are morphisms in C. Then f : A ! B and g : B ! C are functions, f(a) = b and g(b) = c. Thus g ◦ f : A ! C is a function and (g ◦ f)(a) = g(f(a)) = g(b) = c. Moreover, f ◦ α = β ◦ f and g ◦ β = γ ◦ g. Thus (g ◦ f) ◦ α = g ◦ (f ◦ α) = g ◦ (β ◦ f) = (g ◦ β) ◦ f = (γ ◦ g) ◦ f = γ ◦ (g ◦ f): Thus g ◦ f is a morphism in C. 7. Let C be a category and let D and E be subcategories of C. Define the intersection D\E of D and E as follows: (1) The objects of D\E are all A such that A is both an object of D and an object of E. (2) If A and B are objects of D\E, then f : A ! B is a morphism in D\E if it is a morphism both in D and in E. Show that D\E is a well-defined category and a subcategory of C. How about the union of D and E? Solution: Let A be an object in D\E. Then A is an object in D (resp. in E). Since D (resp. E) is a subcategory of C, it follows that 1A is a morphism in D (resp. in E). Thus 1A is a morphism in D\E. Assume next f : A ! B and g : B ! C are morphisms in D\E. Then f : A ! B and g : B ! C are morphisms in D (resp. in E). Thus gf : A ! C is a morphism in D (resp. in E). But then gf is a morphism in D\E.
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