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MA5209 Algebraic Topology Lecture 4. Category Theory and Homotopy Theory (1, 4 September 2009) Wayne Lawton Department of Mathematics National University of Singapore S14-04-04, [email protected] http://math.nus.edu.sg/~matwml Categories A category C consists of the following three mathematical entities: •A class ob(C), whose elements are called objects; •A class hom(C), whose elements are called morphisms or maps or arrows. Each morphism f has a unique source object a and target object b. We write f: a → b, and we say "f is a morphism from a to b". We write Hom(a, b), or Mor(a, b), to denote the hom-class of all morphisms from a to b. •A binary operation, called composition of morphisms, such that for any three objects a, b, and c, we have Hom(a, b) × Hom(b, c) → Hom(a, c). The composition of f: a → b and g: b → c is written as gf and is governed by two axioms: Associativity: If f : a → b, g : b → c and h : c → d then h(gf) = (hg)f, and Identity: For every object x, there exists a morphism 1x : x → x called the identity morphism for x, such that for every morphism f : a → b, we have 1b f f f 1a ob(C) = Sets, Mor(a,b) = { functions f : a b} ob(C) = Groups, Mor(a,b) = { homomorphisms f : a b} ob(C) = Top Spaces, Mor(a,b) = { maps f : a b} ob(C) = Pointed Top Spaces, Mor((X,p),(Y,q)) ={ maps f : a b such that f(p) = q} ob(C) = Top Spaces, Mor(a,b) = [a,b] = homotopy classes of maps f : a b with [f] [g] = [f g] Commutative Diagram for Composition of Morphisms Properties of Morphisms A morphism f : a → b is: a monomorphism (or monic) if f g1 = f g2 implies g1 = g2 for all morphisms g1, g2 : x → a. an epimorphism (or epic) if g1 f = g2 f implies g1 = g2 for all morphisms g1, g2 : b → x. an isomorphism if there exists a morphism g : b → a with f g = 1b and g f = 1a. an endomorphism if a = b. End(a) denotes the class of endomorphisms of a. an automorphism if f is both an endomorphism and an isomorphism. Aut(a) denotes the class of automorphisms of a. Homotopy Equivalence Definition Topological Spaces X and Y are homotopy equivalent if [X, Y] contains an isomorphism. This means that there exist maps f : X Y and g : Y X such that g f 1X and f g 1Y Question 1. Explain what isomorphisms are in the following categories: sets, topological spaces with maps as morphisms, groups. Question 2. Show that in any category the composition of isomorphisms is an isomorphism. Question 3. Use Q2 to prove that homotopy equivalence is an equivalence relation on the class of topological spaces. Question 4. Prove that and are homotopy equivalent. Relative Homotopy Definition Let X, Y be topological spaces and A X. Maps f, g : X Y with f |A g | A are homotopic relative to A, written f g rel A, if there exists a homotopy F : X [0,1] Y from f to g such that F (a, t ) f (a) g (a), (a, t ) A [0,1] Definition A groupoid is a category in which every morphism is an isomorphism. Question 5 Show that a group is a groupoid with exactly one object. Contractible Spaces Definition A topological space X is contractible if it is homotopy equivalent to a single point space. Lemma X is contractible iff there exists x0 X and a map F : X [0,1] X such that F ( x,0) x, F ( x,1) x0 , x X Theorem If X is contractible and f, g : [0,1] X with f(0) = g(0) and f(1) = g(1) then f g rel {0,1} Proof Construct H : [0,1] [0,1] X by t [ 12 ,1] t [0, 12 ] H ( s, t ) F ( f (0),2s), s [0, t ] H (s, t ) F ( f ( 1s2tt ),2t ), s (t ,1 t ) H ( s, t ) F ( g (0),2s), s [0,1 t ] H ( s, t ) F ( f (1),2 2s), s [1 t ,1] H ( s, t ) F ( g (1),2 2s), s [t ,1] H (s, t ) F ( g ( 1s212tt ),2 2t ), s (1 t , t ) Geometry of Theorem f F ( f (0),0) f (0) g (0) F ( f (0), t ) F ( f ( s ), t ) F ( g ( s ), t ) x0 F ( g (0), t ) g f (1) g (1) Connectedness Definition A topological space X is disconnected if any of the following equivalent conditions hold: 1. A, B X A , B , A open, B open, X A B and A B 2. A, B X A , B , A closed , B closed, X A B and A B 3. A X A , A X , A closed and A closed otherwise it is connected. A subset Y of a topological space X is d. / c.. if it is d. / c. when it is given the subspace topology. Result The connected subsets of R with the usual topology are exactly the intervals. The image of a connected set under a map is connected. The product of connected subsets is connected. Path Connectedness Definition In a topological space X a point p is path connected to a point q if there exists a map f : [0,1] X such that f(0) = p and f(1) = q. Lemma Path conn. is an equivalence relation. Proof. p is connected to p by the constant map f : [0,1] X defined by f(x) = p, x in X. If p is connected to q by a path f : [0,1] X then q is connected to p by the path g : [0,1] X defined by g(s) = f(1-s), s in [0,1]. If p is connected to q by a path a and q is connected to r by a path b then p is connected to r by the path ba defined by (ba)( s) a(2s) for s [0, 12 ], (ba)( s) b(2s 1) for s [ 12 ,1]. Definition X is path connected if X is an equiv. class Local Path Connectedness Definition X is locally path connected if for every x in X and every open O containing x there exists an open U containing x and U subset O such that every pair of points in U is path connected. Theorem In a locally path connected space path connected equiv. classes are open and closed. Proof Let E be an equiv. class and x E. Then x X and X is open hence there exists an open set O with x O and each point in O is path connected to x and hence O E E open. If x E there exists a path connected open O with x O hence E O O E x E E closed Corollary If X is locally path connected and connected then X is path connected. Fundamental Groupoid ( X ) of a pathwise connected topological space X is the following category: the objects are the points of X; for p, q in X the morphisms Mor(p,q) are the equivalence classes of maps f : [0,1] X with f(0) = p, f(1) = q with respect to homotopy relative to {0,1}, and for p, q, r in X the composition of [a] in Mor(p,q) with [b] in Mor(q,r) (a, b : [0,1] X with a(0) = p, a(1) = b(0) = q and b(1) = r), is given by [b][a] = [ba] where [b] and [a] denote the homotopy (relative to {0,1}) equivalence classes containing a and b respectively. Question 6. Prove that ( X ) is a groupoid. Remark The objects are not sets! Fundamental Group Fix a topological space X. For every p in X, the set Mor(p,p) together with the composition defined in the groupoid on the previous page, is a group, called the fundamental group of X based at p, and denoted by 1 ( X , p) Question 7. Show that if p,q in X, then every [a] in Mor(p,q) defines an isomorphism [a] : 1 ( X , p) 1 ( X , q) 1 by the formula [a] ([b]) [aba ] Covariant Functors Let C and D be categories. A covariant functor F from C to D is a function that associates to each object X in C an object F(X) and to each morphism f : X Y in C a morphism F(f) : F(X) F(Y) such that if X Y Z F (1X ) 1F ( X ) and F ( g f ) F ( g ) F ( f ) f then g Example Fix a topological space X and let C be the associated groupoid category, let D be the category of groups, and for p, q in X and [a] in Mor(p,q) define F ( p) 1 ( X , p) F ([a]) [a] Mor ( F ( p), F (q)) Question 8. Show that this describes a covariant functor from the category ( X ) to the category of groups. The Fundamental Group Functor Definition The category of pointed topological spaces has objects (X,p) and Mor((X,p),(Y,q)) = { maps f : X Y with f(p) = q }. Consider the rule that associates to each pointed space ( X , p) the group 1 ( X , p) and to each pointed map f : ( X , p) (Y , q) the function 1 ( f ) : 1 ( X , p) 1 (Y , q) defined by 1 ( f )([a]) [ f a]) Question 9. Show that 1 ( f ) is well defined. Question 10. Show that 1 ( f ) is a group homomorphism. Question 11. Show that 1 is a functor. Question 12. Show that if f : X Y is a homotopy equivalence then 1 ( f ) : 1 ( X , x) 1 (Y , f ( x)) is an isomorphism for all x X . Category of Groupoids The objects of this category are groupoids and for every pair of groupoids a and b, Mor(a,b) is the set of (covariant) functors from a to b. Question 13. Explain how to define composition of morphisms in this category and show that it actually produces a category. The Groupoid Functor Definition For each topological space X let ( X ) be its fundamental groupoid and for each map f : X Y let ( f ) be the functor from the category ( X ) to the category (Y ) defined by: ( f )( x) f ( y ), x X . ( f )([ a]) [ f a], [a] Mor ( x, y ) (here a : [0,1] X , a(0) x, a(1) y ) Question 14. Show this defines a functor from the CAT = topological spaces to CAT = groupoids Question 15. Show that f g : X Y functors ( f ) and (g ) are homotopic (naturally equiv.) Simply Connected Spaces Theorem The trivial group is the one having a single element, namely, its identity element. Lemma If a space X is path connected, then 1 ( X , p) 1 ( X , q), p, q X . ( is omorphism) Question 16. Prove this. Definition A space is simply connected if it is path connected and its fundamental group is trivial. Lemma Every contractible space is simply conn. Question 17. Prove this and then show that every convex subspace of an affine space and the n n subspaces S \ { p}, n 0, p S are simp. conn. Simply Connected Spaces Theorem If X is a topological space and U , V X are open subsets that are simply connected (regarded as subspaces) and such that U V is path connected and X U V then X U V . Corollary S n , n 2 is simply connected. n 1 n 1 Proof S { ( x1 ,, xn 1 ) R : k 1 x 1 } is clearly path connected. Let p (0,,0,1), q p and U S n \ { p}, V S n \ {q} then show that n U V S \ { p, q} is path connected. n 2 k Simply Connected Spaces Proof (Theorem) Clearly X is pathwise connected. To show that it is simply connected we choose p X and a loop a : [0,1] X based at p , namely a(0) a(1) p. It suffices to show that there exists an integer n N and loops ai : [0,1] X , i 1,..., n based at p , such that each ai c p rel {0,1} where c p ( s ) p, s [0,1], and a an (an1 ((a3 (a2 a1 )) )), rel {0,1}. Proof By Lebesgue’s lemma there exists 0 t0 t1 tn 1 with each a([t k 1 , t k ]) is contained in U or V . Join p to each a(tk ), k 1,..., n 1 by a path k : [0,1] X in U ,V ,U V if a(tk ) U ,V ,U V respectively. Construct bk : [0,1] X , bk (s) a((tk tk 1 )s tk 1 ) 1 1 1 a b , a ( b ), , a ( b and 1 1 1 2 1 2 2 n 1 n2 n 1 n 1 ), an n 1bn . Covering Spaces Definition Let E and X be topological spaces. A map p : E X is a covering map (and E is a covering space of X) if every x X has an open 1 admissible neighborhood U such that p (U ) is a disjoint union of open sets Si , i I index set such that for every i I the restriction pi p |Si : Si U is a homeomorphism. The S , i I are called sheets over U and the i 1 i maps si p : U Si are (local) sections of p since p si 1U , i I . Covering Spaces Lemma If p : E X is a covering map then 1 1. Each fibre p ( x), x X is a discrete subspace. 2. p is a local homeomorphism. 3. p is surjective and X has the quotient topology. Question 18 Prove these statements. Lemma Let E be a topological group, H E be a discrete subgroup, p : E X E / H be the map p (e) eH , e E onto the set X of left cosets, and give X the quotient topology. Then p is a c.m. Question 19 Prove this lemma. Question 20 Prove that R / Z Tc , SU (2) /{ I } SO(3) Fibre Bundles are 4-tuples ( E , X , F , p) consisting of a total space E , a base space X , a fiber (space) F , and a projection p : E X , such that there exists an open cover of X and for each O 1 a homeomorphism O : O F p (O ) such that p O ( x, y) x, ( x, y) O F . Example (trivial) E X F , p( x, y ) x, ( x, y ) X F n 1 Example E SO(n), X S , F SO(n 1), p(M ) M:,1 Example X connected, P : E X a covering, F p 1 ( x) for any x X . Question 21. Prove this. Lifting Definition If p : E X is a map, Y is a space, ~ then a map f : Y E is a lift of a map f : Y X ~ if p f f or, equivalently, the diagram ~ f Y E f p commutes. X 2 Question 22. Show that if p : SO(3) S is the projection on to the first column of M SO (3) then ~ 2 2 2 of f 1 : S S there exists a lift f : S SO(3) S2 iff there exists a nowhere vanishing tangent 2 S vector field on . Then what can you conclude? Fibrations Definition A surjective map p : E X is a fibration if it satisfies the following homotopy lifting property: ~ If h i0 p f then there exists h that makes Y i0 Y [0,1] commute. f h E Remark: ~ is a homotopy h p from i0 to i1 X where h ik ( y) ( y, k ), k 0,1 ~ and h is a lift of h Fibrations Theorem If ( E , X , F , p) is a fiber bundle and X is paracompact then p : E X is a fibration. Proof Corollary 14 on page 96 in Spanier’s Algebraic Topology (requires much work). Definition A paracompact space is one in which every open cover has a locally finite refinement. Hence every compact space is paracompact Question 23. Show that if F is not discrete then ~ the lift h may not be uniquely determined by h and f . Unique Lifting Theorem If p : E X is a covering, Y is conn.&loc ~ ~ X and conn. and f1 , f 2 : Y E lift a map f : Y ~ ~ ~ ~ y1 Y such that f1 ( y1 ) f 2 ( y1 ) then f1 f 2 . ~ ~ Proof Let Y1 { y Y : f1 ( y ) f 2 ( y )}. Since y1 Y1 and Y is connected, it suffices to show that Y1 is both open and closed. Since we assume that all spaces are Hausdorff, it is closed (why?). For 1 y Y1 choose admissble open U f ( y ) p (U ) is a disjoint union of open sheets and let S be the sheet that contains f1 ( y) f 2 ( y). Y loc. conn. conn. open y Vk f 1 (U ) f k (Vk ) p 1 (U ), k 1,2. 1 W open&closed in p (U ) f k (Vk ) S Hence p |S injective and ~ ~ ~ ~ p f1 p f 2 f f1 f 2 on V1 V2 . Coverings are Fibrations Theorem (Covering Homotopy Theorem 3 on page 66 in Singer and Thorps Lecture Notes …) If p : E X is a covering and Y is compact and ~ h i0 p f then h that makes the diagram commute. Proof. Let U1 ,...,U r be an admissible f Y E open cover of i0 Y [0,1] ~ h p h h(Y [0,1]). Then cover Y X V1 ,...,Vs and find 0 t0 tk 1 such that each h(Va [ti , ti 1 ]) U b for some U b . Then use glueing and induction. Implications of Unique Homotopy Lifting Corollary 1.(p. 67 in S&T) If p : E X is a covering e E 1 ( p) : 1 (E, e) 1 ( X , p(e)) is injective. Corollary 2.(p. 67 in S&T) If p : E X is a covering and a : [0,1] X , e E , p(e) a(0) then there ~ ~ a exists unique lift a : [0,1] E with (0) e. Corollary 3.(p. 68 in S&T) If p : E X is a covering e E there exists a ‘natural’ correspondence 1 between the fiber p ( p(e)) and the set of cosets 1 ( X , p(e)) / 1 ( p) 1 ( E, e) Theorem 4.(p. 71 in S&T) X path.&loc. path. conn. loc. simply conn. H subgroup of 1 ( X , x) covering p : E X 1 ( p)1 ( E, e) H . Galois Theory of Covering Transformations Definition If p : E X a homeomorphism c : E E is called a covering transformation if p c p. Question 24 Show that these form a group G, that this group acts freely on each fiber. Theorem 9.(p. 76 in S&T)* If 1 ( p) 1 ( E, e) is a normal subgroup of 1 ( X , p(e)) then G 1 (E, e) \ 1 ( p) 1 ( E, e) and X is the quotient space of E under the action of G. Remark* E is pathwise connected, p is surjective. Assignments Read Handouts: [1] Appendix: Generators and relations [2] pages 131-140 in Armstrong’s Basic Topology [3] pages 62-77 in Singer and Thorpe’s Lecture Notes on Elementary Topology and Geometry Do Problems 20-24 on page 140 of [2]. The ‘dunce cap’ in problem 22 is the polyhedron of a 2-simplex. Learn how to compute the fundamental group of a simplicial complex using the algorithm in [2] and describe the group by its generators and relations as described in [1]. Learn van Kampen’s theorem. Projects Write a 5-10 page report and give a 20 minute talk on either 5,9 October on 1 of following topics: 1. classification of 2-dimensional manifolds 2. Jordan curve theorem 3. Riemann surfaces 4. Covering spaces (and their Galois theory) 5. Differential topology theory (Milnor’s book) 6. Higher homotopy groups 7. Knot theory 8. Lie group topology 9. Gauss-Bonnet theorem 10. Fibrations and Fiber Bundles11. Graph theory 12. Groupoids and van Kampen’s theorem or another topic related to algebraic topology Homotopy Problem 1. Let X be a topological space, let a : [0,1] X be a path from p a (0) to q a(1), 1 1 define a : [0,1] X by a ( s) a(1 s), and 1 define a a : [0,1] X by 1 a ( 2 s ), s [ 0 , 2] 1 a a ( s) 1 1 a (2s 1) a(2 2s), s [ 2 ,1]. Construct a homotopy H RELATIVE TO {0,1} 1 from a a to the constant path c p : [0,1] X defined by c p ( s ) p, s [0,1]. Homotopy Solution 1. We need to construct a map H : [0,1] [0,1] X with the following 3 properties: 1 H (s,0) a a (s), s [0,1], H ( s,1) c p ( s) p, s [0,1], H (0, t ) H (1, t ) p, t [0,1]. Strategy: construct H as a composition of simple G a maps [0,1] [0,1] [0,1] X where 2s(1 t ), s [0, 12 ], Glueing lemma G and G ( s, t ) (2 2s)(1 t ), s [ 12 ,1]. H a G are continuous. s [0, 12 ] H (s,0) a(2s), s [ 12 ,1] H (s,0) a(2 2s) s [0, 12 ] H (s,1) a(0) p, s [ 12 ,1] H (s,1) a(0) p, H (0, t ) H (1, t ) p, t [0,1]. Covering Maps Problem 2. Let E Tc {3,1,1,3}, X Tc {1,1} have the topology as subspaces of C and define p : E X by c z T p( z 3) z 1, p( z 1) z 1, p( z 1) z 1. 2 2 i. Draw a ‘picture’ of E and X and p. Solution X 1 b E a 1 a1 2 2 0 b2 2 a3 b3 a 0 b p(ai ) a, p(bi ) b, i 1,2,3, p 1 (0) {2,0,2}. Covering Maps Problem 2. Let E Tc {3,1,1,3}, X Tc {1,1} have the topology as subspaces of C and define p : E X by c z T p( z 3) z 1, p( z 1) z 1, p( z 1) z 1. 2 2 ii. Is p a covering map? Solution: p is a 3-1 C.M. b1 E a2 a1 2 0 b2 2 a3 1 b3 X a 0 b p(ai ) a, p(bi ) b, i 1,2,3, p 1 (0) {2,0,2}. Covering Maps 1 ( X ,0). iii. Compute Solution Free group generated by a and b. b1 E a2 a1 2 0 b2 2 a3 1 b3 a X 0 b p(ai ) a, p(bi ) b, i 1,2,3, p 1 (0) {2,0,2}. Covering Maps iv. Compute 1 ( E,2), 1 ( E,0), 1 (E,2). Solution Free groups generated by 1 2 2 3 2 1 2 1 3 1 2 1 3 3 3 2 {a1 , b1b2 , b a a b , b a b a b }, 1 2 1 1 1 1 2 3 1 3 3 2 2 2 1 2 {b b , b a b , a a , a b3a3}, and 1 1 1 1 2 {b , a a , a b b a , a b a b a } respectively. 1 X b1 E a2 a b b 3 a1 2 0 0 2 a3 p(ai ) a, p(bi ) b, i 1,2,3, b2 p 1 (0) {2,0,2}. Covering Maps v. Compute 1 ( p)(1 (E,2)), 1 ( p)(1 (E,0)), 1 ( p)(1 (E,2)). Solution Free subgroups of 1 ( X ,0) generated by 2 1 2 1 1 These 3 subgroups {a, b , b a b, b a bab}, are conjugate to 2 1 2 1 {b , b ab, a , a ba}, and each other! 2 1 2 1 1 {b, a , a b a, a b aba} respectively. b1 E a2 a1 2 0 b2 2 a3 1 b3 a X 0 b p(ai ) a, p(bi ) b, i 1,2,3, p 1 (0) {2,0,2}. Cofibrations Definition A map i : A X is a cofibration if it satisfies the following homotopy extension property: if there exist g, G that makes the solid diagram below commute ~ g then there exists that makes the augmented diagram below 1A{0} A {0} A [0,1] commute. G Compare with i 1{0} i 1[ 0,1] fibrations on Y ~ g g slide 25. X {0} 1X {0} X [0,1] Contravariant Functors Given categories C,D . A contracovariant functor F from C to D is a function that associates to each object X in C an object F(X) and to each morphism f : X Y in C a morphism F(f) : F(Y) F(X) such that if X Y Z f g then F (1X ) 1F ( X ) and F ( g f ) F ( f ) F ( g ) Example Let C be a category, A in Obj(C) and define F ( X ) Mor ( X , A) F ( f )( a) a f , f Mor ( X , Y ), a Mor (Y , A) Question 24. Show this describes a covariant functor from the category C to the category SETS Natural Transformations Given categories C, D and functors F,G from C to D. A natural transformation from F to G is a function T : Obj (C ) Mor ( D) Such that if F, G are covariant (contravariant) the left (right) diagram below commutes for every h Mor ( X , Y ) (h) F ( X ) F F (Y ) T(X ) T (Y ) ( h) G( X ) G G(Y ) F (h) F ( X ) F (Y ) T(X ) T (Y ) G ( h) G( X ) G(Y ) Question 25. Show why NT are also called homotopies. Suggestion. Let I {0,1} be the two object groupoid and derive a correspondence between NT and functors H from C I to D such that H | C{0} F , H | C{1} G.