Introduction to Algebraic Topology. Mitschri der Vorlesung von Dr. M. Michalogiorgaki

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1 Introduction to Algebraic Topology Mitschri der Vorlesung von Dr. M. Michalogiorgaki Tobias Berner Universität Zürich Frühjahrssemester 2009

2 Contents 1 Topology Topological spaces and continuous functions Topological spaces Continuous functions Conectedness & Compactness Connected spaces Compactness Algebraic topology Fundamental group Path homotopy e fundamental group Covering spaces Li ing properties e fundamental group of the circle and applications e Fundamental eorem of Algebra Deformation retracts and homotopy type Seifert von Kampen eorem Direct sums of abelian groups Free abelian groups Free products of groups Free groups e Seifert-van Kampen theorem CW complexes (cell complexes) Surfaces (two-dimensional manifolds) Fundamental group of surfaces Homology of surfaces Classi cation of surfaces Knot theory Classi cation of covering spaces Index 66 2

3 CONTENTS Literature Muncres: Topology 2nd edition Jänich: Topologie Massey: Algebraic Topology: An Introduction Stöcker, Zieschang: Algebraische Topologie Lickorish: An introduction to knot theory Introduction In calculus, you have studied R n, n N, as well as functions f R k R l, k, l N. You have studied notions such as neighbourhood of a point x R n, as well as convergence and continuity of a function f at a point x R k. For this study, you have used the Euclidean metric. For instance if f R R, we say that f is continuous at x R if ε > 0 δ > 0 such that if x x 0 < δ then f(x) f(x 0 ) < ε. We used the metric d R R R, with (x, y) x y. In general the Euclidean metric d R n R n R with (x 1,..., x n ), (y 1,..., y n ) n i=1(x i y i ) 2. In topology, we study notions such as neighbourhood of a point x X, convergence, continuity for a general set X. For this study a metric is not necessary. What we will use is open sets in X. In algebraic topology, we use abstract algebra to study topological properties. 3

4 1 Topology 1.1. Topological spaces and continuous functions Topological spaces Consider a set X and P(X) = {U U X}. De nition 1.1 A topology on X is a collection T P(X) such that 1., X T. 2. If U i T i I, then i I U i T. 3. If U i T, i {1,..., n}, then i {1,...,n} U i T. (X, T ) is called a topological space (sometimes we will only write X). U X is called open, if U T. U X is called closed, if X/U T. Example X some set. T d = {U U X} discrete topology. T t = {, X} trivial topology. 2. X = {x 1, x 2, x 3 }. en the three collections {, X} {, {x 1 }, {x 1, x 2 }, X} {, {x 2 }, {x 1, x 2 }, {x 2, x 3 }, X} are topologies on X. 3. X is a set. en the collections T f = {U X X/U is nite or X/U = X} nite topology. T c = {U X X/U is countable or X/U = X} countable topology. are topologies on X. [Homework] 4

5 1.1 Topological spaces and continuous functions De nition 1.3 Suppose that T and T are topologies on X. If T T, then T is called ner than T. If T T, then T is called strictly ner than T. If T T or T T, then T and T are comparable. De nition 1.4 Let (X, T ) be a topological space and A X. e interior of A is inta = U. U T,U A e closure of A is A = U. X/U T, A U Basis for a topology De nition 1.5 A basis B for a topological space (X, T ) is a collection B of open subsetes X (i.e. B T ), such that U T {B i } i I, B i B, i I, with U = i I B i. Remark 1.6 A basis is not unique. Properties: B basis for (X, T ) then B has the following properties 1. x X B B with x B. PROOF X T and B is a basis X = i I B i, B i B, i I. So x B j for some j I. 2. If x B 1 B 2, B 1, B 2 B then B 3 B such that x B 3 B 1 B 2. PROOF B 1, B 2 B T B 1, B 2 T B 1 B 2 T B 1 B 2 = i I B i, Conversely: If a collection B of subsets of X satis es properties 1. and 2. then there is a unique topology T for which B is a basis. T is called the topology generated by B and it consists of all unions of elements of B. PROOF We prove that T is a topology 1. X T? Property 1 if x X then B x B with x B x X. erefore X = x X B x, i.e. X T. T? is the empty union of elements in B, so T. 2. If U i T i I, does i I U i T? U i T U i = j<inj B ij. So i I U i = i I j J B ij T. 3. If U i T for i {1,..., n}, is i {1,...,n} U i T? We will show that if U 1, U 2 T then U 1 U 2 T. U 1 = λ Λ B λ, U 2 = U k K B k. Consider x U 1 U 2. en x B λx and x B kx, for some λ x Λ and k x K. us x B λx B kx [by property 2.] B x B with x B x B λx B kx U 1 U 2. erefore U 1 U 2 = x U1 U 2 B x T. 5

6 Topology Example X = R, B = {(a, b) a, b R, a < b}. B satis es properites 1. and 2. therefore it generates a topology on R. is is the standard topology on R. 2. X = R, B = {(a, b) a, b Q, a < b} B satis es properties 1. and 2. Claim [Homework]: e topology T generated by B is in fact the standard topology on R. Lemma 1.11 Let B, B be bases for the topologies T and T on X. en the following statements are equivalent: 1. T T. 2. x X, B B with x B there is B B such that x B B. PROOF Consider x X and B B with x B. B T T B T B = λ Λ B λ, B λ B. So x B, therefore x B λ, for some λ x Λ. i.e. we have B λ x B, such that x B λ x B Consider U T and x U. en x U = λ Λ B λ, B λ B. at is x B λx, for some λ x Λ. 2. implies that there exists B λ x B such that x B λ x B λx. en U x U B λ x U U = x U B λ x U T. erefore T T. Product topology Let (X, T x ), (Y, T Y ) be two topological spaces. e product topology is the topology with basis the collection B = {U V U T X, V T Y }. Metric topology X is a set. De nition 1.13 A metric on this set is a function d X X R with 1. d(x, y) 0, and d(x, y) = 0 iff x = y. 2. d(x, z) d(x, y) + d(y, z) x, y, z X. 3. d(x, y) = d(y, x), x, y X. We call the set B d (x, ε) = {y X d(y, x) < ε} the ε-ball centered at x. {B d (x, ε)} x X,ε>0 is a basis for a topology on X, the metric topology. 6

7 1.1 Topological spaces and continuous functions Example 1.14 R with the standard topology (T stand ). Consider R R = R 2. Claim: e product topology on R 2 and the metric topology are the same. For this, one has to show T pr T m By lemma 1.11 one has to show x R 2, B pr B pr, x B pr, B m B m such that x B m B pr. T m T pr Subspace topology De nition 1.15 (X, T ) is a topological space and Y X. e collection T Y = {U Y U T } is a topology on Y, called the subspace topology. Example 1.16 Y = [0, 1] {2} X = R with the standard topology. en the sets (a, b), with a, b [0, 1], [0, b), with b [0, 1], (a, 1], with a [0, 1], {2}, [0, 1] are open sets in the subspace Y Continuous functions De nition 1.17 Let (X, T X ), (Y, T Y ) be topological spaces and f X Y. f is called continuous if f 1 (V ) T X V T Y. Claim f R R (with standard topology). e ε δ de nition of continuity is equivalent to the de nition above. PROOF Suppose that f R R is continuous with the de nition above. Consider x 0 R and ε > 0. en V = (f(x 0 ) ε, f(x 0 ) + ε) T stand, so f 1 (V ) T stand by our de nition. Now x 0 f 1 (V ), so there exists (a, b) B stand with x 0 (a, b) f 1 (V ). Take δ = min{x 0 a, b x 0 }. Clearly δ > 0. en x x 0 < δ x (a, b) f 1 (V ) f(x) V f(x) (f(x) ε, f(x) + ε) f(x) f(x 0 ) < ε. 7

8 Topology eorem 1.19 Let X, Y be topological spaces and f X Y. e following are equivalent 1. f is continuous. 2. for every closed subset of Y the inverse image of it is a closed subset of X. PROOF Let B be closed in Y, then X/f 1 (B) = f 1 (Y /B) which is open in X, i.e. f 1 (B) is closed in X Lemma 1.21 (the pasting lemma) Let X, Y be topological spaces and A, B closed subsets of X with X = A B. Let f A Y, g B Y be continuous with f(x) = g(x) x A B. en h X Y with h(x) = { f(x) x A g(x) x B is continuous. PROOF Let C be a closed subset of Y. en h 1 (C) = f 1 (C) g 1 (C) where f 1 (C) is closed in A and g 1 (C) is closed in B. (a) f 1 (C) is closed in A f 1 (C) = A G, where G closed in X. (b) g 1 (C) is closed in B g 1 (C) = B H, where H closed in X. (a) f 1 (C) is closed in X, (b) g 1 (C) is closed in X. h 1 (C) = f 1 (C) g 1 (C) is closed in X. De nition 1.23 Let X, Y be topological spaces. f X Y is called a homeomorphism if f is bijective, and f, f 1 are continuous. Bijective correspondence not only between X and Y but also between the collection of open sets in X and the collection of open sets Y. us any property of X that is expressed in terms of its open subsets yields via f the corresponding properties for Y. Such a property is called a topological property. Quotient topology Example 1.24 Torus De nition 1.25 Let X, Y be topological spaces, p X Y a surjective map. e map p is called a quotient map if a subset U of Y is open in Y, if and only if p 1 (U) is open in X. 8

9 1.2 Conectedness & Compactness De nition 1.26 Let X be a topological space, Y be some set and p X Y a surjective map. e quotient topology on Y induced by p is de ned as follows: A subset U Y is open if and only if p 1 (U) X is open. e fact that this is a topology follows from p 1 ( ) =, p 1 (Y ) = X, p 1 ( a J U a ) = a J p 1 (U a ), p 1 ( n i=1 U i ) = n i=1 p 1 (U i ). Remark 1.27 e quotient topology on Y is the nest topology that makes p continuous. Example X = {(x, y) R 2 0 x 2π, 0 y 1} = [0, 2π] [0, 1] R 2. Y = {(x, y, z) R 3 x 2 + y 2 = 1, 0 z 1}. f X Y, with (x, y) (cos x, sin x, y). f is surjective. Using f we can de ne the quotient topology in Y. 2. p R {x 1, x 2, c 3 } p(x) = x 1 x > 0 x 2 x < 0 x 3 x = 0 e quotient topology on {x 1, x 2, x 3 } induced by p is {, {x 1 }, {x 2 }, {x 1, x 2 }, {x 1, x 2, x 3 }} 1.2. Conectedness & Compactness In calculus you have studies functions f [a, b] R and you have proved three theorems for continuous functions f [a, b] R. 1. Intermediate Value eorem (IVT), 2. Maximus Value eorem (MVT), 3. Uniform Continuity eorem (UCT). ese theorems rely on the continuity of f and some topological properties of [a, b] R. In particular IVT relies on the connectedness of [a, b]. MVT and VCT rely on the compactness of [a, b] Connected spaces De nition 1.29 Consider (X, T ) topological space. A separation of X is a pair U, V where U, V T, U, V, U V =, and X = U V. X is called connected if there is no separation of X. Remark 1.30 Connectedness is a topological property. Example R, T stand, Q = ((, a) Q) ((a, ) Q), a R Q. 2. R, [a, b], [a, b), (a, b], (a, b) are connected. 9

10 Topology Claim: X is connected iff the only subsets of X that are both open and closed are and X. PROOF Suppose, that A X, A, A X, and A T X, X A T X. en X = A (X A) A, X A is separation of X, contradiction. If U, V is a separation of X. en U T X, X U = V T x, U, X U, U (X U) =. V T X, X U T X, U, U X, contradiction. erefore X is connected. Lemma 1.33 If C, D are a separation of X and Y is a connected subspace of X, then Y C or Y D. PROOF C Y, D Y are open subsets of Y. In addition (C Y ) (D Y ) =. If C Y and D Y were both nonempty, then they would form a separation of Y. But Y is connected. So C D =, or D Y =. Y C or Y D. eorem 1.35 Let {A i } i I be connected subspaces of X and p i I A i. en i I A i is connected subspace of X. PROOF Assume that U i I A i is not connected, and i I A i = C D, C, D a separation of i I A i. en p C D and WLOG we can assume that p C. A i is a connected subspace. [Lemma] A i C or A i D for any i I. p i I A i. p A i i I. A i C i I. i I A i C. D =. Contradiction. eorem 1.37 Let A be a connected subspace of X and A B A. en B is also connected. PROOF Assume that C, D is a separation of B. By Lemma A C or A D. WLOG A C. A C A C. D B A C D C. Claim: D C =. Proof: e closure of C in B is C B, where C is the closure of C in X. C = C B = C (C D) = (C C) (C D) = C (C D). C D =. We have that D C and we showed that D C =. erefore D =, contradiction. ere is no separation of B, i.e. B is connected. eorem 1.39 Let f X Y be a continuous function between topological spaces X and Y. If X is connected, then f(x) is connected. 10

11 1.2 Conectedness & Compactness PROOF Consider f X f(x), x f(x). en f is continuous. Indeed, if U open in f(x), then U = f(x) V for some V open in Y. f 1 (U) = f 1 (f(x) V ) = f 1 (f(x)) f 1 (V ) = X f 1 (V ) = f 1 (V ) open in X. Suppose that f(x) = C D, C, D separation of f(x). en f 1 (C), f 1 (D) are open subsets of X, disjoint and nonempty and X = f 1 (C) f 1 (D). Contradcition (X connected). eorem 1.41 A nite (cartesian) product of connected spaces (in the product topology) is connected. PROOF We start by proving that if X and Y are connected, then X Y is connected. Consider a b in X Y, X b. X b is connected. Similarly x Y is connected for any x X. As a result, T x = (X b) (x Y ) is also connected (by previous theorem, since x b (X b) (x Y )). erefore, x X T x is connected, as the union of connected subspaces, T x with common point a b. Now X Y = x X T x, therefore X Y is connected. By induction, the general proof follows. eorem 1.43 (IVT) Let f X R be a continuous function and X be a connected space. If a, b X and r (f(a), f(b)), then c X with f(c) = r. PROOF A = f(x) (, r), B = f(x) (r, ). A, B are open in f(x), A B =, A, B (as f(a) A, f(b) B). If r f(x), then f(x) = A B, i.e. A, B is a separation of f(x). However f(x) is connected, because X is connected and f is continuous. I.e. such a separation cannot exist. I.e. r f(x), that is, c X such that f(c) = r. Remark 1.45 One can prove an even more generalised form of IVT, where instead of R one considers any ordered set. De nition 1.46 Let X be a topological space and x, y X. A path in X from x to y is a continuous map f [a, b] X with f(a) = x and f(b) = y. X is called path connected, if every pair of points in X can be joined by a path in x. Claim: If X is path connected, then X is connected. PROOF Suppose X = C D, C, D a separation of X. Consider f [a, b] X a path. f([a, b]) is connected, so f([a, b]) C or f([a, b]) D. is implies that there is no path in X joining a point in C to a point in D. X not path connected, contradiction. Example 1.48 (Topologist s sine curve S) Consider S = {(x, sin 1 x ) 0 < x 1} R2. f (0, 1] R 2, x (x, sin 1 ). f is continuous, (0, 1] is connected, therefore S = x f((0, 1]) is connected. S = S {0 [ 1, 1]}. S is connected S is connected. Claim: S is not path connected. Proof: Suppopse f [a, c] S continuous with f(a) = (0, 0), and f(c) S. f 1 (0 [ 1, 1]) is closed in [a, c] therefore it has a largest element b. en f(b) 0 [ 1, 1] 11

12 Topology and f((b, c]) S. Let f(t) = (x(t), y(t)), t [a, c]. en x(a) = 0, x(t) > 0 and y(t) = sin 1 x(t) for t > b. Given n N, choose u n with 0 < u n < x(b + 1 ) such that n sin 1 u n = (±1) n. By the IVT there exists t n (b, b + 1 n ) with x(t n) = u n. en t n b but y(t n ) does not converge. Contradicting the continuity of f. Remarks on yesterday s example If X = R 2 and S X, then S = {limits of convergent sequences of points in S}. More generally, this is true if X is rst countable, e.g. X is a metric space. If f [a, c] S continuous. Of t [a, c] n N, with t n b, b [a, c], then f(t n ) f(b). In general, continuous implies sequentially continuous. De nition 1.49 Given a topological space X, de ne an equivalence relation on X, by setting x y (x, y X) if there exists a connected subspace of X with x, y Y. e equivalence classes are called the connected component of X. De ne another equivalence relation by setting x y (x, y X) if there is a path in X from x to y. e equivalence classes are called path components of X Compactness De nition 1.50 A topological space X is called compact if for any open covering {U λ } λ Λ of X, i.e. U λ open λ Λ and X = λ Λ U λ, there exist nitely many λ 1,..., λ n Λ such that X = n i=1 U λ i. Compactness is a topological property. Example R is not compact, R = r R (r 1 2, r ), or R = n N(n, n + 2). 2. (a, b], a, b R is not compact. (a, b] = n N (a + 1 n, b]. 3. Subspaces of X with nitely many points are obviously compact. 4. X = {0} { 1 n n N} is compact. Consider {U λ} λ Λ an open covering of X and consider λ 0 Λ with 0 U λ0. en there exists N λ0 sucht that 1 n U λ 0 n N λ0. For each n {1,..., N λ0 1} choose U λn containing it. en N λ 0 i=0 U λ i = X. Order topology De nition 1.52 A relation C on a set X is an order relation if it has the following properties: 1. If x, y X, x y then xcy or ycx. 2. For no x X does xcx hold. 3. If xcy, ycz then xcz, x, y, z X. 12

13 1.2 Conectedness & Compactness Example 1.53 X = R, C = <. If X is a set with an order relation < and a, b X, a < b. (a, b) = {x X a < x < b}, (a, b] = {x X a < x b}, [a, b) = {x X a x < b}, [a, b] = {x X a x b}. De nition 1.54 X is a set with an order relation and more than two elements. Let B be the collection of all sets of the following types: (a, b), a, b X, (a, b 0 ], a X, b 0 the largest element (if any) in X, [a 0, b), b X, a 0 the smallest element (if any) in X, B is a basis for a topology, and we call the topology generated by B the order topology. Example 1.55 R 2 with the dictionary order, i.e. a b < c d if a < c or a = c and b < d. R R has no largest or smallest element. [Graph.. open sets... ] An ordered set X has the least upper bound property, if every subset of X that is bounded above has a least upper bound. eorem 1.56 Let X an ordered set with the least upper bound property. en any closed interval [a, b] in X is compact. [Without proof] Corollary 1.57 In R (T stand = T ordered ), [a, b] is compact for any a, b R. eorem 1.58 A closed subspace Y of a compact space X is compact. PROOF Consider {U λ } λ Λ an open covering of Y. en U λ = Y U λ for some U λ open in X, λ Λ. Denote U = {U λ } λ Λ. Now U = U {X Y } is an open covering of X. X is compact, so a nite subcollection of U covers X. If this contains X Y, discard it. If not I leave the subcollection unchanged. What we obtain is a nite subcollection of U, say {U λ } i {1,...,n} with Y n i=1 U λ i. is implies, that {U λi } i {1,...,n} is a nite subcovering of Y. eorem 1.60 X compact topological spaces. Y topological space and f X Y continuous. en f(x) is compact. PROOF Let U = {U λ } λ Λ be an open covering of f(x). en λ Λ U λ = U λ f(x) for some U λ open in Y. Denote U = {U λ } λ Λ. {f 1 (U λ )} λ Λ is an open covering of X. X is compact, hence it has a nite subcovering {f 1 (U λ i )} i {1,...,n}. en {U λi } i {1,...,n} is a nite covering of X. 13

14 Topology eorem 1.62 e product of nitely many compact spaces is compact. [Proof omitted] eorem 1.63 (Extreme value theorem) X is a compact topological space and Y is an ordered set in the order topology. If f X Y is continuous then c, d X with f(c) f(x) f(d) x X. PROOF X is compact and f is continuous, so f(x) is compact. De ne (, y) = {a Y a < y} and note that if Y has no smallest element, then (, y) = a Y (a, y) and a<y if Y has a smallest element a 0, then (, y) = [a 0, y). (, y) is an open set in Y. If f(x) has no largest element, then {(, y) f(x)} y f(x) is open covering of f(x). Since f(x) is compact, there exist y i, i {1,..., n}, such that {(, y i ) f(x)} i {1,...,n} covers f(x). If y j = max{y i } i {1,...,n}, then y j f(x), but y j {(, y i ) f(x)}, i {1,..., n}, contradiction, since {(, y i ) f(x)} i {1,...,n} is an open covering of f(x). Similarly we can prove, that f(x) has a smallest element.f Hausdorff spaces Terminology: X topological space, x X, U open set in X with x U, then U is called a open neighbourhood of x. De nition 1.65 A topological space X is called Hausdorff if for every pair x 1, x 2 X with x 1 x 2 there exist open neighbourhoods U 1, U 2 of x 1, x 2 respectively with U 1 U 2 =. Facts: Every ordered set with the order topology is a Hausdorff space. If X, Y are Hausdorff spaces, then X Y is Hausdorff. A subspace of a Hausdorff space is Hausdorff. Every nite point set in a Hausdorff space is closed. eorem 1.66 Every compact subspace of a Hausdorff space is closed. PROOF X Hausdorff space, Y X compact Y is closed. We will prove, that X Y is open. Consider x X Y. y Y chose U y, V y open in X with x U y, y V y and U y V y = (this can be done, because X is Hausdorff). en {Y V y } y Y is an open covering of Y. Y is compact, so {Y V yi } i {1,...,n} a nite subcovery of Y. Note that Y V y1 V yn = V. U x = U y1 U yn. en U x V =. Indeed, if z U x V, then z U yi i {1,..., n} and z V yj for some j {1,..., n} but V yi U yj =. So U x V =. We have constructed U x open in X with x U x and U x Y U x V =, i.e. U x X Y. So X Y = x X Y U x and X Y is open. 14

15 1.2 Conectedness & Compactness Remark 1.68 (X, T ) topological space, x X. Some authors (e.g. Munkres) de ne: A neighbourhood of X is a set U T such that x U. Other authors (e.g. Janich) de ne: A neighbourhood of X is a set U X such that V T with x V U. eorem 1.69 Let f X Y be a continuous bijective function. If X is compact, and Y is Hausdorff, then f is a homeomorphism. PROOF We have to prove that f 1 is continuous. Equivalently tat if U is open in X, then (f 1 ) 1 (U) is open in Y. Equivalently, that if U is closed in X, then f(x) is closed in Y. Indeed, U is closed in X and X compact, so f(u) Y is compact. Y is Hausdorff, therefore f(u) is closed in Y. Local connectedness and local path connectedness De nition 1.71 X is called locally (path) connected at x X if for every open neighbourhood U of x there is a (path) connected open neighbourhood V of x with V U. If Y is locally (path) connected at every x X, then X is called locally (path) connected. 15

16 2 Algebraic topology Determining whether two spaces are homeomorphic and studying continuous functions between topological spaces are two of the central problems in topology. To show that X, Y are homemorphic, we need to construct f X Y bijective, continuous, with f 1 continuous. To show that X, Y are not homeomorphic, we can for instance show that X has a topological property (e.g. connectendes or compactness) but Y does not have this property. Any continuous map between topological spaces induces a homeomorphism between there fundamental groups. x 0 X π 1 (X, x 0 ) fundamental group. We will show that if X Y then their fundamental group is isomorphic Fundamental group Path homotopy Notation: I = [0, 1]. De nition 2.1 Let f X Y, f X Y be continuous maps. f, f are called homeotopic if there exists a continuous map F X I Y with F (x, 0) = f(x) x X and F (x, 1) = f (x) x X. F is called a homotopy between f and f. If f, f are homotopic, we write f f. If f is homotopic to a constant map then f is called nullhomotopic. We focus on the special case X = [a, b]. en if f [a, b] Y is a continuous function, f is called a path. f(a) is called the initial point, and f(b) the nal point. WLOG we can assume, that the domain of f is [0, 1]. De nition 2.2 Let f I X, f I X two paths in X. f, f are called path homotopic if f(0) = f (0), f(1) = f (1) and there exists continuous map F I I X, with F (x, 0) = f(x) x I and F (x, 1) = f (x) x I and F (0, t) = f(0) t I, F (1, t) = f(1) t I. If f, f are path homotopic, we write f p f. Claim: and p are equivalence relations. PROOF We prove this for p f p f: F (x, t) = f(x).

17 2.1 Fundamental group 2. f p f f p f: If F is a path homotopy from f to f, then G(x, t) = F (x, 1 t) is a path homotopy from f to f. 3. f p f, f p f f p f : If F is a path homotopy from f to f and F is a path homotopy from f to f, then F (x, 2t) t [0, 1/2] G(x, t) = { F (x, 2t 1) t [1/2, 1] is a path homotopy from f to f. G(x, 1/2) = F (x, 1) = F (x, 0) = f (x). i.e. G is well de ned. Is G continuous? G is continuous in I [0, 1/2] and G is continuous in [1/2, 1] [pasting lemma (lemma 1.21] G is continuous. G(0, t) = { G(1, t) = f(1), F (0, 2t) t [0, 1/2] F (0, 2t 1) t [1/2, 1] = { f(0) = f (0) f (0) = f (0) G(x, 0) = f(x) x I (since G(x, 0) = F (x, 0) = f(x) x I. G(x, 1) = f (x) x I (since G(x, 1) = F (x, 1) = f (x) x I. If f is a path, we denote its homotopy class by [f]. Example Let f I R 2, g I R 2, paths with f(0) = g(0) and f(1) = g(1). en F I I R 2 with (x, t) (1 t)f(x) + tg(x) is a path homotopy between f and g. F is called a linear homotopy. 2. f I R 2, s (cos(πs), sin(πs)), g I R 2, t (cos(πt), 3 sin(πt)). f(0) = (1, 0) = g(0), and f(1) = ( 1, 0) = g(1). f, g are path homotopic. Indeed, consider the linear homotopy F described in f I R 2 {(0, 0)} with s (cos(πs), sin(πs)), and g I R 2 {(0, 0)} with t (cos(πt), sin(πt). Are f, g path homotopic? Yes. Prove using the linear homotopy. 4. f I R 2 {(0, 0)}, with s (cos(πs), sin(πs)), and g I R 2 {(0, 0)} with t (cos(πt), 3 sin(πt)). e linear homotopy is not a path homotopy between f and g. Inf fact, there exists no path homotopy between f and g. We de ne the following operation: De nition 2.5 Let f I X path in X with f(0) = x 0, f(1) = x 1, x 0, x 1 X. g I X path in X with g(0) = x 1, g(1) = x 2, x 2 X. We de ne the product f g of f, g as the path h with h(s) = { f(2s) s [0, 1/2] g(2s 1) s [1/2, 1]. 17

18 Algebraic topology Remark 2.6 h is well de ned, since f(1) = g(0) = h(1/2). h is continuous by the pasting lemma (lemma 1.21). e product f g induces a well de ned product on path homotopy equivalce classes de ned by [f] [g] = [f g]. Indeed, if f, f paths in X with [f] = [f ], and g, g paths in X with [g] = [g ]. Consider path homotopies F, G between f and f, g and g respectively. en H(s, t) = { F (2s, t) t [0, 1/2] G(2s 1, t) s [1/2, 1] is a path homotopy between f g and f g (homework). erefore [f g] = [f g ], i.e. [f] [g] = [f ] [g ] and the induced product is wellde ned. Reparametrisation De ne a reparametrisation of a path f to be a composition f φ, where φ I I continuous map, such that φ(0) = 0, and φ(1) = 1. Reparametrising a path preserves its homotopy class (i.e. [f φ] = [f]). Indeed, consider the path homotopy f φ t whereφ t (s) = (1 t)φ(s) + ts. φ 0 (s) = φ(s) f φ 0 = f φ, φ 1 (s) = s f φ 1 = f. Note that (1 t)φ(s) + ts lies between φ(s) and s, hence (1 t)φ(s) + ts lies in I, so f φ t is de ned e fundamental group We restrict our attention to paths f I X with f(0) = f(1). Call x 0 = f(0) = f(1). Such paths are called loops in X at the basepoint x 0. e set {[f] f I X, f(0) = f(1) = x 0 } is denoted by π 1 (X, x 0 ). If [f], [g] π 1 (X, x 0 ), then f, g are loops in X at the basepoint x 0, so f g is de ned. e induced multiplication [f] [g] = [f g] is well-de ned. Proposition 2.7 (π 1 (X, x 0 ), ) is a group. PROOF associativity: [f] ([g] [h]) = ([f] [g]) [h], [f], [g], [h] π 1 (X, x 0 ). [f] [g h] = [f g] [h] [f (g h)] = [(f g) h]. Claim: f (g h) is a reparametrisation of (f g) h by the piecewise linear function φ: where Proof: (f (g h))(t) = { Analogously ((f g) h)(s) = { φ(t) = 1 2 t t [0, 1/2] t 1 t [1/2, 3/4] 4 2t 1 t [3/4, 1] f(2t) t [0, 1/2] (g h)(2t 1) t [1/2, 1] (f g)(2s) s [0, 1/2] h(2s 1) s [1/2, 1] = = f(2t) t [0, 1/2] g(4t 2) t [1/2, 3/4] h(4t 3) t [3/4, 1] f(4s) s [0, 1/4] g(4s 1) s [1/4, 1/2] h(2s 1) s [1/2, 1]

19 2.2 e fundamental group So if one easily checks, that ((f g) h)(φ(t)) = (f (g h))(t). 2. Identity: Consider c I X with c(s) = x 0 s I. en f c is a reparametrisation of f via φ: 2t t [0, 1/2] φ(t) = { 1 t [1/2, 1] I.e. f c = f φ. [f] = [f φ] = [f c]. [f] = [f] [c]. 3. inverse: Consider [f] π 1 (X, x 0 ) and f(s) = f(1 s). Claim: f f c f f. Proof: Consider the homotopy f(2s) s [0, (1 t)/2] H(s, t) = f(1 t) [(1 t)/2, (1 + t)/2] f(2 2s) [(1 + t)/2, 1] H(0, t) = f(0) = x 0, H(1, t) = f(0) = x 0. f(2s) s [0, 1/2] H(s, 0) = { f(2 2s) s [1/2, 1] H(s, 1) = x 0. = f f H is a homotopy between f f and c i.e. [f f] = [c], i.e. [f] [f] = [c]. Analogously [f] [f] = [c]. [f] is the inverse of [f]. De nition 2.9 π 1 (X, x 0 ) is called the fundamental group of X at x 0. Example 2.10 Consider X R n convex. x 0 X. π 1 (X, x 0 ) = {[f] f I X path, f(0) = f(1) = x 0 }. If f 0, f 1 are two paths in X at the basepoint x 0, then f t (s) = (1 t)f 0 (s) + tf 1 (s) is a linear homotopy between f 0 and f 1. Let x 0, x 1 X, π 1 (X, x 0 ), π 1 (X, x 1 ). Proposition 2.11 If x 0, x 1 are in the same path component of X, then π 1 (X, x 0 ), π 1 (X, x 1 ) are isomorphic. PROOF x 0, x 1 X. Consider h I X a path with h(0) = x 0, h(1) = x 1 and h I X, h(s) = h(1 s). De ne ĥ M 1(X, x 1 ) π 1 (X, x 0 ) where [f] [h f f]. ĥ is well-de ned: If [f] = [g] π 1(X, x 1 ) and f t is a homotopy between f and g then h f t h is a homotopy between h f h and h g g i.e. [h f h] = [h g h]. ĥ is a homomorphism: ĥ([f] [g])ĥ([f g]) = [h f g h] = [h f h h g h] = [hfh][hgh] = ĥ([f])ĥ([g]). 19

20 Algebraic topology ĥ is bijective: ĥ = (ĥ) 1. (ĥĥ)[f] = ĥ(ĥ([f])) = ĥ([hfh]) = = [f]. De nition 2.13 X is simply connected if it is path connected and π 1 (X) is trivial. Proposition 2.14 X is simply connected iff there is a unique homotopy class of paths connecting any two points in X. PROOF Path connectedness is the existence of paths connecting every pair of points in X. So we only need to check uniqueness. Suppose π 1 (X) = 0. Let x 0, x 1 X. If f, g are two paths from x 0 to x 1, then f fgg g. [f] = [g]. If there is a unique homotopy class of paths connecting any two points, then there is a unique homotopy class of paths connecting a point to itself. π 1 (X) = 0. Induced homomorphisms X, Y topological spaces, x 0 X, y 0 Y and h X Y a continuous map with h(x 0 ) = y 0. We will denote this by h (X, x 0 ) (Y, y 0 ). If f I X is aloop in X, based at x 0, then h f is a loop in Y based at y 0. De nition 2.16 Let h (X, x 0 ) (Y, y 0 ) continuous. De ne h π 1 (X, x 0 ) π 1 (Y, y 0 ) where [f] [h f]. h is called the homomorphism induced by h relative to x 0. Remark 2.17 h is well-de ned. If f, g loops in X based at x 0 with [f] = [g]. Consider H a homotopy from f to g. en h H is a homotopy from h f to h g. I.e. [h f] = [h g] h ([f]) = h ([g]). Remark 2.18 h is a group homomorphism. Just calculate and compare the following two expressions: h ([f] [g]) = h ([f g]) = [h (f g)] h ([f]) h ([g]) = [h f] h[ g] = [(h f) (h g)]. Remark 2.19 h depends on the basepoint x 0. So strictly speaking we should have written (h x0 ). Properties of induces homomorphisms 1. If h (X, x 0 ) (Y, y 0 ), g (Y, y 0 ) (Z, z 0 ) continuous, then (g h) = g h. 2. If 1 X (X, x 0 ) (X, x 0 ), then (1 x ) = 1 π1 (X,x 0 ). Proposition 2.20 If H (X, x 0 ) (Y, y 0 ) a homoeomorphism,t hen h π 1 (X, x 0 ) π 1 (Y, y 0 ) is a grou isomorphism. 20

21 2.3 Covering spaces PROOF Consider h 1 (Y, y 0 ) (X, x 0 ) and (h 1 ) π 1 (Y, y 0 ) π 1 (X, x 0 ) (h 1 ) h = (h 1 h) = 1 = 1. h (h 1 ) = (h h 1 ) = 1 = 1. π 1 gives a covariant functor from the category with objects topological spases with basepoint and morphism basepoint preserving continuous maps to the category with objects groups and morphisms group homomorphisms. A category D consists of 1. a collection of objects Ob(D), 2. sets of morphisms Mor(X, Y ) for each X, Y Ob(D) with a distinguished identity morphism 1 X in Mor(X, X), 3. a composition of morphisms Mor(X, Y ) Mor(Y, Z( Mor(X, Z) for each tripel X, Y, Z Ob(D) with the properties f 1 = 1 f = f and (f g) h = f (g h). Example Ob(D) = {G G group}. Mor(G, H) = {f G H f group homomorphism}. 2. Ob(D) = {(X, x 0 ) X topological space, x 0 basepoint}. Mor((X, x 0 ), (Y, y 0 )) = {f (X, x 0 ) (Y, y 0 ) f continuous}. A coveriant functor F from a categoriy C to a category D assigns to each X Ob(C) an F (X) Ob(D) to each f Mor(X, Y ) a F (f) such that F (1 X ) = 1 F (X) and F (f g) = F (f) F (g). Example 2.23 π 1 is a covariant functor from category C 2 in example 2 to the category C 1 in example 1. (X, x 0 ) Ob(C 2 ) π 1 π 1 (X, x 0 ) Ob(C 1 ). f Mor((X, x 0 ), (Y, y 0 )) π 1 f Mor(π 1 (X, x 0 ), (Y, y 0 )) (1 (X,x0)) = 1 π1(x,x 0). (f g) = f g Covering spaces useful for computing π 1, algebraic features of π 1 can be translated into geometric features of the spaces. De nition 2.24 Consider p X X a continuous and surjective map. If there exists an open cover {U a } a A of X such that a A p 1 (U a ) = b Ba Va b, where V a b Va b = for any b, b B a, b b. p Va b U a is a homeomorphism b B a and Va b open in X b B a, then p is called a covering map. X is called a covering space of X. Example

22 Algebraic topology 1. S 1 = {(x, y) R 2 x 2 + y 2 = 1}. p R S 1 where s (cos 2πs, sin 2πs). p is a covering map, p is continuous (calculus), and p is surjective. U 1 = {(x, y) S 1 x > 2/2} U 2 = {(x, y) S 1 x < 2/2} {U 1, U 2 } is an open cover of S 1 p 1 (U 1 ) = n Z V n, where V n = (n 3/8, n + 3/8). V n is open in R n Z. V n V n =, for any n, m Z, n m. p V n U 1 is a homeomorphism n Z. We can make the analogous construction for U 2. So p is a covering map. Remark 2.26 You can prove that p is a covering by using any open cover of S 1 by two open subsets of S 1 ( S 1 ). 2. S 1 = {z C z = 1}, n N, n 1. p n S 1 S 1, z z n. p n is a covering map (problem sheet 5). f X Y is called an embedding if f X f(x) is a homeomorphism. Consider the solid torus S 1 D 2 where D 2 = {(x, y) R 2 x 2 + y 2 1}, and its boundary (S 1 D 2 ) = S 1 S 1. Consider f S 1 (S 1 D 2 ) an embedding such that f(s 1 ) wraps around the rst S 1 three times. Lastly consider the projection π S 1 D 2 S 1 {(0, 0)} and restrict it to f(s 1 ). 3. f R + S 1, where s (cos 2πs, sin 2πs) is not a covering map. Consider an open cover {U a } a A of S 1. en U {U a } a A with x U. p 1 (U) = V 0 ( b Ba V b a ). p V 0 U is not a homeomorphism. eorem 2.27 If p X X is a covering map, X0 X, and X 0 = p 1 (X 0 ), then p 0 X0 X 0 obtained by restricting p to X 0 is a covering map. PROOF p 0 is continuous (restricting the domain or the range of a continuous function gives a continuous function). p 0 is surjective. Consider an open cover {U a } a A of X with the properties in the de nition of covering map. 22

23 2.3 Covering spaces en a A p 1 (U a ) = b Ba Va b. If x 0 X 0, then there exists a A with x 0 U a. Now U a X 0 is an open set in X 0. p 1 0 (U a X 0 ) = p 1 0 (U a ) p 1 0 (X 0 ) = p 1 (U a ) X 0 = ( b Ba Va b ) X 0 = b Ba (V b X 0 ). We have V b a X 0 open in X 0, (V b a X 0 ) (V b a X 0 ) = for any b, b B a, b b. p 0 V b a X 0 U a X 0 is a homeomorphism. {U a X 0 } a A is an open cover of X 0 with the desired properties p 0 X0 X 0 is a covering map. a eorem 2.29 If p X X, p X X are covering maps, then p p X X X X is a covering map. Example Consider p R S 1, where s (cos 2πs, sin 2πs). e map p p R R S 1 S 1 is a covering map. 2. X 0 = (S 1 p(0)) (p(0) S 1 ). X 0 = (p p) 1 (X 0 ) = (R Z) (Z R). (p p) 0 X0 X 0 is a covering map (by theorem). Two circles S 1 with a one common point is called wedge of two circles (notation S 1 S 1 ). 3. Other covering spaces of S 1 S 1. b a. x Consider X as in the following picture a b a x 1. x 2 en p X S 1 S 1, where x 1 x and x 2 x. p maps each edge of X to the edge of X with the same label by a map that is a homeomorphism and preserves the orientation. 23

24 Algebraic topology 4. Consider p 1 R+ R R + S 1 R + (where p R S 1 : s (cos 2πs, sin 2πs)) and f S 1 R + R 2 {(0, 0)}, where (x, t) t x. f is a homeomorphism. f (p 1 R +) R R + R 2 {(0, 0)} is a covering map. PROOF p p is continuous (problem sheet 3). p p is surjective (if (x, x ) X X, then x X and x X [p, p is surjective] x X, x X, such that p( x) = x and p ( x ) = x. ( x, x ) X X such that (p p )( x, x ) = (x, x )). Consider {U a } a A, {U c} c C open covers of X and X as in the de nition of covering maps. If (x, x ) X X then x X and x X i.e. U {U a } a A with x U and U {U c } c C with x U. p 1 (U) = b B V b, V b open V b V b = for b b and p V b U homeomorphism b B. (p ) 1 (U ) = d D V d, V d open, V d V d = for d d and p V d U homeomorphism d D. (x, x ) U U, U U is open in X X, (p p ) 1 (U U ) = b B d D p p V b V I.e. {U a U c} a A c C (V b V d ), where V b V d open in X X, V b V d disjoint, d U U is a homeomorphism. is the open cover of X X. at shows that p p is a covering map Li ing properties We will discuss two properties of covering spaces. De nition 2.32 Let p X X be a map. If f Y X is continuous a li ing of f is a map f Y X such that p f = f. Example 2.33 p R S 1, s (cos 2πs, sin 2πs), f I S 1 the path f(s) = (cos πs, sin πs). en f I R, f(s) = s is a li ing of f. 2 Proposition 2.34 (path li ing property) Let p X X be a covering map with p( x0 ) = x 0. If f I X is a path with f(0) = x 0, then! li ing of f to a path f I X with f(0) = x 0. PROOF 1. Consider the open cover {U a } a A of X (as in the de ntion of covering map). Let t (0, 1), and ft) X f(t) U a for some a A. f I X is continuous (a t, b t ) (0, 1) with t (a t, b t ) and f([a t, b t ]) U a. 24 Analogously for t = 0 [0, b 0 ) [0, 1] with f([0, b 0 ]) U a, and for t = 1 (a 0, 1] [0, 1] with f([a 0, 1]) U a. I is compact, and {(a t, b t )} t (0,1) together with [0, b 0 ) and (b 1, 1] is an open cover of I. We can choose a nite subcover, say [0, b 0 ), (a 0, 1], (a 1, b 1 ),..., (a m, b m )}. Consider {a i, b i } i {0,...,m} and orther its elements. (Possibly) rename the elements as follows: 0 < t 1 t 2m+2 < 1. this gives a subdivision 0 = s 0 < s 1 < < s n = 1 of I with the property, that i {0,..., n 1} f([s i, s i+1 ]) U a for some a A.

25 2.4 Li ing properties 2. De ne f(0) = X 0. Suppose that f is de ned in [0, s i ]. De ne f in [s i, s i+1 ] as follows: f([s i, s i+1 ]) U for some U in {U a } a A. Let p 1 (U) = b B V b. Now (p f)(s i ) = f(s i ) U. f(s i ) p 1 (U) = b B V b f(s i ) V 0 for some V 0 {V b } b B. De ne f(s) = (p V0 ) 1 (f(s)), s [s i, s i+1 ]. f is continuous in [s i, s i+1 ], since p V0 is a homeomorphism. us we have de ned f [0, 1] X continuous with f(0) = x Suppose f is another li ing of f with f(0) = x0. en f(0) = f(0) = x0. Suppose that f(s) = f(s) in [0, si ]. p f([si, s i+1 ]) = f([s i, s i+1 ]) U a for some a A. f([si, s i+1 ]) p 1 (U a ) = b B V b. f([s i, s i+1 ]) is connected and f(si ) = f(s i ) V 0. f([si, s i+1 ]) V 0. For s S, p f(s) = f(s) f(s) p 1 (f(s)). Also f V0. So f(s) = (p V0 ) 1 (f(s)) = f(s). I.e. f(s) = f(s). Proposition 2.36 Let p X X be a covering map with p( x0 ) = x 0. Let F I I X be continuous with F (0, 0) = x 0 en! li ing of F to a continuous map F I I X with F (0, 0) = x 0. If F is a path homotopy then F is a path homotopy. PROOF Consider {U a } a A an open cover of X with the properties as in the de nition of covering map. 1. De ne F ((0, 0)) = x 0. Extend F to I {0} and {0} I using the proposition from last time. Extend to I I, as follows: Choose subdivision s 0 < s 1 < < s m, t 0 < t 1 < < t n of I with the property that for each such rectangle I i J j = [s i 1, s i ] [t j 1, t j ], F (I i J j ) U a for some a A. De ne F rst in I 1 I j, then I 2 J 1,..., I m J 1, then I 1 J 2, I 2 J 2 and so on. Consider I i0 J j0 and suppose that F is de ned on the union A of the rectangles I i J j with j < j 0 or j = j 0 and i < i 0. Consider C = A (I i0 J j0 ). Choose U {U a } a A with F (I i0 J j0 ) U. p 1 (U)) b B V b, F is de ned in C, C is connected. So, F (C) is connected, i.e. V 0 such that F (C) V 0. If p 0 = p x0 V 0 U, then p 0 F (x) = p F (x) = F (x) F (x) = (p 1 0 )(F (x)). De ne F (x) = p 1 0 (F (x)) x I i0 J j0. F is continuous, according to the pasting lemma. 2. Check that at every step, there is a unique way to de ne F. 25

26 Algebraic topology 3. If F is a path homotopy, then F ({0} I) = x 0. Also, F ({0} I) p 1 ({x 0 }). p 1 ({x 0 }) has the discrete topology as a subspace of X. Anlose F ({0} I) is connected. us F ({0} I) = { x 0 }. Similarly F ({1} I) is a set with one point. us, F is a path homotopy. Proposition 2.38 (Homotopy li ing property) p X X covering map with p( x0 ) = x 0. Consider f, g the paths in X from x 0 to x 1 and f, g their li ings to paths in X with f(0) = g(0) = x 0. If f g, then f g. PROOF Consider F the path homotopy between f and g. en F ((0, 0)) = x 0. Let F I I X the li ing of f to X with F ((0, 0)) = x 0. en F ({0} I) = { x 0 } and F ({1} I) = { x 1 }. F I {0 } is a path i X starting at x 0 that is a li ing of F I {0}. I.e. FI {0 } = f (uniqueness of path li ings). F I {1} = g. So f(1) = g(1) = x 1 and F I I X is a path homotopy from f to g. De nition 2.40 Let p X X be a covering map with p( x0 ) = x 0. Given a loop f in X based x 0, let f be the li ing of f in X with f(0) = x 0. De ne ϕ π 1 (X, x 0 ) p 1 ({x 0 }), [f] f(1). ϕ is called the li ing correspondence derived from the map p Claim: ϕ is well-de ned. PROOF If [f] = [f ], then f, f are path homotopic. We can li this homotopy to a path homotopy between f and f f(1) = f (1). eorem If p X X be covering map with p( x0 ) = x 0. If X is path connected, then ϕ is surjective. 2. If X is simply connected, then ϕ is bijective. PROOF 1. Let x 1 p 1 ({x 0 }). en since X is path connected, there is a path f I X with f(0) = x 0 and f(1) = x 1. en f = p f is a loop in X based at x 0, since f(0) = p( x 0 ) = x 0, f ( 1) = p( x 1 ) = x 0 and ϕ([f]) = f(1) = x We only need to check that ϕ is injective. Indeed consider ä[f], [g] π 1 (X, x 0 ) with ϕ([f]) = ϕ([g]). en f(1) = g(1), where f, g are the li ings of f, g with f(0) = g(0) = x 0. Since X is simply connected, we have that [ f] = [ g], i.e. there is a path homotopy F between f and g. en F = p F is a path homotopy between f and g, i..e [f] = [g]. 26

27 2.5 e fundamental group of the circle and applications 2.5. e fundamental group of the circle and applications Remark 2.44 S 1 is path connected, so we will write π 1 (S 1 ). eorem 2.45 (π 1 (S 1 ), ) (Z, +). PROOF Let p R S 1, p(t) = (cos 2πt, sin 2πt). Denote p(0) = (1, 0) = x 0. en p 1 ({x 0 }) = Z. Consider ϕ π 1 (S 1 ) Z, where [f] f(1). By exercise 1 of sheet 3 R is simply connected. By the second point of the theorem ϕ is bijective. It remains to show that ϕ is a group homomorphism, i.e. ϕ([f] [g]) = ϕ([f]) + ϕ([g]). ϕ([f g]) = ϕ([f]) + ϕ([g]) f g(1) = f(1) + g(1), where f is the li of f with f(0) = 0 R and g the li of g with g(0) = 0 R and f g the li of f g where f g(0) = 0 R. Let g I R be the path g (s) = g(s) + f(1). en (p g )(s) = p( g(s) + f(1)) = (p g)(s) = g(s). g is a li ing of g with g (0) = f(1). Furthermore, p(( f g )(s)) = (f g)(s) f g is a li ing of f g with f g (0) = 0. ( f g )(+) = f(1) + g(1) f g(1) = f(1) + g(1). Retractions De nition 2.47 A retraction of a X onto A X is a continuous map r X A with r(a) = a a A. If such a map exists A is called a retract of X. Claim: If A is a retract of X and i A X is the inclusion map, then i is injective. PROOF r i A A. r i = 1 A (r i) = (1 A ). r i = 1 π1(a,x 0) i is injective. Proposition 2.49 ere is no retraction of D 2 onto S 1. PROOF If S 1 was a retract of D 2 and i S 1 D 2 the inclusion. en i π 1 (S 1 ) π 1 (D 2 ) would be injective. But π 1 (S 1 ) Z and π 1 (D) 2 is trivial as a convex subset of R 2. Lemma 2.51 Let h S 1 X be a continuous map. e following are equivalent: 1. h is nullhomotopic. 2. ere exists a continuous map h D 2 X with h S 1 = h. 3. h is trivial. PROOF 27

28 Algebraic topology : Consider the homotopy H S 1 I X between h and the constant map c. Let q S 1 I D 2 with (x, t) t x q is continuous, q is open and surjective. Vorlesung von Di 17. März fehlt noch Remark 2.53 Ha the corollary: If S 2 = A 1 A 2 A 3, A 1 closed i {1, 2, 3}, then x S 2, i {1, 2, 3} such that {x, x} A i. Inscribe a sphere in the tetrahedron. Project each face of the tetrahedral radially onto the sphere. en one obtains four closed sets A i, i {1,..., 4} with S 2 = A 1 A 2 A 3 A 4, but none of the A i contains a pair of antipodal points e Fundamental eorem of Algebra (Any non-constant polynomial in C[x] has a root in C). proof in algebra, proof in complex analysis (corollary of Liouville s theorem), proof in topology. eorem 2.54 (Fundamental eorem of Algebra) If a n x n + + a 1 + a 0 C[x], n > 0, then x 0 C with a n x n a 1 x 0 + a 0 = 0. PROOF Let a n x n + + a 1 x + a 0 C[x] and n > We can assume a n = 1 (a n (x n + an 1 a n x n a0 a n )). We can assume n 1 i=0 a i < 1. Indeed choose c R >0 and set x = cy is gives c n y n + + a 1 cy + a 0 = c n (y n + an 1 y n a0 ). Choose c large enough c c n such that a n a 0 < 1. Now if y c c n 0 is a root of y n + an 1 y n a 0, c c n then cy 0 is a root of x n + + a Consider f S 1 S 1, z z n and p I S 1, s (cos 2πs, sin 2πs) = e 2πis. en f π 1 (S 1, (1, 0)) π 1 (S 1, (1, 0)). f ([p]) = [f p], where f p(s) = (cos 2πns, sin 2πns). is shows that f is injective. Also if ι S 1 R 2 {(0, 0)} (inclusion), then ι is injective, since S 1 is a retract of R 2 {(0, 0)}. I.e. ι f = (ι f) is injective, which implies (Lemma??) that ι f is not nullhomotopic (a). 3. Suppose, that our polynomial has no root in D 2. Consider h D 2 R 2 {(0, 0)}, z z n + a n 1 z n a 1 z + a 0. By lemma?? h S 1 is nullhomotopic (b). Furthermore: F S 1 I R 2 {(0, 0)} where (z, t) z n + t(a n 1 z n a 1 z + a 0 ). is is a homotopy between h S 1 and ι f (c).

29 2.5 e fundamental group of the circle and applications F (z, t) = z n + t(a n 1 z n a 0 ) z n t a n 1 z n a 0 1 t n 1 i=0 a i > 0. (a),(b),(c) contradiction, i.e. our polynomial has a root in D 2. Exercise: Any root of such a polynomial z n + + a 0 with a i < 1 is in D Deformation retracts and homotopy type Covering spaces fundamental group computation (e.g. π 1 (S 1 )). De nition 2.56 Let A X, A is called a deformation retract of X, if there is a continuous map H X I < rax with H(x, 0) = x x, H(x, 1) A x X, H(a, t) = a a A, i I. e homotopy H is called a deformation retraction of X onto A. Remark r X A, r(x) = H(x, 1) is a retraction of X onto A. 2. H is a homotopy between 1 X and ι r, where ι A X inclusion map. Example {(x, y, z) R 3 (x, y) (0, 0), z = 0} = A is a deformation retract of {(x, y, z) R 3 (x, y) (0, 0)} = X. H X I X, H(x, y, z, t) = (x, y, (1 t)z). H is continuous, H(x, y, z, 0) = (x, y, z) (x, y, z) X, H(x, y, z, 1) = (x, y, 0) A (x, y, z) X, if (x, y, z) A, then z = 0 and thus H(x, y, z, t) = (x, y, 0) A. 2. Let p, q R 2, p q. en S 1 S 1 is a deformation retract of R 2 {p, q}..p. q 3. p, q R 2, p q, R 2 {p, q} deform retracts onto the theta space:. (Homeomorph to S 1 (0 [ 1, 1])). 4. Möbius band. 29

30 Algebraic topology eorem 2.59 Let A be a deformation retract of X, x 0 A, and ι (A, x 0 ) (X, x 0 ) the inclusion map. en ι (A, x 0 ) π 1 (X, x 0 ) is an isomorphism. PROOF Let r (X, x 0 ) (A, x 0 ), r(x) = H(x, 1) en r ι (A, x 0 ) (A, x 0 ) is the id map 1 A and (r ι) = (1 A ). r ι = 1 π1 (A,x 0 ). Furthermore ι r = (ι r). ι r (X, x 0 ) (X, x 0 ) and there is a homotopy H between 1 X and i r. H(x 0, t) = x 0 t I. If f I X loop, f(0) = f(1) = x 0, then H (f 1 I )) I I X is a path homotopy between 1 X f and (ι r) f. (path homotopy: H (f 1 I )(0, t) = H(x 0, t) = x 0 t I, H (f 1 I )(1, t) = H(x 0, t) = x 0 t I). So [1 X f] = [(ι r) f] (1 (f) = (ι r) (f). 1 π1(x,x 0) = (ι r) ι r = 1 π1(x,x 0). Corollary 2.61 If ι S n R n+1 {0} the inclusion map, x 0 S n, then ι π 1 (S n, x 0 ) π 1 (R n+1 {0}) is isomorphism. PROOF De ne H (R n+1 {0}) I R n+1 {0}, H(x, t) = (1 t)x + t x x. H is continuous, H(x, 0) = x x R n+1 {0} = X, H(x, 1) = x x Sn = A. If x S n, then H(x, t) = (1 t)x + tx = x t I. H is a deformation retraction of R n+1 {0} onto S n. Now use theorem. Example 2.63 We can look at example 2.58 again: 1. π 1 (X) m π 1 (A) Cor π 1 (S 1 ) Z. 2.,3. gure eight space is deformaton retraction of R 2 {p, q} m π 1 (A 1 ) π 1 (R 2 {p, q}), theta space is deformation retraction of R 2 {p, q} m π 1 (A 2 ) π 1 (R 2 {p, q}). π 1 (A 1 ) π 1 (A 2 ). Comment: gure eight and theta space have isomorphic fundamental group, but none is a deformation retract of the other (exercise). De nition 2.64 Let f X Y be a continuous map. If there exists a continuous map g Y X such that g f X X is homotopic to 1 X X X and f g Y Y is homotopic to 1 Y Y Y, then f is called a homotopy equivalence. g is called homotopy inverse of f. X,Y are called homotopy equivalent. If X, Y are homotopy equivalent, we say that X, Y have the same homotopy type. Check: Homotopy equivalence is an equivalence relation. Remark 2.65 Homotopy equivalence is more general than deformation retraction. Suppose that H X I X is a deformation retraction of X onto A. Let ι A X the inclusion map and r X A, r(x) = H(x, 1). en r ι A A is the id map 1 A, and H is a homotopy between ι r and 1 X. 30

31 2.5 e fundamental group of the circle and applications Lemma 2.66 Let h, k X Y homotopic continuous maps with h(x 0 ) = y 0, k(x 0 ) = y 1. If H X Y Y is the homotopy between h and k and if a(t) = H(x 0, t), then k = â h. h π 1 (X, x 0 ) π 1 (Y, y 0 ) k â π 1 (Y 1, y 1 ) PROOF der Beweis ist noch sehr fehlerha, noch fehlerha er als gewohnt :P Let f I X a loop based at x 0. We want to show that k ([f]) = â h ([f]) [k f] = â([h f]) [k f] = [â][h f][a] [a][k f] = [h f][a] [a(k f)] = [(h f)a]. h(x) = H(x, 0) x X, so h(f(s)) = H(f(s), 0) = (H f 0 )(s) where f 0 I X I, f 0 (s) = (f(s), 0). I.e. h f = H f 0. k(x) = H(x, 1) x X, so k(f(s)) = H(f(s), 1) = (H f 1 )(s) where f 1 I X I, f 1 (s) = (f(s), 1). I.e. k f = f 1. Consider c I X I, c(t) = (x 0, t). en H(c(t)) = a(t). F I I X I, F (s, t) = (f(s), t). G 0 γ 1 is homotopic to γ 0 B 1, call G the path homotopy between them. en F G I I X I, continuous, (F G)(s, 0) = F (G(s, 0)) = { F (β 0(s), 0), s [0, 1 2 ] F (1, γ 1 (s)) s [ 1 2, 1] = { (f β 0(s), 0) s [0, 1 2 ] (f(1), γ 1 (s)) s [ 1 2, 1] = { f 0(s) s [0, 1 2 ] (x 0, s) s [ 1 2, 1]. (F G)(s, 1) = = (c f 1 )(s), (F G)(0, t) = F ((0, 0)) = (x 0, 0), (F G)(1, t) = F ((1, 1)) = (x 0, 1) (F G is a path homotopy from c f 1 to c f 1. Lastly, H (F G) is a path homotopy from H(f 0 c) = H(f 0 ) H(c) = (h f) a to H(c f 1 ) = H(c) H(f 1 ) = a (k f) Corollary 2.68 Let h, k (X, x 0 ) (Y, y 0 ) homotopic continuous maps. If H X I Y is the homotopy between h and k and H(x 0, t) = y 0 t [0, 1], then k = h. eorem 2.69 Let f X Y homotopy equivalence with f(x 0 ) = y 0. en f π 1 (X, x 0 ) π 1 (Y, y 0 ) is an isomorphism. PROOF Let g Y X be a homotopy inverse of f, and let g(y 0 ) = x 1 and f(x 1 ) = y 1. (X, x 0 ) f (Y, y 0 ) g (X, x 1 ) f (Y, y 1 ) π 1 (X, x 0 ) (fx 0 ) π 1 (Y, y 0 ) g π 1 (X, x 1 ) (fx 1 ) π 1 (Y, y 1 ) g f x0 is homotopy to 1 X [Lemma] there is a path a I X, such that (g f x0 ) = â (1 X ) = â. 31