INTRODUCTION TO TOPOLOGY

Transcription

1 INTRODUCTION TO TOPOLOGY MARTINA ROVELLI These notes are an outline of the topics covered in class, and are not substitutive of the lectures, where (most) proofs are provided and examples are discussed in more detail. Contents Intro 1 1. Topological spaces Topologies The topology generated by a basis The subspace topology The product topology 15, The quotient topology The topology induced by a metric Continuous functions and Homeomorphisms The universal property of the subspace topology 18.2(b) The universal property of the product topology The universal property of the quotient topology Closed subsets 17 and the pasting Lemma Closed subsets Closure and interior The Pasting lemma Sequences and Hausdorff spaces Sequences in a topological space Hausdorff spaces 17, Connectedness 23,24, Connected spaces 23, Path-connected spaces Compactness 26,27, Compactness 29 1

2 6.2. Local compactness Separability axioms 31,32, Spaces with the same homotopy type 51 (and bits of 58) The fundamental group 51, The fundamental group of the circle Homotopy invariance of the fundamental group Higher homotopy groups [Hatcher 4.1] 42 Intro Topology is concerned with the properties of spaces that are preserved under continuous deformations, such as stretching, crumpling and bending, but not tearing or gluing. Here are some cool theorems that can be proven using topology. Theorem 0.1 (Euler). There exist infinitely many prime numbers. Theorem 0.2 (Fundamental Theorem of algebra). Any polynomial equation with complex coeffients admits at least one solution in the complex numbers. Theorem 0.3 (Hairy ball theorem). Any tangent field on a sphere vanishes at least at one point. You can t comb the hair on a coconut! 1.1. Topologies Topological spaces 12 Notation 1.1. Given a set I, and a set U i for every i I, denote by {U i } i I := {U i i I} the collection of all the U i s, by i I U i the union of all the U i s, and by i I U i the intersection of all the U i s. Definition 1.2. Let X be a set. A topology on X consists of a collection T P(X) of subsets of X such that (t1) X, T. (t2) If {U i } i I T, then i I U i T, i.e., T is closed under union. (t3) If {U i } N i=0 T, then N i=0 U i T, i.e., T is closed under finite intersection. The pair (X, T ) is a (topological) space, and the elements of T are open subsets of X. If U X is open and x U, we say that U is a neighborhood of X. Should think of (X, T ) as a shape, and every open subset as a zoom on a specific region. 2

3 Proposition 1.3 (Characterization of open subsets). Let (X, T ) be a topological space. A subset A X is open if and only if for every point x of A there exists an open U of X such that x U A. For [ ], since A is open and x A, we can take U = A. For [ ], we note that A can be written as A = U, U T,U A which is open, as is a union of open subsets of X. Example 1.4 (Canonical topologies on a set). Let X be a set. The collection T disc := P(X) of all subsets of X defines a topology on X, called discrete topology. The collection T codisc := {, X} defines a topology on X, called codiscrete topology. The collection T fin := { } {A X X \ A is finite} defines a topology on X, called cofinite topology. The collection T count := { } {A X X \ A is countable} defines a topology on X, called cocountable topology. Example 1.5 (Upper and lower topology on R). The collection T up := {, R} {(a, ) a R} defines a topology on R, called the upper topology. Dually, the collection T low := {, R} {(, a) a R} defines a topology on R, called the lower topology. Exercise 1.1. Write down all the topologies on the set with two elements {a, b}, and on the set with three elements {a, b, c}. Proposition 1.6 (Comparison of topologies). Let X be a set, with T and T two topologies. TFAE: (1) T T, i.e., every open of T is an open of T (2) For every x X and U T such that x U there exists U T such that x U U. In this case, we say that T is finer than T, or that T is coarser than T, and that T and T are comparable. If T T, we say that T is strictly finer than T and T is strictly coarser than T. Think that T if finer than T if it has a better resolution! [(1) (2)]: Suppose U T and x U. Since T T, U T and if we set U := U we have that x U U. [(2) (1)]: Suppose U T, and prove that U T. By assumption, for every x U there exists U T such that x U U. By Proposition 1.3, U is open with respect to T, namely U T. Example 1.7 (Comparison of canonical topologies). Given a set X 3

4 (1) Any topology T is coarser than the discrete topology and finer than the codiscrete topology, i.e., T codisc T T disc. (2) The cofinite topology is coarser than the cocountable topology, i.e., T codisc T fin T count T disc. When do these become equalities? Exercise 1.2. The topologies T up and T low on R are not comparable The topology generated by a basis 13. Remark 1.8 (Intersection and union of topologies). Let X be a set with two topologies T and T. The collection T T P(X) defines a topology on X. And for intersection of infinitely many topologies? The collection T T P(X) need not define a topology on X. For example, T up T low does not define a topology on R. When given a collection of subsets that do not form a topology, one can solve the problem by considering the topology generated by this collection. Definition 1.9 (Basis for a topology). Let X be a set. A basis (for a topology) on X consists of a collection B P(X) of subsets of X such that (b1) For every x X there exists B B such that x B. (b2) For any B 1, B 2 B and x B 1 B 2 there exists B 3 B such that x B 3 B 1 B 2. The collection T B := { B A B A B} defines a topology on X, called the topology generated by B. To see that T B indeed defines a topology, need the following. Proposition 1.10 (Characterization of open subsets in terms of a basis). Let X be a set and B a basis for a topology on X. A subset A X belongs to T B if and only if for every point x of A there exists an open U B of X such that x U A. For [ ], suppose that A belongs to T B. Then it is of the form A = B A. In particular, for any x A = B A B there exists B A B such that X B B A = A. For [ ], we note that A can be written as A = B = B T B, B B,U A B B where B := {B B B A}. Exercise 1.3. Use Proposition 1.10 to prove that T B defines a topology on X. For many purposes in topology, it s enough to check what happens on elements of a basis. 4

5 Proposition 1.11 (Comparison of topologies in terms of basis). Let X be a set, B and B two basis for topologies on X. TFAE: (1) T B T B. (2) For every x X and U B such that x U there exists U B such that x U U. Play with the definition of basis! Exercise 1.4 (True or false?). Let X be a set, B and B two basis for topologies on X. (1) T B T B if B B. (2) T B T B only if B B. Many important and interesting topologies on R are defined in terms of a basis. Example 1.12 (Basis for topologies on R). The collection B st := {(a, b) a < b} defines a basis for a topology on R, called the standard topology and denoted T st. The collection B K := B st {(a, b) \ K a < b}, where K := { 1 n n N >0 } defines a basis for a topology on R, denoted T K. The collection B lowlim := {[a, b) a < b} defines a basis for a topology on R, called the lower limit topology and denoted T lowlim. Dually, the collection B uplim := {(a, b] a < b} defines a basis for a topology on R, called the upper limit topology and denoted T uplim. How can we recognize if a basis generates a given topology? Proposition Let (X, T ) be a topological space, and B a basis for a topology. TFAE (1) T B = T. (2) For any x X and U X open in T such that x U there exists B B such that x B U. In this case we say that B is a basis for T. [(1) (2)]: Suppose U T and x U. Since T T B, U is open in T B. By Proposition 1.10 there exists B B such that x B U. [(2) (1)]: Suppose U T, and prove that U T. By assumption, for every x U there exists B T such that x B U. By Proposition 1.10, U is open with respect to T B, namely U T B. A topological space admits several basis! For instance, every topology is a basis for itself, but it s good to identify a basis that is as small as possible. Example 1.14 (Basis for canonical topologies on a set). Let X be a set. 5

6 B disc := {{x} x X} is a basis for the discrete topology T disc on X. T codisc is the only basis for the codiscrete topology T codisc on X The subspace topology 16. There is a canonical way to define a topology on any subset of a topological space. Definition Let (X, T ) be a topological space, and A X a subset. The collection T A := {A U U T } defines a topology on A, called the subspace topology. Example Can put the subspace topology on [0, 1], N, Q,... as a subspace of (R, T st ). When unspecified, it is always assumed that the topology on any subset of R is this one, and is still called standard topology. Proposition 1.17 ( 16.1). Let (X, T ) be a topological space with basis B, and A X a subset. The collection B A := {B A B B} defines a basis for the subspace topology T A. Use Proposition Example The collection {[0, 1] (a, b) 0 a < b 1} defines a basis for the topology on the standard interval [0, 1], seen as a subspace of R with the standard topology. Remark 1.19 ( 16.2). Let (X, T ) be a topological space and A Y X. If A is open in (X, T ), it is also open in (Y, T Y ). And the converse? If Y is open in (X, T ), then A is open in (X, T ) if and only if it is also open in (Y, T Y ). Exercise 1.5. Prove that, in [0, 1] {2} R, {2} and ( 1 2, 1] are open, whereas [0, 1 2 ] and {0} are not The product topology 15,19. Notation Given sets X 1, X 2,..., X n, denote by X 1 X 2 X n their cartesian product. An element consists of an n-tuple (x 1, x 2,..., x n ), where x 1 X 1, x 2 X 2,..., x n X n. There is a canonical way to put a topology on the product of topological spaces. Definition Let (X, T ) and (X, T ) be topological spaces. The collection B T T := {U U U T and U T } defines a basis for a topology on X X, called the product topology and denoted T T. Example Can put the product topology on R 2 = R R as a product of (R, T st ) with itself. When unspecified, it is always assumed that the topology on R 2 is this one, and is still called standard topology. Also get a topology on the circle S 1, as a subspace of R 2. 6

7 Proposition 1.23 (Basis for the product topology). Let (X, T ) and (X, T ) be topological spaces, with basis B and B, respectively. The collection {B B B T and B T } defines a basis for the product topology T T on X X. Follows from the definition of basis. Example 1.24 (Basis for R 2 ). The collection {(a, b) (a, b ) a < b and a < b } defines a basis for the topology on R 2 = R R, seen as a product of R with the standard topology. Proposition 1.25 (Compatibility of subspaces and products, 16.3). Let (X 1, T 1 ) and (X 2, T 2 ) be topological spaces, and A 1 X 1 and A 2 X 2 subsets. The topology (T 1 T 2 ) A1 A 2 on A 1 A 2, induced as a subspace of (X 1 X 2, T 1 T 2 ), coincides with the topology (T 1 ) A1 (T 2 ) A2 on A 1 A 2, induced as a product of (A 1, T 1 ) A1 and (A 2, T 2 ) A2, i.e., (A 1 A 2, (T 1 T 2 ) A1 A 2 ) = (A 1 A 2, (T 1 ) A1 (T 2 ) A2 ). By unpacking the definitions, they are generated by the same basis, i.e., {(A 1 U 1 ) (A 2 U 2 ) U 1 T 1 and U 2 T 2 }. Example Can define the standard topology on the cube [0, 1] [0, 1] in two equivalent ways: as a product of subspaces of the standard topology on R, or as a subspace of the product of the standard topology on R 2. Let s now discuss how to topologize an arbitrary (namely possibly infinite) product of topological spaces. Notation Given any set I, and a set X i for any i I, denote by i I X i their cartesian product. An element consists of a family (x i ) i I with x i X i. Note that, if I = {1, 2,..., n} is finite the previous case is recovered. The naive attempt to generalize the product topology to the case of an arbitrary product would be the following. Definition Let I be a set, and (X i, T i ) a topological space for any i I. The collection B box := { i I U i U i T i } defines a basis for a topology on i I X i, called the box topology and denoted T box. This is a fair definition. However, it will become clear later that this topology is not well-behaved, and the correct topology to put on an arbitrary product is instead the following. Definition Let I be a set, and (X i, T i ) a topological space for any i I. The collection { } B i I T i := U i U i T i and U i X i for finitely many i I i I 7

8 defines a basis for a topology on i I X i, called the product topology and denoted i I T i. The box topology and the product topology coincide if the product is finite. In general, they are comparable but different. Remark Since B i I T i B box, the topologies corresponding topologies are also comparable and i I T i T box. The first interesting case in which they differ is the following. Example 1.31 ( 19, Example 2). Denote by R ω := n N R the product of countably many copies of R. The box topology is strictly finer than the product topology on R ω. Also in the case of an arbitrary product, the product topology and the subspace topology are compatible. Proposition 1.32 (Compatibility of subspaces and products, 19.3). Let I be a set, and (X i, T i ) a topological space with a subset A i X i for any i I. The topology ( i I T i ) i I A on i i I A i, induced as a subspace of ( i I X i, i I T i ), coincides with the topology i I (T i ) Ai on prod i I A i, induced as a product of all the (A i, T i ) Ai s, i.e., ( i I A i, (T 1 T 2 ) A1 A 2 ) = ( i I A i, ( i I T i ) i I A i ) The quotient topology 22. Notation Let X be a set, and R X X an equivalence relation on X. We denote by X/R the quotient set, i.e., the set of all equivalence classes of X with respect to R. The class of x is defined by [x] := {y X xry}. Denote by π : X X/R the projection defined by x π(x) := [x]. There is a canonical way to put a topology on the quotient of any topological space. Definition Let (X, T ) be a topological space, and R X X an equivalence relation on X. The collection T q := {U X/R π 1 (U) T } defines a topology on X/R, called the quotient topology, and denoted T q. Many cool spaces can be built as quotients! Example 1.35 (Topology of a bunch of interesting spaces). Can define a circle as a quotient of [0, 1]. Can you guess what the quotient topology is? Can define a cylinder as a quotient of [0, 1] [0, 1]. Can you guess what the quotient topology is? Can define the Möbius strip as a quotient of [0, 1] [0, 1]. Can you guess what the quotient topology is? 8

9 1.6. The topology induced by a metric 20. Definition Let X be a set. A function d: X X [0, + ) is a metric on X if d(x, y) = 0 if and only if x = y for any x, y X, d(x, y) = d(y, x) for any x, y X, d(x, z) (x, y) + d(y, z) for any x, y, z X. We say that (X, d) is a metric space. For any ɛ > 0 and x X, the ball centered in x of radius ɛ is B d (x, ɛ) := {y X d(x, y) < ɛ}. Every metric induces a topology. Definition Let (X, d) be a metric space. The collection B d := {B d (x, ɛ) x X, ɛ > 0} defines a basis for a topology, called metric topology and denoted T d. Proposition 1.38 (Characterization of open subsets in terms of a metric). Let (X, d) be a metric space. A subset A X is open in T d if and only if for every point x of A there exists ɛ > 0 such that x B d (x, ɛ) A. Use Proposition Example 1.39 (Metrics on R n ). The Euclidean distance d st : R R R 0 given by d st (x, y) := x y defines a metric on R. What is the topology induced by d st? The Euclidean distance d 2 : R n R n R 0 given by d 2 ( x, y) := n i=1 (x i y i ) 2 defines a metric on R n. What is the topology induced by d 2? The function d : R n R n R 0 given by d ( x, y) := max k=1,...,n x k y k defines a metric on R n. What is the topology induced by d? Example 1.40 (Metrics on a set). Let X be a set. The function d disc : X X R 0 given by d disc (x, x) := 0 and d disc (x, y) := 1 if x y defines a metric on X, called the discrete metric. What is the topology induced by d disc? How can we recognize if a metric induces a given topology? Proposition Let (X, T ) be a topological space, and d a metric on X. TFAE (1) T d = T. (2) For every x X and U X open there exists ɛ > 0 such that B d (x, ɛ) U. In this case we say that (X, T ) is metrisable, and that the topology T in induced by the metric d. 9

10 Use Proposition Example The Euclidean metric on R induces the standad topology T st on R. More generally, the Euclidean metric d 2 as well as the uniform metric d from Example 1.39 induce the standard topology on R n. Example If X is any set, the discrete metric d disc from Example 1.40 induces the discrete topology on X. Subsets of metric spaces are metrisable. Proposition Let (X, d) be a metric space, and A X a subset. The restriction d A : A A X X R >0 defines a metric on A which induces the subspace topology on A. Use Proposition Finite products of metric spaces are metrisable. Proposition Let (X, d) and (X, d ) be metric space. The sum d + d : (X X) ( X X ) R >0 defines a metric on X X which induces the product topology on X X. Use Proposition Actually all countable products of metric spaces are metrisable. Theorem 1.46 ( 20.5). Let (X n, d n ) be metric space for any n N. One can use the metrics d n s to build a metric on n N X n which induces the product topology on n N X n. Topological spaces that are metrisable are extremely special. For instance, every two points of a metrisable space can be separated by disjoint open subsets. Proposition Let (X, T ) be a topological space, and d a metric on X. For any points x x in X there exist U, U X open such that x U, x U and U U =. We say that (X, T d ) is Hausdorff. Given x, x different points of x, set ɛ := d(x, x ) > 0. Then x B(x, ɛ 2 ), x B(x, ɛ 2 ), and by the triangle inequality B d(x, ɛ 2 ) B d(x, ɛ 2 ) =. This is the first test to perform on a topological space to tell whether it s metrisable. For instance... Example Given a set X, the topological space (X, T codisc ) is Hausdorff if and only if empty or a singleton. 10

11 2. Continuous functions and Homeomorphisms 18 Continuous functions are the mean for topological spaces to interact with each other. Start with a local notion of continuity. Proposition 2.1 (Local continuity). Let (X, T ) and (X, T ) be topological spaces, x 0 X a point. For a function f : X X, TFAE. (1) For any U T containing f(x 0 ), f 1 (U ) T (and necessarily x 0 f 1 (U )), i.e., the preimage of every open neighborhood of f(x 0 ) is an open neighborhood of x 0. (2) For every open U T containing f(x 0 ) there exists an open U T containing x 0 such that f(u) U. In this case, the function f : (X, T ) (X, T ) is continuous at x 0. Use Proposition 1.3. Continuity can be checked on a basis. Proposition 2.2 (Continuity in terms of basis). Let X and X be sets, B and B basis for topologies, and x 0 X a point. For a function f : X X, TFAE. (1) The function f : (X, T B ) (X, T B ) is continuous at x 0. (2) For every B B containing f(x 0 ) there exists an open B B containing x 0 such that f(b) B. Use Proposition We recover the notion of continuity for metric spaces! Proposition 2.3 (Continuity in terms of metrics). Let (X, d) and (X, d ) be metric spaces, and x 0 X a point. For a function f : X X, TFAE. (1) The function f : (X, T d ) (X, T d ) is continuous at x 0. (2) For every ɛ > 0 there exists δ > 0 such that f(b d (x 0, δ)) B d (f(x 0 ), ɛ), i.e., d(x, x 0 ) < δ d (f(x), f(x 0 )) < ɛ. Use Proposition Can also give a global notion of continuity Proposition 2.4 (Global continuity). Let (X, T ) and (X, T ) be topological spaces. For a function f : X X, TFAE. (1) For any U T, f 1 (U ) T, i.e., the preimage of every open subset is open. 11

12 (2) The function f : (X, T ) (X, T ) is continuous at every x 0 X, i.e., for every x 0 X, and open U T containing f(x 0 ) there exists an open U T containing x 0 such that f(u) U. In this case, the function f : (X, T ) (X, T ) is is said to be (globally!) continuous. Use Proposition 1.3. Remark 2.5. Let X be a set, and (X, T ) any topological space. Every function f : (X, T disc ) (X, T ) is continuous. Every function f : (X, T ) (X, T codisc ) is continuous. Very few functions f : (X, T codisc ) (X, T ) are continuous. Very few functions f : (X, T ) (X, T disc ) are continuous. Recovers several notions of continuity! Example 2.6. Let f : R R be a function (and x 0 R a point). (1) f : (R, T st ) (R, T st ) is continuous (at x 0 ) if and only if it is continuous (at x 0 ) in the sense of calculus. (2) f : (R, T uplim ) (R, T st ) is continuous (at x 0 ) if and only if it is left continuous (at x 0 ). (3) f : (R, T st ) (R, T up ) is continuous (at x 0 ) if and only if it is upper semi-continuous (at x 0 ) (and this is a thing! See the Wikipedia page of Semi-continuity for more details). Proposition 2.7. (1) The identity function id X : (X, T ) (X, T ) is continuous if and only if T T. In particular, id X : (X, T ) (X, T ) is continuous. (2) The composite g f : (X 1, T 1 ) (X 2, T 2 ) (X 3, T 3 ) of continuous functions f : (X 1, T 1 ) (X 2, T 2 ) and g : (X 2, T 2 ) (X 3, T 3 ) is continuous. (3) The function c x0 : (X, T ) (X, T ) constant at x 0 X is continuous. We prove that the preimage of any open is open. (1) For any U T, the preimage is id 1 (U ) = U T T. (2) For any U T, its preimage via g is g 1 (U ) T, and its preimage via g f is (g f) 1 (U ) = f 1 (g 1 (U )) T. (3) For any U T, its preimage is c 1 x 0 (U ) = X T if x 0 U and c 1 x 0 (U ) = T otherwise. Definition 2.8. Let (X, T ) and (X, T ) be topological spaces. A function f : (X, T ) (X, T ) is a homeomorphism if f is bijective, and the inverse function g : (X, T ) (X, T ) is continuous. In = this case we write f : (X, T ) (X, T ) 12

13 The spaces (X, T ) and (X, T ) are homeomorphic if there exists a homeomorphism between them. In this case we write (X, T ) = (X, T ). Big warning: Remark 2.9. Unlike in algebra, it s not enough for a continuous map to be bijective to be a homeomorphism! For instance, consider the identity id: (X, T ) (X, T ) on a set X with two topologies T T. Remark Homeomorphism defines an equivalence relation of topological spaces. (1) (X, T ) = (X, T ) via the identity map. (2) If (X, T ) = (X, T ) via a homeomorphism f, the inverse g is also a homeomorphism and (X, T ) = (X, T ) via g. (3) If (X, T ) = (X, T ) via a homeomorphism f and (X, T ) = (X, T ) via a homeomorphism f, the composite f f is also a homeomorphism and (X, T ) = (X, T ) via f f The universal property of the subspace topology 18.2(b). Idea: It s easy to construct/check continuous maps into a subspace topology! Proposition Let (X, T ) be a topological space and A X a subset. (1) The inclusion ι: (A, T A ) (X, T ) is continuous. (2) A function f : (X, T ) (A, T A ) is continuous if and only if the function is continuous. ι f : (X, T ) (X, T ) (1) The preimage of an open U is ι 1 (U) = A U T A. (2) [ =]: Given U T, the preimage of an open U A T A is f 1 (A U) = (ι f) 1 (A U) = (ι f) 1 (A) (A U) 1 (U) = = X (ι f) 1 (U) = (ι f) 1 (U) T. [= ]: Use (1) and Proposition 2.7(2). Example 2.12 (How to use the universal property of subspace). Let (X, T ) be a topological space, and A B X be subsets. Let s use the universal property of the subspace topology to prove that the inclusion (A, T A ) (X, T B ) is continuous. Since the target is endowed with the subspace topology, by Proposition 2.11(2) we might as well prove that the inclusion (A, T A ) (B, T B ) 13

14 (X, T ) is continuous. This map is precisely the inclusion (A, T A ) (X, T ), which is continuous by Proposition 2.11(1) The universal property of the product topology 19. Idea: It s easy to construct/check continuous maps into a product topology! Notation Given a function f : X X 1 X 2, necessarily f(x) = (f 1 (x), f 2 (x)) where f 1 = pr 1 f : X X 1 X 2 X 1 and f 2 = pr 2 f : X X 1 X 2 X 2. We write f = (f 1, f 2 ). Proposition 2.14 ( 18.4). Let (X 1, T 1 ) and (X 2, T 2 ) be topological spaces. (1) The projections pr 1 : (X 1 X 2, T 1 T 1 ) (X 1, T 1 ) and pr 2 : (X 2 X 2, T 2 T 2 ) (X 2, T 2 ) are continuous. (2) A function f = (f 1, f 2 ): (X, T ) (X 1 X 2, T 1 T 1 ) is continuous if and only if the functions f 1 = pr 1 f : (X, T ) (X 1, T 1 ) and f 2 = pr 2 f : (X, T ) (X 2, T 2 ) are continuous. (1) The preimage of an open U 1 T 1 via pr 1 is pr 1 1 (U 1) = U 1 X 2 T 1 T 2. (2) [ =]: Given U 1 T 1 and U 2 T 2, the preimage of an open U 1 U 2 T 1 T 2 is f 1 (U 1 U 2 ) = f 1 1 (U 1) f 1 2 (U 2) T. [= ]: Use (1) and Proposition 2.7(2). Example 2.15 (How to use the universal property of products). Let s use the universal property of the product topology to prove that the function θ : (R, T st ) (R 3, T st ) defined by θ(t) := (cos(t), sin(t), t) is continuous. Since the target is endowed with the product topology, by Proposition 2.14(2) we might as well prove that each of the three components θ i = pr i θ : (R, T st ) (R, T st ) is continuous. (1) The first component θ 1 : (R, T st ) (R, T st ), which can be computed as t θ(t) = (t, cos(t), sin(t)) cos(t), coincides with the cosine function, which is is known from calculus to be continuous. (2) The second component θ 2 : (R, T st ) (R, T st ), which can be computed as t θ(t) = (cos(t), sin(t), t) sin(t), coincides with the sine function, which is is known from calculus to be continuous. (3) The third component θ 3 : (R, T st ) (R, T st ), which can be computed as t θ(t) = (cos(t), sin(t), t) t = id(t), coincides with the identity function on (R, T st ), which is continuous by Proposition 2.7(1). 14

15 It follows that θ : (R, T st ) (R 3, T st ) is continuous, as desired. Another example where the universal property of products is very useful. Example Let (X, T ) be a topological space, and f, g : (X, T ) (R, T st ) two continuous functions. Let s use the universal property of the product topology to prove that the function f + g : (X, T ) (R, T st ), defined by (f + g)(x) := f(x) + g(x), is continuous. We start by observing that f + g can be written as a composition f + g : (X, T ) (f,g) (R 2 +, T st ) (R, T st ), where +: (R 2, T st ) (R, T st, T st ) is the usual sum of real numbers. By Proposition 2.7(2), we might as well prove that the two functions (f, g) and + are continuous. (1) Since the target of (f, g): (X, T ) (R 2, T st ) is endowed with the product topology, to see that (f, g) is continuous, it s enough to note that each of the two components f, g : (X, T ) (R, T st ) are continuous, which is true by assumption. (2) The sum of real numbers +: (R 2, T st ) (R, T st ) is known to be continuous from calculus. It follows that f + g : (X, T ) (R, T st ) is continuous, as desired. Notation Given a function f : X i I X i, necessarily f(x) = (f i (x)) i I where f j = pr j f : X i I X i X j. We write f = (f i ) i I. The product topology has the correct universal property even for arbitrary products Proposition 2.18 ( 19.6). Let I be a set, and (X i, T i ) a topological space for any i I. (1) For every j I the projection ( ) pr i : X i, i I T i (X j, T j ) i I is continuous. (2) A function ( ) f = (f i ) i I : (X, T ) X i, i I T i is continuous if and only if for every j I the function ( ) f j = pr j f : (X, T ) X i, i I T i (X j, T j ) is continuous. i I 15 i I

16 Improve the argument from Proposition However, the box topology doesn t Example 2.19 ( 19, Example 2). Consider the function f : R R ω given by x (x) n N. The function f : (R, T st ) (R ω, n N T st ) is continuous (using the universal property!). The function f : (R, T st ) (R ω, T box ) is not continuous The universal property of the quotient topology 22. Idea: It s easy to construct/check continuous functions from a quotient topology! Notation Let X, Y be sets, and R an equivalence relation on X. Given a function f : X/R Y, necessarily f([x]) = f π(x) = f(x) where we define f := f π : X X/R Y. Note that f(x) = f(x ) if and only if xrx. All functions f : X/R Y are of the form f π where f has this property. Proposition 2.21 ( 22.2). Let (X, T ) be a topological space and R an equivalence relation on X. (1) The projection π : (X, T ) (X/R, T R ) is continuous. (2) A function f : (X/R, T R ) (X, T ) is continuous if and only if the function is continuous. f π : (X, T ) (X, T ) (1) The preimage of an open W T q is π 1 (W ) T. (2) [ =]: The preimage of an open U T via f = f π is T f 1 (U ) = (f π) 1 (U ) = π 1 (f 1 (U )). Then, by definition of the quotient topology, the preimage f 1 (U ) is open, as desired. [= ]: Use (1) and Proposition 2.7(2). Example 2.22 (How to use the universal property of quotients). Let s use the universal property of the quotient topology to prove that the function f : ([0, 1]/R, T q ) (S 1, T st ) defined by θ([t]) := (cos(2πt), sin(2πt)) is continuous. Since the source is endowed with the quotient topology, by Proposition 2.21(2) we might as well prove that the function f π : ([0, 1], T st ) (S 1, T st ) is continuous. Since the target is endowed with the subspace 16

17 topology, by Proposition 2.11(2) we might as well prove that the function f : ([0, 1], T st ) (R 2, T st ) is continuous. Since the target is endowed with the product topology, by Proposition 2.14(2) we might as well prove that each of the two components is continuous. (1) The first component ([0, 1], T st ) (R, T st ), which can be computed as t f([t]) = (cos(2πt), sin(2πt)) cos(2πt), coincides with the cosine function, which is is known from calculus to be continuous. (2) The second component ([0, 1], T st ) (R, T st ), which can be computed as t f([t]) = (cos(2πt), sin(2πt)) sin(2πt), coincides with the sine function, which is is known from calculus to be continuous. It follows that f : ([0, 1]/R, T q ) (S 1, T st ) is continuous, as desired. 3. Closed subsets 17 and the pasting Lemma Closed subsets 17. Proposition 3.1 (Closed subset). Let (X, T ) be a topological space and A X a subset. TFAE (1) The complement X \ A of A in X is open. (2) For every x A there exists an open U such that x U and U A =. In this case we say that the subset A X is closed. Use Proposition 1.3. Example 3.2 (Closed subsets of the canonical topologies on a set). Let X be a set, and A X a subset. A is always closed wrt T disc. A is closed wrt T count if and only if it is countable or A = X. A is closed wrt T fin if and only if it is finite or A = X. A is closed wrt T codisc if and only if A = X or A =. Closed doesn t mean not open! Example 3.3. Consider the space (R, T st ). The subset [0, 1] is closed and not open. The subset R is closed and open. The subset [0, 1) is not closed and not open. The subset (0, 1) is open and not closed. Proposition 3.4 (Properties of closed 17.1). Let (X, T ) be a topological space. 17

18 (1) X and are closed. (2) If {F i } i I P(X) is a collection of closed subsets, then i I U i is closed, i.e., T is closed under intersection. (3) If {U i } N i=0 P(X) is a finite collection of closed subsets, then N i=0 U i is closed, i.e., T is closed under finite union. Use the properties of complements. Better description of closed in terms of a basis Proposition 3.5 (Closed subset in terms of a basis). Let X be set, B a basis for a topology and A X a subset. TFAE (1) The subset A is closed in T B. (2) For every x A there exists B B such that x B and B A =. Better description of closed in terms of a metric Proposition 3.6 (Closed subset in a metric space). Let (X, d) be a metric space and A X a subset. TFAE (1) The subset A is closed in T d. (2) For every x A there exists δ > 0 such that B d (x, δ) A =. Example 3.7. If (X, d) is a metric space, the closed ball of radius ɛ > 0 and centered at x X, defined by B d (x, ɛ) := {y X d(x, y) ɛ}, is indeed closed in T d. Proposition 3.8 (Closed in subspace topology 17.2). Let (X, T ) be a topological space, and A Y X a subsets. (1) The subset A is closed in (Y, T Y ) if and only if there exists a closed F in (X, T ) such that A = F Y. (2) If Y is closed in (X, T ), then the subset A is closed in (Y, T Y ) if and only if it is closed in (X, T ). (1) For [ ], suppose A Y is closed in T Y. Then Y \ A is open in T Y, and therefore Y \A = U Y for some U T. Set F := X \U. Then F is closed in T and A = F Y, as desired. For [ ], suppose that F is closed in T. Then X \ F is open in T and Y (X \ F ) is open in T A. So the complement F Y = Y \ (Y (X \ F )) is closed in T Y, as desired. (2) Use (1). Proposition 3.9 (Closed in product topology). Let (X, T 1 ) and (X 2, T 2 ) be topological spaces, and A X 1 X 1 a subset. TFAE (1) The subset A is closed in T 1 T 2. (2) For any (x 1, x 2 ) X 1 X 2 such that (x 1, x 2 ) A there exists U 1 T 1 containing x 1 and U 2 T 2 containing x 2 such that A (U 1 U 2 ) =. 18

19 Use Proposition 3.5 Exercise 3.1 (Proof of the Euler Theorem). (1) Show that the collection {az + b a Z \ {0}, b Z} is a basis for a topology on Z. (2) Show that, for any a Z \ {0} et b Z, the subset az + b := {an + b n Z} Z is open and closed. (3) Show that Z \ {±1} = (pz + 0). p prime (4) Conclude that the set of prime numbers is infinite Closure and interior 17. Definition (interior and closure) Let (X, T ) be a topological space and A X a subset. (1) The interior of A is the (necessarily open) subset int(a) := U. U T,U A (2) The closure of A is the (necessarily closed) subset A := cl(a) := F. F closed,f A Remark Let (X, T ) be a topological space and A X a subset. (1) The interior of A is the largest open subset contained in A. Indeed, if U is another open that contains A, then U int(a). In particular, A is open if and only if A = int(a). (2) The closure of A is the smallest closed subset containing in A. Indeed, if F is another closed that contains A, then F A. In particular, A is closed if and only if A = A. Proposition 3.12 (Characterization of closure 17.5(1)). Let (X, T ) be a topological space, A X and x X. TFAE (1) x A. (2) Any open U T containing x intersects A. Version in terms of a basis Proposition 3.13 (Characterization of closure in terms of a basis 17.5(2)). Let X be a set, B a basis for a topology on X, A X a subset and x X. TFAE (1) x A. (2) Any B B containing x intersects A. 19

20 Version for metric spaces Proposition 3.14 (Characterization of closure in terms of a metric). Let (X, d) be a metric space, A X a subset and x X. TFAE (1) x A. (2) For any ɛ > 0, the ball B d (x, ɛ) intersects A. Proposition 3.15 (Closure in a subspace 17.4). Let (X, T ) be a topological space, and A Y X subsets. If cl Y (A) and cl X (A) denote the closures of A in (Y, T Y ) and (X, T ), then cl Y (A) = cl X (A) Y. Proposition 3.16 (closure in a product 19.5). Let (X, T ) and (X, T ) be topological spaces, A X and A X subsets. cl X X (A A ) = cl X (A) cl X (A ) The Pasting lemma 18. Notion of closed useful to discuss continuity Proposition 3.17 (Characterization of continuity 18.1). Let (X, T ) and (X, T ) be topological spaces. For a function f : X X, TFAE (1) The function f is continuous. (2) For every F closed in T, the preimage f 1 (F ) is closed in T. Idea: Continuity is a local property, and it s easy to construct/check continuous maps defined on a finite union of closed or an arbitrary union of open subsets. Proposition 3.18 (Pasting lemma 18.2(f),18.3). Let (X, T ) and (X, T ) be topological spaces, and f : X X a function. (1) If X = A B, where A and B are closed in T, then f : (X, T ) (X, T ) is continuous if and only if the restrictions f A : (A, T A ) (X, T ) and f B : (B, T B ) (X, T ) are continuous. (2) If X = i I U i, where U i T for every i I, then f : (X, T ) (X, T ) is continuous if and only if all the restrictions are continuous. f Ui : (U i, T Ui ) (X, T ) (1) Use Definition 1.2 and Proposition 2.4. (2) Use Proposition 3.4 and Proposition

21 Exercise 3.2. Use the pasting lemma to build a continuous function from X := {0} [0, 1] [0, 1] {0} to {0} [ 1, 1] that doesn t move the vertical leg of X and bends the horizontal leg of X. 4. Sequences and Hausdorff spaces 4.1. Sequences in a topological space. Notation 4.1. Recall that a sequence in a set X consists of a function x: N X, with n x n. We write (x n ) n N. Topology tells you when a sequence is approaching to a point of a space! Definition 4.2 (Converging sequence). Let (X, T ) be a topological space, and (x n ) n N a sequence in X. We say that (x n ) n N converges to x X with respect to T if for any open U T containing x there exists N N such that x n U for n N. In this case we say that x is the limit of (x n ) n N, and we write n x n x or x = lim x n. n Proposition 4.3 (Converging sequence in terms of a basis). Let X be a set, B a basis for a topology, and (x n ) n N a sequence in X. The sequence (x n ) n N converges to x X with respect to T B if and only if for any B B containing x there exists N N such that x n B for n N. Recover the notion from calculus! Proposition 4.4. Let (X, d) be a metric space, and (x n ) n N a sequence in X. The sequence (x n ) n N converges to x X with respect to T d if and only if for any ɛ > 0 there exists N N such that x n B d (x, ɛ) for n N., i.e., n N = d(x n, x) < ɛ. Example 4.5. Let (X, T ) be a topological space. (1) Any definitely constant sequence in (X, T ) converges (to the same value). (2) Any subsequence of a converging sequence in (X, T ) converges (to the same value). The limit is a priori not unique! Example 4.6 (Convergent sequences in canonical examples). Let (X, T ) be a topological space and (x n ) n N a sequence in X. (1) The sequence (x n ) n N always converges with respect to the codiscrete topology to any point of X(!). (2) The sequence (x n ) n N converges with respect to the discrete topology to x X if and only if there exists N N such that x n = x for n N. 21

22 Continuous functions respect converging sequences Proposition 4.7. Let f : (X, T ) (X, T ) be a continuous function, and (x n ) n N a sequence in X. If the sequence (x n ) n N converges to x X with respect to T, then the sequence (f(x n )) n N converges to f(x) X with respect to T. special case Remark 4.8. Let X be a set with two topologies T T. If a sequence converges to x X with respect to T, the sequelce also converges to x X with respect to T. Example 4.9. Consider the sequence ( 1 n+1 ) n N in R. (1) The sequence converges uniquely to 0 with respect to T st. Therefore, the sequence converges (uniquely to 0) with respect to T fin, T up and T codisc. (2) The sequence does not converge with respect to T K. Indeed, if it converged to something, it would be 0, since this is the limit with respect to T st. But there exists a neighborhood of 0, namely R \ K, that does not contain any value of the sequence. Therefore, the sequence does not converge with respect to T uplim and T disc. Converging sequences are topological invariants Proposition Given a homeomorphism f : (X, T ) (X, T ), a sequence (x n ) n N in X converges to x X with respect to T if and only if (f(x n )) n N converges to f(x) with respect to T. Use Proposition 4.7. Proposition 4.11 (Converging sequence in a subspace). Let (X, T ) be a topological space, A X a subspace, and (x n ) n N a sequence in A. The sequence (x n ) n N converges to x A with respect to T A if and only if it coverges to x with respect to T. Proposition 4.12 (Converging sequence in a product). Let (X, T ) and (X, T ) be topological spaces, and (x n, x n) n N a sequence in X X. The sequence (x n, x n) n N converges to (x, x ) X X with respect to T T if and only if i(x n ) n N converges to x X with respect to T and (x n) n N converges to x X with respect to T. also true for arbitrary products 22

23 Proposition 4.13 (Converging sequence in an arbitrary product). Let (X i, T i ) be a topological space for any i I, and (x n,i ) n N,i I a sequence in i I X i. The sequence (x n,i ) n N,i I converges to (x i ) i I i I X i with respect to i I T i if and only if (x n,i ) n N converges to x i X with respect to T i for every i I. Proposition 4.14 (Sequences in a quotient). Let X be a set, R an equivalence relation on X, and (x n ) n N a sequence in A. If the sequence (x n ) n N converges to x X with respect to T, then the sequence ([x n ]) n N converges to [x] X with respect to T q. Use Proposition 4.7. However Example The converse is not true. For instance, consider the equivalence relation on [ 1, 1] that identifies 1 and 1. Then the sequence (( 1) n ) n N in [ 1, 1] does not converge with respect to the standard topology, whereas the sequence ([( 1) n ]) n N in [ 1, 1]/R is constant and converges to [1] = [ 1] with respect to the quotient topology Hausdorff spaces 17,31. Definition A topological space (X, T ) is Hausdorff if for every different points x, x X there exist open U, U T containing respectively x and x such that U U =. Example Let X be a set. (X, T disc ) is always metrisable. (X, T codisc ) is metrisable if and only if X is empty or a singleton. (X, T fin ) is Hausdorff if and only if X is finite. (X, T count ) is Hausdorff if and only if X is countable. In terms of a basis Proposition Let X be a set and B a basis for a topology. The topological space (X, T B ) is Hausdorff if and only if for every different points x, x X there exist B, B B containing respectively x and x such that B B =. All metric spaces are Hausdorff Proposition Let (X, d) be a metric space. Then (X, T d ) is a Hausdorff space. Proposition 4.20 ( 17.10). Let (X, T ) be a Hausdorff space, and (x n ) n N be a sequence in X. If (x n ) n N converges with respect to T, then the limit is unique. 23

24 Example Let X be a set. (X, T disc ) is always Hausdorff. (X, T codisc ) is Hausdorff if and only if X is empty or a singleton. (X, T fin ) is Hausdorff if and only if X is finite. (X, T count ) is Hausdorff if and only if X is countable. Being Hausdorff is a topological property: Proposition Given a homeomorphism f : (X, T ) (X, T ), (X, T ) is Hausdorff if and only if (X, T ) is Hausdorff. Remark Let X be a set with two topologies T T. Hausdorff, then (X, T ) is Hausdorff. If (X, T ) is Proposition 4.24 ( 31.2(a)). Let (X 1, T 1 ) and (X 2, T 2 ) be topological spaces. If (X 1, T 1 ) and (X 2, T 2 ) are Hausdorff spaces, then (X 1 X 2, T 1 T 2 ) is a Hausdorff space. also true for arbitrary products Proposition 4.25 ( 31.2(a)). Let (X i, T i ) be a topological space for any i I. If (X i, T i ) is a Hausdorff space for any i I, then ( i I X i, i I T i ) is a Hausdorff space. Proposition 4.26 ( 31.2(a)). Let (X, T ) be a topological space, and A X a subset. If (X, T ) is a Hausdorff space, then (A, T A ) is a Hausdorff space Connected spaces 23,24. Connectedness: 5. Connectedness 23,24,25 Main ingredient of Intermediate value theorem. Tells whether a space has holes. Global property Useful to prove that certain spaces are not homeomorphic. Proposition 5.1. Let (X, T ) be a topological space. TFAE (1) X and are the only subsets that are both closed and open. (2) If X = U U with U, U T and U U =, then either U = X or U = X. 24

25 (3) If X = F F with F, F closed in T and F F =, then either F = or F =. In this case, (X, T ) is said to be connected. Example 5.2 (Connectedness for canonical topologies). Let X be a set. (X, T disc ) is connected if and only if X is empty or a singleton. (X, T codisc ) is always connected. (X, T fin ) is connected if and only if X is empty, a singleton or not finite. (X, T count ) is connected if and only if X is empty, a singleton or not countable. Intuition for the standard topology: a space is connected if it has no holes. Example 5.3. Consider R endowed with the standard topology. Q is not connected. [0, 1] (300, 301] is not connected. We will see that all intervals are connected. Proposition 5.4 (Connectedness of image 23.5). Let f : (X, T ) (X, T ) be a continuous function. If a subset (X, T ) is connected in (f(x), T f(x) ) is connected. However Remark 5.5. The converse is not true. For instance, the constant function ({0, 1}, T disc ) ({0}, T disc ) is continuous, the singleton {0} is connected, whereas the preimage {0, 1} = {0} {1} is not. Being connected is a topological property: Proposition 5.6. Given a homeomorphism f : (X, T ) (X, T ), (X, T ) is connected if and only if (X, T ) is connected. Use Proposition 5.4. Example 5.7. The letter O is not homeomorphic to the letter Q. Suppose that there exists a homeomorphism f : (O, T st ) = (Q, T st ). Denote by P Q the interesting point of Q, and by P O the preimage P O := f 1 (P Q ). By restricting the source and the target of f suitably, we still have a homeomorphism f : (O \ {P O }, T st ) = (Q \ {P Q }, T st ). However, by straightening O \ {P O }, we see that (O \ {P O }, T st ) = ([0, 1], T st ), which is connected, whereas (Q \ {P Q }, T st ) is not (draw a separation to see this!). 25

26 Remark 5.8. Let X be a set with two topologies T T. If (X, T ) is connected, then (X, T ) is connected. Proposition 5.9 (Connectedness of quotient). Let (X, T ) be a topological space, and R X X an equivalence relation on X. If (X, T ) is connected, then (X/R, T q ) is connected. Special case of Proposition 5.4. Remark 5.10 ( 23.2). Let (X, T ) be a topological space such that X = A B with A B =, and Y X a subset. If (Y, T Y ) is connected, then Y A or Y B. Being connected does not depend on which subspace topology you use Remark Let (X, T ) be a topological space, and Y X. A subset A Y X subsets. The subspace (A, T A ) is connected if and only if (A, (T Y ) A ) is connected. Left as an exercise. Proposition 5.12 (Connectedness of non disjoint union 23.3). Let (X, T ) be a topological space such that X = A A and A A. If the spaces (A, T A ) and (A, T A ) are connected, then (X, T ) is connected. Example Consider R 2 endowed with the standard topology. The subspace Q [0, 1] R {1} is connected. actually more general statement Proposition 5.14 (Connectedness of non disjoint union 23.3). Let (X, T ) be a topological space such that X = i I A i and i I A i. If the space (A i, T Ai ) is connected for any i I, then (X, T ) is connected. Proposition 5.15 (Connectedness of products 23.6). Let (X, T ) and (X, T ) be topological spaces. If (X, T ) and (X, T ) are connected, then (X X, T T ) is connected. same argument works for arbitrary products Proposition 5.16 (Connectedness of arbitrary products). Let (X i, T i ) be a topological space for every i I. If (X i, T i ) is connected for every i I, then ( i I X i, i I T i ) is connected. Example Consider R 2 endowed with the standard topology. ([0, 1] (300, 301]) [0, 1] is not connected [0, 1] (20, 30] is connected 26

27 R 2 is connected. Proposition 5.18 (Connectedness of closure 23.4). Let (X, T ) be a topological space. If a subset A Y is connected in (X, T ), then A is connected in (X, T ). We now prove that intervals (and only intervals!) are connected when endowed with the standard topology. Recall that a subset I R is an interval if and only if x < x and x, x I = [x, x ] I. In particular, [0, 1], [0, 1), (0, 1], [0, 1], R and are intervals. Theorem 5.19 ( ). Let I R be a subset. (I, T st ) is connected if and only if I is an interval. Can now state and prove a more general version of the Intermediate value theorem. Theorem 5.20 (Intermediate value theorem 24.3). Let f : (X, T ) (R, T st ) be a continuous function. If (X, T ) is connected, for every y that lies between f(a) and f(b) for a, b X, there exists c X such that y = f(c). Since (X, T ) is connected, by Proposition 5.4 (f(x), T st ) is connected. By Theorem 5.19, the image f(x) is an interval. By definition of interval, if f(x) contains f(a) and f(b), it contains all the points in between. So if y is between f(a) and f(b), it s in the image of f. This means that there exists c X such that y = f(c). Each space consists of a bunch of connected pieces that cannot be deconnected further. Definition Let (X, T ) be a space. We define a relation on X by declaring that x C y if and only if there exists a subset A X such that (A, T A ) is connected and x, y A. This defines an equivalence relation! Indeed For reflexivity, note that for any x X the singleton {x} is connected and contains x. Symmetry is obvious. For transitivity, use Proposition

28 The equivalence class [x] C of x respect to C is the connected component of x. The connected components of X form therefore a partition of X. This means that They cover X, and Each two connected components concide or they are disjoint. The number of connected components is a topological invariant Proposition Given a homeomorphism f : (X, T ) (X, T ), the spaces (X, T ) and (X, T ) have the same number of connected components. Use this to prove... Exercise 5.1. The letter X is not homeomorphic to the letter Y Path-connected spaces 25. A different and perhaps more intuitive flavor of connectedness. Definition A space (X, T ) is path-connected if and only if for any points x, x of X there exists a continuous function a: ([0, 1], T st ) (X, T ), called path, such that a(0) = x and a(1) = x. Path-connectedness behaves similarly to connectedness: Proposition 5.24 (Path-Connectedness of image). Let f : (X, T ) (X, T ) be a continuous function. If a subset (X, T ) is path-connected in (f(x), T f(x) ) is path-connected. Being path-connected is a topological property: Proposition Given a homeomorphism f : (X, T ) (X, T ), (X, T ) is path-connected if and only if (X, T ) is path-connected. Use Proposition Remark Let X be a set with two topologies T T. path-connected, then (X, T ) is path-connected. If (X, T ) is Proposition 5.27 (Path-Connectedness of quotient). Let (X, T ) be a topological space, and R X X an equivalence relation on X. If (X, T ) is path-connected, then (X/R, T q ) is path-connected. Special case of Proposition Proposition 5.28 (Path-Connectedness of non disjoint union). Let (X, T ) be a topological space such that X = A A and A A. If the spaces (A, T A ) and (A, T A ) are path connected, then (X, T ) is path-connected. 28

29 Proposition 5.29 (Path-Connectedness of products). Let (X, T ) and (X, T ) be topological spaces. If (X, T ) and (X, T ) are path-connected, then (X X, T T ) is path-connected. same argument works for arbitrary products Proposition 5.30 (Path-Connectedness of arbitrary products). Let (X i, T i ) be a topological space for every i I. If (X i, T i ) is path-connected for every i I, then ( i I X i, i I T i ) is path-connected. Also a notion of component: Definition Let (X, T ) be a space. We define a relation on X by declaring that x P y if and only if there exists a subset A X such that (A, T A ) is path-connected and x, y A, or equivalently if there exists a path in X from x to y. This defines an equivalence relation! Indeed For reflexivity, note that for any x X the singleton {x} is pathconnected and contains x. For symmetry, note that if a: ([0, 1], T st ) (X, T ) is a path from x to x, then ã: ([0, 1], T st ) (X, T ), defined by ã(t) := a(1 t) is a path from x to x. For transitivity, use Proposition The equivalence class [x] P of x respect to P is the path-connected component of x. The path-connected components of X form therefore a partition of X. This means that They cover X, and Each two path-connected components concide or they are disjoint. What s the relation between connectedness and path-connectedness? Proposition 5.32 ( 24). If a space (X, T ) is path-connected, it is also connected. The converse is true under a further condition Definition A space (X, T ) is locally connected if for any x X and open U T containing x there exists A U and V T such that x V A U and (A, T A ) is connected. Example Many spaces we worked with are locally connected: (R n, T st ), (X, T disc ),... 29

30 Proposition 5.35 ( 25.5). If a space (X, T ) locally connected, then it is path-connected if and only if it is connected. To look for an example of something connected and not path connected we need to seek a space that is at least not locally connected: Example 5.36 ( 24, Example 7 and 25, Example 2). Let S R 2 be the topologist s sine curve, defined by S := {(x, sin 1 ) x > 0} {0} [0, 1], x endowed with the standard topology. The space (S, T st ) is not locally connected (draw a neighborhood of (0, 1)), is not path connected (can t join (0, 0) to (30, sin( 1 30 ))), is connected (it is the closure of {(x, sin 1 x ) x > 0}, which is homeomorphic to (0, + ), which is connected). 6. Compactness 26,27,29 Extreme Value Theorem Will be able to prove that [0, 1] = (0, 1) 6.1. Compactness. Way to see that [0, 1] is not homeomorphic to [0, 1). Key property for the Extreme value theorem. Need some terminology: Let (X, T ) be a space. A collection {U i } i I P(X) of subsets of X is a covering if i I U i = X; is an open covering if it is a covering and each U i is open; is a finite covering if it is a covering and I is finite. If {U i } i I is a covering of (X, T ) and J I, (U i ) i J is a subcovering of the original covering. The subcovering is finite if J is finite. Definition 6.1. (Compactness) A space (X, T ) is compact if and only if any open covering of (X, T ) admits a finite subcovering, i.e., for any collection {U i } i I of open subsets such that X = i I U i = X there exists a finite subset J I such that X = i J U i. Remark 6.2. The definition of a space (X, T ) being compact does not mean that there exists a finite covering of the space X! This is always true (just take the covering with only one element given by {X}). Example 6.3. Can already prove that some spaces are not compact. (1) R is not compact; consider the covering {( n, n)} n N. 30

31 (2) Q [0, 2] is not compact; choose an increasing sequence (q n ) n N of rational numbers converging to 2. Then consider the covering {[0, q n ) ( 2, 2]} n N, which does not admit finite subcoverings. (3) [0, 1) is not compact; consider the covering {( 1 n )} n N, which does not admit finite subcoverings. Example 6.4. Examples of compact spaces: (1) [0, 1] is compact (will prove later) (2) Finite spaces are always compact. More generally finite topologies are always compact. Example 6.5 (Compactness for canonical topologies). Let X be a set. (X, T disc ) is compact if and only if X is a finite; consider the covering {x} x X, which does not admit finite subcoverings. (X, T codisc ) is always compact; the only possible coverings are {X, } and {X}, which are already finite. (X, T fin ) is always compact; given a covering, any non-empty element of the covering already contains everything but possibly finitely many points. Can then add one by one elements of the covering to make sure you cover the missing points. (X, T count ) is compact if and only if X is finite; if {x n } n N is a countable list of different elements of X, consider the covering given by {X \ {x k k n}} n N, which does not admit finite subcoverings. Remark 6.6 ( 26.1). Let (X, T ) be a space, and Y X a subset. The space (Y, T Y ) is compact if and only if any covering of Y by open subsets in X contains a finite subcollection covering Y. If Y Z X, (Y, T Y ) is compact if and only if (Y, (T Z ) Y ) is compact. Proposition 6.7 (Closed in compact is compact 26.2,26.3). Let (X, T ) be a topological space, and A X a subset. (1) If A is closed and (X, T ) is compact, then (A, T A ) is compact. (2) If (A, T A ) is compact and (XT ) is Hausdorff, then A is closed. Proposition 6.8 (Image of compact is compact 26.5). Let (X, T ), (X, T ) be topological spaces, f : X X a continuus function and A X. If (A, T A ) is compact, then (f(a), T f(a) ) is compact. Proposition 6.9 (Quotient of compact is compact). Let (X, T ) be a topological space, and R X X an equivalence relation. If (X, T ) is compact, then (X/R, T R ) is compact. 31

32 Use Proposition 6.8. Being compact is a topological property: Proposition Given a homeomorphism f : (X, T ) (X, T ), (X, T ) is compact if and only if (X, T ) is compact. Use Proposition 6.8. Remark Let X be a set with two topologies T T. compact, then (X, T ) is compact. If (X, T ) is Proposition ] Let f : (X, T ) (X, T ) be a continuous bijective function. If (X, T ) is compact and and (X, T ) is Hausdorff, then f is a homeomorphisms. Denote by g : X X the inverse of f. We prove that g is continuous, by showing that the preimage via g of any closed subset of X is closed in X. Let F X be a closed subset of X. Then (F, T F ) is compact by Proposition 6.7(1). Then (f(f ), T f(f )) is compact by Proposition 6.8. Then g 1 (F ) = f(f ) is closed in X by Proposition 6.7(2), as desired. Proposition 6.13 (Finite product of compact is compact 26.7). Let (X, T ) and (X, T ) be topological spaces. If (X, T ) and (X, T ) are compact, then (X X, T T ) is compact. Lemma 6.14 (Tube lemma 26.8). Let (X, T ) and (X, T ) be topological spaces, with (X, T ) compact. If x 0 X, and N X X is an open subset that contains a slice {x 0 } X, then N contains a tube W X, where W X is an open subset of X. Proof of the Tube Lemma. Cover the open N with elements of the basis of X X of the form U U (that intersect {x 0 } Y ) N = (U U ) U U U SInce {x 0 } X = X is compact, covered by finitely many, say In particular (1) The subset {x 0 } X (U 1 U 1) (U n U n). is open in X and contains x 0. (2) X is covered by W := U 1 U n, X = U 1 U n. 32

33 We then conclude that W X (U 1 U 1) (U n U n) (U U ) = N, as desired. U U U Can now prove the proposition Proof of the proposition. The analog result for arbitrary products is true, but it is an extremely hard theorem, known as Tychonoff Theorem. The proof involves the axiom choice and is not a simple improvement of the argument used for finite products. Theorem 6.15 (Tychonoff Theorem 5, 37.3). Let (X i, T i ) be a topological space for any i I. If (X i, T i ) is compact for any i I, then ( i I X i, i I T i ) is compact. Can now characterize the compact subsets of R, and more generally R n. Proposition 6.16 ( 27.1). The interval ([0, 1], T st ) is compact. Suppose that U := {U i } is an open covering of [0, 1], and prove that there exist a finite subcovering. Without loss of generality, we can assume that each U i is an interval. Consider the set Y := {x [0, 1] [0, x] is covered by finitely many elements of U}. We note that (0) If x Y and x < x, then x Y. (1) Y contains 0, and is therefore not empty; to see this, note that there exists U U that contains 0, and so [0, 0] = {0} U. (2) Y is open; to see this, consider y Y. Then [0, y] U 1 U n for some U 1,..., U n U. In particular, there exists an interval U j such that y U j. Then U j is an open neighborhood of y contained in Y. (3) Y is closed; to see this, suppose by contradiction that there exists z Y such that any open neighborood of z intersects Y. In particular, given an interval U U that contains y, U intersects Y. Suppose then y U Y. Since y Y, [0, y] U 1 U n for some U 1,..., U n U. Then, [0, z] U 1 U n U 33

34 and therefore z Y, contradiction. Since Y is a non-empty closed and open subset of the connected space [0, 1], Y = [0, 1]. In particular 1 Y, which implies that for some U 1,..., U n U, as desired. [0, 1] U 1 U n Proposition Given A R n, TFAE: (1) A is bounded with respect to the Euclidean metric d 2, i.e., there exists N > 0 such that A B d2 (0, N). (2) A is bounded with respect to the uniform metric d, i.e., there exists N > 0 such that A B d (0, N) = [ N, N] [ N, N]. }{{} n copies In this case we say that A is bounded. Proposition 6.18 (Heine-Borel Theorem, 27.1). Let A R n be a subset. (A, T st ) is compact if and only if A is closed and bounded with respect to the Euclidean metric. For the direct implication, use Proposition 6.7(2) and the fact that any subspace of R n can be covered by open balls. For the converse implication, [book]. Can now prove more general version of the Extreme Value Theorem. Theorem 6.19 (Extreme Value Theorem 27.4). Let f : (X, T ) (R, T st ) be a continuous function. If (X, T ) is compact, f admits extreme values, i.e., there exist a, b X such that f(c) lies between f(a) and f(b) for any c X. Since (X, T ) is compact, by Proposition 6.8 (f(x), T st ) is compact. By Proposition 6.18, the image f(x) is a compact subspace of R, and therefore it admits a maximum M and a minimum m, which have to be of the form f(a) and f(b) for some a and b in X Local compactness 29. Compact spaces are cool. When something is not compact, want to find the smallest approximation of the space to something compact. Example Can you make the following spaces compact by adding one point? (1) [0, 1) [0, 1] (2) (0, 1) S 1 34

35 (3) [0, 1) (1, 2] (20, 21] letter T This is a process known as one-point compactification, or Alexandroff compactification, and makes sense for any Hausdorff space that is locally compact. Definition A space (X, T ) is locally compact at a point x X if there exists an open U and a compact C such that x U C X. The space is locally compact if it is locally compact at every point. Example 6.22 (Exampes and non-examples). (1) [0, 1), (0, 1) and [0, 1) (1, 2] (20, 21] are locally compact. (2) Any compact space is locally compact. (3) Any discrete space is locally compact. (4) Any open or closed subset of R n is locally compact, when endowed with the subspace topology (by Proposition 6.27). (5) Q is not locally compact at any point. Can always construct the one-point compactification of a locally compact Hausdorff space. Theorem 6.23 (Alexandroff compactification 29.1). Let (X, T ) a locally compact Hausdorff space. The collection ˆT := T { { } X \ K K X is compact } defines a topology on ˆX := X { }. Moreover, (1) The space ( ˆX, ˆT ) is compact. (2) The space ( ˆX, ˆT ) is Hausdorff if (X, T ). (3) The topology on X as a subspace of ˆX coincides with the original topology of X, i.e., (X, ˆT X ) = (X, T ). Remark The theorem is actually stronger than this (see the book). For instance The space ˆX is unique with these properties. The space X is locally compact and Hausdorff if and only if there exists a one-point compactification. Will prove the theorem later, first let s apply it. Example Find the one-point compactification: (1) [0, 1] [0, 1] {2} (2) K K {0} (3) N K {0} (4) R S 1 (5) R n S n 35

36 Proposition 6.26 (Characterization of locally compact Hausdorff 29.2). Let (X, T ) be a Hausdorff space. Then (X, T ) is locally compact if and only if for any point x X and open U containing x there exists an open neighborhood V of x such that V is compact and x V V U. See the book. Proposition 6.27 ( 29.3). Let (X, T ) be a space and A X a subset. If (X, T ) is locally compact Hausdorff and A is open or closed in X, then A is locally compact. Can now prove the theorem Proof of Theorem Separability axioms 31,32,24 Definition 7.1. A space (X, T ) is said to satisfy the axiom T 0 is for any two points x, y X there exists either an open neighborhood of x that does not contain y or an open neighborhood of y that does not contain x. Example 7.2. The space R endowed with the codiscrete topology does not satisfy the axiom T 0. Definition 7.3. A space (X, T ) is said to satisfy the axiom T 1 is it satisfy the following two equivalent conditions. (1) For any point x X and any point y X there exists either an open neighborhood of x that does not contain y. (2) All singletons of X are closed in (X, T ). Remark 7.4. A space that satisfies the axiom T 1 satisfies the axiom T 0. Example 7.5. The space R endowed with the upper topology satisfies the axiom T 0 but does not satisfy the axiom T 1. Definition 7.6. A space (X, T ) is said to be Hausdorff, or to satisfy the axiom T 2, if for any points x y X there exists disjoint open neighborhoods U and V of x and y. Remark 7.7. A space that satisfies the axiom T 2 satisfies the axiom T 1. Example 7.8. The space R endowed with the cofinite topology satisfies the axiom T 1 but does not satisfy the axiom T 2. Definition 7.9. A space (X, T ) is said to be regular, or to satisfy the axiom T 3, if it satisfies the axiom T 1 and for any point x X and closed subset F X not containing x there exists disjoint open U and V that contain x and F, respectively. 36

37 Remark A space that satisfies the axiom T 3 satisfies the axiom T 2. Example 7.11 ( 31,Example1). The space R endowed with the topology T K satisfies the axiom T 2 but does not satisfy the axiom T 3. Definition A space (X, T ) is said to be normal, or to satisfy the axiom T 4, if it satisfies the axiom T 1 and for any disjoint closed subset F, G X there exists disjoint open U and V that contain F and G, respectively. Remark A space that satisfies the axiom T 4 satisfies the axiom T 3. Example 7.14 ( 31,Example3). The space R R endowed with the topology T uplim T uplim satisfies the axiom T 4 but is not metrisable. Theorem 7.15 ( 32.2). Any metrisable space is normal. Example 7.16 ( 31,Example2). The space R endowed with the topology T uplim satisfies the axiom T 4 but is not metrisable. Can characterize regular and normal spaces as follows. Proposition 7.17 ( 31.1). Let (X, T ) be a space that satisfies the axiom T 1. (1) (X, T ) is regular if and only if and only if for any point x X and open U containing x there exists an open neighborhood V of x such that x V V U. (2) (X, T ) is normal if and only if and only if for closed subset A X and open U containing A there exists an open V containing A such that A V V U. Proposition 7.18 ( 31.2). Let (X, T ) be a space, and A X a subset. If (X, T ) is regular, then (A, T A ) is regular. Proposition 7.19 ( 31.2). Let (X i, T i ) be a space for i I. If each (X i, T i ) is regular, then ( i I X i, i I T i ) is regular. Last time we saw metrisable T 4 T 3 T 2 T 1 T 0. What further conditions can be added to get the inverse implications? Proposition Any locally compact Hausdorff space is regular. Apply Proposition 6.26 and Proposition

38 Proposition 7.21 ( 32.3). Any compact Hausdorff space is normal. Proposition 7.22 ( 32.1). Any regular space that admits a countable basis is normal. In fact... Theorem 7.23 (Uryshon Metrization Theorem 34.1). Any regular space that admits a countable basis is metrizable. Proven using Theorem 7.24 (Uryshon Lemma 33.1). Let (X, T ) be a normal space. For any A, B X two disjoint closed subsets, there exists a continuous function f : (X, T ) ([0, 1], T st ) such that f(a) = 0 for every a A and f(b) = 1 for every b B. 8. Spaces with the same homotopy type 51 (and bits of 58) From now on, we call map a continuous function, omit the topology, and denote the standard interval [0, 1] by I. Will introduce a notion of equivalence for spaces that is more relaxed than being homeomorphic. Definition 8.1 ( 51). Given two maps f, g : X Y, we say that that they are homotopic if there exists a homotopy from f to g, and in this case we write f g. A homotopy from f to g is a map H : X I Y such that H(x, 0) = f(x) and H(x, 1) = g(x) for all x X. Proposition 8.2 ( 51.1). Homotopy of maps from X to Y is an equivalence relation. Remark 8.3. Homeomorphic spaces are also homotopic, but the converse is not true! Definition 8.4 ( 58). Two spaces X and Y are homotopic, or homotopy equivalent, or have the same homotopy type if there exists f : X Y and g : Y X that are homotopy inverses, i.e., f g id Y and g f id X. In this case we write X Y. Proposition 8.5 ( 58). Having the same homotopy type is an equivalence relation. 38

39 Example 8.6. (1) R 2 { }. (2) R 2 \ {(0, 0)} S The fundamental group 51,52 Definition 9.1. Let X be a space and x 0 a point. A loop based at x 0 is a map α: I X such that α(0) = α(1) = x 0. Definition 9.2. Two loops α, β : I X based at x 0 are homotopic as based loops if there exists a homotopy of based loops from α to β, i.e., a map H : I I X such that H(s, 0) = α(s), H(s, 1) = β(s) and H(0, t) = H(1, t) = x 0. We write α β. Proposition 9.3 ( 51.1). Equivalence of based paths is an equivalence relation. Definition 9.4. Given two loops α, β : I X based at x 0, the concatenation of α with β is the loop α β : I X based at x 0 defined by { α(2s) s [0, 1 α β(s) := 2 ] β(2s 1). s [ 1 2, 1] Proposition 9.5. The concatenation of based loops is well-defined up to homotopy of based loops, i.e., if α α and β β then α β α β. Notation 9.6. Let X be a space and x 0 a point. Denote by ɛ: I X the constant path at x 0 defined by ɛ(s) := x 0. For any loop α: I X based at x 0, denote by α: I X the reverse loop defined by α(s) := α(1 s). Proposition 9.7 ( 51.2). Concatenation of based loops defines a group operation up to homotopy of based loops. Precisely (1) it is associative up to homotopy of based loops, i.e., (α β) γ α (β γ). (2) it has a neutral element up to homotopy of based loops given by ɛ, i.e., α ɛ α ɛ α. (3) it admits inverses up to homotopy of based paths given by the reverse, i.e., α α ɛ α α. 39

40 Definition 9.8. The fundamental group of X at x 0 is the set of homotopy classes of based loops π 1 (X, x 0 ) := {α: I X α(0) = α(1) = x 0 } / endowed with the multiplication given by concatenation of based loops, i.e., [α] [β] := [α β]. The neutral element is the class of the constant path e := [ɛ], and inverses are given by the classes of the reverse paths, i.e.., [α] 1 = [α]. Proposition 9.9 ( 52.1,52.2). If X is a path-connected space, the fundamental group of X does not depend on the base point, i.e., for any points x 0 and y 0 in X there is an isomorphism of groups π 1 (X, x 0 ) = π 1 (X, y 0 ). If α: I X is any path from x 0 to y 0, the assignment [β] ˆα([β]) := [α β α] realizes the desired isomorphism. Proposition Any map of spaces f : X Y induces a group homomorphism defined by f : π 1 (X, x 0 ) π 1 (Y, f(x 0 )) [α] f [α] := [f α]. Proposition 9.11 (Functorial properties of the fundamental group 52.4). For every space X (id X ) = id π1 (X,x 0 ) : π 1 (X, x 0 ) π 1 (X, x 0 ). For any maps f : X Y and g : Y Z (g f) = g f : π 1 (X, x 0 ) π 1 (Z, g(f(x 0 ))). 40

41 9.1. The fundamental group of the circle 54. Need some prelimiary constructions. Consider the helix H := {(cos(2πt), sin(2πt), t) t R}. Note that the projection pr 3 : H R given by pr 3 (cos(2πt), sin(2πt), t) = t realizes a homeomorphism H = R. Denote by π : H S 1 the projection defined by π(cos(2πt), sin(2πt), t) := (cos(2πt), sin(2πt)). For any integer n, denote by P n the point of the helix P n := (cos(2πn), sin(2πn), n). Note that π(p n ) := (1, 0) for any n. For any integer n denote by ω n : I S 1 the loop based at P := (1, 0) defined by Will use the following theorem. ω n (s) := (cos(2πns), sin(2πns)). Theorem 9.12 (Path lifting lemma 53.1,54.1,54.2). For any integer k and for any loop α: I S 1 based at P there exists a unique path α k : I H, called the lifting of α starting at P k, such that α k (0) = P k and π α k = α. Moreover, if α β then α k (1) = β k (1). Example The lifting ω k n : I S 1 of ω n starting at P k is given by Theorem The assignment defines an isomorphism of groups ω k n(s) = (cos(2πns), sin(2πns), nt + k). [α] Φ([α]) := pr 3 ( α 0 (1)) Φ: π 1 (S 1, P ) = Z Homotopy invariance of the fundamental group 58. Lemma 9.15 ( 58.4). If h k : X Y, for any x 0 X there exists a path α: I Y from h(x 0 ) to k(x 0 ) such that k = ˆα h : π 1 (X, x 0 ) π 1 (Y, k(x 0 )). Proposition 9.16 ( 58.7). Any homotopy equivalence f : X Y induces an isomorphism π 1 (X, x 0 ) = π 1 (Y, f(x 0 )). 41

42 Example The fundamental group of the Möbius strip is isomorphic to Z. Proposition For any spaces X and X with base points x 0 and x 0 there is an isomorphism of groups π 1 (X X, (x 0, x 0)) = π 1 (X, x 0 ) π 1 (X, x 0). Example The fundamental group of the torus is isomorphic to Z Z. Example The fundamental group of the filled torus is isomorphic to Z. Definition 9.21 ( 68). Let G and H be groups. Denote by F (G H) := n=0 (G H)n the set of all words of finite length with letters in G and H. Consider on F (G H) the equivalence relation generated by the following two rules: If a word contains 1 G, it is equivalent to the word obtained by removing 1 G. If a word contains g g, it is equivalent to the word obtained by adding a comma between g and g. If a word contains h h, it is equivalent to the word obtained by adding a comma between h and h. The free product of G and H is the quotient G H := F (G H)/. It becomes a group when endowed with multiplication given by concatenation of words. Theorem 9.22 (Simplified version of the Van Kampen Theorem 70.3). Let X be a space, together with open subsets U, V X such that X = U V and x 0 U V. If U, V and U V are path-connected and π 1 (X, x 0 ) = {0} then π 1 (X, x 0 ) = π 1 (U; x 0 ) π 1 (V, x 0 ). Example The fundamental group of the eight-shaped space is Z Z. Theorem 9.24 (Van Kampen Theorem 70.2). Let X be a space, together with open subsets U, V X such that X = U V and x 0 U V. Denote by j U and j V the respective inclusions of U V in U and V. If U, V and U V are path-connected then there is a surjective group homomorphism φ: π 1 (U; x 0 ) π 1 (V, x 0 ) π 1 (X, x 0 ) whose kernel is the smallest normal subgroup containing all the elements of the form (j U (α)) 1 j V (α) for α π 1 (U V, x 0 ). In particular, π 1 (X, x 0 ) = π 1 (U; x 0 ) π 1 (V, x 0 )/ ker φ. 42

43 Example For n > 1, the fundamental group of S n is isomorphic to {0}. Example The fundamental group of the real projective plane is isomorphic to Z/ Higher homotopy groups [Hatcher 4.1] Notation For n 0 denote by I n the unitary cube and by I n its boundary. I n := [0, 1] [0, 1] [0, 1], Notation Given subspaces A X and B Y, we write f : (X, A) (Y, B) to mean that f : X Y is a map such that f(a) B. The following equivalence relation recovers the homotopy equivalence of loops for n = 1, and the equivalence relation of being in the same pathconnected component when n = 0. The corresponding quotient recovers the (underlying set of) the fundamental group when n = 1, and the set of path-connected components when n = 0. Definition Let (X, x 0 ) be a pointed space, and n 0. Two maps α, β : (I n, I n ) (X, x 0 ) are homotopic relatively to I n if there exist a map H : I n I X such that H(s, 0) = α(s), H(s, 1) = β(s) and H( I n I) {x 0 }. We denote the quotient by π n (X, x 0 ) := {α: (I n, I n ) (X, x 0 )}/ I n. The following group structure recovers the multiplication we have seen on the fundamental group when n = 1. Proposition Let (X, x 0 ) be a pointed space, and n 1. π n (X, x 0 ) admits a group structure, defined by [α] [β] := [α 1 β], The set where α 1 β denotes the concatenation along the first coordinate, defined by { α(2s1,, s α 1 β(s 1, s 2,..., s n ) := 2,..., s n ) s 1 [0, 1 2 ] β(2s 1 1, s 2,..., s n ). s 1 [ 1 2, 1] Like for n = 1, the key ingredients for the proof are the following properties of the concatenation. Proposition 10.5 ( 51.2). Concatenation of maps (I n, I n ) (X, x 0 ) defines a group operation up to homotopy relatvely to I n. Precisely (1) it is associative up to homotopy relatvely to I n, i.e., (α 1 β) 1 γ I n α 1 (β 1 γ). 43

44 (2) it has a neutral element up to homotopy of based loops given by the constant map ɛ: (I n, I n ) (X, x 0 ), i.e., α 1 ɛ I n α I n ɛ 1 α. (3) it admits inverses up to homotopy of based paths given by the reverse with respect to the first component, i.e., α 1 α ɛ α 1 α, where α(s 1, s 2,..., s n ) := α(1 s 1, s 2,..., s n ). Proposition Concatenation along a different coordinate induces the same group operation. Proposition Let (X, x 0 ) be a pointed space, and n 2. The group π n (X, x 0 ) is abelian. Proposition Any pointed map f : (X, x 0 ) (Y, y 0 ) induces a group homomorphism f : π n (X, x 0 ) π n (Y, f(x 0 )) defined by [α] f [α] := [f α]. Proposition 10.9 (Functorial properties of the higher homotopy groups). For every pointed space (X, x 0 ) (id X ) = id π1 (X,x 0 ) : π n (X, x 0 ) π n (X, x 0 ). For any pointed maps f : (X, x 0 ) (Y, y 0 ) and g : (Y, y 0 ) (Z, z 0 ) (g f) = g f : π n (X, x 0 ) π n (Z, g(f(x 0 ))). Definition Two pointed maps f, g : (X, x 0 ) are homotopic as pointed maps if there exists a map H : X I Y such that H(x, 0) = f(x), H(x, 1) = g(x) and H(x 0, t) = y 0. We write f g Proposition If two pointed maps f, g : (X, x 0 ) (Y, y 0 ) are homotopic as pointed maps, then for every n 0 f = g : π n (X, x 0 ) π n (Y, y 0 ). 44

45 Definition Two pointed space (X, x 0 ) and (Y, y 0 ) are homotopic as pointed spaces if there exist pointed maps f : (X, x 0 ) (Y, y 0 ) and g : (Y, y 0 ) (X, x 0 ) such that g f Id (X,x0 ) and f g Id (Y,y0 ). Proposition If two pointed spaces (X, x 0 ) and (Y, y 0 ) are equivalent as pointe spaces, then for every n 0 π n (X, x 0 ) = π n (Y, y 0 ). Proposition For any pointed spaces (X, x 0 ) and (Y, y 0 ) there is an isomorphism of groups for any n 1 π n (X Y, (x 0, y 0 )) = π n (X, x 0 ) π n (Y, y 0 ). Example Homotopy groups are in general very hard to compute, and for instance understanding the homotopy groups of spheres is a very hard problem in homotopy theory. The following table displays the homotopy groups of spheres in low degrees. 45