Topologie Luk´aˇs Vokˇr´ınek September 22, 2022 Contents 1. Motivation 1 2. Topological space 1 3. Continuous views 3 4. Subspaces, products 5 5. Separability Axioms 7 6. Compact Spaces 9 7. Connectedness 13 8. Locally Compact Spaces 17 9. Real Functions 19 10. Homotopy, fundamental group, covering 22 11. Simplicial complexes, Brouwer’s theorem, dimension invariance 28 12. Compactly generated Hausdorff spaces 34 13. Algebras of continuous functions 36 14. Topological groups, Pontryagin duality 38 15. Paracompact Spaces 41 16. Uniform Spaces 42 i Introduction This text was created by writing my preparations for lectures and exercises for the subject “Topology”. As a starting text, I used my notes, which I took when the subject was taught by Prof. Rosick´y. It was based on Pultr’s book “Subspaces of Euclidean Spaces”. I expanded or supplemented some parts, I drew mainly from Bredon’s book “Geometry and topology”. This also applies to the chapters I added. In the text, the examples we did in the exercises are marked “cv”; I did not write sample solutions for them. I left the examples marked with “hw” as homework. The parts of the text marked with “∗” are more technically demanding passages that I sometimes didn’t even cover in the lecture, but which I can ask about in the exam. I consider the parts marked with “∗∗” to be unnecessarily difficult or special and will not ask about them. We didn’t do the parts marked “nd”, but I plan to cover them in the future. ii 1. Motivation Topology deals with “topological spaces” – these are roughly metric spaces, we just forget the specific distances between points and only remember which points “are close”. The central concept is continuity, more specifically continuous display. As a result, this means that a square is “the same” as a circle (unlike geometry). This is because there are continuous mutually inverse mappings between the square and the circle – together they specify an isomorphism. There are other kinds of spaces – based on other kinds of views. This is, for example, metric spaces – isometry differentiable varieties – differentiable representations algebraic varieties – polynomial representation PL (piecewise linear) variety – piecewise linear representation polyhedra – affine mapping Algebraic topology goes much further in non-geometry (square = circle), which declares Rn and the space consisting of a single point to be equal spaces, since Rn can be “continuously deformed” to a point. 2. Topological space In the metric space M we define an open sphere around x with radius ε > 0 as Bε(x) = {y ∈ M | dist(x, y) < ε}. We say that a subset U ⊆ M is open if for every x ∈ U there exists ε > 0 such that Bε(x) ⊆ U. In short, we say that U contains some of its neighborhoods with every point. Definition 2.1. A Topology on a set X is a system of subsets X ⊆ P(X) satisfying the following conditions (1) ∅, X ∈ X, (2) Ui ∈ X, i ∈ I ⇒ i∈I Ui ∈ X, (3) Ui ∈ X, i ∈ I, I final ⇒ i∈I Ui ∈ X. A Topological space is the set X together with the topology X on X . We call the elements of X wiped subsets of X. Remark. Condition (0) follows from the other two, since ∅ is the union of the empty system of subsets and X is the intersection of the empty system. Examples 2.2. 1. metric spaces (we can do this in more detail over time), 2. for any set X we define a discrete topology on X as X = P(X) (everything is open, it is specified by the metric dist(x, y) = 1), 3. for any set X we define a trivial topology on X as X = {∅, X} (“nothing” is open, it is given by the pseudometric dist(x, y) =0), 4. for any set X we define the topology of finite complements on X as X = {U ⊆ X | X ∖ U final} ∪ {∅}, 1 5. if X is any (pre)ordered set, it ishw 1 X = {U ⊆ X | U meets ∀x ∈ U ∀y ≤ x : y ∈ U} topology. Conversely, if X is an arbitrary topology satisfying (2) even for infinite index sets I , then there is a preorder on X specifying this topology. We will now try to prove that the open sets specified by the metric really define the topology. The following concept will be useful for this. Definition 2.3. A system of sets S ⊆ P(X) is called a subbase of a topology X if X is the smallest topology containing S. A system of sets B ⊆ P(X) is called a base of the topology X if X are precisely all unions of the elements of B. In other words, X = {A ⊆ X | ∀x ∈ A ∃U ∈ B: x ∈ U ⊆ A} (because then A = {U ∈ B | U ⊆ A}). Lemma 2.4. It holds that B is a basis of some topology (by definition, however, the only one), if and only if the following conditions hold 1. X = B a 2. for each U, V ∈ B and x ∈ U ∩ V there exists a W ∈ B such that x ∈ W ⊆ U ∩ V . Prove the previous lemma.cv Example 2.5. Let M be a metric space. Then the system of spheres B = {Bε(x) | x ∈cv M, ε > 0} is the basis of the topology. (First prove that they are based on some topology, then identify it as a canonical topology on a metric space.) Now if S ⊆ P(X) is an arbitrary system of subsets, B = {finite intersections of elements of S} satisfies the conditions of the lemma and therefore the topology generated by S right now X = {unification of finite intersections of elements S}. Definition 2.6. A subset F ⊆ X is called closed if X ∖ F is open. For example, in the space of finite complements precisely the finite sets and X are closed. For X = C, equivalently closed sets can be described as null sets of polynomials – this example has a generalization to Cn, see algebraic geometry. Remark. For closed sets, “dual” axioms apply to topology axioms. Equivalently, the topology can be specified by a system of closed sets that satisfy these axioms. Definition 2.7. Closure A of a subset A ⊆ X is the smallest closed subset containing A, i.e. A = A⊆F uz. F. Example 2.8. Prove the following properties of the closurecv 1. ∅ = ∅, 2. A ∪ B = A ∪ B, 2 3. A ⊆ A, 4. A = A. Furthermore, show that A ∩ B = A ∩ B does not hold in general. Using a closure (or better said a closure operator), the topology can be reconstructed as follows: a subset A ⊆ X is closed if and only if A = A. Remark. It is true that the topology can be equivalently specified by a closure operator (ie operator P(X) → P(X) satisfying axioms (1)–(4)). Lemma 2.9. It holds A = {x ∈ X | ∀U open, x ∈ U : A ∩ U ̸= ∅}. We refer to the points on the right as the limit points of A (and we do not need to say what the limit of the sequence is). Proof. x /∈ A holds if and only if there exists a closed F ⊇ A not containing x. Going to complements, that is if and only if there exists an open U = X ∖F, A∩U = ∅ and containing x. But that is exactly x /∈ RHS. Definition 2.10. “Dually” we define the interior of A as the largest open set contained in A, i.e. ˚A = A⊇U rot. U. We say that the interior points of A are those that fit into A with some of their surroundings. We will define the surroundings more precisely later. 3. Continuous maps Definition 3.1. A mapping f : X → Y between two topological spaces is called continuous if for every open U ⊆ Y there is also f−1(U) ⊆ X dad Exercise 3.2.cv 1. It is enough to verify the continuity for U from some (arbitrary) subbase of the topology on Y . 2. A representation of f is continuous if and only if the pattern of every closed set is closed. end of 1. lecture Definition 3.3. A subset N ⊆ X is called a neighborhood of a point x ∈ X if there exists an open set U with property x ∈ U ⊆ N. In particular, the open neighborhood of x is the same as the open set containing x. Open sets can be characterized by neighborhood as those that are neighborhoods of all their points. Definition 3.4. We say that the mapping f : X → Y is continuous at a point x ∈ X if for every neighborhood N of the point f(x) there is also f−1(N) around the point x. Exercise 3.5. Prove that a mapping f : X → Y is continuous if and only if it is continuoushw 2 at every point x ∈ X. 3 Definition 3.6. We say that the system N of the neighborhood of x is a neighborhood basis of x if every neighborhood of x contains as a subset some element of N. (I worked for regulars later.) Example 3.7. In metric space, the open sphere Bε(x), ε > 0, centered at x forms the neighborhood of x. Alternatively, B1/n(x), n ∈ N forms the basis of the neighborhood of the sphere. This set is countable. Therefore, every metrizable space, i.e. a space whose topology is specified by some metric, must have a countable basis around each point – it is so-called “first countable”. Exercise 3.8.nd 1. Continuity at the point x ∈ X is sufficient to verify on the neighborhoods of f(x) from some (arbitrary) basis of the neighborhood. 2. A map f : M → N between metric spaces is continuous if and only if it satisfies the ε-δ-definition of continuity. Proof. Part 1. is elementary. Part 2. follows from the fact that the open sphere Bδ(x), δ > 0, forms the basis of the neighborhood of x and Bε(f(x)), ε > 0, forms the basis of the neighborhood of f(x). Exercise 3.9. Prove that the mapping f : X → Y between preordered sets X, Y is continuous∗ if and only if it is isotonic. Lemma 3.10. The following conditions on the mapping f : X → Y are equivalent∗∗ 1. f is continuous, 2. f−1(B) ⊆ f−1(B) for any subset of B ⊆ Y , 3. f(A) ⊆ f(A) for any subset of A ⊆ X. The last condition is a generalized statement that xn → x implies f(xn) → f(x) (however, sequences may not be sufficient for general topological spaces). Proof. We will first show the equivalence of 1. and 2. Since f−1(B) is a closed subset containing f−1(B), it must also contain f−1(B). In the opposite direction, for closed F ⊆ Y f−1(F) ⊆ f−1(F) = f−1(F) and thus f−1(F) is closed. Now we show 2. ⇒ 3. We want A ⊆ f−1(f(A)), while obviously A ⊆ f−1(f(A)) and thus A ⊆ f−1(f(A)) 2. ⊆ f−1 (f(A)). It remains to show 3. ⇒ 2. We want f(f−1(B)) ⊆ B, while obviously f(f−1(B)) ⊆ B and thus f(f−1(B)) 3. ⊆ f(f−1(B)) ⊆ B. Definition 3.11. A mapping f : X → Y is called a homeomorphism if f is a bijection and both mappings f, f−1 are continuous. Example 3.12. 1. The interval (0, 1) is homeomorphic to R; for example, the homeomorphism (0, 1) → R is the mapping t → tg(πt − π/2). 4 2. The mapping id: Xdisc → Xtriv is a continuous bijection, but its inverse id: Xtriv → Xdisc is not continuous ; see next example. 3. Consider when the mapping id: (X, X0) → (X, X1) is continuous.nd 4. The mapping [0, 1) → S1, t → e2πit is a continuous bijection, but its inverse is not continuous. 5. In fact, there is no homeomorphism [0, 1) ∼= −→ S1. This is best demonstrated by finding some “invariant” that distinguishes the two spaces. In this case, for example, we can say (in time we will be able to formulate it precisely) that removing any point from S1 does not disintegrate the space, while removing the point t ̸= 0 from [0, 1) disintegrates this interval . Another such invariant is compactness. 6. For m ̸= n there is no homeomorphism Rm ∼= −→ Rn. For n = 1 this can be seen, similar to the previous example, using point removal and context. For higher n , “higher connectedness” is needed. Example 3.13. Describe continuous mappings from a trivial space and continuous mappingscv into a discrete space. Remark. The topology is non-algebraic (a continuous bijection is not necessarily a homeomorphism). From one of the examples, we see that the completeness of a metric space is not a topological concept, i.e. there are homeomorphic spaces, one of which fulfills this property and the other does not. On the other hand, compactness is a topological concept, later we characterize it purely in the language of open sets. 4. Subspaces, products Definition 4.1. Let X be a topological space and let A ⊆ X be its subset. We define on A the subspace topology as {A ∩ U | U ⊆ X open}. We call the set A together with the topology of the subspace subspace X. An important property of a subspace is that the embedding i: A → X is continuous and has the following universal property: the mapping f : T → A is continuous if and only if if : T → X is continuous . T if 22 f GG A _ i  X (both follow from the fact that A ∩ U = i−1(U)). The proof can be summarized in the observation: the topology of a subspace is the smallest one for which the inclusion i is continuous. Lemma 4.2. For the subset B ⊆ A holdshw 3 clA B = A ∩ B, where clA B denotes the closure of B in the subspace A. Remark. Nothing similar applies to the interior. 5 Definition 4.3. A system A ⊆ P(X) of sets is called a cover of the space X if A = X. Exercise 4.4. Let U be an open covering of the space X. Prove that the map f : X → Y iscv continuous if and only if every taper of f|U : U → Y , U ∈ U, is continuous. Similarly, prove the same for a finite closed coverage of F. Exercise 4.5. Prove that the square is homeomorphic to the circle.hw 4 Definition 4.6. Let X, Y be topological spaces. We define on X × Y the product topology generated by the basis {U × V | U ∈ X, V ∈ Y}. We call the set X × Y together with the product topology the product of the topological spaces X, Y . An important property of the product is that the projections p: X×Y → X, q: X×Y → Y are continuous and have the following universal property: the mapping f = (g, h) colT → X × Y is continuous if and only if its components pf = g : T → X and qf = h: T → Y are continuous. X X T pf FF qf II f GG X × Y p XX q 66 ≡ T g FF h II (g,h) GG X × Y p XX q 66 Y Y (both follow from the fact that U × V = p−1(U) ∩ q−1(V )). A little more complicated is the product of infinitely many topological spaces, where the clue to the correct definition is just the previous universal property and its proof. Let’s mark pj : i∈I Xi → Xj projection onto the j-th component. Definition 4.7. Let Xi, i ∈ I, be topological spaces. We define on i∈I Xi the product topology generated by the subbase {p−1 j (U) | j ∈ I, U ⊆ Xj open}. We call the set i∈I Xi together with the product topology product of the topological spaces Xi, i ∈ I. Exercise 4.8. Prove that the product i∈I Fi of closed sets Fi ⊆ Xi is closed.cv end of 2. lecture 6 5. Separability Axioms Remark. There is a separability axiom T0. Definition 5.1. A topological space X is called T1if for every two points x, y ∈ X, x ̸= y there exists an open neighborhood U ∋ x disjoint with y, i.e. y /∈ U. Lemma 5.2. A topological space X is T1if and only if all its one-point subsets are closed. Proof. “⇐”: It is enough to choose U = X ∖ {y} in the definition. “⇒”: Let y ∈ X. Then for any x ̸= y there exists an open Ux ∋ x not containing y. Therefore, x̸=y Ux = X ∖ {y} is open and {y} is therefore closed. Definition 5.3. A topological space X is called T2 (Hausdorff) if for every two points x, y ∈ X, x ̸= y there exist disjoint open neighborhoods U ∋ x, V ∋ y, i.e. U ∩ V = ∅. Example 5.4. The space of finite complements is T1but not Hausdorff (unless the supportcv set is finite). Lemma 5.5. A topological space X is Hausdorff if and only if ∆X ⊆ X × X is a closed subset. Here ∆X = {(x, x) | x ∈ X} is the “diagonal”. Proof. “⇒”: We show that X × X ∖ ∆X is open. Let (x, y) ∈ X × X ∖ ∆X, i.e. x ̸= y. By definition, U ∋ x, V ∋ y exist disjoint open. Then (x, y) ∈ U × V ⊆ X × X ∖ ∆X, where U × V is basic open. “⇐”: By analogy; let x, y ∈ X, x ̸= y, i.e. (x, y) ∈ X × X ∖ ∆X. Since X × X ∖ ∆X is open, there exists a basic open subset U × V with property (x, y) ∈ U × V ⊆ X × X ∖ ∆X. Therefore, x ∈ U, y ∈ V , and U ∩ V = ∅. Alternatively, one can prove “⇒” using ∆X = {(x, y) | p(x, y) = q(x, y)}.nd Corollary 5.6. Let f, g: X → Y be two continuous mappings and Y be Hausdorff. Then {x ∈ X | f(x) = g(x)} is a closed subset of X. Proof. The representation (f, g): X → Y × Y is continuous, whereas {x ∈ X | f(x) = g(x)} = (f, g)−1 (∆Y ). Example 5.7. The orthogoal group O(n) ⊆ GL(n) is closed. (similarly SL(n))cv Theorem 5.8. 1. Subspaces of Hausdorff spaces are Hausdorff. 2. Products of Hausdorff spaces are Hausdorff. Proof. Let x, y ∈ A be separated in X by open sets U, V . Then A ∩ U, A ∩ V are open sets in A separating x from y. Let (xi), (yi) ∈ i∈I Xi be different points. Then there exists an index j ∈ I such that xj ̸= yj. Since Xj is Hausdorff, there exist U ∋ xj, V ∋ yj, U ∩ V = ∅. Then p−1 j (U), p−1 j (V ) separate (xi) from (yi). 7 Definition 5.9. A T1-space X is called T3 (regular) if for each of its points x ∈ X and a closed subset F ⊆ X not containing x there are open disjoint neighborhoods U ∋ x, V ⊇ F, i.e. U ∩ V = ∅. Lemma 5.10. The topological space X is regular if and only if for every point x ∈ X the closed neighborhood x forms a basis of the neighborhood, i.e. for every neighborhood N ∋ x there exists a closed neighborhood F ∋ x satisfying N ⊇ F. Proof. “⇒”: For every open neighborhood W ∋ x, it is enough to find a closed subneighborhood. By definition, one can separate x from X ∖ W, i.e. x ∈ U, X ∖ W ⊆ V , U ∩ V = ∅. In other words, x ∈ U ⊆ X ∖ V ⊆ W, so X ∖ V is a closed subregion of x. “⇐”: Let x /∈ F, i.e. x ∈ X ∖ F form an open neighborhood. By assumption, there exists x ∈ G ⊆ X ∖ F, where G is a closed neighborhood of x, i.e. x ∈ U ⊆ G, F ⊆ X ∖ G = V . Example 5.11. Every metric space M is regular – closed spheres form the basis of the neighborhood of every point. In a moment we will prove an even stronger claim in another way. Theorem 5.12. 1. Subspaces of regular spaces are regular. 2. Products of regular spaces are regular. Proof. Let F ⊆ A be closed not containing x ∈ A. Then A ∩ F = F, so x /∈ F and can be separated in X by U, V ; in A they can then be separated by A ∩ U, A ∩ V . An alternative proof leads via the previous lemma. Let (xi) ∈ i∈I Xi, (xi) ∈ U be an open neighborhood. Then there exist j1, . . . , jn ∈ I and open sets Uk ⊆ Xjk such that (xi) ∈ p−1 j1 (U1) ∩ · · · ∩ p−1 jn (Un) ⊆ U. Let xjk ∈ Fk ⊆ Uk be closed subregions. Then (xi) ∈ p−1 j1 (F1) ∩ · · · ∩ p−1 jn (Fn) closed neighborhood ⊆ U. Definition 5.13. A T1-space X is called T4 (normal) if for each of its two disjoint closed subsets F, G ⊆ X there exist open disjoint neighborhoods U ⊇ F, V ⊇ G. The analogy of the previous theorem does not hold – see the proof: if F, G ⊆ A are disjoint closed subsets, it is not necessarily true that F, G are disjoint (except for the case where A is closed). So it should not be hard to believe that there are normal spaces whose subspaces and products are not normal. Example 5.14. Every metric space M is normal. This is because for any A ⊆ M the function dist(A, −): M → R is continuous (does not shorten distances). If we now put f(x) = dist(F, x) dist(F, x) + dist(G, x) , is this function defined and continuous everywhere, im f ⊆ [0, 1]. Meanwhile, f(x) = 0 on F and f(x) = 1 on G, so one can choose U = f−1 [0, 1/2), V = f−1 (1/2, 1). The phenomenon from the previous example is separation using continuous functions. We will return to it later. 8 6. Compact Spaces Definition 6.1. A topological space X is called compact if a finite subcovering can be selected from any of its openopencoverings. In what follows, we will very often use the following interpretation of the compactness of the subspace A ⊆ X. If U is a system of open sets in X such that A ⊆ U, then there exist finitely many U1, . . . , Un ∈ U such that A ⊆ U1 ∪ · · · ∪ Un. Example 6.2. R is not compact.cv Example 6.3. (a, b) is not compact (no coverage found).cv Theorem 6.4. The closed interval [a, b] is compact. Remark. The previous theorem uses the completeness of real numbers – it does not hold over Q. The interval [a, b]Q = Q ∩ [a, b] is not compact: let c ∈ (a, b) be irrational. Then [a, b]Q = n∈N [a, c − 1/n)Q ∪ (c + 1/n, b]Q. Proof. Let U be an open covering of [a, b]. Let’s consider T = {t ∈ [a, b] | the interval [a, t] can be covered by finitely many elements U}. Obviously a ∈ T and hence T ̸= ∅. So we can put t0 = sup T. We first show that t0 ∈ T. Indeed, there exists U ∈ U such that t0 ∈ U and therefore there exists some t1 < t0 such that the entire interval [t1, t0] ⊆ U. Since t1 ∈ T, [a, t1] ⊆ U1 ∪· · ·∪Un holds. Therefore, [a, t0] ⊆ U1 ∪ · · · ∪ Un ∪ U. Now we show by argument that t0 = b. If t0 < b, we get t0 ∈ U ∈ U again, and T also contains some t1 ∈ U, t1 > t0. This is a dispute with t0 = sup T. Theorem 6.5. A closed subspace of a compact space is compact. Proof. Let F ⊆ X be closed and U be some system of open sets with property U ⊇ F. Then V = U ∪ {X ∖ F} is an open covering of X. Due to the compactness of X , it is X = U1 ∪ · · · ∪ Un ∪ (X ∖ F) and therefore F ⊆ U1 ∪ · · · ∪ Un. end of 3. lecture Theorem 6.6. A compact subspace of a Hausdorff space is closed. Proof. Let C ⊆ X be compact, x /∈ C. We want to find some U ∋ x, U ∩ C = ∅. Let y ∈ C. Then there exist disjoint Uy ∋ x, Vy ∋ y. The system {Vy | y ∈ C} forms an open covering of C. For compactness, a finite subcover C ⊆ Vy1 ∪ · · · ∪ Vyn can be selected from it. Then x ∈ Uy1 ∩ · · · ∩ Uyn = U is an open neighborhood of x and U ∩ C = ∅ because U ∩ Vyk ⊆ Uyk ∩ Vyk = ∅. 9 Corollary 6.7. In a compact Hausdorff space, closed sets are precisely compact. Theorem 6.8 (about the product). The product X × Y of two compact spaces X, Y is compact. We will prove the theorem later. Corollary 6.9. A subset Rn is compact if and only if it is closed and bounded. Proof. “⇒”: Closedness follows from the Hausdorffness of Rn, boundedness follows from the coverage of Bk(0), k ∈ N. “⇐”: From the boundedness A ⊆ [−k, k]n, while the cube [−k, k]n is compact by the product theorem. Therefore, even its closed subset A is compact. Theorem 6.10. A continuous image of a compact space is compact. Proof. Let f be a continuous mapping f : X → Y from a compact space X and let U be an open covering of f(X) . Then f−1(U) = {f−1(U) | U ∈ U} is an open covering of X and thus X = f−1(U1) ∪ · · · ∪ f−1(Un), or f(X) ⊆ U1 ∪ · · · ∪ Un. Theorem 6.11. A continuous bijection f : X → Y from a compact space X to a Hausdorff space Y is a homeomorphism. Proof. It suffices to show that f−1 is continuous, i.e. that for a closed F ⊆ X, f(F) ⊆ Y is also closed. However, F is compact, so f(F) is also compact and therefore closed. Definition 6.12. Let X be a topological space and let ∼ be an equivalence relation on X. Let us denote the projection p: X → X/∼. Let us define the quotient (identification) topology on the decomposition of X/∼ as {V ⊆ X/∼ | p−1 (V ) ⊆ X open}. We call the decomposition of X/∼ together with the quotient topology the quotient of X according to the relation ∼. A fundamental property of the quotient is that the mapping f : X/∼ → Y is continuous if and only if fp: X → Y is continuous, X fp GG p  Y X/∼ f aa In connection with the previous sentence, some quotients can be described very concretely. Examples 6.13. 1. Describe the topology of the quotient R/Q of the group R by its subgroup Q. It’s notcv even T1 even though R is even T4. 2. ([0, 1] × Sn−1)/({0} × Sn−1) ∼= Dn; here X/A = X/∼, where a ∼ a′ for any a, a′ ∈ A.cv (The display required is “polar coordinates”.) 10 3. Dn/Sn−1 ∼= Sn. (The necessary representation is given by orbiting around Sn alongcv major circles, for which there is a simple formula.) 4. S1 × S1 ∼= [0, 1]2/∼ (∼= R2/Z2 – this is the group quotient).cv 5. Figure illustrating other areas as quotients of polygons; a note on hyperbolic tiling. 6. SSn−1 ∼= Sn∗∗ 7. A line with double origin R × {−1, 1}/∼ where (x, −1) ∼ (x, 1) at any time x ̸= 0 – is∗∗ not Hausdorff, although it is “locally Hausdorff”. 8. Each retract is both a quotient and a subspace.nd 9. R2/∼, x ∼ y ⇔ |x| = |y|, is homeomorphic to R≥0hw 5 Now we can prove the product theorem. Let us first state the lemma. Lemma 6.14 (tube lemma). Let X be a compact space, Y any, W ⊆ X × Y an open set containing X × {y}. Then there exists an open neighborhood V ∋ y such that X × V ⊆ W. Proof. From the definition of the product topology, for each (x, y) there exist open neighborhoods Ux ∋ x and Vx ∋ y such that Ux × Vx ⊆ W. A finite subcover Ux1 , . . . , Uxn can be selected from the coverage {Ux | x ∈ X}. Let us set V = Vx1 ∩ · · · ∩ Vxn . Then Uxi × V ⊆ Uxi × Vxi ⊆ W and thus also X × V = Uxi × V ⊆ W. Example 6.15. Prove that the lemma is equivalent to the following statement: the projectionhw 6 X × Y → Y is closed, i.e. the image of a closed set is closed. Proof of the Product Theorem. Let W be the open coverage of X × Y . For each y ∈ Y , consider a subspace X × {y} that is homeomorphic to X and therefore compact. Since W is its open coverage, W1,y, . . . , Wn,y ∈ U covering X × {y} can be selected. By the lemma, W1,y ∪ · · · ∪ Wn,y contains a subset of the form X × Vy. Thus, we see that it suffices to cover Y with finitely many Vy since each X × Vy is covered with finitely many elements of W . But since {Vy | y ∈ Y } is an open coverage of Y , this follows from the compactness of Y . Our next goal will be the proof of Tikhonov’s theorem about infinite products of compact spaces. We can first prove the so-called Alexander’s lemma. Lemma 6.16 (Alexander). Let X be a topological space. If there exists a subbase S such that a finite subcover can be selected from every open coverage U ⊆ S , then X is compact. Proof. The proof is based on the axiom of choice, specifically on the principle of maximality, or as it is called in Czech. Assume that X is not compact and choose a maximal open covering U that has no finite subcovering (the assumptions of Zorn’s lemma are easily verified).hw 7 Let x ∈ X be an arbitrary point. Since U is a coverage, there exists x ∈ U ∈ U. Since S is a subbase, then there exist S1, . . . , Sn ∈ S such that x ∈ S1 ∩ · · · ∩ Sn ⊆ U. We now show by argument that some Si is an element of U. If Si /∈ U, by the maximality of U there exists a finite Ui ⊆ U such that {Si} ∪ Ui is a covering, i.e. Ui covers X ∖ Si. But then U1 ∪ · · · ∪ Un covers (X ∖ S1) ∪ · · · ∪ (X ∖ Sn) = X ∖ (S1 ∩ · · · ∩ Sn) ⊇ X ∖ U 11 and thus {U} ∪ U1 ∪ · · · ∪ Un covers X , which is a dispute. Denoting the corresponding Si ∈ U as Sx, we have x ∈ Sx ∈ S ∩ U. But since {Sx | x ∈ X} ⊆ S is an open coverage of the elements of S, a finite subcover can be selected from it by assumption. But this will also be the final sub-covering of U, dispute. Theorem 6.17 (Tikhonov). The product of any number of compact spaces is compact. Proof. Let X = i∈I Xi where Xi is compact. We show that the subbasis {p−1 j (Uj) | j ∈ I, Uj ⊆ Xj open} satisfies the conditions of Alexander’s lemma. Let U be an open covering of subbasic sets. Let us define Uj as the set of those open Uj ⊆ Xj such that p−1 j (Uj) ∈ U. Assume that no Uj is covered. Then there exists, for each j ∈ I, a point xj ∈ Xj such that xj /∈ Uj. But then the point with components (xj)j∈I does not lie in U, which contradicts the fact that U is a cover. Therefore, there is some Uj covering, and due to compactness, a finite subcovering U1, . . . , Un can be selected from it. Then apparently p−1 j (U1), . . . , p−1 j (Un) is a finite subcovering of U. end of 4. lecture Theorem 6.18. A compact Hausdorff space is normal. Proof. Let F ⊆ X be closed, x /∈ F. For any y ∈ F, there exist Uy ∋ x, Vy ∋ y open disjoint. Since F is compact, there are finitely many y1, . . . , yn ∈ F such that F ⊆ Vy1 ∪ · · · ∪ Vyn , x ∈ Uy1 ∩ · · · ∩ Uyn ; these are open and disjoint. The implication of T3⇒T4 is for homework.hw 8 A mass point of a sequence (xn) is a point x such that for every open neighborhood U ∋ x there are infinitely many terms xn ∈ U. Our goal will now be to show that a metric space is compact if and only if every sequence has a mass point. Let’s call this property sequential compactness for now. Let us define the diameter diam A = sup{dist(x, y) | x, y ∈ A}. Theorem 6.19 (Lebesgue’s lemma). Let U be an open covering of a (sequentially) compact metric space M. Then there exists ε > 0 such that every subset A ⊆ M of diameter diam A ≤ ε lies in some U ∈ U. We call the number from the theorem the Lebesgue number coverage of U. Proof. Obviously, it suffices to find ε such that every closed sphere of radius ε lies in some U ∈ U. Assume that no such ε exists and choose, for each n ∈ N, a sphere B1/n(xn) that does not fit into any U ∈ U. Passing to the subsequence, we can assume that xn → x. Since U is an open coverage, there exists δ > 0 such that B2δ(x) ⊆ U ∈ U. For n ≫ 0, dist(xn, x) < δ and 1/n < δ and therefore B1/n(xn) ⊆ B2δ(x) ⊆ U, dispute. Theorem 6.20. A metric space is compact if and only if it is sequentially compact. 12 Proof. The “⇒” direction is simple. Let xn be a sequence that has no mass point. Then for each x ∈ M there exists some ball Bεx (x) containing only a finite number of terms of the sequence. By choosing a finite subcovering, we get that there are only finitely many points of the sequence in the entire space, disputes. For the opposite direction “⇐”, let ε > 0 be the Lebesgue number U. Since every sphere of radius ε fits into some U ∈ U, it suffices to cover M with finitely many spheres of radius ε. Let us successively choose a sequence of points that are at least ε away from each other. Such a sequence must necessarily be finite, since no subsequence of it is Cauchy and therefore cannot converge; let us denote it by x1, . . . , xn. Then M = Bε(x1) ∪ · · · ∪ Bε(xn). Question. Say something about networks, their convergence and the equivalence of compactness with the existence? of a convergent subnet? Say something about filters, ultrafilters and compactness equivalence with every ultrafilter converging (to at least one point)? 7. Connectedness Classically, the empty topological space is considered connected, but for various reasons it is better not to consider it connected. We avoid this dilemma by limiting ourselves to non-empty spaces. Definition 7.1. Let X be a nonempty topological space. We say that X is continuous if the only subsets of A ⊆ X that are both open and closed are ∅ and X. Subsets from the definition (i.e., those that are both open and closed) are called ambiguous, in English clopen. If U is dual and V = X ∖ U is its complement, then the entire space X is a disjoint set of X = U ⊔ V . In the section on separation axioms, we used disjoint open sets to separate subsets of X, and this decomposition then corresponds to the fact that the entire space consists of two separate parts. Lemma 7.2. A nonempty space X is continuous if and only if every continuous mappingcv χ: X → {0, 1} is constant. Theorem 7.3. The closed interval [a, b] is continuous. Remark. This interval property depends, like compactness, on the completeness of the real numbers. Specifically, [a, b]Q is not continuous: we choose an arbitrary irrational c with property a < c < b, then [a, b]Q = [a, c)Q cup(c, b]Q. Proof. Assume that U ⊆ [a, b] is dual, ∅, X ̸= U. By possibly passing to the complement, we can assume that a ∈ U. Let’s mark T = {t ∈ [a, b] | [a, t] ⊆ U} We want b ∈ T. Obviously a ∈ T and therefore t0 = sup T exists. We get t0 ∈ T from the closedness of U, and t0 + ε ∈ T from the openness, except for the case t0 = b. This is a conflict with U ̸= X. (Otherwise: if the continuous mapping χ: X → {0, 1} were to take on the values 0 and 1, it would have to take on the values 1/2 as well, dispute.) Theorem 7.4. A continuous image of a continuous space is continuous. 13 Proof. Let f : X → Y be a continuous mapping where X is continuous. Let χ: f(X) → {0, 1} be an arbitrary continuous function. Then χf : X → {0, 1} is also continuous and therefore constant. Since f : X → f(X) is surjective, χ is also constant. (Otherwise: if U ⊆ f(X) is dual, f−1(U) ⊆ X is also dual.) Theorem 7.5. The closure of a continuous subset is continuous. Proof. If χ: A → {0, 1} is a continuous function, then its taper to A is constant, say χ|A = 0. Meanwhile, χ−1(0) is a closed set containing A and therefore χ−1 = A, so χ is constant. Theorem 7.6. Let M be a system of connected subsets in X such that A ∩ B ̸= ∅ for every two A, B ∈ M. Then M is continuous. Proof. Let f : M → {0, 1} be a continuous mapping. Then its narrowing to every A ∈ M is constant, due to the connection of A. At the same time, this constant value must be the same for all A ∈ M, due to the non-emptiness of the intersections. So f is constant. Corollary 7.7. The real axis R = n∈N[−n, n] is continuous. Example 7.8. The intervals [a, b], [a, b), (a, b) are continuous – prove. Further prove thatcv they are not homeomorphic by removing points. Example 7.9. Prove that space∗∗ {(x, sin π x ) | x > 0} ∪ {(0, y) | −1 ≤ y ≤ 1} is continuous but not arc-continuous. Definition 7.10. A Component of a nonempty space X is a maximal continuous subset. Theorem 7.11. Any topological space is a disjoint union of its components. These components are closed. Proof. Let x ∈ X and consider the system M = {A ⊆ X | A continuous, x ∈ A}. Then M is a continuous subset containing x, obviously maximal. If two different components had a non-empty intersection, their union would be continuous, which would be a contradiction of maximality. Closedness follows from the fact that the closure of a continuous subset is continuous and from maximality. Definition 7.12. We say that a topological space X is totally discontinuous if its components are one-point. Example 7.13. Prove that Q is totally disjoint.hw 9 The set of dual sets together with the inclusion, (Ob(X), ⊆), forms a Boolean algebra (the dual sets are closed for finite union, finite intersections, and complements). This way we get all Boolean algebras (except for isomorphism). After narrowing down to compact Hausdorff totally disjoint spaces, so-called Stone spaces, we get a unique correspondence, so-called Stone’s duality. Conversely, let B be a Boolean algebra. Consider the set S(B) of all homomorphisms ofnd 14 Boolean algebras φ : B → 2.1 Thus, we can equip S(B) ⊆ 2B with a subspace topology, where 2B has a product topology ( 2 has a discrete topology). As a subspace of the product of Hausdorff totally disjoints, is Hausdorff and totally disjoint. We now show that S(B) is closed, i.e., compact. Let φ : B → 2 not be a homomorphism of Boolean algebras. For example, suppose φ(b) = 0 = φ(c) but φ(b ∧ c) = 1. If we denote the projection 2B → 2 onto the bth component as pb, then φ ∈ p−1 b (0) ∩ p−1 c (0) capp−1 b∧c(1), with no mapping from this open neighborhood lying in S(B). Theorem 7.14 (Stone). The above constructions specify the (contravariant) equivalence be-nd tween Stone spaces and Boolean algebras. Contravariance means that homomorphisms of algebras B → B′ correspond to continuousnd mappings S(B′) → S(B). Proof. The homeomorphism X → S(Ob(X)) sends the point x ∈ X to the mapping U →∗∗ (x ∈ U), where x ∈ U needs to be understood as logical value, i.e. element 2. The isomorphism B → Ob(S(B)) sends an element b ∈ B to the dual set p−1 b (1) = {φ ∈ S(B) | φ(b) = 1} ⊆ S(B). end of 5. lecture Example 7.15 (Path filling a square). Let’s start with the path shown in the first figure, which we traverse at a constant speed, denote it by γ1. In the next steps, we replace all sections of γn that look like γ1 with corresponding sections that look like γ2. All paths are traversed at constant speed. γ1 γ2 γ3 γ4 Let us set γ = limn→∞ γn. Since γn+1(t) and γn(t) lie in the same square of side (1/2)n−1, this sequence is uniformly convergent and therefore γ continuous. It remains to show that it is surjective. Let x be any point of the square and write it as the intersection of the sequence of squares with sides (1/2)n−1 shown in the figures. In each such square lies some point γn(tn) and therefore x = limn→∞ γn(tn). Passing to the convergent subsequence, we can assume tn → t and then γ(t) = limn→∞ γ(tn) = limn→∞ γn(tn) = x from uniform convergence. (Otherwise 0, x1x2x3x4 . . . → (0, x1x3 . . . ; 0, x2x4 . . .).) Theorem 7.16. The product of two continuous spaces is continuous. 1 An alternative characterization of such homomorphisms is using φ−1 (0) – these are maximal ideals – or using φ−1 (1) – these are ultrafilters. (A filter in B is an upwardly closed set F ⊆ B, closed on finite suprems. An ultrafilter is a maximal filter not containing 0.) 15 Proof. Let f : X×Y → {0, 1} be a continuous mapping and let (x, y) and (x′, y′) be two points of X × Y . Then f(x, y) = f(x′, y) from the relation X × {y} ∼= X and f(x′, y) = f(x′, y′) from the context {x′} × Y ∼= Y . Thus, f is constant. Example 7.17. The spaces R and Rn, where n > 1, are not homeomorphic – again using point removal. For this, it is necessary to prove that Rn ∖{0} is continuous. It can be written as the union of continuous sets of the form Ri ×R± ×Rj, where R± is either a set of positive or negative numbers and i+1+j = n. Although these sets do not have non-empty intersections, this only happens for the pairs Ri × R+ × Rj and Ri × R− × Rj and those have a non-empty intersection with any other set from the system (n > 1). Theorem 7.18. Any product of continuous spaces is continuous. Proof. Again, let f : i∈I Xi → {0, 1} and let x = (xi) ∈ f−1(0). Since f is continuous, it takes the value 0 also on some neighborhood p−1 j1 (Uj1 ) ∩ · · · ∩ p−1 jn (Ujn ) of the point x. In particular, f takes on the value 0 at all points of y satisfying xj1 = yj1 , . . . , xjn = yjn . In the previous proof, we showed that f has the same values at points differing in finitely many components. Together, f is constant. From now on, we will denote I = [0, 1]. Definition 7.19. A Path in X is a continuous mapping γ : I → X. We say that γ connects the points γ(0), γ(1). Definition 7.20. A non-empty space X is called path-continuous (traditionally arc-continuous) if every two of its points can be connected by a path. Theorem 7.21. Any path-continuous space is continuous. Proof. If γx is a path connecting some selected point x0 ∈ X to a point x , then X = x∈X im γx , where each im γx is continuous (as the image of I ) and all intersect at x0 . In the opposite direction, the theorem does not always apply, but only under certain limiting conditions. We say that a space X is locally path-continuous if a path-continuous neighborhood forms a basis of the neighborhood at every point, i.e. if for every neighborhood N ∋ x there exists a path-continuous neighborhood O satisfying N ⊇ O ∋ x. Example 7.22. Open subsets of Euclidean spaces are locally path-connected. General sub-cv sets need not be locally path-connected (see Example 7.9). Lemma 7.23. If x can be connected to y and y can be connected to z by a path, then x can also be connected to z by a path. Proof. Let γ be the path connecting x to y and let δ be the path connecting y to z. Let’s putcv (γ ∗ δ)(t) = γ(2t) 0 ≤ t ≤ 1/2 δ(2t − 1) 1/2 ≤ t ≤ 1 Since the intervals [0, 1/2] and [1/2, 1] form a finite closed coverage and γ ∗ δ is continuous on each, it is continuous on all of [0, 1] . Meanwhile, (γ ∗ δ)(0) = γ(0) = x and (γ ∗ δ)(1) = δ(1) = z. 16 Theorem 7.24. If X is both continuous and locally path-connected, then it is path-connected. Proof. Let x ∈ X and consider C(x) = {y ∈ X | x can be connected by path s y}. According to the local path connection assumption, C(x) is open – whenever x can be connected to y and O is any path-connected neighborhood of y, then x can be connected to any point O . Since X is a disjoint union of C(x), where x ∈ X, the complement of C(x) is also open and therefore C(x) is dual. At the same time, x ∈ C(x), so C(x) = X follows from the connection, and therefore x can be connected to every point of X. Remark. For a general space X , the set C(x) is called a path component from the previoushw 10 proof, and similarly, it can be shown that for a locally path-connected X , its components coincide with the path components. Question. Mention locally contiguous spaces, that their components are open? Say something about continua (compact,? continuous metric spaces) and the Hahn–Mazurkiewicz theorem (continuous images of an interval are precisely locally continuous continua)? 8. Locally Compact Spaces Definition 8.1. A Hausdorff space X is called locally compact (Hausdorff) if a compact neighborhood forms the basis of the neighborhood of every point, i.e. if every neighborhood N ∋ x contains a compact subneighborhood N ⊇ C ∋ x . Question. Consistently say locally compact Hausdorff.? Example 8.2. 1. Every compact Hausdorff space is locally compact because it is regular (the uclosedneighborhood is a basis of the neighborhood) and closed=compact. 2. Every Euclidean space is locally compact – closed spheres form the basis of the neighborhood and are compact. 3. Discrete spaces are locally compact. 4. The rational numbers Q are not locally compact – no neighborhood is compact. Probably skip the next one, the proof is pretty useless; otherwise, of course, a picture will help - I used round=open, square=closed watering cans. Theorem 8.3. If every point of the Hausdorff space X has some compact neighborhood, then∗∗ it is locally compact. Proof. Let C be a compact neighborhood of x and let N be any of its neighborhoods. Then C ∩ N is a neighborhood of x in the compact Hausdorff space C. Therefore, there exists a compact subneighborhood x ∈ D ⊆ C ∩ N. Since D is a neighborhood of x in C and C is a neighborhood of x in X , D is also a neighborhood of x in X . Theorem 8.4. The product of two locally compact spaces is locally compact. Proof. Obviously, it suffices to find a compact neighborhood of (x, y) ∈ X × Y that fits into U × V ∋ (x, y). Let U ⊇ C ∋ x and V ⊇ D ∋ y. Then C × D is the sought-after compact neighborhood. 17 Construction 8.5 (one-point compactification). Let X be a locally compact space and ∞ /∈ X. Let’s put X+ = X ∪ {∞}. Let us define a topology on X+ as follows: U ⊆ X+ is openif and only ˆ ∞ /∈ U and U is open in X or ˆ ∞ ∈ U and X ∖ U is compact. This space is called a one-point compactification of the space X. Example 8.6. Prove that this is indeed a topology (for this you will need to prove that thehw 11 final union of compact subsets is compact). Theorem 8.7. The space X+ is a compact Hausdorff space of which X is a subspace. Proof. Obviously, all “tracks” X ∩ U of open U ⊆ X+ are open (this requires Hausdorffness), so indeed X is a subspace of X+. Let U be an arbitrary open covering of X+. Then some U0 ∈ U contains ∞ and from openness X ∖ U0 is compact. Therefore, finitely many U1, . . . , Un covering X ∖ X0 can be selected from U . Together, U0, U1, . . . , Un cover X+. The points of X can be separated by open subsets in X. So it remains to separate x ∈ X and ∞. Due to local compactness, there exists a compact neighborhood C ∋ x. From the definition of the neighborhood of C ⊇ U ∋ x and then U ∋ x and X+ ∖C ∋ ∞, the separating open sets are sought. Example 8.8. The topology on X+ is uniquely determined by the above requirements.cv Example 8.9. Describe the one-point compactification of Rn. (describe the stereographiccv projection Sn → Rn, x → x−e0 x0−1, show that it is continuous together with its inverse v → −1+|v|2 1+|v|2 · e0 + 2 1+|v|2 · v – both are given by the formula (my universal reasoning) – and use the previous example). The formula for the inversion can be left behind HW, with the fact that I would deduce for them that it must be of the form e0 + k(v − e0). Example 8.10. Describe the one-point compactification of a compact Hausdorff space.hw 12 end of 6. lecture Proposition 8.11. A continuous bijection f : X → Y between locally compact spaces is a homeomorphism if and only if it is “regular” (proper, i.e., the pattern of a compact set is compact). Proof. By the definition of topology on one-point compactification, f+ : X+ → Y + is continuous if and only if f is regular. But then it is a continuous bijection between compact Hausdorff spaces and thus a homeomorphism. Its contraction f must then also be a homeo- morphism. This can (maybe) be applied to the example of the stereographic projection Sn ∖ {e0} → Rn – but one needs to prove regularity. Theorem 8.12. Locally compact spaces are just open subspaces of compact Hausdorff spaces. Proof. “⇒”: every locally compact space X is an open subspace of X+. “⇐”: every compact Hausdorff space is locally compact and these are obviously closed on open subspaces. 18 Remark. It also holds that a closed subset of a locally compact subspace X is locally compact: compact neighborhoods of x in F can be obtained as intersections of F with compact neighborhoods of x in X . Even more generally, a subset A ⊆ X is locally compact if and only if it is the intersection∗∗ of an open and a closed subset (specifically, A is open in A). 9. Real Functions Definition 9.1. A Compactification of a space X is an embedding of X → K of the space X into some compact Hausdorff space K as a subspace such that X = K holds. A basic example is the one-point compactification mentioned above. The opposite extreme is the so-called Stone-ˇCech compactification, which adds as many points as possible. This compactification works for any completely regular spaces – we’ll deal with those in this chapter. Definition 9.2. Let X be a T1 topological space. We say that X is T31 2 (completely regular) if for each of its points x ∈ X and a closed subset F ⊆ X not containing x there exists a continuous function f colX → [0, 1] such that f(x) = 0 and f|F = 1. Question. Exercise: every locally compact Hausdorff space is completely regular (this also follows from one-point com-? pactification). Example 9.3. Every metric space is completely regular. We even proved that it is “completely normal”, i.e., that any two closed sets can be separated by a function. We will see in a moment that this condition is equivalent to normality. Theorem 9.4. Completely regular spaces are closed to subspaces and products. Proof. If A ⊆ X is an arbitrary subspace and x ∈ A, F ⊆ A is closed in A and does not contain x, then x and F are also disjoint in X. Therefore, there exists f : X → [0, 1] separating x from F . Its narrowing to A separates x from F. Now let Xi be completely regular, i ∈ I, and consider the product X = i∈I Xi. If x ∈ X and F ⊆ X is closed not containing x, then x ∈ X ∖ F (open) and by the definition of the product topology x ∈ p−1 j1 (U1) ∩ · · · ∩ p−1 jn (Un) ⊆ X ∖ F for some open Uk ⊆ Xjk . By passing to the complements Fk = Xjk ∖ Uk we get p−1 j1 (F1) ∪ · · · ∪ p−1 jn (Fn) ⊇ F and the closed set on the left still does not contain x. Therefore, it is enough to separate it from x. Let us choose continuous functions fk : Xjk → [0, 1] separating pjk (x) from Fk and set f(x) = max{f1(pj1 (x)), . . . , fn(pjn (x))}. Exercise 9.5. Prove that for continuous functions f, g: X → R, max{f, g} is also continuous.cv Theorem 9.6. Completely regular spaces are precisely the subspaces of the cube [0, 1]S. 19 Proof. Every subspace [0, 1]S is completely regular according to the previous theorem. On the contrary, let X be completely regular and let us set S = {f : X → [0, 1] | f is continuous}. We will write the components t ∈ [0, 1]S as tf = pf (t). Let us define the mapping h: X → [0, 1]S in terms of its components X h GG f 33 [0, 1]S pf  [0, 1] thus h(x) = (f(x))f∈S. According to the universal product property, h is continuous. Next, we show that is injective and, finally, that it is a homeomorphism to its image. Let x, y ∈ X be two distinct points and let f : X → [0, 1] be a continuous function separating x from y . Then (h(x))f = 0 and (h(y))f = 1, therefore h(x) ̸= h(y). It remains to show that the image of the closed set F ⊆ X is closed in h(X). Therefore, let’s choose an arbitrary point h(x) /∈ h(F) and look for its neighborhood disjoint with h(F). By injectivity, x /∈ F holds and therefore there exists f : X → [0, 1] separating x from F . But then (h(x))f = 0, while pf |h(F) = 1. Therefore, (pf )−1[0, 1) is the searched open neighborhood of h(x) disjoint with h(F). Corollary 9.7. A topological space is compactified if and only if it is completely regular. Proof. If X has a compactification, it is a subspace of the compact Hausdorff space, which we proved to be normal. We will see in a moment that T4 ⇒ T31 2 (Uryshohn’s theorem). Conversely, let X be completely regular. Then the mapping h : X → h(X) from the previous proof is a compactification. Definition 9.8. To embed h: X → [0, 1]S from the previous proof, let us set β(X) = h(X). This is a compactification of the space X and it is called the Stone- ˇCech compactification. Remark. The Stone-Cech compactification has the following universal property: if X → K is an arbitrary compactification, then there exists a single continuous extension β(X) → K (it simply extends to the mapping β(X) → β(K) ∼= K). For this reason, this is the “biggest” possible compactification. Remark. The continuous functions f : β(X) → R are in unique correspondence with thend bounded continuous functions X → R. Because of the compactness of β(X), f(β(X)) is bounded and more so f(X). Conversely, if g: X → R is an arbitrary bounded function, then we can understand it as g: X → [a, b] and get f : β(X) → β([a, b]) ∼= [a, b]. Question. If X is a discrete topological space, then β(X) should be the set of all ultrafilters on X (it is clear that each? ultrafilter determines one point of β(X); but why can’t two ultrafilters enter the same point?). This is related to Stone’s duality, see elsewhere. Theorem 9.9. A completely regular topological space with a countable topology basis is metriz-∗ able. Proof. Analyzing the proof of the cubing theorem, one can easily arrive at the following observation. Let S0 ⊆ S = {f : X → I continuous}. Then the mapping h0 : X → IS0 with components h0 = (f)f∈S0 is an embedding if for every closed set F and a point x /∈ F there exists f ∈ S0 such that f(x) = 0, f|F = 1. 20 Next, we find a countable set S0 with this property. Then h0 : X → Iω and there exists a metric on Iω dist(x, y) = 1 2n |xn − yn|. Prove that this metric specifies on Iω the product topology Iω = ∞ n=1 I. The metric onhw 13 X ∼= h0(X) is then obtained by narrowing the metric on Iω. It remains to find S0. Let U ⊆ V be basic open sets. If there is a continuous f : X → I with property f|U = 0, f|X∖V = 1, then we choose one and denote it FU,V . Let’s put S0 = {fU,V | U ⊆ V basic such that fU,V exists}. The condition needs to be verified. Let x /∈ F, so x ∈ X ∖ F. By the definition of a topology base, there exists a base V with property x ∈ V ⊆ X ∖ F. Due to complete regularity, there exists f : X → I such that f(x) = 0 and f|X∖V = 1. By appropriate reparameterization φ(t) = 0 t ≤ 1 2 2t − 1 t ≥ 1 2 we get a function φf which is zero on the neighborhood f−1[0, 1 2) ∋ x. Again, there exists a basis U with property f−1[0, 1 2) ⊇ U ∋ x. Therefore, the function fU,V exists; may not be equal to φf , but nevertheless any fU,V separates x from F as we require. In particular, a compact Hausdorff space is metrizable if and only if it has a countablend topology basis. Indeed, we saw in the proof of Theorem 6.20 that every compact metric space X can be covered by a finite system Bn of spheres of radius 1/n for every n ∈ N. It then simply turns out that n∈N Bn forms the basis of the topology: if U is open and x ∈ U, then one can find B2/n(x) ⊆ U. In Bn then one can find a ball containing x and it will necessarily lie in B2/n(x). Theorem 9.10 (Urysohn). Let X be a normal space and let F, G ⊆ X be its two disjoint∗ closed subsets. Then there exists a continuous f : X → [0, 1] such that f|F = 0 and f|G = 1. Proof. Let us set F0 = F, U1 = X ∖G. So we are looking for a function f with the properties f|U0 = 0, f|X∖U1 = 1. By normality, there exist F0 ⊆ U1/2 ⊆ F1/2 ⊆ U1 (since X ∖ F1/2 is a neighborhood of X ∖ U1 disjoint with U1/2). In the next step we get F0 ⊆ U1/4 ⊆ F1/4 ⊆ U1/2 ⊆ F1/2 ⊆ U3/4 ⊆ F3/4 ⊆ U1. and inductively then the system of open sets Um/2n and closed sets Fm/2n satisfying Ur ⊆ Fr and for r < s also Fr ⊆ Us . Let f(x) = inf{r = m/2n ∈ [0, 1] | x ∈ Ur}, where f(x) = 1 if x does not lie in no Ur. In particular, then, f|X∖U1 = 1; apparently also f|F0 = 0. The continuity of the display f follows from the easily verifiable formula f−1(a, b) = a |an−1| + · · · + |a1| + |a0|, R ≥ 1. Then for |z| = R it holds |g(z)| ≥ |zn | − |an−1zn−1 + · · · + a1z + a0| ≥ Rn − |an−1|Rn−1 + · · · + |a1|R + |a0| ≥ Rn−1 R − (|an−1| + · · · + |a1| + |a0|) > 0 and thus g has all its roots inside a circle of radius R. For the same reason, for t ∈ I the polynomial zn + t(an−1zn−1 + · · · + a1z + a0) has roots only inside the circle of radius R . We consider again the loop γ(t) = Re2πit. Then the above homotopy of polynomials determines the homotopy gγ ∼ fnγ of loops in R2 ∖ {0}. We calculated that the second loop represents the element n ∈ Z ∼= π1(R2 ∖ {0}, 1) and the same is therefore true for gγ. Now suppose that g has no roots. Then any constant-loop homotopy h : γ ∼ ε specifies the homotopy of loops gγ ∼ ε in R2 ∖ {0}. But this is only possible for n = 0. 27 Remark. In the proof, we “counted” only the roots inside a sufficiently large circle. Similarly, one can count the roots of g inside any region bounded by the curve γ : I → C as the homotopy class of the loop gγ in R2 ∖{0}. The “continuous dependence” of the roots on the polynomial also follows from this principle. If z0 is a root of the polynomial g and g′ is a polynomial close to g, then g′ has a root close to z0. end of 9. lecture Example 10.29. Prove that π1(Sn) = {e} for every n > 1. (Hint: cut each path intocv connecting paths that lie in the complement of the north/south pole and change each by homotopy to a path with “small image”.) Example 10.30. Let us consider Pn = P(Rn+1) the topology of the quotient Sn/∼, x ∼ −x.cv Prove that the mapping Sn → Pn, x → [x], is a covering and compute π1(Pn), n > 1. (Hint: if the open set U lies inside one hemisphere, then for the canonical projection p: Sn → Pn it holds that p: U ∼= −→ p(U); due is that p is open.) Example 10.31. Generalize the previous to “properly discontinuous” actions (see Bredon,hw 18 but the name does not fit the classical definition), i.e. actions satisfying: for each point x ∈ X there exists a neighborhood U ∋ x such that gU ∩U = ∅ for g ̸= e. In that case, the projection X → G\X is a covering, and if X is simply continuous, then π1(G\X) ∼= G. Example 10.32. Describe the fundamental group of the Klein bottle as G\R2.hw 19 Example 10.33. Prove that [S1, X] ∼= π1(X)/conj holds for a path-connected space X.∗∗ 11. Simplicial complexes, Brouwer’s theorem, dimension invariance Next, we prove Brouwer’s theorem in the general dimension. Theorem 11.1 (Brouwer’s theorem). Every continuous mapping Dn → Dn has a fixed point. First, let’s think about what it says in the 1 dimension. We have D1 = [−1, 1] and thus we claim that every mapping f : [−1, 1] → [−1, 1] has a fixed point. But this follows from the fact that f(−1) ≥ −1, f(1) ≤ 1 and therefore somewhere in the interval [−1, 1] must be f(x) = x. We will prove Brouwer’s theorem combinatorially. Therefore, we first need to replace the disk Dn with some combinatorial object. We will use the following theorem for this, in which ∂X = X ∖ ˚X is the boundary of X. Theorem 11.2. Let X ⊆ Rn be a compact convex subset with nonempty interior. Then there exists a homeomorphism h: Dn → X such that h(Sn−1) = ∂X. Proof. First, we can achieve that the origin of 0 is an interior point of X by possibly shifting X, which is a homeomorphism. Let us define the representation d: Sn−1 → R+ as d(v) = max{t ∈ R+ | tv ∈ X}. 28 Since 0 is an interior point, the above set is nonempty and, due to compactness, also bounded and closed; therefore the maximum element exists. An important step will be to show the continuity of the display d. Then we define h′ : I × Sn−1 → X, h′ (t, v) = td(v)v; this apparently sends {0} × Sn−1 to 0 and thus induces the display h: Dn ∼= (I × Sn−1 )/({0} × Sn−1 ) → X, which is continuous and by definition also a bijection between compact Hausdorff spaces. This is the desired homeomorphism. We will first show that tv ∈ ˚X holds for t ∈ R+ and v ∈ Sn−1 if and only if t < d(v). This exactly corresponds to the condition h(Sn−1) = ∂X. Obviously, for an interior point tv, t < d(v). Conversely, an isosceles centered at d(v)v converting 0 to tv sends some sphere Bε(0) ⊆ X to some sphere Bε′ (tv) ⊆ X and therefore tv is internal. Now we prove the continuity of the mapping d. Let vn → v be a convergent sequence. Since d(Sn−1) is bounded, it suffices to prove that every convergent subsequence of d(vn) converges to d(v). Assume for simplicity that d(vn) itself converges to some t. Because d(vn)vn → tv and the sequence on the left lies in X, it must also be tv ∈ X and therefore t ≤ d(v). Now suppose that t < d(v). Then tv is an interior point of X and therefore also d(vn)vn is an interior point for n ≫ 0. According to the previous paragraph, however, these are border points, a dispute. We now describe our combinatorial Dn disk model. We say that the points A0, . . . , Ak ∈ Rn are affine independent if A1 −A0, . . . , Ak −A0 are linearly independent vectors. A simplex of dimension k (also k-simplex) is called a convex hull s = [A0, . . . , Ak] = {ξ0A0 + · · · ξkAk | ξi ≥ 0, ξ0 + · · · + ξk = 1}, of affinely independent points A0, . . . , Ak. Example 11.3. A standard simplex ∆k of dimension k is defined as a convex hull ∆k = [e0, . . . , ek] of standard basis vectors Rn+1. If s = [A0, . . . , Ak] is any other k-dimensional simplex, then there exists a single affine mapping ∆k → s sending ei to Ai and it is homeomorphism. Thus, any two simplexes of dimension k are homeomorphic. According to the previous theorem, any simplex of dimension n is homeomorphic to Dn – since every two simplexes of dimension n are homeomorphic, this follows from the case of the convex hull of the n-tice of affinely independent points in Rn (it has a non-empty interior). The Wall of the simplex s is [Ai0 , . . . , Aiℓ ], where 0 ≤ i0, . . . , iℓ ≤ k (and we can assume that these indexes are mutually different and ordered). The combinatorial interior of s is intc s = {ξ0A0 + · · · ξkAk | ξi > 0, ξ0 + · · · + ξk = 1} and the combinatorial boundary then ∂cs = s ∖ intc s. Simplicial complex K in Rn is a finite set of simplexes such that 29 ˆ every simplex face of K lies in K, ˆ if s, t are two simplexes from K, then s ∩ t is their common face. The body of the complex K is the set |K| = K = s∈K s. A subset P ⊆ Rn is called a polyhedron if there exists a simplicial complex K such that P = |K|. The simplicial complex K is then called the triangulation of the polyhedron P. Example 11.4. A square has a triangulation consisting of two triangles. The cube is more complicated – it has a triangulation consisting of six tetrahedra. Subdivision L of the complex K is a simplicial complex such that |L| = |K| and every simplex s ∈ L lies in some simplex t ∈ K, i.e. s ⊆ t. Barycentric subdivision sd K is defined as follows: sd[A0, . . . , Ak] is given by all faces of k-simplexes [Ai0 , 1 2(Ai0 + Ai1 ), . . . , 1 k+1(Ai0 + · · · + Aik )], where i0, . . . , ik is an arbitrary permutation of 0, . . . , k; the barycentric subdivision of sd K is the union of all sd s, s ∈ K. The Fineness of triangulation K is µ(K) = max{diam s | s ∈ K} = max{dist(A, B) | [A, B] ∈ K}, i.e. the largest diameter of the simplex K or, equivalently, the largest distance of points connected by an edge. Lemma 11.5. µ(sd K) ≤ dim K dim K+1µ(K). Proof. In the proof, we will use the following observation several times: the greatest distance of a point from a point of a simplex is always realized in some vertex of this simplex.2 It is apparently enough to prove the statement for the simplex s = [A0, . . . , Ak]. Let t be the simplex of its barycentric subdivision. The diameter of t is the maximum length of an edge between its vertices. If this edge does not contain 1 k+1(A0 + · · · + Ak), it is a simplex barycentric subdivision of some wall s and we can use induction with respect to dimension k. The maximum distance 1 k+1 (A0 +· · ·+Ak) from the vertices of the barycentric subdivision occurs for some vertex Ai and thus µ(sd s) ≤ max{dist(Ai, 1 k+1(A0 + · · · + Ak)) | i = 0, . . . , k}, where dist(Ai, 1 k+1 (A0 +· · ·+Ak)) = k k+1 dist(Ai, 1 k (A1 +· · · Ai · · ·+Ak)) and this is bounded by k k+1 times the distance Ai from one of the vertices Aj, j ̸= i. Theorem 11.6 (Sperner’s lemma). Let T be the triangulation of the simplex [A0, . . . , An] and let φ be the mapping of the set of vertices T into the set {0, . . . , n} such that for B ∈ [Ai0 , . . . , Aik ] is φ(B) ∈ {i0, . . . , ik}. Then the number of simplexes [B0, . . . , Bn] ∈ T such that φ{B0, . . . , Bn} = {0, . . . , n} is odd. end of 10. lecture 2 The greatest distance of a point X from s is the same as the greatest distance of X′ from s, where X′ is the projection of X into the linear subspace L generated by s. By the proof of the previous theorem ( s has a nonempty interior inside L ), this is maximal for points from the boundary. Next, induction is used. 30 Proof. By induction with respect to n. Let’s mark S0 set of n-simplexes whose vertices are marked by exactly all 0, . . . , n − 1, p0 = #S0; S1 set of n-simplexes whose vertices are marked by exactly all 0, . . . , n, p1 = #S1; R the set of (n − 1)-simplexes whose vertices are marked by exactly all 0, . . . , n − 1, x = #R. Let us count the number of (s, r) pairs in two ways, where s is an n-simplex and r is its face of dimension n − 1 with property r ∈ R. Let us first deal with how many ways s can be chosen. The latter obviously lies either in S0 or in S1. In the first case, the wall r can be selected in exactly two ways, in the second case, in exactly one way. Therefore, the number of (s, r) pairs is equal to 2p0 + p1. Now let’s divide the same number according to the options to choose the simplex r. This can be chosen in exactly x ways, while each such simplex lies in exactly two3 n-simplexes s with the exception of those r that lie on the boundary of the simplex [A0, . . . , An]. Thanks to the condition on marking the vertices in the walls, r then lies in the wall [A0, . . . , An−1] and according to the induction assumption, the number y of such simplexes is odd. So the number of pairs is equal 2p0 + p1 = 2x − y and thus p1 = 2(x − p0) − y is odd. Proof of Brouwer’s Theorem. According to the previously performed reduction, it suffices to show the non-existence of the retraction Dn on Sn−1. Thanks to ∆n ∼= Dn, converting ∂c∆n to Sn−1, then it suffices to show the non-existence of the retraction r: ∆n → ∂c∆n. From the possible retraction, we will construct a vertex label sdN ∆n, N ≫ 0, which will contradict Sperner’s lemma. The label φ(B) of the vertex B ∈ sdN ∆n is given by the index i of any ei for which in the expression r(B) = ξ0e0 + · · · + ξnen is ξi the largest of all (there may be more of them – in that case we choose any of them; the vertex ei is closest to the point r(B)). The idea of the proof is that, with sufficiently fine triangulation, the images of simplexes will be small and cannot have vertices marked by all 0, . . . , n. We now define the open coverage U0, . . . , Un of the boundary ∂c∆n by the prescription Ui = {ξ0e0 + · · · ξnen | ξi < 1 n+1} Since ξj are non-negative and their sum is 1, the only point ∆n not belonging to U0 ∪· · ·∪Un is the barycenter 1 n+1(e0 + · · · + en), which, however, does not lie on the boundary. At the same time, it is also obvious that for r(B) ∈ Ui, ei is not the closest vertex to r(B). By Lebesgue’s lemma, there exists ε > 0 such that every subset of mean ε lies in some of r−1(U0), . . . , r−1(Un). Let us choose N ≫ 0 such that µ(sdN ∆n) ≤ ε. Then, for s ∈ sdN ∆n, 3 Formally, this follows from the fact that every n-simplex s having r as a wall lies in exactly one of the half-spaces determined by r. Thus, if there is only one such s, r cannot lie inside [A0, . . . , An]. If two such n-simplexes s, s′ lie in the same half-space, they intersect at some interior point; therefore, this possibility cannot occur in the simplicial complex. 31 r(s) ⊆ Ui will hold for some i and, in particular, all vertices of s will be evaluated by numbers from the set {0, . . . , i − 1, i + 1, . . . , n}. To dispute Sperner’s lemma, it is sufficient to verify the boundary condition. So let B ∈ sdN ∆n be a vertex lying in the wall [ei0 , . . . , eik ]. Then r(B) = B = ξ0e0 + · · · + ξnen with coefficients ξj = 0 for all j /∈ {i0, . . . , ik}; in particular φ(B) ∈ {i0, . . . , ik}. Theorem 11.7 (on dimension invariance). If Rn ∼= Rm are homeomorphic, then n = m. Let’s start with a simple reduction. If Rn ∼= Rm, then the one-point compactifications, Sn ∼= Sm, will also be homeomorphic. In what follows, we will show that spheres of different dimensions are not even homotopically equivalent. We say that a mapping f : X → Y is trivial if it is homotopy to a constant mapping. Otherwise, we say it is substantial. Theorem 11.8. Let Y be a topological space. The continuous mapping f : Sn → Y is trivial just as it can be extended continuously to Dn+1. Proof. If f can be extended to g : Dn+1 → Y , the homotopy of f with constant mapping iscv h(t, x) = g(tx); namely h(0, x) = g(0) = const and h(1, x) = g(x) = f(x). Conversely, let h : I × Sn → Y be a homotopy between the constant mapping and f. Since h(0, x) is independent of x, we obtain from the universal property of the quotient a continuous representation h′ : Dn+1 ∼= (I × Sn )/∼ → Y, where (0, x) ∼ (0, x′). That’s the extension you’re looking for. Theorem 11.9. Every homeomorphism f : Sn → Y is essential. Proof. This is a direct consequence of Brouwer’s theorem and the previous theorem. A possible extension of g: Dn+1 → Y would give a retraction Dn+1 g −→ Y f−1 −−→ Sn . In particular, no sphere Sn is contractible, i.e., homotopically equivalent to a one-point space. This is equivalent to the fact that identity is irrelevant. Example 11.10. Prove that every homotopy equivalence f : Sn → Y is essential.∗ In what follows, we show that every mapping f : Sn → Sm, n < m, is trivial. According to the previous, then it cannot be a homeomorphism, which proves the dimension invariance theorem. Definition 11.11. Let K, L be two simplicial complexes. We say that the representation f : |K| → |L| is simplicial with respect to the triangulations K, L, if for any simplex s = [A0, . . . , Ak] inK holds [f(A0), . . . , f(Ak)] ∈ L and on s f is affine, i.e. holds f(ξ0A0 + · · · + ξkAk) = xi0f(A0) + · · · + ξkf(Ak). Let us emphasize that the vertices f(A0), . . . , f(Ak) do not have to be different, thus we get a simplicial representation of the triangle on the segment. This doesn’t quite correspond to the combinatorial definition of a simplicial complex as a set of simplexes of different dimensions that satisfy something – one would expect a simplicial representation to send k-simplexes to 32 k-simplexes. However, a degree of generality of the definition is needed – otherwise there would be no simplicial mapping of a nontrivial polyhedron to a point.4 Theorem 11.12. (on simplicial approximation) Let K, L be two simplicial complexes and let f : |K| → |L| be a continuous mapping. Then there exists a subdivision K′ of the triangulation K such that f is a homotopy mapping g : |K′| → |L| that is simplicial with respect to the triangulations K′, L. Before the actual proof of the simplicial approximation theorem, let’s prove the dimensional invariance theorem. Proof of the dimension invariance theorem. As said, it suffices to show that every mapping f : Sn → Sm, n < m, is trivial. Thanks to the homeomorphisms Sn ∼= ∂∆n+1, Sm ∼= ∂∆m+1, it suffices that every mapping f′ : ∂∆n+1 → ∂∆m+1, n < m, is irrelevant. By the simplicial approximation theorem, f′ is homotopy to the simplicial mapping g′ : |K| → |L|. Since g′ is affine on every simplex, its image is the union of simplexes of dimension at most n and, in particular, g′ is not surjective (since L has a larger dimension). The backward transition to spheres is f a homotopy mapping of g which is not surjective and hence g: Sn → Sm ∖ {P} → Sm . Since Sm ∖ {P} ∼= Rm is contractible, the first mapping is homotopy to constant and the same is therefore also true for the composition g. We now return to the proof of the simplicial approximation theorem. Let us denote the vertex A of the triangulation K by its open star st(A) = [A,A1,...,Ak]∈K intc[A, A1, . . . , Ak], i.e. union of the interiors of all simplexes containing A. Prove that st(A) ⊆ |K| is an open neighborhood of A.hw 20 Since k i=0 st(Ai) is the union of the interiors of those simplexes that contain all the vertices A0, . . . , Ak, this intersection is nonempty if and only if [A0, . . . , Ak] ∈ K. This will come in handy in the proof. Proof of the simplicial approximation theorem. The simplistic mapping g is uniquely determined by its values at the vertices of the triangulation K′, which will be a multiple barycentric subdivision, K′ = sdN K. The homotopy between f and g will be linear – so we need that for each x ∈ |K| f(x), g(x) lie in the same simplex L. Let N ≫ 0 be such that the open star of each vertex A ∈ K′ = sdN K is mapped by f to the open star of some vertex B ∈ L. This is possible because {st(B) | B ∈ L} is an open covering of |L|, so U = {f−1(st(B)) | B ∈ L} open coverage |K|, and st(A) ⊆ Bµ(K′)(A). So it is enough to choose N so that the fineness of µ(K′) is smaller than the Lebesgue number of the open coverage U. 4 Formally, this “problem” can be circumvented by also considering formal “degenerate” k-simplexes [A0, . . . , Ak] for which we do not require the vertices to be different (but we still want the set {A0, . . . , Ak} to be affinely independent). These considerations lead to the so-called simplicial sets. 33 Now we can define g(A). Choose a vertex B ∈ L arbitrarily such that f(st(A)) ⊆ st(B) and set g(A) = B. Let [A0, . . . , Ak] ∈ K′. Then f(intc[A0, . . . , Ak]) ⊆ k i=0 st(g(Ai)) and the intersection on the right is therefore nonempty; but this means that [g(A0), . . . , g(Ak)] ∈ L and g is really simplicial. Let x lie inside [A0, . . . , Ak]. According to the previous one, then f(x) lies in the interior of some simplex s containing g(A0), . . . , g(Ak). Since g(x) lies inside the simplex [g(A0), . . . , g(Ak)], which is the wall of s, the segment connecting f(x), g(x) lies in s ⊆ |L| and the linear homotopy between f and g really has values in |L|. end of 11. lecture 12. Compactly generated Hausdorff spaces Definition 12.1. A topological space X is called compactly generated Hausdorff (CGH) if it is Hausdorff and for a subset A ⊆ X: if for every compact C ⊆ X the intersection C ∩ A open in C, then A is open. In what follows, we will call the set A for which C ∩ A ⊆ C is open compactly open. A compactly closed set is defined analogously. Thus, X is compactly generated if every compactly open set is open. Example 12.2. Every locally compact Hausdorff space X is compactly generated: let U be compactly open and x ∈ U. There exists a compact neighborhood C ∋ x and by the definition of compact openness, C ∩ U ⊆ C is open, that is, by the intersection of C ∩ V with the open set V ⊆ X. Since both C, V are neighborhoods of x, C ∩ U = C ∩ V is also a neighborhood of x and even more U ⊇ C ∩ U. Let X be a Hausdorff space. Let kX denote the set X together with the topology given by the system of compact open subsets. Exercise 12.3. The mapping f : kX → Y is continuous if and only if f : X → Y is continuouscv on every compact subset. Lemma 12.4. The space kX is a compactly generated Hausdorff space. Proof. Since there are more open sets in kX, it is apparently a Hausdorff space. We show that it has the same compact subspaces. The identical mapping kX → X is continuous and therefore every compact subspace of kX is also compact in X. Conversely, let C ⊆ X be compact. According to the exercise, the composition C → X → kX is continuous (since id : kX → kX is continuous), so its image is a compact set. The two possible topologies on C as a subspace of X and as a subspace of kX are identical because both identities on C are continuous. Therefore, the compactly open sets X and kX are the same and therefore kX is compactly generated (a compactly open subset of kX is compactly open in X, thus open in kX). 34 Definition 12.5. Let Y , Z be topological spaces. On a set of continuous views ZY = {f : Y → Z | f continuous} let’s define a compact-open topology using a subbase M(C, U) = {f ∈ ZY | f(C) ⊆ U}, where C ⊆ Y is compact and U ⊆ Z is open. Example 12.6. Let Y be a compact Hausdorff space and Z a metric space. Let us define∗ the Uniform Convergence Metric on ZY using a prescription dist(f, g) = max{dist(f(y), g(y)) | y ∈ Y }. This metric specifies exactly the compact-open topology on ZY . More generally, for a locally compact Hausdorff space Y , the compact-open topology on ZY is given by uniform convergence on compact subsets. To represent f : X × Y → Z, let us define f♭ : X → ZY , f♭(x)(y) = f(x, y). Conversely, for g: X → ZY , let us define g♯ : X × Y → Z, g♯(x, y) = g(x)(y) . We define X ×k Y = k(X × Y ). The importance of this construction lies in the following sentence. Theorem 12.7 (about adjunction). Let X, Y be compactly generated Hausdorff spaces and Z an arbitrary space. Then the mapping f : X ×k Y → Z is continuous if and only if the mapping f♭ : X → ZY is continuous. Proof. The continuity of f♭ just needs to be verified on each compact subset of C ⊆ X and can be expressed as follows. Let M(D, U) be a subbasic set. Then {x ∈ C | ∀y ∈ D: f(x, y) ∈ U} is open in C. This follows from the continuity of f : X × Y → Z on C × D and from the compactness of D using the “tube lemma”. The continuity of f is equivalent to the continuity of f : X × Y → Z on every compact set C × D ⊆ X × Y . Let U ⊆ Z be open and let f(x, y) ∈ U. Since f(x, −) : Y → Z is continuous and D ⊆ Y is locally compact, there exists a compact neighborhood y ∈ D′ ⊆ D such that f(x, D′) ⊆ U. This means that x ∈ (f♭)−1(M(D′, U)) and from the continuity f♭ there is a neighborhood x ∈ C′ ⊆ C such that f♭(C′) ⊆ M(D′, U), i.e. f(C′ ×D′) ⊆ U. Thus, f is continuous on C × D. Remark. The continuity f♭ is equivalent to the continuity f♭ : X → k(ZY ). This means that the category of compactly generated Hausdorff spaces is Cartesian closed (since X ×k Y is a product in this category and k(ZY ) is an object of functions). Let ∼ be an equivalence relation on a compactly generated Hausdorff space X such that X/∼ is again Hausdorff. Then the Hausdorff is compactly generated. This follows from the fact that the projection X → X/∼ induces a continuous mapping X = kX → k(X/∼). But since X/∼ is the largest topology for which this mapping is continuous, and k(X/∼) has more open sets, k(X/∼)mustbe = X/∼, i.e. X/∼ is compactly generated Hausdorff. 35 Corollary 12.8. Let ∼ be an equivalence relation on a compactly generated Hausdorff space X such that X/∼ is also Hausdorff. Then there is a homeomorphism (X ×k Y )/∼ ∼= (X/∼) ×k Y. Proof. The continuous mapping (X×k Y )/∼ → (X/∼)×k Y is equivalently given as X×k Y → (X/∼) ×k Y respecting the relation. So just take p × id, where p: X → X/∼ is the canonical projection. It then follows from the continuity that (X ×k Y )/∼ is also a Hausdorff space. In the opposite direction, the continuous mapping (X/∼) ×k Y → (X ×k Y )/∼ is equivalently given as X/∼ → ((X ×k Y )/∼)Y , thus as the relation-respecting mapping X → ((X ×k Y )/∼)Y , and thus as the mapping X ×k Y → (X ×k Y )/∼ respecting the session. Just take the canonical projection. An important special case is when Y is locally compact. Proposition 12.9. Let X be a compactly generated Hausdorff, Y a locally compact Hausdorff. Then X ×k Y = X × Y . Proof. Let A ⊆ X × Y be compactly open and (x0, y0) ∈ A. Then also ({x0} × Y ) ∩ A is compactly open in {x0} × Y ∼= Y and due to the compact generation of Y is open. Thus, there exists a compact neighborhood y0 ∈ D ⊆ Y with property {x0} × D ⊆ A. Consider a set U = {x ∈ X | {x} × C ⊆ A} ⊆ X. We show that U is compactly open, that is, open – if C ⊆ X is compact, then (C × D) ∩ A is open in C × D; C ∩ U is then open according to the “tube lemma”. Therefore (x0, y0) ∈ U × D ⊆ A. Since (x0, y0) was arbitrary, A is open. An alternative proof consists of the following: the representation in: X → (X ×k Y )Y is continuous by the adjoint theorem and the evaluation ev : ZY × Y → Z, (f, y) → f(y), is continuous thanks to a very simple argument (if U ∋ f(y) is open, then from the continuity of f and the local compactness of Y there exists a compact neighborhood C ∋ y; then M(C, U) × C is the neighborhood of (f, y) that appears in U). Therefore, the composition is also continuous X × Y in × idY −−−−−→ (X ×k Y )Y × Y ev −→ X ×k Y and therefore X × Y is compactly generated. 13. Algebras of continuous functions Recall that the (associative, with unit) C-algebra A is a vector space over C together with the bilinear mapping A × A → A that makes A Circle. The mapping C → A, z → z1, is then a circle homomorphism. Conversely, every homomorphism of circuits ι: C → A imposes on A the structure of the vector space over C by za = ι(z) · a. One simply verifies that it is a C-algebra if and only if the image ι lies at the center of the circle A. In particular, the commutative C-algebra is exactly a homomorphism of the circles C → A. 36 The C-algebra homomorphism Φ : A → B is a homomorphism of circuits that is also linear. Equivalently, it commutes the diagram C ÕÕ %% A Φ GG B Let X be a compact Hausdorff space. Let’s define C(X) = {f : X → C continuous}. Together with addition and multiplication functions, this is a circuit. The embedding of constant functions is a homomorphism C → C(X) and is therefore a C-algebra. Theorem 13.1. There is a natural bijection between the points of X and the maximal ideals of C(X). Proof. We first describe the maximal ideals corresponding to the points X. Let x ∈ X. We define mx = {f ∈ C(X) | f(x) = 0}. Since mx is the kernel of a surjective homomorphism of C-algebras (especially circles) evx : C(X) → C, f → f(x) and C is a solid, is mx = ker evx really a maximal ideal. Show that the assignment x → mx is injective.hw 21 It remains to show that every maximal ideal is of the form mx for some x ∈ X . Suppose by argument that I ⊆ C(X) is a maximal ideal different from mx. Then there exists fx ∈ I ∖mx. By multiplying by the complex combined function fx, we get the non-negative function gx = fxfx ∈ I with the property gx(x) > 0. Let us set Ux = {y ∈ X | gx(y) > 0}. We thus obtain the open coverage U = {Ux | x ∈ X}. Thanks to its compactness X = Ux1 ∪ · · · ∪ Uxn and the function g = gx1 + · · · + gxn is positive on all X. Therefore g−1 exists and I contains 1 = g−1g and cannot be maximal. Remark. The previous theorem does not hold without the compactness condition of X. Ideal I = {f : Rn → C | ∃C is compact: f = 0 on Rn ∖ C} does not lie in any maximal ideal mx. Thus, it must lie in some maximal ideal different from mx and, in particular, such maximal ideals of m exist. Note also that the dimension C(X)/m will always be infinite. (Assume this dimension is finite and set f(x) = |x|. Then [1, f, f2, . . .]∗∗ has infinite dimension – anfn(x) = 0 only for f(x) with a root of anzn, there are finitely many of them and therefore the equality cannot hold for all x. Therefore, there must be some h = anfn inm and the zero set Z(h) is compact; we proceed as in the proof of the theorem.) 37 Thus, X can be reconstructed as a set from the algebra C(X); let us now show how the topology can be reconstructed. If I ⊆ C(X) is an arbitrary ideal, is a set Z(I) = f∈I f−1 (0) = {x ∈ X | ∀f ∈ I : f(x) = 0} closed (this is the set of common zeros of the ideal I). Show that every closed set arises in this way from some ideal.hw 22 Theorem 13.2. There is a natural bijection between continuous mappings φ : X → Y and C-algebra homomorphisms Φ: C(Y ) → C(X). Proof. Let φ : X → Y be a continuous mapping. Let us define the homomorphism Calgebra φ∗ : C(Y ) → C(X) by the prescription φ∗(f) = f ◦ φ. We now show that every homomorphism C-algebra Φ: C(Y ) → C(X) is of this form. Let x ∈ X and consider the ideals mx ⊆ C(X), Φ−1(mx) ⊆ C(Y ). The induced mapping C(Y )/Φ−1(mx) → C(X)/mx is clearly injective. Both quotients contain C as a subbody, and this is fixed by this representation. Since C(X)/mx is equal to C, this is an isomorphism. Therefore, Φ−1(mx) is also maximal and Φ−1(mx) = mφ(x) for some φ(x) inY . This determines the representation of φ: X → Y . end of 12. lecture It remains to show that φ is continuous and that Φ = φ∗. Let x0 ∈ X, y0 = φ(x0); then my0 = Φ−1(mx0 ), so Φ(my0 ) ⊆ mx0 . Let’s do the math Φ(f) = Φ(f(y0) const +(ff(y0))) = f(y0) + Φ(ff(y0) ∈my0 ) ∈ f(y0) + mx0 , i.e. Φ(f)(x0) = evx0 Φ(f) = evx0 f(y0) = f(y0) = f(φ(x0)) = ( varphi∗f)(x0). Since x0 ∈ X was arbitrary, we have Φ(f) = φ∗f, i.e. Φ = φ∗. For continuity, just note that Z(I) = {y ∈ Y | I ⊆ my}. Then Z(Φ(I)) is the set of those x ∈ X for which Φ(I) ⊆ mx ⇔ I ⊆ Φ−1 (mx) = mφ(x) ⇔ φ(x) ∈ Z(I), thus Z(Φ(I)) = φ−1(Z(I)) and, in particular, φ−1(Z(I)) is closed. Since every closed set is of the form Z(I), φ is continuous. 14. Topological groups, Pontryagin duality Definition 14.1. The topological group G is a Hausdorff topological space and at the same time a group in such a way that group operations µ: G × G → G, µ(x, y) = xy; ν : G → G, ν(x) = x−1 are continuous. Let us define a left translation λy : x → yx and a right translation ρy : x → xy. Both are homeomorphisms since (λy)−1 = λy−1 and (ρy)−1 = ρy−1 . Similarly, the inversion ν and conjugation homeomorphisms are x → yxy−1. 38 Lemma 14.2. Every open subgroup is closed at the same time. In particular, a subgroup containing some unit neighborhood e contains the entire unit component. Proof. The complement of a closed subgroup H ⊆ G is the union G ∖ H = x/∈H xH, where xH = λx(H) is open – λx is a homeomorphism and H is open; thus G ∖ H is also open and H is indeed closed. A unit component Ge is a continuous closed subgroup – the images Ge · Ge = µ(Ge × Ge), G−1 e = ν(Ge) are also continuous and contain e , therefore must lie in Ge. If H ⊆ G is an arbitrary subgroup containing some neighborhood U ∋ e, then it is clearly open – for every x ∈ H also contains some neighborhood xU ∋ x. Therefore, the intersection of Ge ∩ H is an open subgroup of the continuous group Ge and must therefore be equal to Ge. Lemma 14.3. A subgroup closure is a subgroup. The closure of a normal subgroup is a normal subgroup. Proof. If H ⊆ G is a subgroup, then H × H ⊆ µ−1(H) holds. It is not difficult to verify5 that H × H = H × H and therefore also H × H ⊆ µ−1(H), i.e. H · H ⊆ H. Together with H ⊆ ν−1(H), i.e. H −1 ⊆ H, this means that H is a group. Normality flows in a similar way using conjugations. Example 14.4. 1. Prove that the Hausdorffness of the topological group follows from the weaker require-hw 23 ment T1, in fact from the closure of {e}. (Hint: if U, V are two neighborhoods of e and x, y are two points of G, then xU ∩ yV = ∅, if and only if x−1y /∈ U · V −1.) 2. Every topological group is a regular topological space.∗∗ Proposition 14.5. The quotient G/H of a topological group G by a closed normal subgroup H ⊆ G is a topological group. (Here G/H is equipped with a quotient topology.) Proof. Thanks to the previous example, it suffices to show that multiplication and inversion on G/H are continuous, and that G/H is T1. Let p : G → G/H denote the canonical projection. Any point G/H is closed because its pattern is the class xH = λx(H). First, let us note that the projection p is open – for any open U ⊆ G, p(U) ⊆ G/H is also open – namely p−1(p(U)) = y∈U yH = U · H = x∈H Ux. The continuity of multiplication follows from the following diagram G × G µ GG p×p  G p  G/H × G/H µ′ GG G/H If W ⊆ G/H is open, then (µ′)−1(W) = (p × p)(µ−1(p−1(W))) open due to the openness of the view p × p. Inversion continuity is similar but simpler. 5 It is true that A × B is the set of mass points A × B, i.e. those (x, y) whose every neighborhood intersects A timesB. Obviously, it suffices to restrict oneself to an arbitrary basis of the neighborhood, e.g. to the neighborhood of the form U × V . Then the intersection condition of A × B is exactly A ∩ U ̸= ∅ & B ∩ V ̸= ∅. This is equivalent to x ∈ A & y ∈ B. 39 The quotient of the group G under the (closed) nonnormal subgroup H is just a set, in our case a Hausdorff topological space. We call it homogeneous space. Homogeneous spaces are characterized by the following theorem in the case of the compact group G. There is also an extension of this theorem to locally compact groups, but it is technically more demanding. Proposition 14.6. Let G be a compact topological group having a continuous action on the Hausdorff space X. Then display G/Gx → G(x), gGx → gx is a homeomorphism of the quotient G/Gx according to the stabilizer x to the orbit G(x) passing through x. Proof. The continuity of the mapping G/Gx → G(x) follows from the universal property of the quotient. Since it is also a bijection and G/Gx is compact and G(x) is Hausdorff, it is a homeomorphism. Example 14.7. 1. Show that GL+(n), SO(n) are continuous. (This can also be shown via SO(n +cv 1)/ SO(n) ∼= Sn and due to the connection Sn – it is convenient that the projection SO(n + 1) → Sn is open.) 2. O(n + 1)/ O(n) ∼= Sn.cv 3. O(n)/({E} × O(nk)) ∼= Vk(Rn).cv 4. O(n)/(O(k) × O(nk)) def = Gk(Rn).cv Let G be a locally compact abelian group and define Γ = G = hom(G, T) ⊆ TG, i.e., the space of continuous homomorphisms G → T into complex units T = R/Z. Again, this is a locally compact abelian group – its elements are called characters. Let us now consider the second dual Γ. There is a natural display E : G → Γ, x → (evx : χ → χ(x)). The essence of Pontryagin duality is that E is an isomorphism of topological groups. Example 14.8. Valid R = R, T = Z, Z = T. Then on G there exists a measure µ defined on the set of Borel subsets E ⊆ G, i.e. the smallest σ-algebra containing closed sets, with the following properties 1. is regular, µ(E) = sup{µ(C) | C ⊆ E compact} = inf{µ(U) | U ⊇ E open}, 2. is translation invariant, µ(xE) = µ(E), 3. is not identically zero. Such a measure is called the Haar measure, it exists and is unique up to a multiple. In the case G = R, the Lebesgue measure is the Haar measure. For a general G , the construction of the Haar measure is performed as follows: an appropriate continuous linear form Cc(G) → C is constructed, where Cc(G) are compact carrier functions; according to Riesz’s representation theorem, this linear form corresponds to a single measure, while the properties of the measure are derived from the properties of this functional. 40 The Fourier transform is then defined L1 (G) → C(Γ), f(χ) = G f(x)χ(x)dµ, where L1(G) is the space of absolutely integrable functions. The inverse Fourier transform is given by L1 (Γ) → C(G), ˇg(x) = Γ g(χ)χ(x)dν, where ν is a certain “dual” measure on Γ. These transformations are mutually inverse on the space of functions that are absolutely integrable even with their square and specify an isometry L2 (G) ∼= −→ L2 (Γ) (the so-called Plancherel theorem). From these considerations, Pontryagin’s duality follows quite simply. end of 13. lecture 15. Paracompact Spaces∗∗ Definition 15.1. Let U be the coverage of the space X. We say that the coverage of V is a refinement of the coverage of U if every element of V ∈ V lies in some U ∈ U. We say that a coverage of V is locally finite if every x ∈ X has a neighborhood of N ∋ x that intersects only finitely many V ∈ V. We say that a Hausdorff topological space X is paracompact if every open covering of it has a locally finite open refinement. Lemma 15.2. The union of a locally finite system of closed sets is closed. Proof. Let F be a locally finite system of closed sets and x /∈ F. Then some of its neighborhoods N ∋ x intersect only finitely many elements of F and thus N ∩ F is closed in N and does not contain x, i.e. N ∖ mcF is a neighborhood of x and X ∖ F is open. Lemma 15.3. A closed subspace of a paracompact space is paracompact. Proof. similar to compact. Proposition 15.4. Every paracompact space is normal. Proof. We can prove regularity, then normality can be proved in the same way. Let x /∈ F where F ⊆ X is closed. For each y ∈ F, we choose an open neighborhood Uy ∋ y such that x /∈ Uy; we thus obtain an open coverage {Uy | y ∈ Y } of the set F. Since this is paracompact by the previous lemma, there exists a locally finite open refinement V of it. Then V ∈V V is a closed (due to local coequality) neighborhood (containing V) of the set F that does not contain x. 41 Definition 15.5. The Carrier of a continuous function f : X → R is the set supp f = f−1(R ∖ {0}). Let U be an open covering of X. We say that the system of functions fλ : X → I, λ ∈ Λ, is a unit decomposition subordinate to U if {supp fλ midλ ∈ Λ} locally finite refinement of U and holds λ∈Λ fλ = 1. The sum in the definition makes sense because the system of carriers is locally finite, i.e. around every point this sum is finite. For the same reason, such a sum is always a continuous function. Theorem 15.6. Let U be an open covering of the paracompact space X. Then there exists a unit decomposition subordinate to U. Proof. We can assume that U is a locally finite coverage (eventual transition to refinement). Let U = {Uλ | λ ∈ Λ} and choose a good ordering on the index set Λ. We will construct the decomposition of the unit by transfinite induction. Obviously, it suffices to construct a system of functions fλ such that supp fλ ⊆ Uλ and f = λ∈Λ fλ > 0 – such a system then it is enough to normalize, i.e. replace each fλ with the quotient fλ/f. For the already constructed functions fλ, let us denote Vλ = f−1 λ (0, 1]. By induction, we will assume that for i < λ Vi subseteqUi and i<λ Vi ∪ i≥λ Ui = X. For this reason, Fλ = i<λ (X ∖ Vi) ∩ i>λ (X ∖ Ui) ⊆ Uλ and we choose fλ : X → I arbitrarily such that it is 1 on Fλ and has a carrier inside Uλ – this is possible due to the normality of X. It is straightforward to verify that λ∈Λ fλ > 0 as we wish. It holds that every locally compact Hausdorff space with a countable topology basis is paracompact (the proof is not difficult). Also, every metric space is paracompact, but the proof of this claim is quite demanding. 16. Uniform Spacesnd Uniform space is another abstraction of metric space. In the topological space, we can compare the proximity of points to the given point x – they are similarly close if they belong to some neighborhood of x. However, we cannot compare the proximity of arbitrary pairs of points as in metric space. The basic “topological” concept in this direction is not a continuous representation, but a uniformly continuous representation. The formalization of this concept is the so-called uniform spaces. Definition 16.1. Let X be a set. Uniformity on X is a system of sets U ⊆ P(X × X) satisfying 1. ∀U, V ∈ U : U ∩ V ∈ U, 2. ∀U ∈ U, ∀V ∈ P(X × X): V ∈ U, 3. ∀U ∈ U : ∆X ⊆ U, 4. ∀U ∈ U : U−1 ∈ U, 5. ∀U ∈ U : ∃V ∈ U : V ◦ V ⊆ U. 42 We call the set X together with the uniformity uniform space. Since every U ∈ U is a relation on X, we will write xUy instead of (x, y) ∈ U. The first three conditions say that U is in some sense a neighborhood system of ∆X. It is more precisely expressed in the following construction. Let X be a uniform space and x ∈ X. We say that N is a neighborhood of the point x if there exists U ∈ U such that N is equal to xU = {y ∈ X | xUy}. According to property 5. let us choose a sequence Un ∈ U such that U0 = U, Un ◦ Un ⊆ Un−1. Then V = U1 ∪ U1U2 ∪ U1U2U3 ∪ · · · is an element of U (for example, because it contains U1 ∈ U) and for y ∈ xV xU1 · · · Uny holds for some n and then for z ∈ yUn+1 holds xU1 · · · UnyUn+1z, i.e. xV z. Therefore, xV includes the point y and its neighborhood yUn+1 and xV is therefore open, while xV ⊆ xU. Let us now deal with the uniformity relation U to the induced topology on X in more detail. We first show that each U ∈ U is a neighborhood of ∆X ⊆ X × X in the product topology. Let (x, x) ∈ ∆X. Let us choose V using properties 4. and 5. such that V −1 ◦V ⊆ U. Then yV −1xV z ⇒ yUz holds for (y, z) ∈ xV × xV , i.e. (y, z) ∈ U. Example 16.2. A typical example of a uniform space is the topological group G. Let N ∋ e be the unit neighborhood. Let us define the appropriate neighborhood of ∆G as UN = {(x, y) | y−1x ∈ N}. Then let us set U = {UN | N ∋ e unit neighborhood}. Example 16.3. Let X be a compact Hausdorff space. Let us define uniformity on X using the system of all neighborhoods of ∆X. The only axiom that is not obvious is 5. First, let’s realize that due to normality, closed neighborhoods form the basis of all ∆X neighborhoods. So let U be an arbitrary open neighborhood of ∆X. It is easy to show that {V ◦ V | V is the neighborhood of ∆X} = ∆X, for each x ̸= y it is enough to choose a closed neighborhood N ⊇ ∆X not containing (x, y); then V ◦ V ̸∋ (x, y) holds for V = N ∖ (xN × {y}). In other words, the union of U and all complements of sets of the form V ◦ V is the whole X × X. By compactness, X × X is the union of U and finally many such complements, or (V1 ◦ V1) ∩ · · · ∩ (Vn ◦ Vn) ⊆ U. Let’s put V = V1 ∩ · · · ∩ Vn, then V ◦ V ⊆ U. uniformly continuous mapping, topological groups, compact T2, uniformizability is the same as T31 2 . 43