INTRODUCTION TO TOPOLOGY

LÁSZLÓ FEHÉR

October 18, 2002

Preface

These notes are based on a course given at the Budapest Semester in Mathematics in 2002. It is intended for a one semester undergraduate course. For readers who are interested in more

details I suggest the Topology of Munkres which covers much more material than can be covered in a one semester course. The * sign at some homework problem indicates that I

think it is more difficult. It may not apply for you.

Contents

Preface 1 1. Introduction—What is Topology? 2 2. Continuity and Metric Spaces 2 3. Continuity and Topological Spaces 5 3.1. Examples of Topological Spaces 6 4. Further Examples of Topological Spaces 7 5. Intermediate Value Theorem and Connectedness 10 6. Separation Axioms 12 6.1. Limit of sequences 12 7. Countability Axioms 13 8. Compactness 14 9. The notion of Homotopy 17 10. The Fundamental Group 17 11. Induced homomorphisms and Functors 19 12. Basic properties of the Fundamental Group 21 13. Other constructions for Topological Spaces 22 14. Manifolds 24 15. Homotopy Equivalences 24 16. Simply Connected Spaces 25 17. The Fundamental Group of the Circle 26 18. Covering Spaces and Maps 27 19. Compact surfaces 29 20. Group Actions 30 21. Seifert-van Kampen Theorem 31 22. The Dirac experiment and the no-wind theorem. 35

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1. Introduction—What is Topology?

Topology is elastic (rubber-sheet) geometry. It studies properties of shapes unchanged under deformations. As the joke says: Topologist is the person who cannot distinguish the teacup

from the doughnut. I list some problems leading to the field of Topology:

Problems 1.1.

1. The two-pancake problem: Suppose you have two pancakes on a plate. Find a straight cut halving both.

2. The sleeping snake: Our snake sleeps in a tubelike bed having the same length as the snake has. The snake goes to sleep straight but has nightmares and wakes up all twisted. Show that there is a point on the snake which wakes up at the spot it went to sleep.

3. A polynomial of odd order has a root.

4. Is there a good map of Earth? If it is one piece, then it is extremely distorted at the poles. Otherwise you need two pieces.

5. When can you untie a knot? For a mathematician a knot has no ends: a knot is the image of a circle in space without self-intersections.

6. How to distinguish knots?

7. Can you brush a hedgehog? In—not yet defined—mathematical terms: Is there a nowhere zero vector field on the sphere? An other equivalent question: Is it possible that the wind blows at every point on the surface of Earth at the same time? We assume that the Earth is a big sphere and the direction of the wind is always tangential (horizontal).

8. The Dirac experiment: If you rotate an object 360^{∘} then things (or your hand) get twisted. Do it once more and the strings get untwisted.

Homework for Section 1.

1. Solve the 1-pancake problem.

2. * Solve the 2-pancake problem.

3. Solve the sleeping snake problem and formulate the sleeping amoeba problem.

4. Prove that x^{5} − 7x^{3} + x^{2} − 2 has a root.

5. * Prove that any polynomial of odd order has a root.

6. Find a wind distribution on Earth with two (one) no-wind spot.

2. Continuity and Metric Spaces

The single most important notion underlying all these problems is continuity. We are surrounded by continuous functions. That is why our life is reasonably safe and predictable.

What do we mean by continuous in everyday life?

1. There are no jumps: if I buy a little more gas it costs a little more.

2. Small changes have small effect. More precisely small change in the input results in a small change in the output: if I throw a ball with a little different speed or angle, it lands at a little different position.

3. No missing value: If someone crosses a bridge then at a certain moment he has to be at the middle of the bridge.

Exercise 2.1. You may try to find discontinuous functions in everyday life.

To define continuity we need a notion of “small”. Traditionally it is done by measuring. It leads to the notion of metric spaces: We want to measure the distance of two points.

Definition 2.2. A metric space is a pair (X, d), where X is a set and d : X × X → ℝ is a distance function, i.e. it satisfies the following properties:

1. (positivity) d(x, y) ≥ 0 and d(x, y) = 0 ⟺ x = y.

2. (symmetry) d(x, y) = d(y, x)

3. (triangle inequality) d(x, z) ≤ d(x, y) + d(y, z).

Exercise 2.3. You may try to find examples where one of the three conditions are not satisfied.

Examples 2.4. (of metric spaces)

1. The set of real numbers ℝ with the distance function: d(x, y) = |x − y|. This is the most important example. Positivity and symmetry are clear, for |x − z| ≤ |x − y| + |y − z| notice that if y is in the middle then we have an equality, otherwise |x − z| ≤ |x − y| or |x − z| ≤ |y − z|.

2. The discrete metric spaces: X is any set and

d(x, y) = \begin{cases}1 & if x ≠ y \\ 0 & if x = y. \end{cases}

3. The real plane ℝ^{2} with the distance function:

d(P_{1}, P_{2}) = \sqrt{(x_{1} − x_{2})^{2} + (y_{1} − y_{2})^{2}},

where P_{1} = (x_{1}, y_{1}) and P_{2} = (x_{2}, y_{2}) are points of the plane. This is sometimes called the round metric. (Check the triangle inequality!).

4. The real plane ℝ^{2} with the distance function:

d(P_{1}, P_{2}) = |x_{1} − x_{2}| + |y_{1} − y_{2}|,

the taxicab metric: the shortest way in NY city.

5. The real plane ℝ^{2} with the distance function:

d(P_{1}, P_{2}) = max{|x_{1} − x_{2}|, |y_{1} − y_{2}|},

the maximum or square metric.

6. X = {f : [0, 1] → ℝ continuous} and d(f_{1}, f_{2}) = ∫_{0}^{1} |f_{1}(t) − f_{2}(t)|dt (the so called L^{1}- metric).

7. X = {f : [0, 1] → [0, 1]} and d(f_{1}, f_{2}) = sup_{0≤t≤1} |f_{1}(t) − f_{2}(t)| (the so called supremum metric).

Constructions 2.5. We can build new metric spaces out of old ones:

1. Subspace:

Take any subset Y ⊂ X of a metric space (X, d) and define a metric on Y by restricting the metric d to Y: d_{Y} := d|_{Y}. (Show that the discrete metric space with 3 elements is a subspace of (ℝ^{2}, d_{round}).)

2. Product of two metric spaces:

If (X, d_{X}) and (Y, d_{Y}) are metric spaces then we can put a metric on the set X × Y. In fact we have several choices:

(a) d_{round}((x_{1}, y_{1}), (x_{2}, y_{2})) = \sqrt{d_{X}(x_{1}, x_{2})^{2} + d_{Y} (y_{1}, y_{2})^{2}},

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(b) d_{cab}((x_{1}, y_{1}), (x_{2}, y_{2})) = d_{X}(x_{1}, x_{2}) + d_{Y} (y_{1}, y_{2}),

(c) d_{max}((x_{1}, y_{1}), (x_{2}, y_{2})) = max{d_{X}(x_{1}, x_{2}), d_{Y} (y_{1}, y_{2})}.

3. Product of infinitely many metric spaces:

The previous construction can be generalized to the case of countably many metric spaces with some boundedness assumption. (Try to define d_{round} and d_{cab}.) The third metric works even for uncountably many spaces:

Suppose that you have metric spaces {(X_{i}, d_{i}) : i ∈ I} with the property that d_{i}(x, y) ≤ 1 for all i ∈ I and x, y ∈ X_{i}. Then

d_{sup}(f, g) = sup_{i∈I}{d_{i}(f(i), g(i))}

is a metric on the product set ∏_{i∈I} X_{i}. Recall that an element of the product set ∏_{i∈I} X_{i} is a function on the index set I such that f(i) ∈ X_{i} for all i ∈ I. Notice that the function space in Example 2.4.7 is a product space.

These are powerful constructions. It is not difficult (but tricky) to prove that every bounded metric space—i.e. d(x, y) < K for some constant K—can be obtained as a subspace of some

product of copies of the interval [0, K].

The paradox of measuring: We want to use measuring—the distance function—to define continuity but “small” cannot be defined rigorously. But we are not far from resolving this

conflict by defining topological space. Nevertheless this paradox makes the definition of continuity quite awkward: If you tell what you mean on “small” in the range of the function

then I tell what I mean on “small” in the domain of the function.

Definition 2.6. A map f : X → Y of two metric spaces (X, d_{X}) and (Y, d_{Y}) is continuous at the point x_{1} ∈ X if for every given ϵ > 0 there is a δ > 0 such that d_{X}(x_{1}, x_{2}) < δ implies that d_{X}(f(x_{1}), f(x_{2})) < ϵ. f is said to be continuous if it is continuous at every point x_{1} ∈ X.

There is a more geometric way to think about continuity.

Definition 2.7. The open ball centered at x with radius r in the metric space (X, d_{X}) is

B_{r}(x) = {z ∈ X : d_{X}(x, z) < r}.

Remark 2.8. So f is continuous at x iff for every open ball B_{ϵ}(f(x)) around f(x) there is an open ball B_{δ}(x) around x such that f(B_{δ}(x)) ⊂ B_{ϵ}(f(x)).

Remark 2.9. The name “ball” might be misleading. Try to draw a picture of a ball in the different metrics on ℝ^{2} defined in Examples 2.4. This explains the names of these metrics.

Homework for Section 2.

1. Check the triangle inequality for the round metric 2.5.2a.

2. Check the triangle inequality for the taxicab metric 2.5.2b.

3. Check the triangle inequality for the maximum metric 2.5.2c.

4. Show that d_{min}((x_{1}, y_{1}), (x_{2}, y_{2})) = min{d_{X}(x_{1}, x_{2}), d_{Y} (y_{1}, y_{2})} is not a metric.

5. * Show that that every bounded—i.e. d(x, y) < K for some constant K—metric space can be obtained as a subspace of some product of copies of the interval [0, K] (use the supremum metric!).

6. Find a subspace of ℝ^{3} which is isometric to the discrete metric on four points.

7. Describe the open ball B_{1}(0) for the metrics d_{round}, d_{cab} and d_{max}.

8. Let (X, d_{X}), (Y, d_{Y}) and (Z, d_{Z}) be metric spaces and f : X → Y, g : Y → Z continuous functions. Show that the composition g ∘ f : X → Z is continuous.

9. Show that the function f : ℝ → ℝ defined by f(x) = x^{2} is continuous at x = 1 by finding a δ for any given ϵ.

3. Continuity and Topological Spaces

You can rephrase the paradox in a different way: We want elastic geometry, so size shouldn’t

matter. Then why do we use it?

Now we try to find a better definition for continuity. We try to replace balls with a less metric concept. Let (X, d_{X}) be a metric space what we denote by X for short.

Definition 3.1. The set U ⊂ X is open if it is the union of open balls.

Proposition 3.2. U ⊂ X is open iff for all x ∈ U there is an ϵ > 0 such that B_{ϵ}(x) ⊂ U.

We may call a set satisfying this condition safe since if we are standing in U we always have a safety zone: if we move less than ϵ distance we are still in U. So we show that a set is open iff

it is safe.

Proof. ⇐ x ∈ B_{ϵ}(x) so U is the union of these open balls. ⇒ Enough to show that an open ball is safe. So let y ∈ B_{ϵ}(x). Then by the triangle inequality B_{δ}(y) ⊂ B_{ϵ}(x) if δ = ϵ − d_{X}(x, y).

Theorem 3.3. Let (X, d_{X}) and (Y, d_{Y}) be metric spaces. Then a map f : X → Y is continuous iff for all U ⊂ Y the preimage f^{−1}(U) ⊂ X is open.

Proof. ⇒ Let x ∈ f^{−1}(U). Then f(x) is in U which is open by assumption. So there is an open ball B_{ϵ}(f(x)) ⊂ U. Continuity of f implies that there is an open ball B_{δ}(x) such that f(B_{δ}(x)) ⊂ B_{ϵ}(f(x)) so B_{δ}(x) ⊂ f^{−1}(U). Since x was arbitrary we showed that f^{−1}(U) is open. ⇐ Since any open ball is open by the assumption f^{−1}(B_{ϵ}(f(x)) is open, too. x ∈ f^{−1}(B_{ϵ}(f(x)), so there is a δ such that B_{δ}(x) ⊂ f^{−1}(B_{ϵ}(f(x)) in other words f(B_{δ}(x)) ⊂ B_{ϵ}(f(x).

Proposition 3.4. Union of open sets is open and if U_{1} and U_{2} are open then U_{1} ∩ U_{2} is open, too.

Proof. The first statement is obvious, for the second let x ∈ U_{1} ∩ U_{2} then we have two balls centered at x, one in U_{1} and the other one in U_{2}. The one with the smaller radius is in U_{1}∩U_{2}.

Notice that it was the triangle inequality which made it possible to show that the intersection

of two open sets is open. Now we are in the position to get rid of the metric and make the open sets the fundamental objects.

Definition 3.5. Let U be a collection of subsets of the set X. They define the open sets of a topology on X if the following conditions are satisfied:

1. ∅, X ∈ U.

2. U_{1}, U_{2} ∈ U implies that U_{1} ∩ U_{2} ∈ U.

3. if U_{i} ∈ U for every i ∈ I then ⋃_{i∈I} U_{i} ∈ U.

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The collection U is usually called topology on X and a pair (X, U) is called a topological space.

Definition 3.6. Let f : X → Y is a map between topological spaces (X, U_{X}), (Y, U_{Y}). f is continuous if for every U ∈ U_{Y} we have f^{−1}U ∈ U_{X}. (In other words if preimage of open sets is open.)

Definition 3.7. The topological spaces (X, U_{X}) and (Y, U_{Y}) are homeomorphic ( (X, U_{X}) ≅ (Y, U_{Y}) in notation) if there is a continuous bijective function f : X → Y such that the inverse function f^{−1} is continuous, too. Such a function is called homeomorphism.

Homeomorphic spaces are indistinguishable in Topology. One main objective of Topology is to decide whether two topological spaces are homeomorphic or not. We can define continuity at a point, we imitate the ϵ − δ-definition:

Definition 3.8. N ⊂ X is a neighborhood of x ∈ X if there is an open set U ∈ U_{X} such that x ∈ U ⊂ N.

Warning: Munkres defines neighborhood what others—and these notes—call open neighborhood.

Definition 3.9. f : X → Y is continuous at the point x ∈ X if for every neighborhood N of f(x) there is a neighborhood M of x such that f(M) ⊂ N.

Exercise 3.10. Show that f : X → Y is continuous iff it is continuous at every point x ∈ X.

3.1. Examples of Topological Spaces. The main examples for topological spaces come from metric spaces. We have defined open sets in a metric space in Definition 3.1. To avoid confusion we can call them metrically open sets. Proposition 3.4 shows that metrically open sets in a metric space satisfy the axioms in Definition 3.5. For a metric space (X, d) we denote

the topological space defined by the metrically open sets by (X, U_{d}).

Not every topological space comes from a metric space. One traditional question of topology was to find sufficient and necessary conditions for a topological space to be “metrizable”. I

will give a partial answer to this question.

Examples 3.11.

1. The trivial topological space on a set X: U_{triv} = {∅, X}.

2. The discrete topological space on a set X: U_{dis} = P(X) i.e. every set is open.

Different topologies on the same set. We have already found three different topologies on the set ℝ. It is easy to see that U_{triv} ⊂ U_{d} ⊂ U_{dis}. So we say that U_{triv} is smaller or coarser

than the usual topology U_{d} and U_{dis} is bigger or finer than U_{d}.

Exercise 3.12. Show that if U_{1}, U_{2} are topologies on the set X then U_{1} is bigger than U_{2} iff the identity map is continuous from (X, U_{1}) to (X, U_{2}).

Closed sets. Defining closed sets in a topological space is very easy:

Definition 3.13. Let (X, U) be a topological space. Z ⊂ X is called closed if X \ Z ∈ U.

We usually denote the set of closed sets by Z.

Proposition 3.14. Let Z be the collection of closed subsets of the topological space X. Then the following conditions are satisfied:

1. ∅, X ∈ Z.

2. Z_{1}, Z_{2} ∈ Z implies that Z_{1} ∪ Z_{2} ∈ Z.

3. if Z_{i} ∈ Z for every i ∈ I then ⋂_{i∈I} Z_{i} ∈ Z.

Proof. Apply the De Morgan identities:

X \ (⋃_{i∈I} Z_{i}) = ⋂_{i∈I}(X \ Z_{i}), X \ (⋂_{i∈I} Z_{i}) = ⋃_{i∈I}(X \ Z_{i}).

This allows us to give an alternative definition of topological space on a set X by giving a collection Z of subsets of X satisfying properties (1) to (3) of Proposition 3.14. Also

continuity can be described via closed sets:

Proposition 3.15. f : X → Y is continuous iff preimage of closed sets is closed.

Proof is left to the reader. Homework for Section 3.

1. Prove Proposition 3.15.

2. List all topological spaces on the set X = {1, 2}. In other words list all collections of subsets of X, which satisfy the axioms in 3.5.

3. Show that (ℝ^{2}, d_{round}) and (ℝ^{2}, d_{max}) have the same open sets, i.e. the topologies U_{d_{round}} and U_{d_{max}} are the same.

4. Show that (X, U_{triv}) is not metrizable if X has at least 2 points.

5. Define Ā = ⋂{Z ∈ Z : Z ⊃ A}. The subset Ā is called the closure of A. Show that (a) Ā is the smallest closed set containing A.

(b) ¯ Ā = Ā.

(c) A ⊂ B implies Ā ⊂ B̄.

6. Show that x ∈ Ā ⇔ every neighborhood of x intersects A.

7. Prove Exercise 3.10.

8. Prove Exercise 3.12.

4. Further Examples of Topological Spaces

Generating a Topological Space. Suppose that S ⊂ P(X) and you want to know in which

topologies these sets are open. They are certainly open in U_{disc} so the natural question is which is the smallest topology in which they are open.

Proposition 4.1. For any S ⊂ P(X) the smallest topology U_{S} such that S ⊂ U_{S} is

U_{S} = {⋃_{i∈I} B_{i} : B_{i} ∈ B_{S}} where B_{S} = {⋂_{i=1}^{n} S_{i} : S_{i} ∈ S}.

Remark 4.2. You may wonder why X ∈ U_{S}. We can agree that the intersection of no sets is the whole space X, or if you don’t like this convention, just simply add the condition that ⋃{S ∈ S} = X i.e. S is a cover of X.

Proof. The axioms of topology in Definition 3.5 implies that any topology U ⊃ S has to be bigger than U_{S} so we only have to verify that U_{S} is a topology.

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1. ∅, X ∈ U_{S} by the convention on empty union and intersection.

2. Suppose that U_{k} ∈ U_{S} for k ∈ K. Then U_{k} = ⋃_{i∈I_{k}} B_{i} for B_{i} ∈ B_{S} so ⋃_{k∈K} U_{k} = ⋃_{i∈I} B_{i} where I = ⋃_{k∈K} I_{k}.

3. Let U, V ∈ U_{S}. Then U = ⋃_{i∈I} B_{i} and V = ⋃_{j∈J} B_{j} where B_{i}, B_{j} ∈ B_{S}. By the distributivity for ∪ and ∩ we get:

U ∩ V = ⋃_{i∈I,j∈J} B_{i} ∩ B_{j}.

It is clear from the definition of B_{S} that B_{i} ∩ B_{j} is also in B_{S}, so U ∩ V ∈ U_{S}.

Definition 4.3.

1. A set S ⊂ P(X) generating the topology U is called a subbasis for U if S is a cover of X (See Remark 4.2).

2. A set B ⊂ P(X) is called a basis for the topology U if U = {⋃_{i∈I} B_{i} : B_{i} ∈ B}.

Exercise 4.4. B ⊂ P(X) is a basis for a topology iff B_{1}, B_{2} ∈ B implies the existence of B_{i} ∈ B for i ∈ I such that B_{1} ∩ B_{2} = ⋃_{i∈I} B_{i} and B is a cover of X.

Defining topologies by giving a basis or a subbasis for them is a useful method:

Examples 4.5.

1. Metric topologies: If (X, d) is a metric space then B := {B_{ϵ}(x) : x ∈ X, ϵ > o}. Proposition 3.4 shows that B is a basis for the metric topology on (X, d).

2. Discrete topology: {{x} : x ∈ X} is a basis for U_{dis}.

3. {U ⊂ X : X \ U is a finite set} is a basis for the cofinite topology U_{cof}.

4. {U ⊂ X : X \ U is a countable set} is a basis for the cocountable topology U_{coco}.

5. {(a, b] : −∞ ≤ a < b ≤ ∞} is a basis for the left topology U_{left} on the real line ℝ (also called the Sorgenfrey line).

6. Let (X, <) be an ordered set, and let (a, ∞) = {x ∈ X : a < x} and (−∞, a) = {x ∈ X :

a > x}. Then

{(a, ∞), (−∞, a) : a ∈ X} ∪ {X}

is a subbasis for the order-topology U_{<}.

Subbasis is also a useful tool when you want to check the continuity of a function:

Proposition 4.6. Suppose that S is a subbasis for the topological space Y. then f : X → Y is continuous iff f^{−1}(S) is open for all S ∈ S.

The proof is left to the reader.

Constructing new topologies from old ones.

Definition 4.7. Subspace topology: If (X, U) is a topological space and A ⊂ X then U_{A} = {U ∩ A : U ∈ U} is a topology on A.

Exercise 4.8. Show that U_{A} is the smallest topology for which the inclusion i : A → X is continuous.

Definition 4.9. Product topology: If (X, U_{X}) and (Y, U_{Y}) are topological spaces then let U_{X×Y} be the smallest topology on the product set X ×Y for which the projection maps π_{X} : (x, y) ↦ x and π_{Y} : (x, y) ↦ y are continuous. In other words U_{X×Y} is generated by the subbasis

{π_{X}^{−1}(U) : U ∈ U_{X}} ∪ {π_{Y}^{−1}(U) : U ∈ U_{Y} }.

Proposition 4.10. The function f : Z → X × Y is continuous iff the coordinate functions π_{X} ∘ f and π_{Y} ∘ f are continuous.

Proof. ⇒ follows from the definition since the projections π_{X} and π_{Y} are continuous. ⇐ By Proposition 4.6 it is enough to check the preimages of elements of a subbase to prove continuity. But

f^{−1}(π_{X}^{−1}(U)) = (π_{X} ∘ f)^{−1}(U)

and the right hand side is open by assumption. Replacing X with Y in the equation above we finished the proof.

Proposition 4.10 can be used to show the following universal property of the product:

Theorem 4.11. Let X and Y be topological spaces. Then there exists a space T unique up to homeomorphism such that there are continuous maps t_{X} : T → X and t_{Y} : T → Y such that for any continuous maps f_{X} : Z → X, f_{Y} : Z → Y there is a unique continuous map h : Z → T with the property: f_{X} = t_{X} ∘ h and f_{Y} = t_{Y} ∘ h.

The statement of the Theorem can be summarized with a so called commutative diagram:

\xymatrix{X & T \ar[l]_{t_{X}} \ar[r]^{t_{Y}} & Y \\  & Z \ar@{-->}[ul]^{f_{X}} \ar@{-->}[ur]_{f_{Y}} \ar@{-->}[u]^{h}}

The commutativity of such a diagram means that if there are two ways of getting from one

point to the other following the arrows then the corresponding compositions of maps are equal.

Proof. By Proposition 4.10 T = X × Y, t_{X} = π_{X} and t_{Y} = π_{Y} have the required property so the existence part is proved. Proof of unicity is left as an exercise.

This theorem provides us an alternative definition of the product which is very useful in

applications since it reflects the basic property of the product.

Similarly you can define the product of any number of topological spaces by looking for the

smallest topology on the product set for which all the projection maps are continuous.

Homework for Section 4.

1. What are the continuous maps from (ℝ, U_{triv}) to (ℝ, U_{d})?

2. Show that every map is continuous from (X, U_{dis}) and every map is continuous to (X, , U_{triv}).

3. Prove Exercise 4.4: B ⊂ P(X) is a basis for a topology iff it is closed for finite intersection i. e. B_{1}, B_{2} ∈ B implies the existence of B_{i} ∈ B for i ∈ I such that B_{1} ∩ B_{2} = ⋃_{i∈I} B_{i}.

4. Find a basis for the order-topology in Example 4.5.

5. Which previously defined topologies on the real line are smaller than (ℝ, U_{cof})?

6. Which previously defined topologies on the real line are smaller than (ℝ, U_{left})?

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7. Prove Exercise 4.8: Show that U_{A} is the smallest topology for which the inclusion i : A → X is continuous.

8. Show that the open rectangles {U × V : U ∈ U_{X}, V ∈ U_{Y} } provide a basis for the product topology U_{X×Y}.

9. Show that the metric topology on the real plane ℝ^{2} coincides with the product topology on ℝ × ℝ. (More generally metric product is compatible with topological product.)

10. * Show that the product of finitely many discrete top spaces is discrete, but ∏_{i=1}^{∞} X_{i} where X_{i} is the discrete topology on {0, 1} is not discrete.

11. Show that (0, 1) ≅ ℝ.

12. Prove Exercise 4.6, that it is enough to check continuity for a subbasis.

13. Prove the unicity part of Theorem 4.11.

14. Let Y_{x} = {(x, y) : y ∈ Y } ⊂ X × Y (usually called a slice) Show that Y_{x} ≅ Y.

5. Intermediate Value Theorem and Connectedness

One of the properties of continuous function we listed at the beginning of this course was the “no missing value” property. One of the first theorems in calculus is the

Theorem 5.1 (Intermediate Value Theorem ). If f : [0, 1] → ℝ is a continuous function, then [f(0), f(1)] ⊂ f([0, 1]).

There is a straightforward way to generalize this property:

Definition 5.2. The topological space (X, U) has the Intermediate Value (IMV) property if for any continuous function f : X → ℝ the image f(X) is an interval (possibly infinite).

Exercise 5.3. Find examples of topological spaces with and without the IMV property.

We can give a simpler description of the IMV property:

Definition 5.4. The topological space (X, U) is connected if for any nonempty U, V ∈ U the property U ∪ V = X implies U ∩ V ≠ ∅.

Exercise 5.5. Show that (X, U) is connected iff for any nonempty F, G closed sets if F ∪G = X then F ∩ G ≠ ∅.

Exercise 5.6. Show that (X, U) is connected iff the only open and close sets (sometimes called clopen) are ∅ and X.

Theorem 5.7. The topological space (X, U) has the IMV property iff it is connected.

Proof. ⇒ Suppose that X doesn’t have the IMV property. Then ∃ a continuous function f and real numbers a < z < b such that a, b ∈ f(X) but z ∉ f(X). Then the open decomposition X = f^{−1}((−∞, z)) ∪ f^{−1}((z, ∞)) shows that X is not connected. ⇐ Suppose we have a decomposition X = U_{1} ∪ U_{2} to nonempty disjoint open sets. Then

f(x) = \begin{cases}1 & if x ∈ U_{1} \\ 2 & if x ∈ U_{2} \end{cases}

is a continuous function demonstrating that X doesn’t have the IMV property.

Theorem 5.8. The closed interval [0, 1] is connected.

Proof. Suppose we have a decomposition [0, 1] = U ∪ V to disjoint open sets. We can assume that 1 ∈ V. Let x = sup U. Then x ∈ U since U = [0, 1] \ V is closed as well. But U is open, too, so it has to contain a small interval around x. So x has to be equal to 1. Contradiction.

In fact the connectedness of [0, 1] is equivalent with the existence of supremum. Consequently at the definition of real numbers we can replace the Cantor axiom with declaring [0, 1] to be

connected.

Exercise 5.9. Show that the connected subsets of ℝ are the intervals (possibly infinite).

The most general and also the simplest version of the IMV-type theorems is the following:

Theorem 5.10. The continuous image of a connected space is connected.

Proof. The preimage of a clopen subset is clopen.

Definition 5.11. Let a ∼ b if a and b are elements of the topological space X and they belong to a connected subset of X.

The following exercise shows that ∼ is an equivalence relation (you can use it to prove the transitivity of ∼):

Exercise 5.12. Suppose that A_{i} ⊂ X are connected and the pairwise intersections A_{i} ∩ A_{j} are not empty. Prove that ⋃ A_{i} is connected.

Definition 5.13. The equivalence classes of ∼ are called the connected components of the topological space X.

Exercise 5.12 implies that the connected components are indeed connected. There is a similar concept we will use more frequently later:

Definition 5.14. The topological space (X, U) is path connected (or arcwise connected) if for any two points a and b in X there is a continuous map f : [0, 1] → X (called path) such that f(0) = a and f(1) = b.

Proposition 5.15. A path connected space is connected.

Proof. The unit interval is connected so by Theorem 5.10 f([0, 1]) is also connected. So if (X, U) is path connected then X has only one connected component so X is connected.

Example 5.16. The following subspace of the plane is connected but not path connected:

X := {(0, 0)} ∪ {x, sin(1/x) : x ∈ (0, π]}.

Hint: Show that any path f : [0, 1] → X connecting (0, 0) with (π, 0) contains all the points of

X. Then check continuity of f at 0.

Homework for Section 5.

1. Show that if A ⊂ X is connected then \overline{A} is connected, too.

2. Show that the connected components are closed.

3. Show that ℝ ≅ ℝ̸^{2} with their usual topology.

4. Show that ℝ ≅ (0, 1].

5. What are the continuous maps from (ℝ, U_{d}) to (ℝ, U_{dis})?

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6. Which previously defined topologies are connected on ℝ?

7. Prove Exercise 5.5: Show that (X, U) is connected iff for any nonempty F, G closed sets if F ∪ G = X then F ∩ G ≠ ∅.

8. Prove Exercise 5.6: Show that (X, U) is connected iff the only clopen are ∅ and X.

9. Suppose that A and B are connected and A ∩ B is not empty. Prove that A ∪ B is connected.

10. Prove Exercise 5.9: Show that the connected subsets of ℝ are the intervals (possibly infinite).

11. * Prove the existence of supremum for subsets of the closed interval using that it is connected.

12. * Show that the space of Example 5.16 is connected but not path connected.

13. Show that the space X is connected iff every continuous function f : X → {1, 2} (with the discrete topology on {1, 2}) is constant.

14. Show that the product of two connected spaces is connected and the product of two path connected spaces is path connected. (Hint: You can use the slice lemma Homework (14) in Section 3.)

6. Separation Axioms

If we want to deal with only “reasonable” (for example metrizable) topological spaces we need some restrictions.

Definition 6.1. (X, U) is T_{0} if for any two points x ≠ y there is an open set containing one of them but not the other.

Exercise 6.2. The topological space (X, U) is T_{0} if every continuous function f : Y → X— where Y is a trivial topological space with at least two elements—is constant.

Definition 6.3. The topological space (X, U) is T_{1} if for any two points x ≠ y there is an open set containing x but not y and another open set containing y but not x.

Exercise 6.4. (X, U) is T_{1} iff every point is closed.

Definition 6.5. The topological space (X, U) is T_{2} or Hausdorff if for any two points x ≠ y there are disjoint open sets U and V with x ∈ U and y ∈ V.

Exercise 6.6. Every metric space is Hausdorff.

Hausdorff spaces behave well because sequences behave well:

6.1. Limit of sequences.

Definition 6.7. Let {x_{n} : n ∈ ℕ} is a sequence in the topological space (X, U). We say that {x_{n}} converges to x ∈ X (x_{n} → x or x is the limit of the sequence {x_{n}}) if for every open set U containing x the set {n ∈ ℕ : x_{n} ∉ U} is finite.

Exercise 6.8. If x_{n} → x and \{x_{n} : n ∈ ℕ\} ⊂ A then x ∈ \overline{A}.

Proposition 6.9. In a Hausdorff space a sequence has at most one limit.

The proof is left to the reader. In applications you hardly ever meet topological spaces which are not Hausdorff. One notable exception is in algebraic geometry. Let X = ℝ^{n} (or ℂ^{n}) and

let Z ⊂ X be closed if there are polynomials {p_{i} : i ∈ I} of n variables such that Z = {x ∈ X : p_{i}(x) = 0 for all i ∈ I}. Such sets are usually called algebraic sets. The

corresponding topology is called the Zariski topology.

Exercise 6.10. Show that the algebraic sets in X satisfy the axioms for closed sets in a topology.

Homework for Section 6.

1. Prove Exercise 6.2 that the topological space (X, U) is T_{0} if every continuous function f : Y → X—where Y is a trivial topological space with at least two elements—is constant.

2. Prove Exercise 6.4 that (X, U) is T_{1} iff every point is closed.

3. Describe the sequences which have a limit in U_{dis}, U_{triv}, U_{cof} and U_{coco}.

4. Prove that if f : X → Y is continuous then x_{n} → x implies f(x_{n}) → f(x).

5. Find an example for a space in which limits are unique, but it is not Hausdorff.

6. Show that the product of Hausdorff spaces is Hausdorff.

7. Prove that X is Hausdorff iff the diagonal Δ = {(x, x) : x ∈ X} ⊂ X × X is closed.

8. Let f : X → Y be a continuous map. Show that if Y is T_{1} then the level set

(f = c) = {x ∈ X : f(x) = c}

is closed.

9. Let f, g : X → Y be continuous maps. Show that if Y is Hausdorff then E = {x ∈ X : f(x) = g(x)} is closed in X.

10. Show that Id : (ℝ, U_{cof}) → (ℝ, U_{dis}) is not continuous but x_{n} → x implies f(x_{n}) → f(x).

11. Prove Exercise 6.8: If x_{n} → x and \{x_{n} : n ∈ ℕ\} ⊂ A then x ∈ \overline{A}.

12. Prove Exercise 6.10 on the Zariski topology.

13. Prove that the standard topology on X = ℝ^{n} (or ℂ^{n}) is bigger than the Zariski topology.

7. Countability Axioms

We know that for real functions continuity can be defined via convergent sequences:

Theorem 7.1. f : ℝ → ℝ is continuous iff for every sequence x_{n} → x implies f(x_{n}) → f(x).

Homework 4 in 6 says that ⇒ is always true, but not the converse by Homework 10. So we introduce a class of topological spaces where the converse holds, too:

Definition 7.2. The topological space (X, U) is M_{1} or first countable if for every x ∈ X there exist open sets {U_{n} : n ∈ ℕ} containing x such that for all V ∋ x open there is an n such that U_{n} ⊂ V.

Theorem 7.3. Suppose that (X, U) is first countable. Then f : X → Y is continuous if for every sequence x_{n} → x implies f(x_{n}) → f(x).

To simplify the language we introduce a definition:

Definition 7.4. {N_{i} : i ∈ I} ⊂ P(X) is a neighborhood basis for x ∈ X if all N_{i} is a neighborhood of x and for all N neighborhood of x there is an i such that N_{i} ⊂ N.

So X is first countable iff every point has a countable neighborhood basis.

Proof of Theorem 7.3. Suppose that f is not continuous. Then there is an open set U such that f^{−1}(U) is not open, i.e. there is an x ∈ f^{−1}(U) which is in the closure of the complement i.e. for all V ∋ x open V \ f^{−1}(U) is not empty. Let {U_{n} : n ∈ ℕ} be a countable neighborhood basis of x. Then we can choose elements x_{n} ∈ U_{n} \ f^{−1}(U). Then

1. x_{n} → x since x_{n} ∈ U_{n} so for any V ∋ x open there is an n such that U_{n} ⊂ V so all but the first n − 1 elements of the sequence {x_{n}} is in V.

2. f(x_{n}) ↛ f(x) since f(x_{n}) ∉ U.

Proposition 7.5. Every metric space is first countable.

Proof. {B_{1/n}(x) : n ∈ ℕ} is a neighborhood basis of x. (For specialists: we used the Archimedean axiom here ;-)

Definition 7.6. The topological space (X, U) is M_{2} or second countable if it has a countable basis.

Homework for Section 7.

1. If X is first countable then Ā = {x ∈ X : ∃a_{n} ∈ A a_{n} → x}.

8. Compactness

Compactness is probably the most frequently used topological property. The basic example for compact space is the unit interval [0, 1]. Let us list some properties related to compactness

shared by [0, 1]:

1. Every continuous function f : X → ℝ is bounded.

2. Every continuous function f : X → ℝ attains its maximum.

3. Every sequence in X has a convergent subsequence.

4. Every decreasing sequence of closed subsets of X has a nonempty intersection. (The Cantor Axiom)

5. Every open cover of X has a finite subcover. Here open cover means a collection C of open sets such that ∪C = X and a finite subcover is a

finite subset C′ of C such that ∪C′ = X.

For general topological spaces these properties are not equivalent.

Definition 8.1. The topological space X is compact if every open cover of X has a finite subcover.

Exercise 8.2. Show that for a topological space X a subset A ⊂ X is compact in the subspace topology iff for every collection C of open subsets of X such that ∪C ⊃ A there is a finite subset C′ of C such that ∪C′ ⊃ A. So we use the name open cover for C and the name finite subcover for C′ in this situation as well.

Proposition 8.3. Let Y be a Hausdorff space. Then every compact subset of Y is closed.

INTRODUCTION TO TOPOLOGY 15

Proof. Fix a x ∈ X \ K. Then for any k ∈ K let U_{k} ∋ k and V_{k} ∋ x disjoint open sets provided by the Hausdorff property. Then ⋃_{k∈K} U_{k} is an open cover of K so it has a finite subcover {U_{k_{1}}, …, U_{k_{n}}}. Then

W_{x} = ⋂_{i=1}^{n} V_{k_{i}} ⊂ (X \ K)

is open (being the finite intersection of open sets). Construct such a W_{x} for all x ∈ X \ K. Then X \ K = ⋃_{x∈X\K} W_{x} is open. Exercise 8.4. Suppose that X is compact and A ⊂ X is closed. Then A is compact. Exercise 8.5. If K is a compact subspace of a metric space then K is bounded (i.e. there is a ball B_{N}(x) ⊃ K). Exercise 8.6. It is not true that a closed and bounded subspace of a metric space is compact. Find an example. Proposition 8.7. The unit interval [0, 1] is compact. Proof. Let C be an open cover of [0, 1] and let

S = {s ∈ [0, 1] : there is a finite subcover C′ ⊂ C for [0, s]}.

Let b = sup S. There is a U ∈ C such that b ∈ U. The set U is open, so there is an ϵ > 0 such that (b − ϵ, b + ϵ) ⊂ U. Since b − ϵ/2 ∈ S there is a finite subcover C′ covering [0, b − ϵ/2] so C′ ∪ {U} is a finite subcover for [0, b + ϵ/2]. So b had to be 1. Proposition 8.8. If f : X → Y is continuous and X is compact then f(X) is compact. Proof. Preimage of an open cover is an open cover. Corollary 8.9. If X is compact then every continuous function f : X → ℝ is bounded. Proof. f(X) is compact by 8.8 therefore bounded by 8.5. Corollary 8.10. If X is compact then every continuous function f : X → ℝ attains its maximum. Proof. f(X) is compact therefore closed by 8.3, so sup f(X) ∈ f(X). Proposition 8.11. If X is compact and M_{1} then every sequence in X has a convergent subsequence.

We prepare the proof with a definition and two lemmas:

Definition 8.12. x ∈ X is an accumulation point of A ⊂ X if for every open set U ∋ x the set (U \ x) ∩ A is not empty.

The set of accumulation points of A ⊂ X is denoted by A′.

Exercise 8.13. Show that Ā = A ∪ A′. Lemma 8.14. If X is compact then every infinite set has an accumulation point.

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Proof. Suppose that A ⊂ X has no accumulation point. Then for all x ∈ X there is an open set U_{x} ∋ x such that (U_{x} \ x) ∩ A is empty. {U_{x} : x ∈ X} is an open cover of X so it has a finite subcover {U_{x_{1}}, …, U_{x_{n}} : x_{i} ∈ X}. Then A ⊂ {x_{1}, …, x_{n}} so A is finite.

Lemma 8.15. If X is M_{1} and x ∈ X is an accumulation point of A ⊂ X then there is a sequence {a_{n}} ⊂ A such that a_{n} → x.

Proof. Let {U_{n} : n ∈ ℕ} be a countable neighborhood basis of x. Since x is an accumulation point of A we can pick points a_{n} ∈ U_{n} ∩ A. As in the proof of Theorem 7.3 we can see that a_{n} → x.

These lemmas immediately imply Proposition 8.11. A space X with the property that every sequence in X has a convergent subsequence is called sequentially compact. So Proposition

8.11 says that a compact M_{1} space is sequentially compact.

Proposition 8.16. If X is compact then every decreasing sequence of closed subsets of X has a nonempty intersection.

The proof is left to the reader.

Theorem 8.17. The product of compact spaces is compact.

The proof uses the Axiom of Choice and not so easy so we omit it.

Theorem 8.18. If X is compact Hausdorff and M_{2} then metrizable.

This is a special case of Urison’s metrization theorem. We omit the proof.

Homework for Section 8.

1. Prove Exercise 8.4: Suppose that X is compact and A ⊂ X is closed. Then A is compact.

2. Prove Exercise 8.5: If K is a compact subspace of a metric space then K is bounded.

3. Prove Exercise 8.6: find a subspace of a metric space which is closed and bounded but not compact.

4. Prove the Basis theorem for compactness: Suppose that B is a basis for the topological space X. then X is compact iff for any cover C ⊂ B there is a finite subcover. The theorem is true for subbasis, too. It is called Alexander subbasis theorem. The proof is more difficult and uses the Axiom of Choice.

5. * Prove Theorem 8.17 for two spaces. Hint:

(a) Use Homework (8) in Section 3 that the ”rectangles”

{U × V : U open in X and V open in Y }

give a basis for the topology on X × Y.

(b) Use the slice lemma Homework (14)in Section 3 that the “slice” Y_{x} := {(x, y) : y ∈

Y } ⊂ X × Y is homeomorphic to Y.

6. * Let X_{S} = {0, 1} with the discrete topology for S ∈ P(ℕ). Then X = ∏_{S∈P(ℕ)} X_{S} is compact by Theorem 8.17. Let

x_{n}(S) = \begin{cases}0 & if n ∈ S \\ 1 & if n ∉ S \end{cases} .

Show that {x_{n}} doesn’t have a convergent subsequence.

7. Prove that if X is T_{1} then x is an accumulation point of A ⊂ X iff for every neighborhood N of x the intersection A∩N is infinite. Find a T_{0} space which shows that the T_{1} condition above is necessary.

8. Let f : X → Y be a continuous bijection where X is compact and Y is Hausdorff. Show that f is a homeomorphism.

9. The notion of Homotopy

Till now we treated functions in a stricter way than spaces. Two spaces are considered to be the same if you can “deform” one into the another. We can do something similar to functions

as well. We say that f : X → Y is homotopic to g : X → Y if there is a “continuous” deformation {f_{t} : t ∈ I} (here I denotes the unit interval) such that f_{0} = f and f_{1} = g. The problem is that we haven’t defined a topology on the space of continuous functions C(X, Y)

yet. It would be possible but we take a slightly different approach:

Definition 9.1. H : X × I → Y is a homotopy between f and g if H_{0} = f and H_{1} = g where H_{s} = H ∘ i_{s} and i_{s} : X → X × I such that i_{s}(x) = (x, s). If there is a homotopy between f and g then we say that f and g are homotopic or homotopy equivalent and we write f ∼ g.

Exercise 9.2. Show that ∼ is an equivalence relation.

As I mentioned before, the mapping spaces C(Z, X) and C(X, Z) are fundamental in studying topological invariants of X. But studying only the set of homotopy equivalence

classes is an easier task. Homotopy theory studies the sets [Z, X] = C(Z, X)/ ∼. We will look

at the case of Z =circle in details. As we will see it provides us a strong and calculable topological invariant. It also helps us to solve almost all the problems listed in Section 1.

10. The Fundamental Group

For technical reasons we think a map of the circle to the space X as a map—called loop—f : I → X such that 0 and 1 are mapped to the same point. In fact we assume that

f(0) = f(1) = x_{0} where x_{0} is a distinguished point of X called the base point. This seemingly unnecessary restriction becomes important when we want to define a group structure on loops

to enter the field of Algebraic Topology:

Definition 10.1. Suppose that f, g : I → X are loops. Then

f · g = \begin{cases}2t & if 0 ≤ t ≤ 1/2 \\ 2t − 1 & if 1/2 ≤ t ≤ 1 \end{cases}

is the concatenation of f and g.

We can see that f · g is a loop, we just have to check that f · g is continuous. That is a consequence of the following fact called the Pasting Lemma.

Lemma 10.2. Suppose that X = A∪B and A, B ⊂ X are closed. If f : X → Y is continuous restricted to A and restricted to B then f is continuous.

Proof. Exercise.

18 LÁSZLÓ FEHÉR

Concatenation resembles to a group multiplication on the set of loops. But it is easy to see that there is not even a unit element in this structure. However, moving to the homotopy

level fixes all the problems.

Definition 10.3. Suppose that f, g : I → X are loops. Then H : I × I → X is a (based) homotopy between f and g if H_{0} = f and H_{1} = g and H(0, s) = H(1, s) = x_{0} for all s ∈ I.

Definition 10.4. Let π_{1}(X, x_{0}) be the set of homotopy equivalence classes of loops in (X, x_{0}).

Theorem 10.5. Concatenation defines a group structure on π_{1}(X, x_{0}).

Proof. This proof is typical in homotopy theory. We work with equivalence classes so with every definition first we have to check whether it makes sense on the level of equivalence classes. In this particular case we have to check the following:

If f_{0} ∼ f_{1} and g_{0} ∼ g_{1} then f_{0} · g_{0} ∼ f_{1} · g_{1}.

So we have a homotopy F between f_{0} and f_{1} and a homotopy G between g_{0} and g_{1}. Then the concatenation F · G will give a homotopy between f_{0} · g_{0} and f_{1} · g_{1} as soon as we define it:

Definition 10.6. Suppose that F, G : I × X → Y are maps. Then

F · G(s,) = \begin{cases}(2t, x) & if 0 ≤ t ≤ 1/2 \\ (2t − 1, x) & if 1/2 ≤ t ≤ 1 \end{cases}

is the concatenation of F and G.

Remark 10.7. The pasting lemma 10.2 shows that F · G is continuous.

You can see that Definition 10.1 is a special case of Definition 10.6. It is also easy to see that if F and G are based homotopies than F · G is a based homotopy, too. So we showed that concatenation defines a binary operation on π_{1}(X, x_{0}). Now we have to check whether the group axioms

1. there is a unit e such that γe = eγ = γ for all γ ∈ G,

2. for all γ ∈ G there is an inverse γ^{−1} ∈ G such that γγ^{−1} = e,

3. the associativity α(βγ) = (αβ)γ holds are satisfied for G = π_{1}(X, x_{0}).

1. It is clear that the unit in this group should be the constant loop \underline{x}_{0}. So we have to show that f · \underline{x}_{0} ∼ f for all loops f, which can be seen from the following:

Lemma 10.8 (on reparametrization). Let h : I → I such that h(0) = 0 and h(1) = 1. Then for any loop f : I → (X, x_{0}) we have f ∼ f ∘ h.

Proof. We use the trick of convex combination (and this is not the last time ;-). Let H(t, s) = f(sh(t) + (1 − s)t).

Exercise 10.9. Show that H is indeed a homotopy between f and f ∘ h.

2. If γ = [f] (i.e. f is an element of the equivalence class γ), then let γ^{−1} = [\overline{f}] where \overline{f}(t) = f(1 − t).

Exercise 10.10. Use the following generalization of the reparametrization lemma 10.8 to show that \overline{f} · f ∼ \underline{x}_{0}: Let h_{1}, h_{2} : I → I such that h_{1}(0) = h_{2}(0) and h_{1}(1) = h_{2}(1). Then for any map f : I → X such that f(h_{1}(0)) = f(h_{1}(1) = x_{0} the loops f ∘ h_{i} are homotopic.

3. Associativity can be checked with an other application of the reparametrization lemma 10.8 to

h(t) = \begin{cases}2t & if 0 ≤ t ≤ 1/4 \\ t + 1/4 & if 1/4 ≤ t ≤ 1/2 \\ t/2 + 1/2 & if 1/2 ≤ t ≤ 1 \end{cases} .

This group π_{1}(X, x_{0}) is called the first homotopy or fundamental group of (X, x_{0}).

11. Induced homomorphisms and Functors

As I mentioned before maps are frequently even more important than the objects. That makes the following simple fact very important:

Proposition 11.1. Suppose that f : (X, x_{0}) → (Y, y_{0}) is a base point preserving map (i.e. f(x_{0}) = y_{0}). Then

1. f induces a homomorphism f_{∗} : π_{1}(X, x_{0}) → π_{1}(Y, y_{0}) where f_{∗}[l] = [f ∘ l].

2. (f ∘ g)_{∗} = f_{∗} ∘ g_{∗} for any g : (Z, z_{0}) → (X, x_{0}) base point preserving map.

Proof. Left to the reader.

An other, less popular notation for f_{∗} is π_{1}(f). The advantage of the latter is that it reminds us that π_{1} is a functor. In other words it is a map from the category of topological spaces to the category of groups. Though I am not going to use category theory in these notes I give

some basic definitions because I believe thinking in these terms helps to get a wider perspective.

Definition 11.2. A category is a pair C = (O, M) together with a lot of maps. Elements of O are called objects and elements of M are called morphisms. For every A, B ∈ O there is a set Mor_{C}(A, B) ⊂ M called the set of morphisms from A to B. We can think of Mor_{C}(A, B) as the set of maps from A to B with a certain property. We also have maps describing the composition of these morphisms: Comp = Comp_{A,B,D} : Mor_{C}(A, B) × Mor_{C}(B, D) → Mor_{C}(A, D). We can denote Comp(α, β) by α ∘ β or simply αβ. We require two axioms to be satisfied:

1. Associativity of composition (try to formalize this using the maps Comp).

2. The existence of identity: For every A ∈ O there is an element Id_{A} ∈ Mor(A, A) such that α Id_{A} = α and Id_{A} β = β for any morphisms for which the composition is defined.

In most examples Comp is indeed the composition of certain maps, so the axioms are automatically satisfied.

Exercise 11.3. Check that the following examples are categories (i.e. composition of morphisms are morphisms).

1. Sets and maps.

2. Metric spaces and isometries.

3. Topological spaces and continuous maps.

20 LÁSZLÓ FEHÉR

4. Topological spaces with base point and base point preserving continuous maps.

5. Groups and homomorphisms.

6. Vector spaces and linear maps.

7. Ordered sets and order preserving maps.

Definition 11.4. A functor F : C_{1} → C_{2} is a collection of maps: a map F_{O} : O_{1} → O_{2} and maps F_{A,B} : Mor_{C_{1}}(A, B) → Mor_{C_{2}}(F (A), F (B)) such that F respects composition. (i.e. F (αβ) = F (α)F (β)).

So Proposition 11.1 shows that π_{1} is a functor from the category of Topological spaces with

base point and base point preserving continuous maps to the category of Groups and homomorphisms.

Exercise 11.5. Find other functors among the categories above (an important example is the forgetful functor, give a definition!).

Category theory helps to apply a construction in one category into an other category. For example you are encouraged to define the product of two objects in an arbitrary category in analogy with the product in topology (hint: use Theorem 4.11). What will be the product in

the category of groups and homomorphisms? Warning: not every category has a product.

Proposition 11.6. If f, g : (X, x_{0}) → (Y, y_{0}) are (based) homotopic, then f_{∗} = g_{∗}.

Proof. Left to the reader.

Theorem 11.7 (Brouwer). Every continuous map ϕ : D^{2} → D^{2} has a fixed point ϕ(x) = x.

Sketch of Proof. Suppose that there is no fixed point. Then for every x ∈ D^{2} there is a unique line l_{x} through ϕ(x) and x. Let p : D^{2} → S^{1} be defined by the following rule: p(x) is the intersection point of the line l_{x} and the unit circle S^{1} which is closer to x than to ϕ(x). p is continuous and p|_{S^{1}} = Id_{S^{1}} (Check these!). In other words i ∘ p = Id_{S^{1}} where i : S^{1} → D^{2} is the inclusion of the unit circle into the unit disc. Then by the functorial property of π_{1}:

i_{∗}p_{∗} = Id_{π_{1}(S^{1})} .

Assuming that π_{1}(S^{1}) = ℤ and knowing that π_{1}(D^{2}) = 0 we get a contradiction.

Remark 11.8. Of course you can give a more down to earth argument if you dig into the definition of induced homomorphism. Let f : I → S^{1} is a loop where the base point of S^{1} is (1, 0). Then i ∘ f is a loop in (D^{2}, (1, 0)). So there is a homotopy H showing that f ∼ \underline{(1, 0)}. Since p ∘ i ∘ f = f then p ∘ H is a homotopy connecting f and \underline{(1, 0)} inside S^{1} which would imply that π_{1}(S^{1}) = 0.

Homework for Section 11.

1. Prove Proposition 11.1.

2. Give 3 examples of functors.

3. Show that π_{1}(X, x_{0}) = 0 ∀x_{0} ∈ X ⟺ ∀f : S^{1} → X extends to a map F : D^{2} → X (i.e. F |_{S^{1}} = f).

4. If f : S^{1} → S^{1} is nulhomotopic (i.e. homotopic to a constant map) then f has a fixed point.

12. Basic properties of the Fundamental Group

Proposition 12.1. π_{1}(X × Y, (x_{0}, y_{0})) = π_{1}(X, x_{0}) × π_{1}(Y, y_{0})

Proof. Enough to show that f × g ∼ \underline{(x_{0}, y_{0})} ⟺ f ∼ \underline{x}_{0} and g ∼ \underline{y}_{0}, which is left to the reader.

A fancy way to interpret this result is to say that the functor π_{1} preserves product. Now we turn to to a seemingly technical but important issue: the choice of the basepoint.

Proposition 12.2. If X is path connected then π_{1}(X, x_{0}) ≅ π_{1}(X, x_{1}) for any x_{1} ∈ X.

Proof. Let p : I → X be a path connecting x_{0} and x_{1}. Then we define a map

p̂: π_{1}(X, x_{0}) → π_{1}(X, x_{1}) by p̂[f] := [(\overline{p} · f) · p],

where \overline{p}(t) = p(1 − t). Though p is not a loop, we defined concatenation for homotopies in Definition 10.6. We have to check that this definition is correct i.e. if f ∼ g then (\overline{p} · f) · p ∼ (\overline{p} · g) · p. But if H is a homotopy between f and g, then (\underline{\overline{p}} · H) · \underline{p} is a homotopy between (\overline{p} · f) · p and (\overline{p} · g) · p (Here \underline{p} denotes the constant homotopy \underline{p}(t, s) = p(t)). We leave it to the reader to check that ˆ \overline{p} is an inverse for p̂, therefore p̂ is an isomorphism between π_{1}(X, x_{0}) and π_{1}(X, x_{1}).

So for a path connected space X we can use the notation π_{1}(X) not mentioning the basepoint. However it is an important fact that the isomorphism p̂ depends on the path p:

Proposition 12.3. Let p, q : I → X be paths connecting x_{0} and x_{1}. Then

q̂^{−1}p̂(α) = [q · \overline{p}]α[q · \overline{p}]^{−1}.

Notice that q̂^{−1}p̂ is an automorphism of π_{1}(X, x_{0}) and that q · \overline{p} is a loop based at x_{0} so

[q · \overline{p}] ∈ π_{1}(X, x_{0}).

Proof. Let f be a loop representing α. Then

q̂^{−1}p̂(α) = ˆ \overline{q}p̂(α) = ˆ \overline{q}[(\overline{p} · f) · p] = [(q · ((\overline{p} · f) · p)) · \overline{q}],

but as we have seen in the proof of Theorem 10.5 the bracketing don’t matter up to homotopy so:

q̂^{−1}p̂(α) = [((q · \overline{p}) · f) · (p · \overline{q})].

We can finish the proof by noticing that \overline{q · \overline{p}} = p · \overline{q}.

Remark 12.4. Proposition 12.3 says that the deviation of the isomorphism q̂ from the isomorphism p̂ is measured by conjugation with [p · \overline{q}] ∈ π_{1}(X, x_{0}). It is easy to see that a group is commutative iff every conjugation is trivial, which implies that p̂ is independent of the path p iff π_{1}(X, x_{0}) is a commutative group. So it would be interesting to find a space with non commutative fundamental group:

Exercise 12.5. Try to guess what π_{1}(ℝ^{2} \ {a, b}) is and give examples of loops such that the corresponding elements in π_{1}(ℝ^{2} \ {a, b}) don’t commute.

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13. Other constructions for Topological Spaces

Definition 13.1. Let f : X → Y be a map and U_{Y} be a topology on Y. Then the pull back of U_{Y} is

U\overleftarrow{f} = \{f^{−1}(U) ⊂ X : U ∈ U_{Y} \}.

Remarks 13.2. 1. It is easy to see that U\overleftarrow{f} is a topology on X.

2. If f is injective we get back the definition of subspace topology. If f is not injective then U\overleftarrow{f} is not even T_{0}.

3. An alternative definition would be that U\overleftarrow{f} is the smallest topology on X such that f : X → Y is continuous.

There is a natural dual of this construction.

Definition 13.3. Let f : X → Y be a map and U_{X} be a topology on X. Then the push forward of U_{X} is

U\overrightarrow{f} = \{U ⊂ Y : f^{−1}(U) ∈ U_{X}\}.

In other words U\overrightarrow{f} is the biggest topology on Y for which f is continuous. (X, U\overrightarrow{f}) restricted

to the complement of f(X) has the discrete topology (Show it!) so the interesting case is when f is surjective. In this case U\overrightarrow{f} is called the quotient topology by the following reasons: If f : X → Y is a surjective map then we can define an equivalence relation on X by taking

x_{1} ∼ x_{2} if f(x_{1}) = f(x_{2}). On the other hand if ∼ is an equivalence relation on X then we can

take Y := X/ ∼ (the set of equivalence classes) and we can define f : X → Y to be the surjective map sending x ∈ X to its equivalence class [x] ∈ Y. Y is usually called the quotient

of X by ∼.

Exercise 13.4. Describe the following quotient topologies:

1. X := ℝ^{2} and (x_{1}, x_{2}) ∼ (y_{1}, y_{2}) if x_{1} = y_{1}.

2. X := ℝ and x ∼ y if there is an integer n ∈ ℤ such that x = y + n.

3. X := [0, 1] and 0 ∼ 1. All other point are equivalent only to themselves.

4. X := \overline{B_{1}(0)} = D^{2} ⊂ ℝ^{2} the closed unit disc. Let (x_{1}, x_{2}) ∼ (y_{1}, y_{2}) if x_{1}^{2} +x_{2}^{2} = y_{1}^{2} +y_{2}^{2} = 1. All other point are equivalent only to themselves.

At the last two cases you can use the following fact called the Quotient Lemma:

Lemma 13.5. Suppose that f : X → Y is a continuous surjection, X is compact and Y is T_{2}. Then U_{Y} ≅ U\overrightarrow{f}.

Proof. f is continuous so U_{Y} ⊃ U\overrightarrow{f}, i.e. Id_{Y} : (Y, U\overrightarrow{f}) → (Y, U_{Y}) is continuous. (Y, U\overrightarrow{f}) is the continuous image of X therefore compact. So we can apply Proposition 8 saying that a continuous bijection from a compact space to a T_{2} space is a homeomorphism.

A special case of the quotient topology is the glueing of two spaces. The simplest is when we don’t glue at all:

Definition 13.6. The disjoint union of the topological spaces (X, U_{X}) and (Y, U_{Y}) is X ∐ Y (the disjoint union of the two sets) with the topology generated by the basis B = U_{X} ∪ U_{Y}.

Disjoint union has the following universal property:

Theorem 13.7. Let X and Y be topological spaces. Then there exists a space T unique up to homeomorphism such that there are continuous maps t_{X} : X → T and t_{Y} : Y → T such that for any continuous maps f_{X} : X → Z, f_{Y} : Y → Z there is a unique continuous map h : T → Z with the property: f_{X} = h ∘ t_{X} and f_{Y} = h ∘ t_{Y}.

The statement of the Theorem can be summarized with the commutative diagram:

\xymatrix{X \ar[r]^{t_{X}} \ar[dr]_{f_{X}} & T \ar@{-->}[d]_{h} & Y \ar[l]_{t_{Y}} \ar[dl]^{f_{Y}} \\  & Z}

Notice that we simply reversed the arrows in the commutative diagram of the product (Theorem 4.11). In other words this is a ”dual” construction. In an arbitrary category this

construction is called sum. We leave the proof as an exercise that disjoint union is the sum in the category of topological spaces and continuous maps.

Definition 13.8. Let (X, U_{X}) and (Y, U_{Y}) be topological spaces, g_{X} : A → X and g_{Y} : A → Y continuous. Then we can glue X and Y together along g_{X} and g_{Y}:

X ∪_{g_{X},g_{Y}} Y := X ∐ Y/ ∼

(where g_{X}(a) ∼ g_{Y} (a) for all a ∈ A and all other point are equivalent only to themselves) with the quotient topology. g_{X} and g_{Y} are called the glueing maps.

In most applications A ⊂ X and g_{X} is the inclusion map of A into X. Then we use the notation X ∪_{g_{Y}} Y. Of course you can glue a space to itself: If A ⊂ X and g : A → X continuous then you can take the quotient topology on X_{g} = X/ ∼ where a ∼ g(a) for all

a ∈ A and all other point are equivalent only to themselves. Notice that the last two examples in Exercise 13.4 can be obtained via glueing.

Remark 13.9. You can define glueing with a universal property corresponding to the commutative diagram:

\xymatrix{ & A \ar[dl]_{g_{X}} \ar[dr]^{g_{Y}} \\ X \ar[r]^{t_{X}} \ar[dr]_{f_{X}} & T \ar@{-->}[d]_{h} & Y \ar[l]_{t_{Y}} \ar[dl]^{f_{Y}} \\  & Z}

so glueing can be defined in any category (it may not exist of course).

Exercise 13.10. We noticed that the edge E of the Moebius band Mb is homeomorphic to the unit circle S^{1}, i.e. we have a homeomorphism h : E → S^{1}. How does Mb ∪_{f} D^{2} look like?

Homework for Section 13.

1. Solve Exercise 13.4.

2. Give a glueing function f such that I_{f}^{2} ≅ Mb the Moebius band.

3. Show that the quotient topology on X/ ∼ is T_{1} iff the equivalence classes of ∼ are closed in X.

24 LÁSZLÓ FEHÉR

4. Show that if A ∪ B = X and i_{A} : A ∩ B → A and i_{B} : A ∩ B → B are the corresponding inclusions then X ≅ A ∪_{i_{A},i_{B}} B if both A and B are open in X or both A and B are closed in X but not in general. Hint: for A and B closed this is the pasting lemma 10.2.

14. Manifolds

A particularly interesting class of topological spaces is the class of manifolds. They look like a Euclidean space locally. So they have only “global” topological properties.

Definition 14.1. The topological space M is an n-dimensional manifold if ∀m ∈ M ∃U ∋ m open such that U ≅ ℝ^{n} and M ∈ T_{2}, M_{2}.

Remarks 14.2. 1. The assumption M ∈ T_{2} is important if we want to avoid strange spaces: Let X = (0, 2), Y = (4, 6), A = (0, 2) \ {1} and f : A → Y, f(a) = a + 4. The space Z = X ∪_{f} Y is locally homeomorphic to ℝ but not Hausdorff. Z is called the interval with a double point (why?).

2. The assumption M ∈ M_{2} is less important for us. We are mainly interested in compact manifolds and for compact spaces the two other conditions imply M_{2}.

Theorem 14.3. The dimension of a manifold is a topological invariant.

This is a consequence of the deep fact that ℝ^{m} ≅ ℝ^{n} implies m = n. We will prove a special case later.

Examples 14.4.

1. 1-dimensional manifolds are called curves. The only connected compact curve is S^{1}. The only connected non compact curve is ℝ. The proof is not difficult but technical.

2. 2-dimensional manifolds are called surfaces. Examples of surfaces are ℝ^{2} and the annulus (0, 1) × S^{1}. Examples of compact surfaces are the sphere S^{2} and the torus (or donut) T^{2} = S^{1} × S^{1}. Later we give a complete list of connected compact surfaces.

3. In higher dimensions no complete list is known. This is a very active area of modern mathematics. Examples of higher dimensional manifolds: ℝ^{n}, the n-dimensional sphere S^{n}, the n-dimensional torus T^{n}. Taking products you can get more examples.

Exercise 14.5. Show that the product of two manifolds is a manifold.

Homework for Section 14.

1. Show that S^{n} is an n-dimensional manifold.

2. Prove Exercise 14.5 that the product of two manifolds is a manifold.

15. Homotopy Equivalences

Definition 15.1. Two spaces X and Y are homotopy equivalent and denoted by X ≃ Y if there are continuous maps f : X → Y and g : Y → X such that f ∘ g ≃ Id_{Y} and g ∘ f ≃ Id_{X}.

Homotopy equivalent spaces can look very different however their homotopic properties are the same.

Exercise 15.2. Find explicit maps and homotopies showing that ∗ ≃ ℝ^{n}, S^{n} ≃ ℝ^{n+1} \ {0}

and 8 ≃ Θ,

INTRODUCTION TO TOPOLOGY 25

Definition 15.3. If a space is homotopy equivalent to the one point space ∗ then is is called contractible. Theorem 15.4. Let X and Y be path connected and X ≃ Y. Then π_{1}(X) ≅ π_{1}(Y). The proof is not difficult but tricky because the homotopies involved don’t have to preserve a

base point. We prove a weaker statement instead.

Proposition 15.5. Suppose that (X, x_{0}) and (Y, y_{0}) are “based” homotopy equivalent: there are base point preserving maps f : (X, x_{0}) → (Y, y_{0}), g : (Y, y_{0}) → (X, x_{0}) and base point preserving homotopies H_{X}, H_{Y} connecting f ∘ g to Id_{Y} and g ∘ f to Id_{X}. Then π_{1}(X) ≅ π_{1}(Y). Proof. By Proposition 11.6 g_{∗} ∘ f_{∗} = (g ∘ f)_{∗} = (Id_{X})_{∗} = Id_{π_{1}(Y)}.

Homework for Section 15.

1. Prove Exercise 15.2.

2. * Prove Theorem 15.4.

16. Simply Connected Spaces

Definition 16.1. X is simply connected if path connected and π_{1}(X) = 0. Definition 16.2. D ⊂ ℝ^{n} is convex if for any x, y ∈ D the line segment \overline{xy} connecting x and y is in D. Proposition 16.3. If D ⊂ ℝ^{n} is convex then simply connected. Proof. The trick of the trade is the parametrization of \overline{xy}: Let f(t) = (1 − t)x + ty. The path f(t) shows that D is path connected. The point (1 − t)x + ty for t ∈ [0, 1]is called a convex combination of x and y. Now let l : I → D is a loop in (D, x_{0}) (where x_{0} is a base point for D). Then H(s, t) = (1 − s)l(t) + sx_{0} is a nulhomotopy.

Proposition 15.4 implies that a contractible space has trivial fundamental group. So the

following exercises give an alternative proof for Proposition 16.3:

Exercise 16.4. Show that a contractible space is path connected. Exercise 16.5. Show that if D ⊂ ℝ^{n} is convex then contractible. Theorem 16.6. The spheres S^{n} for n > 1 are simply connected. Proof. The proof is based on the following Exercise 16.7. S^{n} \ {N} ≅ R^{n} where N = (0, …, 0, 1) is the North pole. So we will show that any loop f in S^{n} can be homotoped to a loop which avoids a point of S^{n}. Then f is nulhomotopic and we are done. It is difficult even to imagine a loop covering the whole sphere however such monsters exist: Theorem 16.8 (Peano). There is a continuous surjective map p : [0, 1] → . Using such a p—called Peano curve—you can construct curves covering the cube, or S^{n} for any n (try!). Our major tool is the following

26 LÁSZLÓ FEHÉR

Lemma 16.9 (Lebesgue lemma for [0, 1]). Let l : I → X be a continuous map and let C = {U_{α} : α ∈ A} be an open cover of X. Then there are t_{0} = 0, t_{1}, …, t_{n} = 1 ∈ [0, 1] such that for every i ≤ n there is an α ∈ A with l([t_{i}, t_{i+1}]) ⊂ U_{α}.

Proof. Let C′ = {(a, b) ⊂ [0, 1] : ∃α ∈ Al(a, b) ⊂ U_{α}}. C′ is an open cover of [0, 1]. The compactness of the unit interval implies that C′ has a finite subcover {[0, b_{0}), (a_{1}, b_{1}), …, (a_{n}, 1]}. We can assume that a_{1} < b_{0} < a_{2} < b_{1} < ⋯ so we can pick the t_{i} from the ”overlap” (a_{i+1}, b_{i}).

Continuation of the proof of Theorem 16.6: Let U_{S} = S^{n} \ {N} and U_{N} = S^{n} \ {S}. Then C = {U_{S}, U_{N}} is an open cover of S^{n}. So if l : [0, 1] → S^{n} is a loop then by the Lebesgue lemma for [0, 1] we have t_{0} = 0, t_{1}, …, t_{n} = 1 ∈ [0, 1] such that l([t_{i}, t_{i+1}]) ⊂ U_{N} or l([t_{i}, t_{i+1}]) ⊂ U_{S}. The paths l|_{[t_{i},t_{i+1}]} are homotopic (keeping endpoints fixed) to a path c_{i} not containing the North pole N. With the usual pasting of these homotopies we can see that l is homotopic to the concatenation c = c_{0} · c_{1} ⋯ c_{n}. So l is nulhomotopic.

Homework for Section 16.

1. Prove Exercise 16.4.

2. Prove Exercise 16.5.

3. Prove Exercise 16.7

4. Prove the Lebesgue number lemma: If (X, d) is a compact metric space and C ⊂ U_{d} is an open cover of X then there is a δ > 0—called the Lebesgue number of the cover C—such that if A ⊂ X has diameter less then δ then there is a U ∈ C with A ⊂ U.

5. Use the Lebesgue number lemma to give an alternative proof of the Lebesgue lemma for [0, 1].

17. The Fundamental Group of the Circle

Theorem 17.1. π_{1}(S^{1}) ≅ ℤ.

This is a consequence of a general theorem about covering maps we are going to prove in the

next section. As a warm up I give a sketch of the proof of π_{1}(S^{1}) ≅ ℤ:

As we saw S^{1} ≅ ℝ/ ∼ where a ∼ b if a = b + k for k ∈ ℤ. In other words S^{1} can be identified with the modulo 1 real numbers. (From this point of view a natural base point for S^{1} is 0=1 ;-) So if we have a map f : X → S^{1} we can construct a ”lift” f̃: X → ℝ such that p ∘ f̃ = f where p is the quotient map. Such a lift is not unique and sometimes it is impossible to find a

continuous lift even if f was continuous. However we have the following:

Lemma 17.2 (path lifting). Let f : [0, 1] → S^{1} be a continuous path such that f(0) = 0. Then for any k ∈ ℤ there is a unique lift f̃: [0, 1] → ℝ such that f̃(0) = k.

We will prove this in the next section. An intuitive argument would be the following: Think ℝ as an infinite helix above S^{1} and think f as the driving instructions for our car standing at 0. We don’t see whether our car is on S^{1} or on the helix (locally they look the same). So the

directions work in both case giving f and f̃ respectively.

Lemma 17.2 allows us to assign an integer to any loop l : [0, 1] → S^{1}:

Definition 17.3. Φ(l) := l̃(1) ∈ ℤ where l̃: [0, 1] → ℝ is the unique lift of l such that l̃(0) = 0.

Proposition 17.4. Φ induces a map ϕ : π_{1}(S^{1}, 0) → ℤ.

Proof. We have to show that if l_{0} ∼ l_{1} then l̃_{0}(1) = l̃_{1}(1). For this we need the following

Lemma 17.5 (homotopy lifting). If H : → S^{1} is a based homotopy (i.e. H(s, 0) = H(s, 1) = 0 for all s ∈ [0, 1]) then there is a unique lift H̃ : → ℝ such that H̃(0, 0) = 0.

Notice that the unicity follows from the path lifting lemma applied to the paths H_{s}. We will prove the continuity part in the next section. So if H is a homotopy connecting l_{0} and l_{1}, then H̃ connects l̃_{0} and l̃_{1}. In particular H̃(s, 1) is a path connecting l̃_{0}(1) and l̃_{1}(1). Since H̃ is a lift H̃(s, 1) has to be entirely in p^{−1}(0) = ℤ ⊂ ℝ so it is the constant path: l̃_{0}(1) = l̃_{1}(1).

Exercise 17.6. Show that ϕ is a homomorphism.

Proposition 17.7. ϕ : π_{1}(S^{1}, 0) → ℤ is an isomorphism.

Proof.

1. ϕ is surjective: Let l_{k}(t) = [kt] ∈ S^{1} for k ∈ ℤ then l̃_{k}(t) = kt so ϕ[l_{k}] = l̃_{k}(1) = k.

2. ϕ is injective: Since ϕ is a homomorphism it is enough to show that if ϕ[l] = 0 then l ∼ \underline{0}. But ϕ[l] = 0 means that l̃(1) = 0. Let H̃ : → ℝ be defined by H̃(s, t) = sl̃(t). H̃ is nulhomotopy for l̃ so H := p ∘ H̃ is a nulhomotopy for l.

18. Covering Spaces and Maps

We used some special properties of the quotient map p : ℝ → S^{1}. The proper generalization is the notion of covering map. We follow Munkres: Topology (p. 336) for a while:

Definition 18.1. Let p : X̃ → X be a continuous surjective map. An open set U ∈ X is evenly covered by p if p^{−1}(U) is the disjoint union of open subsets V_{α} of X̃ such that for each α the restriction p|_{V_{α}} : V_{α} → U is a homeomorphism.

Definition 18.2. Let p : X̃ → X be a continuous surjective map. If X is covered by open subsets evenly covered by P then p is called a covering map and X̃ is said to be a covering space of X.

Exercise 18.3.

1. Show that the quotient map p : ℝ → S^{1} is a covering map.

2. Show that if p : X̃ → X is a covering map then for all x ∈ X the preimage p^{−1}(x) has the discrete topology.

3. Show that if p : X̃ → X is a covering map and X is connected then p^{−1}(x) has the same number of points for all x ∈ X. This number is called the number of sheets of p.

4. Find a 3-sheeted covering of S^{1}. OK, find an other one ;-)

5. Find a covering map p : ℝ^{2} → S^{1} × S^{1}.

Definition 18.4. Let p : X̃ → X be a map. If f : Z → X is a continuous map then a lift of f is a map f̃: Z → X̃ such that p ∘ f̃ = f.

Lemma 18.5 (path lifting). Let p : X̃ → X be a covering map such that p(x̃_{0}) = x_{0}. Then any path f : [0, 1] → X beginning at x_{0} has a unique lift f̃: [0, 1] → X̃ beginning at x̃_{0}.

28 LÁSZLÓ FEHÉR

Proof. Existence of a lift: Apply the Lebesgue lemma for [0, 1] to the open cover of X provided by the definition of covering space. Let t_{0} = 0, t_{1}, …, t_{n} = 1 ∈ [0, 1] such that f([t_{i}, t_{i+1}]) ⊂ U_{i} for an evenly covered open set U_{i} ⊂ X. Let f̃(0) = x̃_{0}. Then suppose that f̃ is defined for 0 ≤ t ≤ t_{i}. By assumption p^{−1}(U_{i}) is the disjoint union of open sets V_{α} such that for each α the restriction p|_{V_{α}} : V_{α} → U is a homeomorphism. f̃(t_{i}) lies in one of these, say in V_{0}. Define f̃(t) := p|_{V_{0}}^{−1}(f(t) for t ∈ [t_{i}, t_{i+1}]. Continuing this way we define f̃ on the whole interval. By the pasting lemma f̃ is continuous. Unicity: Suppose we have an other lift g̃ of f. Then let t_{i} be the largest of the division points such that g̃(t) = f̃(t) for t ∈ [0, t_{i}]. But then g̃ and f̃ has to agree on [t_{i}, t_{i+1}] as well since the image should stay in the same connected component of p^{−1}(U_{i}).

Lemma 18.6 (homotopy lifting). If H : → X is a based homotopy (i.e. H(s, 0) = H(s, 1) = x_{0} for all s ∈ [0, 1]) then there is a unique lift H̃ : → X̃ such that H̃(0, 0) = x̃_{0}.

Notice that the unicity follows from the path lifting lemma applied to the paths H_{s}. We will prove the continuity part using a version of the Lebesgue lemma for [0, 1]:

Lemma 18.7 (Lebesgue lemma for ). Let H : I × I → X be a continuous map and let C = {U_{α} : α ∈ A} be an open cover of X. Then there are s_{0} = 0, s_{1}, …, s_{m} = 1 ∈ [0, 1] and t_{0} = 0, t_{1}, …, t_{n} = 1 ∈ [0, 1] such that for every i ≤ m, j ≤ n there is an α ∈ A with H([s_{i}, s_{i+1}] × [t_{j}, t_{j+1}]) ⊂ U_{α}.

Proof. Let

C′ = {(a_{1}, b_{1}) × (a_{2}, b_{2}) ⊂ [0, 1] : ∃α ∈ A such that H((a_{1}, b_{1}) × (a_{2}, b_{2})) ⊂ U_{α}}.

C′ is an open cover of [0, 1]. The compactness of the unit square implies that C′ has a finite subcover. Slightly decrease the size of the rectangles occuring in the finite subcover. Let s_{0} = 0, s_{1}, …, s_{m} = 1 ∈ [0, 1] be the possible s-coordinates for the corners of these rectangles and t_{0} = 0, t_{1}, …, t_{n} = 1 ∈ [0, 1] be the possible t-coordinates for the corners of these rectangles.

Exercise 18.8. Explain what ”slightly” means exactly in the previous proof.

Proof of Lemma 18.6. Apply the Lebesgue lemma for to the open cover of X provided by the definition of covering space. Let R_{i} the i^{th} rectangle, where we ordered the rectangles column by column. We can define the lift H̃ by induction if we notice that L_{n+1} = R_{n+1} ∪ ∩_{i=1}^{n}R_{i} is connected (one or two sides of R_{n+1}) so if U ⊃ H(R_{n+1}) is evenly covered i.e. p^{−1}(U) = ∐ V_{α} then H̃(L_{n+1}) ⊂ V_{α} for some α so we can define H̃|_{R_{n+1}} by (p|_{V_{α}})^{−1} ∘H|_{R_{n+1}}. Continuity follows from the pasting lemma.

Now we actually finished the proof of that π_{1}(S^{1}) ≅ ℤ, but we can get a more general statement. For this we generalize the map ϕ of the previous section:

Proposition 18.9. Let p : (X̃, x̃_{0}) → (X, x_{0}) be a covering map and l : [0, 1] → X be a loop. Let Φ(l) = l̃(1) ∈ p^{−1}(x_{0}) where l̃: [0, 1] → X̃ is the unique lift of l such that l̃(0) = x̃_{0}. Then Φ induces a map ϕ : π_{1}(X, x_{0}) → p^{−1}(x_{0}) called the lifting correspondence.

Proof is the same as of Proposition 17.4.

Theorem 18.10. Let p : (X̃, x̃_{0}) → (X, x_{0}) be a covering map and ϕ : π_{1}(X, x_{0}) → p^{−1}(x_{0}) be the lifting correspondence. If X is path connected then ϕ is surjective. If X is simply connected then ϕ is bijective.

Proof is the same as of Proposition 17.7.

19. Compact surfaces

Definition 19.1. A surface is non-oriented if it has a subspace homeomorphic to a Moebius band and oriented if it doesn’t have.

The projective plane is the first example of a compact non oriented surface. Difficult to imagine since it cannot be embedded into ℝ^{3}.

Definition 19.2.

ℝℙ^{2} := {lines in ℝ^{3} through origin}.

ℝℙ^{2} is a metric space called projective plane with the metric d(l_{1}, l_{2}) = ∢(l_{1}, l_{2}) where ∢(l_{1}, l_{2}) is measured between 0 and π/2.

Exercise 19.3. Show that (ℝℙ^{2}, U_{∢}) is homeomorphic to 1. S^{2}/ ∼ where a ∼ −a (think S^{2} ⊂ ℝ^{3} the unit sphere).

2. D^{2}/ ∼ where a ∼ −a if |A| = 1.

3. Mb ∪_{ϕ} D^{2} where ϕ : ∂Mb → S^{1} ⊂ D^{2} is a homeomorphism.

4. S^{1} ∪_{ψ} D^{2} where ψ : S^{1} → S^{1}, ψ(t) = [2t].

Proposition 19.4. π_{1}(ℝℙ^{2}) ≅ ℤ_{2}.

Proof. The quotient map p : S^{2} → ℝℙ^{2} is a 2-sheeted covering map, S^{2} is simply connected so Theorem 18.10 implies that π_{1}(ℝℙ^{2}) has two elements. The only group with two elements is ℤ_{2}.

All compact connected surfaces can be obtained from the sphere S^{2}, the torus T^{2} and the projective plane ℝℙ^{2} using connected sum:

Definition 19.5. If M and N are n-dimensional manifolds then their connected sum

M#N = M \ B^{n} ∪_{ϕ} N \ B^{n},

where B^{n} is a small open disc and ϕ is a homeomorphism between the boundaries of the two discs (they are both homeomorphic to S^{n−1}).

Theorem 19.6.

1. A compact connected oriented surface is homeomorphic to the sphere S^{2} or to Σ_{g} =

\overbrace{T^{2}# ⋯ #T^{2}}^{g} and these are all non-homeomorphic surfaces.

2. A compact connected non-oriented surface is homeomorphic to K_{g} = \overbrace{ℝℙ^{2}# ⋯ #ℝℙ^{2}}^{g} and these are all non-homeomorphic surfaces.

Remark 19.7. We will show later that these surfaces all have different fundamental groups, so they are not even homotopy equivalent. We will not prove however that this is a complete list.

Example 19.8. K_{2} = ℝℙ^{2}#ℝℙ^{2} is called the Klein bottle. In other words K_{2} = Mb ∪_{ϕ} Mb where ϕ glues the two Moebius bands together along their boundary.

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Homework for Section 19

1. Prove Exercise 19.3.

2. What is / ∼ if (a) (s, 0) ∼ (s, 1) and (0, t) ∼ (1, t),

(b) (s, 0) ∼ (s, 1) and (0, t) ∼ (1, 1 − t),

(c) (s, 0) ∼ (1 − s, 1) and (0, t) ∼ (1, 1 − t).

3. What is S#S^{2} if S is any surface?

4. Where is ℝℙ^{2}#T^{2} on the list?

20. Group Actions

We say that a group acts on the set X if for any g ∈ G we have a map ρ(g) : X → X such that ρ(e) = Id_{X} and ρ(g_{1}g_{2}) = ρ(g_{1})ρ(g_{2}). Usually the action ρ is not mentioned and we think

g as a self map of X. If there is a topology on X then we also require the maps ρ(g) to be continuous (therefore homeomorphism). So we can say that ρ is a homomorphism from the

group G to Homeo(X) the group of self homeomorphisms of X.

If G also admits a topology then you want ρ to be continuous. Since we haven’t define topology on Homeo(X) we can use the following:

Definition 20.1. Let G be a group with a topological structure and X be a topological space. An action of G on X is a continuous map μ : G × X → X such that μ(e, x) = x and μ(g_{1}, μ(g_{2}, x)) = μ(g_{1}g_{2}, x).

We are mainly concerned about the case when the topology on G is discrete. In this case the two definitions are equivalent.

Definition 20.2. Suppose that G acts on X. Then we can define an equivalence relation on X by x ∼ y if ∃g ∈ G such that y = g(x). (Check that this is indeed an equivalence relation.) The equivalence classes of ∼ are called orbits and X/ ∼ is called the orbit space and denoted by X/G. If X was a topological space then we have the quotient topology on X/G .

Exercise 20.3.

1. Find a ℤ-action and a ℤ_{2}-action on ℝ.

2. Find a ℤ_{n}-action on S^{1} and on S^{2}.

We look at closely to two examples:

1. ℝ/ℤ ≅ S^{1}, the quotient map is exactly the covering map we studied to calculate π_{1}(S^{1}) ≅ ℤ.

2. S^{2}/ℤ_{2} ≅ ℝℙ^{2}, the quotient map is exactly the covering map we studied to calculate π_{1}(ℝℙ^{2}) ≅ ℤ_{2}.

You may wonder whether this is a coincidence that the group acting is the same as the the fundamental group. Of course not, but we need that the quotient map has to be a covering

map as well.

Definition 20.4. The action of G on X is called a covering action (also called properly discontinuous) if for all x ∈ X there is an open set U ∋ x such that g(U) ∩ U = ∅ if g ≠ e.

Proposition 20.5. The quotient map q : X → X/G is a covering map for any covering action.

Proof. Let y ∈ X/G and x ∈ X such that q(x) = y. Then take V = q(U) for the U ∋ x in the definition of covering action. We claim that V is evenly covered. It follows from the equality

q^{−1}(V) = q^{−1}(q(U)) = ∐_{g∈G} g(U)

and the definition of quotient topology since all the sets g(U) are open (g^{−1} is continuous!).

Remark 20.6. It is easy to see that a covering action is always a free action, i.e. g(x) = x implies g = e. So a covering action defines a |G|-sheeted covering. Applying Theorem 18.10 we get that if X is simply connected then there is a bijection between the groups π_{1}(X/G) and G.

Theorem 20.7. Suppose that the action of the group G is a covering action on the simply connected space X. Then π_{1}(X/G) ≅ G.

Proof. Pick base points x_{0} ∈ X and y_{0} = q(x_{0}) ∈ X/G. Since the action is free we have a bijection m : q^{−1}(y_{0}) → G such that m(x) = g if x = g(x_{0}). We claim that h = m ∘ ϕ : π_{1}(X/G, y_{0}) → G is a homomorphism therefore an isomorphism where ϕ : π_{1}(X/G, y_{0}) → q^{−1}(y_{0}) is the lifting correspondence. To see that it is enough to show that if a, b : [0, 1] → X/G are loops with lifts ã, b̃ : [0, 1] → X, then ã · h[a](b̃) is a lift for ã · b̃. Here h[a](b̃) denotes the path t ↦ h[a](b̃(t)). By the definition of h we have that h[a](x_{0}) = ã(1) which shows that the concatenation ã · h[a](b̃) is continuous. It is a routine to check that ã · h[a](b̃) is a lift of a · b, so ϕ([a][b]) = (ã · h[a](b̃))(1) = h[a](b̃(1)) = h[a](ϕ([b])) = h[a](h[b](x_{0})) which means that h is a homomorphism.

Every reasonable path connected space X has a simply connected cover X̃ called the universal cover. Moreover there is a covering action of π_{1}(X) on X̃. This gives a way to determine the fundamental group of any reasonable space. However the construction of the universal cover

is not too constructive so we need an other method to calculate fundamental groups.

Homework for Section 20

1. Show that the formula (k, l)(x, y) = (k + x, l + y) defines a covering action of ℤ × ℤ on ℝ^{2}. Use this result to calculate the fundamental group of something.

2. Let X = D^{2} ∐ D^{2} ∐ D^{2}/ ∼ where all three boundary circles are identified. You can call X the 3-layered disc or the pita bread with two pockets. Define a covering action of ℤ_{3} on X. (Later we will see that X is simply connected, so X/ℤ_{3} has fundamental group ℤ_{3}.)

21. Seifert-van Kampen Theorem

Goal: Calculate π_{1}(A ∪ B) where A and B are open in A ∪ B if we know π_{1}(A), π_{1}(B) and π_{1}(A ∩ B). This way you can calculate the fundamental group of any space ”built up” from

simpler pieces.

Idea: By the Lebesgue lemma for [0, 1] any loop γ in A ∪ B is homotopy equivalent to a “factorized” loop γ_{1} · γ_{2} ⋯ γ_{n}, where γ_{i} : [0, 1] → A or γ_{i} : [0, 1] → B. So π_{1}(A ∪ B) is going to

be a group generated by elements of π_{1}(A) and π_{1}(B) subject to the condition that loops in A ∩ B are considered to be the same no matter if we think them as elements of π_{1}(A) or π_{1}(B).

The algebraic construction: By Homework 4 in Section 13 we can consider A ∪ B as A and B glued together along A ∩ B. So we can expect that π_{1}(A ∪ B) is going to be the

32 LÁSZLÓ FEHÉR

glueing of π_{1}(A) and π_{1}(B) in the category of groups and homomorphisms called

amalgamation or amalgamated product. The Seifert-van Kampen theorem says that under some conditions this is indeed the case:

Definition 21.1. Suppose we have groups G, H, S and homomorphisms g : S → G, h : S → H. Then we denote by G ∗_{S} H the amalgamation i.e. the unique group satisfying the universal property corresponding to the commutative diagram:

\xymatrix{ & S \ar[dl]_{g} \ar[dr]^{h} \\ X \ar[r]^{t_{g}} \ar[dr]_{f_{g}} & G ∗_{S} H \ar@{-->}[d]_{α} & Y \ar[l]_{t_{h}} \ar[dl]^{f_{h}} \\  & Z}

It would be more precise to have the maps g and h in the notation but this is the traditional notation.

Theorem 21.2 (Seifert-van Kampen). Let X = A ∪ B s.t. A, B are open in X and A, B, A ∩ B are all path connected. Let x ∈ A ∩ B and suppose that x is the base point for all fundamental groups appearing. Then

π_{1}(A ∪ B) ≅ π_{1}(A) ∗_{π_{1}(A∩B)} π_{1}(B),

where we use the maps a_{∗} : π_{1}(A ∩ B) → π_{1}(A) and b_{∗} : π_{1}(A ∩ B) → π_{1}(B) induced by the inclusions a : A ∩ B → A and b : A ∩ B → B.

We defined amalgamation using the universal property of glueing. It is not easy to see how

this group looks like. Let me list some group theoretic facts—mainly without proof—to understand the structure of amalgamation better. The main tool will be presenting a group

with generators and relations. Generators and relations:

Examples 21.3.

1. ℤ_{2} = ⟨g : g^{2} = 1⟩,

2. ℤ × ℤ = ⟨g, h : ghg^{−1}h^{−1} = 1⟩,

3. D_{3} = ⟨g, h : g^{3} = 1, h^{2} = 1, hgh^{−1}g = 1⟩, where D_{3} is the symmetry group of the equilateral triangle.

A formal definition would be:

Definition 21.4. Let g_{1}, …, g_{n} be different letters and r_{1}, …, r_{k} are words using the letters g_{1}, …, g_{n}, g_{1}^{−1}, …, g_{n}^{−1}. The group G = ⟨g_{1}, …, g_{n} : r_{1} = 1, …, r_{k} = 1⟩ is the set of equivalence classes of words using the letters g_{1}, …, g_{n}, g_{1}^{−1}, …, g_{n}^{−1}. Two such words w_{1}, w_{2} are equivalent if w_{1} can be transformed into w_{2} using finitely many steps of the following types:

1. a “syllable” of the form g_{i}g_{i}^{−1} can be deleted or inserted anywhere.

2. a “syllable” of the form r_{i} can be deleted or inserted anywhere.

Remark 21.5. The number of generators or relations can be infinite as well.

INTRODUCTION TO TOPOLOGY 33

Theorem 21.6. Concatenation induces a group structure on G. Remark 21.7. If there are no relations, the group F_{n} = ⟨g_{1}, …, g_{n}⟩ is called the free group with n generators. F_{1} ≅ ℤ, the other free groups are not Abelian. Every group can be described with generators and relations. A not so economic way is to list

all elements as generators and all products which happen to be 1 in this group as relations. Theorem 21.8. G = ⟨g_{1}, …, g_{n} : r_{1} = 1, …, r_{k} = 1⟩ ≅ F_{n}/N, where N is the normal subgroup generated by the relations r_{1}, …, r_{k}. Corollary 21.9. Every group is a factor of a free group.

Suppose now that we have homomorphisms g, h from a group S to

G = ⟨g_{1}, …, g_{n} : r_{1} = 1, …, r_{k} = 1⟩ and H = ⟨h_{1}, …, h_{m} : q_{1} = 1, …, q_{l} = 1⟩.

Theorem 21.10. The amalgamation G ∗_{S} H of G and H over S can be presented as G ∗_{S} H = ⟨g_{1}, …, g_{n}, h_{1}, …, h_{m}|r_{1} = 1, …, r_{k} = 1, q_{1} = 1, …, q_{l} = 1, g(s)h(s)^{−1} = 1 : s ∈ S⟩.

Sum in the category of groups and homomorphisms is called free product.

Corollary 21.11. The free product G ∗ H of G = ⟨g_{1}, …, g_{n} : r_{1} = 1, …, r_{k} = 1⟩ and H = ⟨h_{1}, …, h_{m} : q_{1} = 1, …, q_{l} = 1⟩ is

G ∗ H = ⟨g_{1}, …, g_{n}, h_{1}, …, h_{m} : r_{1} = 1, …, r_{k} = 1, q_{1} = 1, …, q_{l} = 1⟩.

So you can see that free groups are free products of copies of the group ℤ. This corollary also

shows that any element of G ∗ H can be written as a product g_{1}h_{1} ⋯ g_{n}h_{n}.

Definition 21.12. A word using letters which are non unit elements of G or H is called reduced word if no consecutive letters are from the same group. Theorem 21.13. If a ∈ G ∗ H then there is a unique reduced word representing a.

The theorem generalizes to the free product of more that two groups.

Corollary 21.14. The free group F_{2} is not commutative. Proof. xy and yx are different reduced words in F_{2} = ⟨x⟩ ∗ ⟨y⟩. Exercise 21.15. Show that ℤ_{2} ∗ ℤ_{2} is the semidirect product of ℤ_{2} and ℤ. Example 21.16. Let G ≅ ℤ, H ≅ 1 (the trivial group), S ≅ ℤ, g(n) := 7n and h is the trivial map. Then G ∗_{S} H ≅ ℤ_{7}. Similarly you can show that if H is trivial then G ∗_{S} H ≅ G/⟨Im g⟩, where ⟨Im g⟩ denotes the

normal subgroup generated by the image of g.

Remark 21.17. Im g is usually not a normal subgroup. Exercise 21.18.

1. G ≅ F_{2}, H ≅ 1, S ≅ ℤ and g(1) = aba^{−1}b^{−1}. G ∗_{S} H ≅?

2. G ≅ H ≅ ℤ, g(1) = m and h(1) = n. G ∗_{S} H ≅? After this detour in algebra we turn to the proof of the Seifert-van Kampen theorem:

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Proof of Theorem 21.2. We want use the universal property of amalgamation, so suppose we have homomorphisms f_{A} : π_{1}(A) → Z and f_{B} : π_{1}(B) → Z to an arbitrary group Z. Then we have to show that there is a unique homomorphism α : π_{1}(A ∪ B) → Z such that the following diagram commutes:

\xymatrix{ & π_{1}(A ∩ B) \ar[dl]_{a_{∗}} \ar[dr]^{b_{∗}} \\ π_{1}(A) \ar[r]^{t_{A}} \ar[dr]_{f_{A}} & π_{1}(A ∪ B) \ar@{-->}[d]_{α} & Y \ar[l]_{t_{B}} \ar[dl]^{f_{B}} \\  & Z}

where t_{A} and t_{B} are induced by the inclusions A → A ∪ B and B → A ∪ B respectively. First we show the uniqueness of α:

Lemma 21.19. The image of t_{A} and t_{B} generates π_{1}(A ∪ B).

Proof. The Lebesgue lemma for [0, 1] implies that any loop γ in A ∪ B has a factorization γ = p_{1} · p_{2} ⋯ p_{n} where p_{i} are paths in A or B. Using the path connectedness of A ∩ B we can choose paths q_{i} connecting the base point x with the division points. Inserting “round trips” back to x into γ along the paths q_{i} we get a factorization of [γ] ∈ π_{1}(A ∪ B): We get that [γ] = [γ_{1}][γ_{2}] ⋯ [γ_{n}] where γ_{i} are loops in A or B.

So the only possible choice is α([γ]) = f_{A}([γ_{1}])f_{B}([γ_{2}]) ⋯ f_{B}([γ_{n}]). Here for simplicity we assumed that γ_{1} is a loop in A, γ_{2} is a loop in B and so on. The difficult part of the proof is to show that this is a correct definition of α, i.e. if we choose an other factorization of γ or an other representant of [γ] then we get the same result. Independence of factorization: There are two things to check. One is what happens if γ_{i} is a loop in A ∩ B and we switch from f_{A}([γ_{i}]) to f_{B}([γ_{i}]) in the product. But the diagram implies that f_{A}([γ_{i}]) = f_{B}([γ_{i}]). The other is what happens if we add an extra division point. I leave this case as an (easy) exercise. So we have now a well defined map α : {loops in A ∪ B} → Z, and we have to check that α depends only the homotopy class of the loops: Independence of the choice of representant: Suppose that [δ] = [γ] ∈ π_{1}(A ∪ B) i.e. δ ∼_{H} γ. Apply the Lebesgue lemma for to the homotopy H : [0, 1] × [0, 1] → A ∪ B. So we get a subdivision of [0, 1]×[0, 1] to smaller rectangles {R_{i} : i = 1, ⋯, k}. Using this subdivision we can factorize H, i.e. we construct a sequence of loops γ(0) = \underline{x} · γ, γ(1), ⋯, γ(k) = δ · \underline{x} (I denoted the constant loop by \underline{x}) such that γ(i) is the restriction of H to the piecewise linear curve “cutting down the first i rectangles”, see figure:

It is easy to see that α(γ) = α(\underline{x} · γ) and α(δ) = α(δ · \underline{x}). What left is to show that α(γ(i)) = α(γ(i + 1)). We can choose a factorization of γ(i) and γ(i + 1) such that all factor are the same except one, the different factor is homotopic in A (or B) so the value of f_{A} (or f_{B}) is the same. And we finished the proof of the Seifert-van Kampen theorem.

22. The Dirac experiment and the no-wind theorem.

The Dirac experiment. Moving an object like the coat hanger in space defines a path of isometries of the 3-dimensional space ℝ^{3}. If we keep one point of the object fixed (think of it

to be the origin of a 3-dimensional vector space) then the path is in the space of linear isometries of ℝ^{3}: This is the group usually denoted by O(3) (Check that this is a group

indeed). Elements of O(3) can be represented by 3 × 3 matrices, so O(3) ⊂ ℝ^{9} and admit a natural topology. In fact by moving an object we can only realize orientation preserving isometries: your left shoe never becomes a right shoe though they are isometric. It means

that our path is in the group of orientation preserving linear isometries denoted by SO(3). If at the end the coat hanger arrives to the same position, then this path is a loop. So the Dirac

experiment suggests that π_{1}(SO(3)) has an order two element.

Theorem 22.1. Any isometry X ∈ SO(3) is a rotation along an axis through origin.

The proof can be found in most geometry or linear algebra textbook. So we can describe an element of SO(3) by giving an axis and an angle. Let

c : B̄_{π}(0) ≅ D̄^{3} → SO(3) defined by sending v ∈ B̄_{π}(0) to the rotation around the axis spanned by the vector v rotating with angle |v|. It is not difficult but technical to see that c is

continuous so we omit the proof.

Exercise 22.2. Show that c is one to one in the interior B_{π}(0) but sends antipodal points on the boundary sphere to the same element in SO(3).

Since B̄_{π}(0) is compact and SO(3) is T_{2} we can apply the quotient lemma 8 to show that SO(3) ≅ D̄^{3}/ ∼ where x ∼ −x on the boundary sphere.

Corollary 22.3. SO(3) ≅ ℝℙ^{3}.

The proof is left to the reader. Since π_{1}(ℝℙ^{3}) ≅ ℤ_{2}, the next exercise gives the explanation for the Dirac experiment.

Exercise 22.4. Show that making a full rotation along a fixed axis defines a loop in SO(3) which represents the non-trivial element in π_{1}(SO(3)) ≅ ℤ_{2}.

The no-wind theorem. Suppose there is a wind blowing everywhere on Earth. It means that we can assign a nonzero tangent vector to every point of the sphere S^{2} continuously. By normalizing these vectors we can assume that these are unit vectors. Such an object is usually

called a unit tangent vector field.

Theorem 22.5 (No-wind theorem). There is no continuous unit tangent vector field on the sphere.

Let T_{1}S^{2} denote the space of unit tangent vectors (define a topology on this set!). An element of T_{1}S^{2} can be described by a point p in S^{2} and a direction in the tangent plane of S^{2} at p. If a unit tangent vector field is given then this direction can be described via the angle between

36 LÁSZLÓ FEHÉR

the direction and the vector given by the vector field. In other words we get a map

T_{1}S^{2} → S^{2} × S^{1}. It is not difficult to see that this map would be a homeomorphism. We know that π_{1}(S^{2} × S^{1}) ≅ ℤ so to prove the no-wind theorem it is enough to show that π_{1}(T_{1}S^{2}) ≅ ℤ̸.

Theorem 22.6. SO(3) ≅ T_{1}S^{2}.

Proof. The first observation is that the image of the standard basis of ℝ^{3} uniquely determines a linear transformation. Since the image of the standard basis under an orientation preserving isometry is a right handed orthonormal basis (RHONB) we get that SO(3) ≅ {RHONB’s}. Now we define a map h : T_{1}S^{2} → {RHONB’s}: If (p, v) ∈ T_{1}S^{2} then let b_{1} = p, b_{2} = v and b_{3} = p × v the unique unit vector forming a RHONB with p and v. It is not difficult to check that h is a homeomorphism. So we showed that π_{1}(T_{1}S^{2}) ≅ ℤ_{2} and proved the no-wind theorem.