MAS435 / MAS6370 Algebraic Topology Part A: Semester 1 Lecturer: Miss Magdalena Kedziorek

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1 MAS435 / MAS6370 Algebraic Topology Part A: Semester 1 Lecturer: Miss Magdalena Kedziorek Notes by Dr E Cheng Weekly tests Once a week (every Monday) at the beginning of the lecture there will be a quick test of some definitions, theorems, examples and counterexamples from the previous week. This will count as 20% of your final mark. The exam counts as 80%. Homeworks Weekly exercises are available online. Contents 1 Introduction and Motivation Topology and algebra Algebraic topology Homotopy Examples of spaces we will consider Examples of groups we will be thinking about Questions about the fundamental group The eventual goals of algebraic topology Reminder on metric spaces Basic definitions Topological spaces Spaces and continuous maps Examples of continuous maps Paths and loops Homotopy 25

2 CONTENTS 2 5 The fundamental group Definition Dependence on basepoint Categories Homotopy invariance The circle Covering spaces Definitions The classification of covering spaces Universal covering spaces Covering space constructions Van Kampen s Theorem Free product of groups Glueing spaces Equivalence relations on groups Van Kampen s Theorem Generators and relations Cell complexes The classification of surfaces Applications The fundamental theorem of algebra The Brouwer fixed point theorem Plan Week 1: Section 1 Week 2: Sections 2 and 3 Week 3: Section 4 Week 4: Section 5 (first half) Week 5: Section 5 (second half) Week 6: Section 6 (first half) Week 7: Reading Week Week 8: Section 6 (second half) Week 9: Section 7 Week 10: Section 8 Week 11: Revision Each week you should study the previous week s notes and the next week s notes.

3 1 Introduction and Motivation 3 1 Introduction and Motivation The idea of algebraic topology is to study topology using algebra. Many of the greatest advances in mathematics have come from finding a link between two apparently different subjects enabling us to use our understanding of one to help our understanding of the other. Often the flow of understanding starts in one direction and but is later discovered to go in both directions. Topology Algebra 1.1 Topology and algebra Topology Topology is about space, shape and form. You ve already seen metric spaces before, but we want to move away from the notion of distance. For example: How do you get from A to B? Do you need to fly? This is a separate issue from how far it is. Think about things made of plasticine that we can stretch and shrink but we don t want to make new holes or stick things together. Stretching changes the distance between points but we don t care. We re not worried about size, e.g. are fundamentally the same. We re not worried about curvature, so is the same as We need a better notion of space that doesn t depend on having a ruler.

4 1.2 Algebraic topology 4 Example 1.1. A city is made up of neighbourhoods, but these are not defined by neat circular areas around points. Algebra Algebra is about operations on things. Here we will mainly be thinking about groups but later in the subject there are modules, chain complexes, groupoids, and more. 1.2 Algebraic topology The idea is to find a way of relating topology to algebra, and then, amazingly, we can use algebra to help us understand space. Some examples of applications include robotic arms computers crashing polymers ties. Perhaps even more amazingly, we can go the other way and actually use space to help us understand algebra. For example we will prove the Fundamental Theorem of Algebra. There are many ways to convert space into algebra. The main methods used in algebraic topology are called homotopy, homology, and cohomology. We will almost exclusively think about homotopy in this part of the course (Part A).

5 1.3 Homotopy Homotopy Example 1.2. What if someone told you that you were living on the surface of a torus? Here are some questions we can ask ourselves about the space we inhabit. 1. Can we get from A to B without flying? 2. How many different ways are there of getting from A to B? 3. Are those ways really different? 4. If we all hold hands in a circle can we all walk to the same place? That last one might sound a bit silly in this context, but it turns out that considering loops like this tells us tons of stuff about the space we re in. This idea turns out to be the same as the idea of bending and stretching spaces around, and this is the notion of homotopy. The key to using these loops is: The loops form a group. This is called the fundamental group and helps us tell spaces apart. However, this is only the beginning. For example, it can t tell higher-dimensional spheres apart we need higher-dimensional methods to to that, and it s very hard. However, the fundamental group is extremely powerful. For example, it enables us to classify surfaces completely. Surfaces are one of the most important examples of spaces that we ll look at. Exercise 1.3. Which of the following spaces are the same in the sense that a doughnut is the same as a teacup? 1. 2.

6 1.4 Examples of spaces we will consider Any of the above spaces and (a) a pair of trousers (b) a cardigan (c) a sweater (d) a skirt (e) a woolly hat (f) a Klein bottle 1.4 Examples of spaces we will consider 1. Discs/balls D n or B n : not very interesting as they re the same as a point. But they are very useful building blocks. 2. The unit interval I = [0, 1]. Also the same as a point, but also a crucial building block. If you were stuck on a desert island with nothing but a unit interval, you could (in some sense) built any topological space, provided you knew how to make products, disjoint unions and quotients. 3. Circles/spheres S n : these are very interesting and to date their homotopy is still not completely understood in high dimensions. 4. The Möbius strip (also known as Möbius Band): it s somehow the same as a circle, and yet The torus and its many-holed cousins. These are the orientable surfaces. 6. Non-orientable surfaces such as the Klein bottle and more complicated versions. 7. We can cut holes in a surface and glue on a Möbius Band or handle. We get the following mysterious equation handle + MB = 3MB = KB + MB.

7 1.4 Examples of spaces we will consider 7 8. Letters of the alphabet: A D B C E F 9. We can stick circles together e.g. 10. We can also stick circles together in a pathological way e.g. 11. We can stick spheres together e.g. 12. We can mix things up, and stick anything to anything else e.g. 13. Real projective space is defined by RP n = S n / where is the equivalence relation that sticks antipodal points together. This is hard to visualise, especially for n 3. The formal methods of studying spaces really come into their own when we re dealing with spaces we can t draw/visualise, so we can t rely on our spatial intuitions any more.

8 1.4 Examples of spaces we will consider We can take products e.g. R R = R 2 I I = I 2 B 2 I I I = I 3 B 3 I R I S 1 cylinder S 1 S 1 torus S 1 R 15. Knots and links can be studied via their complements e.g. The complement of is like The complement of is like The complement of is like

9 1.4 Examples of spaces we will consider We can take unions and intersections. This might not sound so interesting, but will be very important later. 17. We can take a polygon and glue its edges together according to some scheme this is a form of quotient e.g. 18. Other quotients e.g. We can take I I and identify the entire boundary with a point. What does this give us? We can take D 2 D 2 and identify the boundaries. What does this give us? We can take MB D 2 and identify the boundaries. We can quotient R by the relation x y y x Z. We can quotient R R by the relation (x, y) (x, y) x x Z.

10 1.5 Examples of groups we will be thinking about Examples of groups we will be thinking about 1. Z 2. Z n = Z/nZ 3. Z Z, Z n = Z Z 4. Z Z, Z Z 5. quotient groups 6. abelianisation. 1.6 Questions about the fundamental group 1. How do we actually calculate it? We will build up techniques to help us understand groups in terms of simpler ones, for example: covering spaces: wrapping spaces up with other spaces Van Kampen s theorem: glueing spaces together to make other spaces The crucial starting point will be to understand the fundamental group of the circle. 2. How does the fundamental group help us? We can t classify all homotopy types using the fundamental group, for example for all n 2 the fundamental group of S n is the trivial group. However, we can classify all surfaces using the fundamental group. That is, we can show that any surface is entirely determined up to homotopy by its fundamental group. However, for spaces in general we need something much more complicated: homotopy equivalence same fundamental group but not the converse. This is why further methods are introduced such as higher homotopy groups, and homology/cohomology in all dimensions.

11 1.7 The eventual goals of algebraic topology The eventual goals of algebraic topology Arguably, the eventual goal of algebraic topology is to use algebra to classify space completely, up to homotopy. (Some would say the real goal is to classify space up to homeomorphism but that is much harder, possibly unrealistically so.) That is, the aim is to find some algebraic structures that correspond to topological spaces in an extremely comprehensive way. This is a field of research that is still wide open. In the last 15 years or so, a whole new type of algebra called ω-categories has been developed to try to achieve this goal, and many arguments are going on about what will be the most fruitful way to do this.

12 2 Reminder on metric spaces 12 2 Reminder on metric spaces 2.1 Basic definitions Exercise 2.1. What can you remember about metric spaces? What do you think we might retain if we ditch the notion of metric? Definition 2.2. A metric space is a set X equipped with a metric such that d : X 2 R 0 1. d(x, y) = 0 x = y, 2. for all x, y X, d(y, x) = d(x, y), and 3. for all x, y, z X, d(x, z) d(x, y) + d(y, z) triangle inequality. One thing we definitely want to retain even without metrics is the notion of continuous map. Can you remember how to define continuous maps by convergent sequences: by ɛ and δ: In fact we have the following three equivalent ways of characterising continuity. Definition 2.3. Let f : X Y be a function between metric spaces. Then f is continuous if any of the following equivalent conditions holds: 1. Given any sequence {x n } such that x n x, we have fx n fx. 2. a X ɛ > 0 δ > 0 s.t. x X, d(a, x) < δ d(f(a), f(x)) < ɛ 3. Given any open set U Y, f 1 U is open in X. Before we go any further, we d better remind ourselves of what the various parts of these definitions mean. For definition (1) we need to remember convergent sequences.

13 2.1 Basic definitions 13 Definition 2.4. Given a sequence {x n } in a metric space X, we say {x n } converges to x, written x n x if and only if ɛ > 0 N N s.t. n N d(x n, x) < ɛ. Definition (2) speaks for itself, but for definition (3) we need to remind ourselves about images and pre-images, and open sets. Definition 2.5. Let f : X Y be any function, A X, B X. The image of A under f is f(a) = {f(a) a A} Y. The pre-image of B under f is f 1 B = {x X f(x) B} X Definition 2.6. Let X be a metric space, a X. The open ball of radius r centred at a is defined by Br (a) = {x X d(a, x) < r}. The closed ball of radius r centred at a is defined by B r (a) = {x X d(a, x) r}. Definition 2.7. Let X be a metric space and A X. Then A is open iff a A δ > 0 s.t. Bδ (a) A. The only definition of continuity that has a hope of working without a metric is the last one, but to do that we have to somehow think about open sets without using open balls. This brings us to the definition of topological space.

14 3 Topological spaces 14 3 Topological spaces We are going to concentrate on the idea of open sets. But we need a notion that doesn t use distance. How is it possible? We re going to do something mind-boggling: we re just going to say what the open sets have to satisfy, and we re not actually going to say what the open sets have to be. We will define a topological space to be a set equipped with some subsets called open and then specify some conditions that ensure that we have made a sensible choice of open set. Question 3.1. What do open sets look like in R 2? Note: we re not actually going to worry about the definition of topological space very much. We will hardly even use it. We ll probably never prove that something actually is a topological space using this definition. Likewise the definition of continuity. You might wonder how and why we avoid the definition of topological space in the study of topological spaces. One possible answer is that this definition is sort of disgusting to work with, and somehow doesn t capture what is really going on with topological spaces. So we run away from the definition as fast as possible and use algebra to study spaces instead. The important thing we ll need to know is: There is some notion of space a bit like metric space and some notion of continuous map and some things are true about them. For example: we can compose maps, take products, do quotients Spaces and continuous maps Recall that a metric space is a set equipped with a metric; similarly a topological space is a set equipped with a topology, as follows. Definition 3.2. A topology on a set X is a collection of subsets of X called open, satisfying: 1. and X are open, 2. any union of open sets is open, and 3. any finite intersection of open sets is open.

15 3.1 Spaces and continuous maps 15 A topological space is a set X equipped with a topology. Note that we only ask for finite intersections it is possible to have an infinite intersection of open sets that isn t open, e.g. ( 1 n, 1 n ) = n Example 3.3. Any metric space is a topological space, with the metric topology in which the open sets are the ones we defined before (Definition 2.7). In fact many of the spaces we consider can be thought of as metric spaces, but in a manner that might just not be very helpful. For example when we do funny glueings like we just don t want to be thinking about how far apart points are when we re done. Note that a given set X might have many different possible topologies on it. Here are a couple of examples that sound a bit silly but are actually quite important. Example 3.4. Any set X can be given the following two topologies. These may be thought of as maximal and minimal. 1. Discrete: every set is open. 2. Indiscrete: only and X are open. We will later see (Exercises #4) that every map out of a discrete space is continuous, and every map into an indiscrete space is continuous. Definition 3.5. A subset A of a topological space X is called closed if X \ A is open. For example and X are both open and closed. Note that we can state the definition of topology using closed sets instead of open (Exercises #3). Now that we don t have a notion of distance any more, we don t have a notion of convergent sequence. The important thing is that we do have a notion of

16 3.1 Spaces and continuous maps 16 continuity. Which definition of continuity do we use? There is clearly only one of our three that is still feasible. Definition 3.6. A continuous map of topological spaces f : X Y is a function such that whenever U is open in Y, f 1 U is open in X. Note that this is very different from saying that f maps open sets to open sets. This condition on pre-images tells us nothing about images. If f is continuous then: the image of an open set does not have to be open e.g. R 0 R (0, 1) [0, 0] the image of a closed set does not have to be closed e.g. under the map R R x e x the image of the subset N R is a sequence whose limit is 0 as n. Since this limit is not actually attained the image f(n) is not closed, but N is closed. The following result is important, and is also a good exercise in manipulating pre-images. Proposition 3.7. If f : X Y and g : Y gf : X Z is continuous. Z are continuous then Proof. We need to check that if U is open in Z, (gf) 1 (U) is open in X. Now (gf) 1 (U) = {x X g(f(x)) U} = {x X f(x) g 1 (U)} = f 1 (g 1 (U)). Since g is continuous, g 1 (U) is open in Y, and thus since f is continuous, f 1 (g 1 (U) is open in X. We now come to two crucial constructions that we will use heavily, for building up spaces: products and quotients. Products go pretty much as you would expect, but perhaps quotients take a bit more thought.

17 3.1 Spaces and continuous maps 17 Definition 3.8. If X and Y are topological spaces then X Y is a topological space with the product topology whose open sets are: U V where U is open in X and V open in Y, and all unions of these. Exercise 3.9. Why don t we need to specify that all intersections of these are open as well? Note that given continuous maps X 1 map f X 2 and Y 1 X 1 Y 1 f g X 2 Y 2. g Y 2 we get a continuous Definition If X is a topological space and is an equivalence relation on the set X then X/ is a topological space with the quotient topology, in which a subset of equivalence classes is open in X/ if their union is open in X. The resulting space is called a quotient space. For example a subset of equivalence classes might look like {[x], [y], [z]} and to check if this is open in X/ we look at the subset [x] [y] [z] X that is, the subset of X consisting of all the points equivalent to x, y or z under. Example We define an equivalence relation on R by x y y x Z and form the quotient. We can depict this as:

18 3.1 Spaces and continuous maps 18 Note that the quotient topology is defined precisely to make the quotient map continuous. That is, the map: X X/ x [x]. Quotients are dealt with much better by topological spaces than by metric spaces. Quotients of metric spaces are basically disgusting. We can think of this as one of the major limitations of metric spaces, as glueing things together is one of our most important tools for understanding complicated spaces. 1 Exercise Think about the way we glue the ends of the interval I together to make a circle. How would that work if we thought of them as metric spaces? We can now state the most straightforward notion of sameness of topological spaces. Definition A homeomorphism is a continuous map with a continuous inverse. It is very important to stress the fact that the inverse must be continuous. A map can certainly be continuous and bijective but not have a continuous inverse. That is, it would have an inverse as a map between sets, but it does not have an inverse as a continuous map between topological spaces. (This is different from, say, groups if a group homomorphism is bijective then its inverse is automatically a group homomorphism. However the same problem arises if we 1 This fact is expressed categorically by the fact that the category of metric spaces does not have coequalisers.

19 3.2 Examples of continuous maps 19 think about smooth maps a smooth map can certainly have an inverse that isn t smooth.) Examples Consider the map [0, 1) f S 1 t (cos2πt, sin2πt). This is bijective, so it has an inverse, but its inverse is not continuous. For example, [0, 1 4 ) is open in [0, 1) but its pre-image under the inverse of f is not open in S 1 (which you should prove properly in Exercises #4). 2. However, if we consider the quotient space I/ under the relation 0 1, then we have a well-defined map I/ f S 1 t (cos2πt, sin2πt) and it is a homeomorphism. The above problem has gone, because the subset like [0, 1 4 ) is not open in I/. To see this, we have to remember how the quotient topology goes, and [0, 1 4 ) 1 is not open in I. 3. The map D 2 {0} is quite obviously not a homeomorphism, but we want to think of these spaces as being somehow the same. So we need more subtle ways of telling spaces apart. Note that from now on whenever we say space we will mean topological space. 3.2 Examples of continuous maps 1. For any space X we have the identity map 1 X : X X.

20 3.2 Examples of continuous maps There are things that are a bit like identities inclusions/embeddings e.g. (a) R R 2 embeds in many ways. (b) I 2 I 3 (c) S n S n+1 (d) S 1 I S 1 x (t, x) we have one of these for each t I. We can do the same to embed X I X. (e) S 1 MB by inclusion into the boundary. (f) B n S n on the north or south hemisphere. Write B n = {(x, y) R 2 x 2 + y 2 1} S n = {(x, y, z) R 3 x 2 + y 2 + z 2 = 1} Then the map is given by 3. Glueing e.g. (x, y) (x, y, 1 x 2 y 2 ).. (a) I 2 = MB. (b) I 2 = torus. (c) All our other polygon glueing examples. (d) We can glue the ends of an interval to make a circle: I α S 1 t (cos2πt, sin2πt)

21 3.2 Examples of continuous maps 21 (e) The above map induces a map for glueing the edges of a square to make a cylinder: 1 α (f) D 2 D 2 S 2 I 2 S 1 I (t, z) (cos2πt, sin2πt, z) N.B. we naturally get two maps D 2 D 2 D 2 S 2 for the upper and lower hemisphere. We have a diagram of continuous maps: boundary S 1 D 2 D 2 S 2 This diagram commutes, which means that the different ways of following the arrows round the diagram produce the same composite. (g) Paths and loops we will be studying these a lot. A path in X from a to b is a continuous map γ : I X such that γ(0) = a and γ(1) = b. A loop at a in X is a path that starts and ends at a, i.e. continuous map γ : I X a such that γ(0) = γ(1) = a.

22 3.3 Paths and loops 22 (h) There is a type of continuous map called a retraction e.g. R 2 \ {0} S 1 see Exercises #3. Exercise How many different maps S 1 S 1 can you think of? 3.3 Paths and loops Definition A path from a to b (written a b) in a space X is a continuous map γ : I X such that γ(0) = a and γ(1) = b. A loop at a X is a continuous map γ : I X such that γ(0) = γ(1) = a. Paths tells us some things immediately e.g. in the following space is there a path from a to b? Definition Given a space X we can put an equivalence relation on it by x y there is a path in X from x to y. The equivalence classes are called the path components of X, written π 0 X. X is called path-connected if there is only one path componenet; equivalently x, y X a path in X from x to y. Exercise Is that really an equivalence relation?

23 3.3 Paths and loops 23 Definition Let X be a space. 1. Given a point a X we define the constant path at a by cst a : I X t a. 2. Given a path γ : a b in X we define the reverse path γ : b a by γ : I X t γ(1 t). We can check that γ (0) = γ(1) and γ (1) = γ(0). 3. Given paths a γ1 b γ2 c in X we define the concatenation γ 2 γ 1 : a c by γ 1 (2t) 0 t 1 2 (γ 2 γ 1 )(t) = 1 γ 2 (2t 1) 2 t 1 So 1. gives reflexivity a a, 2. gives symmetry a b b a, and 3. gives transitivity a b and b c a c. so it really did make sense for us to take path-components π 0 X. π 0 is a rather crude way of classifying spaces, for example it makes these three spaces all the same. We want a better test instead of just saying Is there a path from a to b? we want to say something like:

24 3.3 Paths and loops 24 How many different paths are there from a to b? For example these paths are different : whereas these should count as the same : So we need a way of saying when paths are the same really. Note that, as you saw in Exercises #2, just having the same image is not subtle enough going round a circle once or twice gives the same image but should definitely count as different paths. So in the next section we introduce the notion of homotopy.

25 4 Homotopy 25 4 Homotopy Example 4.1. Imagine policemen searching an area that happens to include a pond. Each individual can get to the other side, but they ll have to break their line in the middle somewhere. Consider two paths f, g : I Y. We want to say what it means to deform f into g. At a first guess we might say x I we have a path f(x) px g(x) but that isn t quite enough. We also need to say something like For each t I we have a path consisting of all the p x (t) s.

26 4 Homotopy 26 The easiest way to do this is to think of mapping an entire square: which is sort of what our diagram looked like anyway. If we demand that this map is continuous it ensures that we didn t try and deform anything past a hole. More generally we can do this for any continuous maps f, g : X Y. This is the notion of homotopy. Note that from now on whenever we say map we will mean continuous map. Definition 4.2. Let X, Y be spaces and f, g : X Y continuous maps. Then a homotopy α : f g or f X α Y g is a continuous map α : I X Y such that for all x X α(0, x) = f(x) also written α(0, ) = f α(1, x) = g(x) α(1, ) = g Note in particular this means: for all x X we have a path p x : f(x) g(x) in Y given by I Y t α(t, x) i.e. p x = α(, x), and

27 4 Homotopy 27 for all t I we have a continuous map α t : X Y x α(t, x). We can think of this as continuously deforming f into g via all these intermediate maps α t. Definition 4.3. If there is a homotopy f g we write f g and say f is homotopic to g. We will show that this is an equivalence relation. Example 4.4. This example makes precise the idea that the map from the circle to itself doing nothing is more or less the same as the map that rotates it by an angle φ. This is simplest to write down in polar coordinates, with the points of the circle given as (1, θ). We have two maps f, g : S 1 S 1 where f is the identity and g(1, θ) = (1, θ + φ). We can now define a homotopy α : f g by α : I S 1 S 1 (t, (1, θ)) (1, θ + tφ) Example 4.5. This example makes precise the idea that if we map the circle into R 2 it s more or less the same as mapping it to a point, because we can shrink the circle down to a point in the plane. Again, this is simplest in polar coordinates. We have two maps f, g : S 1 R 2 where f is the embedding and g sends everything to the origin, i.e. for all θ f(1, θ) = (1, θ) g(1, θ) = (0, 0) We can now define a homotopy α : f g by α : I S 1 R 2 (t, (1, θ)) (1 t, θ) Example 4.6. The boring but important example. map f : X Y there s an identity homotopy given by For every continuous I X Y (t, x) f(x) t

28 4 Homotopy 28 Example 4.7. This example is about changing the speed we travel along a path. Given a path f from a to b in X there s a new one which we can think of as doing f twice as fast and then just sitting still at b for the rest of the time. Formally, this is given by f(2t) 0 t 1 2 g(t) = 1 b 2 t 1 We can now define a homotopy α : f g by α : I I X (t, s) f( 2s 2 t ) b 0 s 2 t 2 2 t 2 s 1 Example 4.8. This example constructs a reverse homotopy very much like a reverse path. Informally, we know that a homotopy f g gives for each x a path f(x) g(x). We we might wonder if we can reverse all these paths and form a homotopy g f. Formally we do it like this. Given maps f, g : X Y and a homotopy α : f g we define a homotopy α : g f by α : I X Y (t, x) α(1 t) Example 4.9. This example constructs a concatenation of two homotopies very much like a concatenation of paths. Suppose we have homotopies f X g α β Y h As above, we know informally that a homotopy α : f g gives for each x a path α(, x) : f(x) g(x). Now here we also have for each x a path β(, x) : g(x) h(x). The idea is to concatenate these paths to get a path f(x) g(x) h(x)

29 4 Homotopy 29 for each x. Formally we define the homotopy by: β α : I X Y α(2t, x) 0 t 1 2 (t, x) 1 β(2t 1, x) 2 t 1 and then we must check that it is continuous. Proposition For any maps f, g : X Y write f g if f is homotopic to g. Then is an equivalence relation. Proof. Reflexivity is Example. Symmetry is Example. Transitivity is Example. Example This example is a slightly more complicated way of sticking homotopies together. Suppose this time we have two homotopies f h X α Y β Z g k Informally we know that α gives us, for each t I a continuous map α(t, ) : X Y, and β gives us β(t, ) : Y Z. The idea is to compose these so that for each t we have a map β(t, ) α(t, ) : X Z, and make this into a homotopy. Formally: β α : I X Z (t, x) β ( t, α(t, x) ) Now that we have defined homotopy we are finally ready to make precise the idea of spaces being more or less the same given some squashing and/or stretching. The idea is that we want continuous maps between our two spaces that are not exactly inverse to one another, but only up to homotopy.

30 4 Homotopy 30 Definition A continuous map f : X Y is called a homotopy equivalence i there is a continuous map g : Y X such that g f 1 X and f g 1 Y Then the spaces X and Y are called homotopy equivalent, written X Y. Equivalently, X and Y are said to have the same homotopy type. Example The following spaces are homotopy equivalent: So are the following spaces: Proposition Homotopy equivalence is an equivalence relation. Proof. This is mostly left as an exercise. The only part that isn t straightforward is transitivity. We ll see some more examples of homotopy equivalence in a minute, but one particularly important kind of homotopy equivalence is when something is homotopy equivalent to a point: Definition A space is called contractible if it has the homotopy type of a point, that is, it is homotopy equivalent to a point. Lemma A space X is contractible if and only if there is a homotopy cst a id X for some point a X. Here cst a is the map X X that sends everything to a. Proof. X is contractible if and only if we have maps f : X and g : X together with homotopies gf id X and fg id.

31 4 Homotopy 31 Now fg = 1 so the second homotopy can be taken to be the identity. For the first homotopy, first observe that a map g : X simply picks out a point in X, so the composite gf sends everything to this point g( ). So it is cst a, where we write a = g( ). Thus a homotopy gf id X is just a homotopy cst a id X, hence the result. Example R 2 is contractible. We take a to be the origin and construct a homotopy id R 2 cst a by I R 2 R 2 (t, (x, y)) ((1 t)x, (1 t)y) Exercise Show that the following two spaces are homotopy equivalent: Example Suppose we have maps f, g : X Y where Y is a convex subset of a vector space. Then f g via α : I X Y (t, x) (1 t).f(x) + t.g(x) We can check that α(0, x) = f(x) and α(1, x) = g(x). This is called a straight line homotopy. For example, any two paths in R n are homotopic via a straight line homotopy. Note that if we have two paths with the same endpoints, i.e. f, g : a b R n then the straight line homotopy keeps the endpoints fixed: α(t, 0) = (1 t).a + t.a = a α(t, 1) = (1 t).b + t.b = b This last example leads us into a special notion of homotopy for paths. Definition Let f, g : I X be paths from a to b in X. Then f and g are homotopic or equivalent if they are homotopic through paths from a to b. That is, we have a map α : I I X such that α(0, t) = f(t) α(1, t) = g(t)

32 4 Homotopy 32 and for all s I α(s, 0) = a α(s, 1) = b This is called a path homotopy. Similarly we have the notion of loop homotopy. Definition As before, path homotopy and loop homotopy are equivalence relations. Given a path (or loop) f we write [f] for the homotopy class of f. Note that being path homotopic is a stronger condition than being just homotopic. From now on we will assume that if we are dealing with paths we are using path homotopy. We are now ready for the next section, in which we will make the homotopy classes of loops into a group: the fundamental group.

33 5 The fundamental group 33 5 The fundamental group So far we have defined a way of going from the world of spaces to the world of sets, using path components. π 0 This is useful, but rather crude, and ignores the differences between many very different spaces. So now we re going to be a bit more subtle about it, and map into groups instead: π 1 We are going to define, for each space X, a group π 1 X called the fundamental group of X, which captures much more information than π 0 X did. We have already informally seen some fundamental groups: π 1 (S 1 ) = Z π 1 ( ) = 0 π 1 (R 2 ) = 0 π 1 ( ) = free group on two generators (see Exercises #4) First we ll show how to construct the fundamental group formally, and then we ll ask ourselves some questions about this construction: 1. Is π 1 (X Y ) related to π 1 (X) and π 1 (Y )? 2. What about for X Y? What about for X/? 3. What happens to maps X Y? 4. What kind of map is π 1 itself? 5. Can we go backwards: given a group can we make a space? Does every group arise as the fundamental group of some space?

34 5.1 Definition If two spaces have the same fundamental group, what does that tell us? 7. What does π 1 (X) = 0 tell us? 8. What do subgroups of π 1 (X) tell us? What about cosets? Homomorphisms? 5.1 Definition Theorem 5.1 (Fundamental group construction). Write π 1 (X, a) for the set of equivalence classes of loops based at a. This can be given the structure of a group as follows: multplication: [f].[g] = [f g] inverses: [f] 1 = [f ] identity: e = [cst a ] Proof. We need to check that: the operation is well-defined, the operation is associative, e really acts as a unit, and every element has an inverse. This is an exercise.

35 5.2 Dependence on basepoint 35 Definition 5.2. The group π 1 (X, a) is the fundamental group of X based at a. 5.2 Dependence on basepoint When we defined the fundamental group, we had to choose a basepoint. What would have happened if we had chosen a different one? Obviously if we chose one in a different connected component then we d be a bit doomed e.g. However if there is a path a γ b then it will turn out okay: every loop at a corresponds to a loop at b via γ: Proposition 5.3. Let γ be a path a b in X. Then there is a group isomorphism given by ˆγ : π 1 (X, a) π 1 (X, b) Its inverse is given by ˆγ. [f] [γ.f.γ ] Proof. First we need to show that ˆγ is a group homomorphism, i.e. ˆγ([f].[g]) = ˆγ[f].ˆγ[g]

36 5.2 Dependence on basepoint 36 that is γfgγ γfγ γgγ. We can construct such a homotopy as indicated by the following diagram: We now need to show that ˆγ is really inverse to ˆγ, i.e. f γ γfγ γ and f γγ fγγ. This follows from the fact that γγ cst b and similarly γ γ cst a. So now we know that if X is path-connected then π 1 does not depend on the choice of basepoint. Note however that π 1 (X, a) and π 1 (X, b) are not canonically isomorphic we had to choose a path a b which gave us a group isomorphism. A non-equivalent path might give us a different isomorphism. The next thing we might ask is: what happens to maps? If we have a map φ : X Y do we get a group homomorphism π 1 X π 1 Y? And the answer is yes. Any path γ : I X becomes a path in Y by I γ X φ Y. If γ is a path from a to b then the new path will go from φ(a) to φ(b), so in particular if γ is a loop then φ γ is also a loop:

37 5.2 Dependence on basepoint 37 We would like this to give us a homomorphism of fundamental groups as well; we will have to be a bit careful about homotopy classes. Proposition 5.4. Given a continuous map φ : X Y, we get a group homomorphism φ : π 1 (X, a) π 1 (Y, φ(a)) [γ] [φ γ] Proof. We need to show that this map is well-defined, and that φ ([γ 1 ].[γ 2 ]) = φ [γ 1 ].φ [γ 2 ]. This is an exercise. We would now like to show that this process satisfies all sorts of good properties. This is best summed up using the notions of category and functor.

38 5.3 Categories Categories Definition 5.5. A category C consists of: a collection obc of objects, and for every pair X, Y obc, a collection C(X, Y ) of morphisms f : X Y, equipped with for each X obc, an identity morphism 1 X C(X, X), and for all X, Y, Z obc, a composition map m XY Z : C(Y, Z) C(X, Y ) C(X, Z) (g, f) g f satisfying the following axioms: unit laws given f : X Y we have 1 Y f = f = f 1 X, and associtiavity given X f Y g Z h W we have h (g f) = (h g) f. A category is said to be small if obc and all of the C(X, Y ) are not just collections but actually sets, and locally small if each C(X, Y ) is a set. Here are some examples of categories. Examples 5.6. Large categories of mathematical structures You may or may not have met these mathematical structures before, but it doesn t particularly matter for now. 1. Set of sets and functions. This is in many ways the prototype category. Set has many, many wonderful features that make it a good place to start doing mathematics, and we would like to know what other categories have these features. 2. Categories derived from or related to Set:

39 5.3 Categories 39 Rel of sets and relations; a morphism A B is a subset of A B giving the pairs (a, b) such that a b. Given a relation on A and B, and a relation on B and C we produce the composite relation on A and C which has a c b B such that a b and b c. Set of pointed sets and basepoint-preserving functions. A pointed set is just a set with a chosen element as its basepoint ; the morphisms are functions which send basepoints to basepoints. 3. Algebraic structures and structure-preserving maps: Grp of groups and group homomorphisms; Ab of abelian groups and group homomorphisms; Ring of rings and ring homomorphisms; Vect k of vector spaces over a chosen field k; R-Mod of R-modules and their homomorphisms, for some fixed ring R. Field of fields and field homomorphisms. This actually turns out not to be a very well-behaved category, and it is often better to restrict to fields of a fixed characteristic. 4. Categories of topological spaces: these were among the earliest motivating examples of categories. Using this framework gives very concise statements of results in algebraic topology. Top of topological spaces and continuous maps; actually unlike Set and Vect, Top is not a well-behaved category. For this reason we often end up working in some related category of better spaces, to eliminate various pathologies that can otherwise arise and ruin everything. Haus of Hausdorff spaces and continuous maps; CHaus of compact Hausdorff spaces and continuous maps; this is a much better category than Top. Top of based spaces, that is topological spaces equipped with a chosen point called the basepoint; morphisms are the continuous maps that preserve the basepoint. We have to work with this category when we do the fundamental group, because our loops have to be based somewhere. Met of metric spaces and uniformly continuous maps; Htpy of topological spaces and homotopy classes of maps; this category is sometimes preferable to Top because we are often interested in things up to homotopy.

40 5.3 Categories 40 Examples 5.7. Very small categories When categories are very small indeed we can quite simply and vividly draw them on the page using points for objects and arrows for morphisms. 1. There is a category with one object and one morphism, which has to be the identity. We can draw a picture of this category: x 1 x This category is often referred to as There is an empty category. It has no objects and no morphisms, so must surely be the tiniest and most stupid category. Still, it s not completely worthless and is often called 0, and after all, the number 0 is very far from useless. 3. There is a category that we might draw like this: which means there are two objects and one morphism between them, as shown. Really this category has three morphisms and could be drawn like this: 1 y 1 x x y 4. Here s a category with some non-trivial composition: 5. There is a category defined by the commutative square: a f 1 b g 1 c f 2 g 2 d. Saying the square commutes means that going round the square in different ways gives the same answer, that is, that the two composites are equal: f 2 f 1 = g 2 g 2.

41 5.3 Categories 41 Examples 5.8. Curious categories The following examples are a bit curious and might surprise you a bit. But they re a central part of category theory and yield all sorts of interesting constructions. 1. Every set is a category with only identity morphisms. 2. Every group is a category with only one object. That is to say we can express a group G as a category: it has only one object, which we might call ; the morphisms are precisely all the elements of G; composition is group multiplication, and the identity 1 x in the category is the unit element of the group G. In the resulting category every morphism is invertible; the next example is more general. 3. A category with only one object is precisely a monoid. 4. A category in which every morphism is invertible is a groupoid. 5. We can generate a category by the diagram x f where f : x x is not supposed to be the identity. additive monoid N. The result the 6. We could do something similar but impose the relation f f = 1. This would turn our category into the one-object category corresponding to the cyclic group of order Every poset is a category. Given a poset P with ordering, we make a category with objects the elements of P and precisely one morphism x y when x y, and none otherwise. 8. There is a curious category Mat of matrices. The objects are the natural numbers; the morphisms n m are all the n m matrices (in some fixed field, k, say). Composition is given by matrix multiplication.

42 5.3 Categories 42 One of the basic principles of Category Theory is that we should study structures together with their structure-preserving maps. We should look at the totality of these structures and their maps, rather than just looking at individual structures in isolation. The structure-preserving maps of categories are called functors. Definition 5.9. Let C and D be categories. A functor F : C D associates: to every object X C an object F X D, and to every morphism f C(X, Y ) a morphism F f D(F X, F Y ) such that for all morphisms X f Y for all X C, F (1 X ) = 1 F X. g Z, F (g f) = F g F f, and The conditions in this definition are generally referred to as functoriality conditions, and should distinctly remind you of the axioms for a group homomorphism. Examples Functors on other mathematical structures expressed as categories 1. If C and D are groups (expressed as one-object categories) then a functor C D is precisely a group homomorphism. 2. If C and D are monoids, that is, one-object categories, then a functor C D is precisely a monoid morphism. 3. If C and D are posets (expressed as categories) then a functor C D is precisely an order-preserving map. Examples Forgetful functors Forgetful functors forget all or part of the structure in question. 1. There is a forgetful functor Gp Set which sends a group to its underlying set; the action on morphisms is induced in the obvious way.

43 5.3 Categories Similarly there are forgetful functors Ring Set Vect Set Ab Set Top Set Mnd Set Poset Set Set Set and many more. 3. Here are some forgetful functors that only forget part of the structure: Ring Ab Ring Mnd Vect Ab Top Top 4. Here are some examples of functors that just forget some property: Examples Free functors Ab Gp Haus Top Free functors go in the opposite direction of forgetful functors: they take something with little or no structure, and create something with more structure. The new extra structure is added in freely. 1. There is a free functor Set Mnd that takes a set X and produces the free monoid on it. The result is a set X which consists of words in X, that is, finite lists of elements of X, possibly with repetitions. 2. We can do something similar but harder to get a free functor Set Gp. The reason it s harder is that we have to make sure everything has an inverse and this complicates matters rather.

44 5.3 Categories Similarly we have a free functor Set Ring but we also have a more subtle free functor Mnd Ring. Examples Mathematical constructions as functors 1. If G is a group expressed as a one-object category, a functor G Set is a set with a G-action, also known as a G-set. 2. If G is a group expressed as a one-object category, a functor G Vect is a representation of G. 3. We also get functors for the constructions of fundamental groups, higher homotopy groups, homology, cohomology, geometric realisation, sheaves, compactification, parallel transport, topological quantum field theory... The fundamental group functor This brings us to the functor that launched us into this whole discussion about categories in the first place. That is, I launched into it because I wanted to be able to say that the fundamental group construction gives us a functor as follows. Theorem There is a functor π 1 : Top Gp given by on objects: π 1 (X, x) is the fundamental group, on morphisms: given a morphism (X, x) φ (Y, y) we put π 1 (φ) = φ. Proof. We need to check that this satisfies the axioms for a functor. First, given maps (X, x) φ ψ (Y, y) (Z, z) we need to check that i.e. π 1 (ψ φ) = π 1 (ψ) π 1 (φ) (ψ φ) = ψ φ.

45 5.3 Categories 45 We also need to check that for all (X, x) we have π 1 (1 (X,x) ) = 1 π1(x,x). This is an exercise Remark We have seen something about the fundamental group being independent of the choice of basepoint. Does this mean we have a functor Top Gp? We would now like to know that the fundamental group is homeomorphism invariant that is, if two spaces are homeomorphic then they have the same fundamental group (or rather, isomorphic ones). This now turns out to be immediate from the fact that π 1 is a functor, because all functors preserve isomorphisms: Lemma Let F : C D be a functor and f : X Y C an isomorphism with inverse g : Y X, so fg = 1 Y and gf = 1 X. Then F f is an isomorphism. Proof. See Exercises #6. Corollary π 1 is homeomorphism invariant. More precisely, if φ : (X, x) (Y, y) is a homeomorphism then φ : π 1 (X, x) π 1 (Y, y) is a group isomorphism.

46 5.4 Homotopy invariance 46 It is crucial that π 1 be homeomorphism invariant, otherwise it would be a rather stupid invariant of topological spaces. However, we would also like it to be homotopy invariant and that is the subject of the next section. 5.4 Homotopy invariance A property or calculation is said to be homotopy invariant if any homotopy equivalent spaces give the same answer. Of course, we ll need to say this more precisely every time we come to it. We start with π 0. Proposition π 0 is homotopy invariant. That is, if X and Y are homotopy equivalent then we have π 0 X = π 0 Y. In fact we prove something stronger that a homotopy equivalence f induces an isomorphism on path components. Proof. We have maps X f Y g and homotopies α : gf = 1 X β : fg = 1 Y In particular we have x X a path gfx αx x, and y Y a path fgy βy y. We now show that f induces an isomorphism on path components. injective: we need to show [fx 1 ] = [fx 2 ] = [x 1 ] = [x 2 ] i.e. fx 1 fx 2 = x 1 x 2

47 5.4 Homotopy invariance 47 γ i.e. if there is a path fx 1 fx 2 then there was already a path x 1 x 2. We construct this path by: x 1 surjective: we need to show α x gfx 1 gfx 2 y Y x X s.t.fx y that is, there is a path fx y. We do this by: with x = gy. fgy y α x x2 We now prove the analogous result for π 1. Proposition π 1 is homotopy invariant. That is, if X and Y are homotopy equivalent then π 1 (X) = π 1 (Y ). More precisely, given a homotopy equivalence f : X Y, the map is a group isomorphism. π 1 f : π 1 (X, x) π 1 (Y, f(x)) Proof. (Sketch) We have homotopies α and β as in the previous proof. We now construct a left inverse to π 1 f as π 1 (Y, f(x)) π1g αˆ π 1 (X, gf(x)) x π1 (X, x). and similarly we construct a right inverse to π 1 g. This is enough as... Definition We call a space X simply-connected if it is path-connected and π 1 (X) = 0. Intuitively, this means that the space has no holes its fundamental group is trivial so every loop is homotopic to the constant loop at a point. We really need to start calculating some fundamental groups that aren t trivial! The best building block to start with is the circle, but actually it s going to be easier for us to look at that using the theory of covering spaces. However, before we move onto that section it s worth getting a rough idea how we want to think about the circle, so that when we go onto covering spaces properly we have some pictures in our head.

48 5.5 The circle The circle The fundamental group of the circle is Z, but we haven t proved it yet. This result is intuitively clear, but the proof involves a lot of technicalities. Here are some things that follow from knowing this result. The fundamental theorem of algebra Every non-constant polynomial with coefficients in C has a root in C. The Brouwer fixed point theorem Every continuous map D 2 D 2 has a fixed point. The Borsuk-Ulam theorem in dimension 2 For every continuous map f : S 2 R 2 there is a pair of antipodal points x, x S 2 such that f(x) = f( x). Corollary If S 2 is expressed as the union of three closed sets then at least one must contain a pair of antipodal points. Question When you re walking up the stairs in the Hicks Building, is it easier to count how many times you ve gone round, or to look at the signs to see what floor you ve reached?

49 5.5 The circle 49 Sketch proof that π 1 (S 1 ) = Z We get a bit confused about going round our circle multiple times, so let s unwind our circle: R p S 1 Now, instead of counting how many times we went round, we count count how many floors up (or down) we ve gone. Moreover we can choose the map p so that the preimages of the basepoint x are precisely the integers. So the number of times we ve gone round the circle is precisely counted by what number we land on in the spiral version. This is very convenient! Now the question is, can we somehow: 1. turn this into a proof that π 1 (S 1 ) = Z, and 2. generalise this idea for other spaces? We are looking for an isomorphism [loops in S 1 ] p 1 (x) and we ll do this in steps. The idea is that loops downstairs correspond to paths upstairs, so we can fix the starting point of the path and see where the other end of is. So we start by choosing a lift of the basepoint x i.e. x p 1 (x). 1. Every loop γ downstairs lifts to a unique path γ upstairs starting at x i.e. such that p γ = γ and γ(0) = x.

50 5.5 The circle So we get a function loops in S 1 p 1 (x) γ γ(1) which is a good start, but we want a function from homotopy classes of loops to p 1 (x). So we need to check that this function is homotopy invariant. 3. Now, any loop homotopy downstairs lifts to a path homotopy upstairs, so we do indeed get a function [loops in S 1 ] p 1 (x). 4. We check that it is surjective, that is, that every pre-image of x upstairs gets hit this is true because R is path-connected. So given any a p 1 (x) we just taken a path x f a in R, and applying p must give a path p( x) p(a) in S 1, i.e. a loop γ at x such that γ(1) = b. 5. We check that it is injective, that is γ 1 (1) = γ 2 (1) = γ 1 γ 2. This is true because R is simply-connected, so all paths with the same endpoints are homotopic. Thus we have a homotopy γ 1 γ 2 and this must map via p to a homotopy γ 1 γ 2. This sketch proof has highlighted several of the important properties of covering spaces that we are going to look at: 1. path lifting, 2. the lifting correspondence, and 3. homotopy lifting. The construction we did where we unwound all the loops around the circle produced was is called the universal covering space. We could have just unwound some of the loops and not others e.g.

51 5.5 The circle 51 In this case in our proof we would still have surjectivity (4) but not injectivity (5) some different loops downstairs would lift to the same path upstairs. Which? Another important question is: what would the induced map on fundamental groups be? These are all important questions we will consider in the next section, on covering spaces.

52 6 Covering spaces 52 6 Covering spaces So far we don t have many ways of calculating fundamental groups. Covering spaces give us some ways of doing this, but also much more it s a very specific manifestation of the correspondence between algebra and topology fundamental group covering spaces Recall that we tried to calculate the fundamental group of S 1 by thinking about a helix mapping onto the circle: This is a good prototype example of a covering space. We are unwinding the loops in the space. Here are some more covering spaces of S 1 :

53 6 Covering spaces 53 And some other covering spaces can you see how the first two cover the space? Here are some of the main ideas about covering spaces. A covering space wraps itself evenly around the base space. The fibre over any point has n points in it, for some fixed n. (It might be infinite.) We would then call this an n-sheeted cover. We get neighbourhoods whose pre-image splits into a disjoint union of homeomorphic copies. There are various symmetries interchanging the sheets. The above are all geometric features, and we will see how these correspond to algebraic features of the fundamental group: subgroups, and group actions on sets or spaces. We will ask ourselves questions like: can we find all the covering spaces of a space? How do the geometric and algebraic features correspond? The main answer is a correspondence connected subgroups of the covering spaces fundamental group

54 6.1 Definitions 54 This might remind you of Galois theory, in which you have a correspondence between field subgroups of the extensions Galois group A covering space sort of unloops some of the loops in your space. The corresponding subgroup of the fundamental group consists of the loops that you haven t unlooped yet. Example 6.1. You walk around a tree and you think you re back to where you started but you have a layer of mud on your shoes. Example 6.2. Pooh, Piglet and the Heffalump. So as your covering space gets bigger and bigger, your corresponding subgroup gets smaller and smaller, and finally you get the universal cover corresponding to the trivial subgroup. This is where everything has been maximally unwound you no longer have any loops left. Exercise 6.3. Look at page 58 of Hatcher, where he gives a page of lovely diagrams of some covering spaces of (he refers to this space as S 1 S 1, a piece of notation we ll get to later). 6.1 Definitions Recall that a neighbourhood of a point x X is an open set containing x. Definition 6.4. A covering space of a space X is a space X equipped with a map X X with the following property: every x X has a neighbourhood U for which p 1 (U) is a disjoint union of open sets, each of which is mapped by p homeomorphically to U. p

55 6.1 Definitions 55 Example 6.5. We can also say this in terms of open covers of the space X. Definition 6.6. An open cover of a space X is a collection of open sets {U i } such that U i = X. i Definition 6.7. A covering space of a space X is a space X together with a map X X where p evenly covers X, that is, there is an open cover {U i } of X such that for each i, p 1 (U i ) is a disjoint union of open sets of X, each mapped by p homeomorphically to U i. p

56 6.1 Definitions 56 Example 6.8. Example 6.9. We can cover the torus with R 2. Example We define a subspace of R 3 called a helicoid surface: {(rcos2πt, rsin2πt, t) r (0, ), t R} This covers R 2 \ {0}, with the covering map p given by: (x, y, z) (x, y) Example The identity always gives us a covering space, trivially. This amounts to unwinding no loops, so should correspond to the biggest possible subgroup of the fundamental group the whole thing. Exercise Can you think of a (non-trivial) cover of RP 2? S 2?

57 6.1 Definitions 57 Proposition Crucial properties of covering spaces Let X X p be a covering space. 1. Path lifting Given any path f in X starting at x and any lift x of x, i.e. p( x) = x there is a unique lift of f to a path f in X starting at x i.e. a unique path I f X such that f(0) = x and the following diagram commutes: I f f X X p 2. Homotopy lifting Given any homotopy α : I A X starting at φ : A X and any lift φ of φ i.e. A φ φ X X p there is a unique lift of α to a homotopy α starting at φ i.e. a unique homotopy I A and the following diagram commutes: α X such that α(0, ) = φ I A α α X X p 3. Lifting correspondence Given any lift x of x we have a map π 1 (X, x) p 1 (x) [γ] γ(1)

58 6.1 Definitions 58 and furthermore if X is path-connected then this map is surjective, and if X is simply connected then this map is also injective. 4. Sheets If X is path-connected then the cardinality of p 1 (x) is constant. This is called the number of sheets of the cover. 5. Important consequence The group π 1 ( X, x) embeds as a subgroup of π 1 (X, x) via the homomorphism i.e. π 1 (p) is injective. π 1 ( X, x) π 1(p) π 1 (X, x) Idea: π 1 ( X, x) is smaller because we have unwound some loops. Proof. 1. Path lifting We start with a path f in X and construct a lift as required. Idea: If p were a homeomorphism it would be easy, but p is only locally a homeomorphism on the U α s exhibiting the covering space so we edge our way along, staying in one U α at a time We use the definition of covering space, and compactness of I every cover has a finite subcover.

59 6.1 Definitions 59 Step 1 By the definition of covering space we have a cover {U α } of X such that for all α, p 1 (U α ) is a disjoint union of open sets in X, each of which maps homeomorphically to U α under p. Now f : I X is continuous so {f 1 (U α )} is an open cover of I. So for all t I, there is some α with t f 1 (U α ) and there is an open ball Br (t) f 1 (U α ). This gives a cover of I, so by compactness of I there is a finite subcover [a 0, b 0 ), (a 1, b 1 ), (a 2, b 2 ),..., (a m, b m ] where a 0 = 0, b m = 1 and without loss of generality i.e. a 0 < a 1 < b 0 < a 2 < b 1 < < b m Now put t i = a i + b i i m So we have 0 = t 0 < t 1 < < t m < t m+1 = 1 such that for each i, f[t i, t i+1 ] U i, say, where U i {U α }. Step 2 We now lift the path f on each of these sub-intervals one at a time. Consider the first patch [0, t 1 ], with f[0, t 1 ] U 0. We have f(0) = x and we fix f(0) = x. Now x p 1 (U 0 ) so must lie in one homeomorphic copy of U 0 we call it Ũ0. We can now define f on [0, t 1 ] by [0, t 1 ] f U 0 p 1 Ũ0 X.

60 6.1 Definitions 60 We can then proceed by induction we repeat this process for each patch This constructs a lift f of f. [t 1, t 2 ], [t 2, t 3 ],..., [t m, 1]. Step 3 Finally we must check that this lift is unique, by essentially the same argument. Suppose we have lifts f, f : I X, so p f = p f = f, and f(0) = f (0) = x. As above, we choose a partition 0 = t 0 < t 1 < < t m+1 = 1 such that for each i, f[t i, t i+1 ] some U i. Now we edge our way along : f is continuous and [0, t 1 ] is connected, so f[0, t 1 ] is connected in X so f[0, t 1 ] lies in one homeomorphic copy of U 0, say Ũ0. Furthermore f [0, t 1 ] must lie in the same copy as f(0) = f (0). But p gives a bijection Ũ0 U 0 and we also know p f = p f so we must have f = f on [0, t 1 ]. We can then proceed by induction and patch all the way along. 2. Homotopy lifting This is basically similar to path lifting but slightly harder. 3. Lifting correspondence We need to show that this is well-defined, i.e. γ 1 γ 2 = γ 1 = γ 2. This follows from the homotopy lifting property:

61 6.2 The classification of covering spaces 61 Given a homotopy α : γ 1 γ 2 fixing the endpoints, we get a lift which must also fix the endpoints. The last part is on Exercises #8. α : γ 1 γ 2 4. Sheets Idea: the cardinality of the pre-image must be constant on each U α and we can then patch since it must agree on intersections. 5. Important consequence We want to show that π 1 (p) is injective, i.e. [pγ 1 ] = [pγ 2 ] = [γ 1 ] = [γ 2 ] i.e. pγ 1 pγ 2 = γ 1 γ 2. This follows from the homotopy lifting property. 6.2 The classification of covering spaces Slogan: Connected covering spaces of X correspond to subgroups of π 1 (X). To make this more precise: we must consider covering spaces together with the covering map ( X, x) p (X, x) we must think of two covering spaces as the same if there is a homeo-

62 6.2 The classification of covering spaces 62 morphism f making the following diagram commute f ( X, x) ( X, x ) p (X, x) p we must restrict to nice spaces: X must be path-connected, locally path-connected, and semi-locally simply-connected. We won t really go into what this last part means, but for example we want to exclude the Hawaiian earring. Here are the definitions, for completeness: Definition A space X is called locally path-connected if for any point x X and any open set U containing x, there is a path-connected open set V U also containing x. Definition A space X is called semi-locally simply-connected if for any point x X there is an open set U containing x in which every loop is nullhomotopic. Examples The quasi-circle is path-connected but not locally path-connected. 2. Let X R 2 be the space formed by taking each point (0, 0) and ( 1 n, 0) for n N and joining it to the point (0, 1). Then X is path-connected but not locally path-connected.

63 6.2 The classification of covering spaces The Hawaiian earring is locally path-connected but not semi-locally simply connected. Theorem Let X be path-connected, locally path connected and semilocally simply connected. Then there is a bijection between isomorphism classes of connected covering spaces ( X, x) and subgroups of π 1 (X, x). p (X, x) The correspondence is given by ( X, x) Remark p (X, x) Im π 1 p Note that π 1 p is a group homomorphism π 1 ( X, x) π 1 (X, x). We know it is injective (by part 5 of Proposition 6.13), so by the First Isomorphism Theorem we have However π 1 ( X, x) could embed as a subgroup of π 1 (X, x) in many different ways. For example when we consider covering spaces of the circle, we consider

64 6.2 The classification of covering spaces 64 all the different embeddings Z Z. So in the correspondence above, we say Im π 1 p and not just π 1 ( X, x). Example Connected covering spaces of S 1. We have seen the following covering connected covering spaces of S 1 : For each n N S 1 p n S 1 (1, θ) (1, nθ) and R p S 1 θ (1, θ) We know π 1 (S 1 ) = Z so we should get one connected covering space for each subgroup of Z, i.e. for each nz for n N including 0. Now, if we consider the covering space S 1 S 1 p n we get and Im π 1 p n is nz Z. N.B. This shows why we can t just get a correspondence using π 1 of the covering

65 6.2 The classification of covering spaces 65 space in this case we would just keep on getting Z. Finally for R we get p S 1 giving the subgroup: This gives all subgroups of Z so we must have found all connected covering spaces of S 1. Example Connected covering spaces of RP 2. We can cover RP 2 by S 2 by identifying antipodal points: S 2 1 RP 2 x 1 [x] = {x, x} We will later see that π 1 (RP 2 ) = Z 2. So we get giving the subgroup: Z 2 has no non-trivial subgroups, so this must be the only connected covering space of RP 2. (Of course, we also have the covering of RP 2 by itself, the trivial cover, giving the subgroup Z 2.) Note that there are many disconnected covering spaces.

66 6.2 The classification of covering spaces 66 Example Connected covering spaces of S 2. We know that π 1 (S 2 ) = 0, which has no non-trivial subgroups. So the only connected covering space of S 2 is the trivial one S 2 1 S 2 Example Conected covering spaces of RP 2 RP 2. We will see that π 1 preserves products, so π(rp 2 RP 2 ) = π1 (RP 2 ) π 1 (RP 2 ) = Z 2 Z 2 which has subgroups 0 and Z 2 Z 2, and subgroups of order 2: (0, 1), (1, 0), (1, 1) (each isomorphic to Z 2 ) We have a covering space S 2 S 2 RP 2 RP 2 (α, β) ([α], [β]) and π 1 (S 2 S 2 ) = π 1 (S 2 ) π 1 (S 2 ) = 0 so this corresponds to the trivial subgroup. We also have and RP 2 S 2 RP 2 RP 2 S 2 RP 2 RP 2 RP 2 ([α], β) ([α], [β]) (α, [β]) ([α], [β])

67 6.2 The classification of covering spaces 67 corresponding to (1, 0) and (0, 1). Finally we seek the covering space corresponding to (1, 1). It may help at this point to think of RP 2 S 2 as a quotient of S 2 S 2 by the relation (α, β) ( α, β) and similarly S 2 RP 2 is a quotient of S 2 S 2 by the relation (α, β) (α, β). It is now clearer what the last covering space should be X = (S 2 S 2 )/ where is the equivalence relation defined by (α, β) ( α, β). Note that this is not the same as RP 2 RP 2 = S 2 / S 2 / = (S 2 S 2 )/ with the equivalence relation (α, β) (±α, ±β). We have covering map X RP 2 RP 2 [(α, β)] ([α], [β]) and this covering space corresponds to (1, 1). This gives us all the connected covering spaces. Note that we have a diagram of quotients: S 2 S 2 4 sheets S 2 RP 2 X RP 2 S 2 2 sheets RP 2 RP 2 1 sheet Question Are the upper level maps covering maps as well?

68 6.3 Universal covering spaces Universal covering spaces A consequence of the classification theorem is that we must have a connected covering space corresponding to the trivial subgroup 0 (provided X is a nice space). Thus we get a simply-connected covering space, that is, a connected covering space ( X, x) with π 1 ( X, x) = 0. This is called the universal covering space. Remarks This is the covering space in which all loops have been unwound. Or: we are stupid forever, and never notice that we ve come back to where we started! 2. This has a universal property it is a covering space of every other covering space. 3. Actually if we really proved the classification theorem we would start by constructing the universal cover, and then quotient it to make all the other covers, rather like in the RP 2 RP 2 example. 6.4 Covering space constructions 1. Universal covering spaces We want to unwind all loops. We start with (X, x), and we make a new space by taking the points of X together with the path we took to get there. More precisely, we can put a topology on {[γ] γ is a path in X starting at x}. For example on S 1 if we go round the circle once we will no longer consider ourselves back where we started because that path is not homotopic to the constant path at x. 2. Non-universal covering spaces Given a covering space ( X, x) p (X, x)

69 6.4 Covering space constructions 69 this corresponds to the subgroup of π 1 (X, x) of those loops that are still loops upstairs. Every loop upstairs maps to a loop downstairs, but some loops downstairs lift to paths upstairs that are not loops. We make non-universal covering spaces by quotienting the universal covering space just to make those loops we want still to be loops in X. Comments on the classification of covering spaces 1. In fact the most rigorous statement of the classification theorem involves a category of covering spaces and a category of subgroups of π 1 (X, x). We then have an equivalence of these two categories. 2. If ( X, x) corresponds to the subgroup H G = π 1 (X, x) then the number of sheets of the cover is the index of H in G i.e. G H. This can be seen in Example 6.22, the covering spaces of RP 2 RP 2. Wedge sums Definition Given based spaces (X, x) and (Y, y), their wedge sum is defined by glueing X and Y at the basepoint. (X, x) (Y, y) = (X Y )/x y. When it doesn t matter what the basepoints are we just write X Y and understand that we have glued at one point. For example S 1 S 1 is the familiar. S 1 S 2 is. Here is an informal recipe for making the universal covering space of (X, x) (Y, y).

70 6.4 Covering space constructions Make the universal covers X of X and Ỹ of Y. 2. Find all the preimages of the basepoint x in X, and glue on a copy of Ỹ at each one. 3. Find all the preimages of the basepoint y in each copy of Ỹ, and glue on a copy of X at each one. 4. Iterate... Examples S 1 S 2 RP 2 RP 2 S 1 S 1

71 6.4 Covering space constructions 71 Non-example Let X be the space given by glueing two copies of R at the origin. We have a map to S 1 S 1 but this is not a covering space. We can see this by looking at neighbourhoods of the basepoint this shows that in general to make a universal cover of X Y we can t just stick X to Ỹ, but rather we have to stick a copy of Ỹ to every preimage of the basepoint in X, and then repeat for Ỹ and then iterate. Non-example This is not a covering space; again we can see this by looking at neighbourhoods of the basepoint.

72 6.4 Covering space constructions 72 Non-example We have seen how covers by looping the middle circle around twice so that the outer two circles land on top of each other. But why can t we do it by just folding the whole thing in half, so that the middle circle folds in half, and then glueing the two points where we folded? This is an interesting non-example because it does cover the space evenly but still isn t a covering space. Why? Non-example [0, 1) S 1 t (cos2πt, sin2πt) This does have the property that the cardinality of p 1 (x) is constant, but this is not a covering space; we can see this by looking at neighbourhoods of the point (1, 0).

73 7 Van Kampen s Theorem 73 7 Van Kampen s Theorem This theorem is about sticking spaces together and finding the fundamental group of the resulting space. For example S 1 S 1 S 1 S 1 MB glue MB KB MB glue RP 2 Key building block examples 1. Take any space X, cut out a disc and glue handle. 2. Take any space X, cut out a disc and glue a Möbius Band. 3. Take a wedge of n circles labelled a 1,..., a n and a word in the letters a 1,..., a n e.g. a 2 3a 5 a 4 1, then glue a disc onto the loops following this word. Key consequences (1) and (2) enable us to classify all 2-dimensional surfaces. (3) enables us to construct a space with any given fundamental group we like.